M.Nelkon&R Parker
Advanced
Level
Physics
M.Nelkon&R Parker
Advanced
Level
Physics
Third Edition With SI Units
*£§iP
'
HEB
Advanced
Level
Physics
Books by M. Nelkon
Published by Heinemann
ADVANCED LEVEL PRACTICAL PHYSICS (with J. Ogbom)
SCHOLARSHIP PHYSICS (Si)
LIGHT AND SOUND (Si)
MECHANICS AND PROPERTIES OF MATTER (Si)
PRINCIPLES OF ATOMIC PHYSICS AND ELECTRONICS (Si)
REVISION NOTES IN PHYSICS (Si)
Book I, Heat, Light, Sound
Book II. Electricity, Atomic Physics, Mechanics, Properties of Matter
GRADED EXERCISES IN PHYSICS (Si)
TEST PAPERS IN PHYSICS (Si)
REVISION BOOK IN ORDINARY LEVEL PHYSICS (Si)
elementary physics, Book I and II (with A. F. Abbott)
AN INTRODUCTION TO THE MATHEMATICS OF PHYSICS
(with J. H. Avery)
GENERAL SCIENCE PHYSICS
electronics and radio (with H. I. Humphreys)
SOLUTIONS TO ADVANCED LEVEL PHYSICS QUESTIONS (Si)
SOLUTIONS TO ORDINARY LEVEL PHYSICS QUESTIONS (Si)
Published by Chatto & Windus
FUNDAMENTALS OF PHYSICS (O-Level, Si)
EXERCISES IN ORDINARY LEVEL PHYSICS (with SI units)
C.S.E. PHYSICS
REVISION BOOK IN C.S.E. PHYSICS
SI UNITS: AN INTRODUCTION FOR ADVANCED LEVEL
Published by Edward Arnold
ELECTRICITY AND MAGNETISM
electricity (Advanced Level, si)
Published by Blackie
heat (Advanced Level, SI)
Books by P. Parker
Published by Heinemann
intermediate heat
electricity and magnetism
Published by Arnold
electronics
Advanced
Level
Physics
Third edition with SI units
M. NELKON, M.Sc.(Lond.), F.Inst. P., A.K.C.
formerly Head of the Science Department,
William Ellis School, London
P. PARKER, M.Sc, F.lnst.P., A.M.I.E.E.
Late Senior Lecturer in Physics,
The City University, London
HEINEMANN EDUCATIONAL
BOOKS LTD LONDON
Heinemann Educational Books Ltd
LONDON EDINBURGH MELBOURNE TORONTO
SINGAPORE JOHANNESBURG AUCKLAND
IBADAN HONG KONG NAIROBI
NEW DELHI
ISBN 435 68636 4
M. Nelkon and Mrs. P. Parker 1958, 1961, 1964, 1965
1 966, 1 968, 1 970
First published as one volume 1958
Reprinted 1 959, 1 960, 1 961 , 1 962
Second edition 1964
Reprinted with additions 1965
Reprinted with additions 1966
Reprinted with corrections 1968
Third edition (SI) 1970
Reprinted 1 970
Published by
Heinemann Educational Books Ltd
48 Charles Street, London WIX 8AH
Parts 1 , 2 and 4 Phototypeset by
Keyspools Ltd., Golborne, Lanes.
Illustrations drawn by
Bucken Ltd., Wealdstone, Middlesex
Printed and bound in Great Britain
by Jarrold and Sons Ltd, Norwich
Preface to Third Edition
This edition covers the new syllabus of the examining boards and
is written in SI units to conform to their use in all future Advanced
level examinations.
The main change in the text has been in the sections in Electricity
on magnetic and electric fields and their associated phenomena. In
the treatment, (i) magnetic flux density or induction B and electric
intensity E have been used in preference to H and D— this follows the
recommendations of the 1966 report of the Association for Science
Education, (ii) magnetic flux density has been defined from the relation
F = BIl. A new section on electromagnetic waves has been added.
Other changes are as follows: Waves. This has now been treated
generally. Optics. The sections on interference and diffraction have
been expanded. Sound. An account of recording on tape and film has
been added. Heat . The joule has been used as the unit of heat and van
der Waals' equation has been discussed. Properties of Matter . The
repulsive and attractive forces between molecules have been em-
phasised. Mechanics. There are further discussions on angular momen-
tum and the dynamics of a rigid body. Throughout the text, worked
examples in SI units have been given to assist the student and the
exercises at the end of chapters contain recent questions from examining
boards. It is hoped that this SI edition will continue to assist students
of Advanced level standard.
We are very much indebted to M. V. Detheridge, Woodhouse
Grammar School, London, for his valuable co-operation in the writing
and the preparation of the new electricity sections and for his generous
assistance with this edition. We are also grateful to Rev. M. D. Phillips,
O.S.B., Ampleforth College, York; S. S. Alexander, Woodhouse
Grammar School, London ; C. A. Boyle, William Ellis School, London ;
S. Freake, Queen's College, Cambridge, and R. P. T. Hills, St. John's
College, Cambridge, for reading parts of the work; and to Prof.
M. L. McGlashan, Exeter University, and M. Sayer, Chetham's
Hospital School, Manchester, for advice on SI units.
I am grateful to the following for permission to include photographs
in this book. To the Head of the Physics Department, the City
University, London, for Newton rings, Fresnel biprism interference
bands, Diffraction rings and Diffraction bands; to the late Sir J. J.
Thomson for Positive Rays photographs; to the National Chemical
Laboratory, for X-Ray diffraction rings; to Lord Blackett of the
Imperial College of Science and Technology, for Transmutation of
Nitrogen ; to Professor George Thomson and the Science Museum, for
Electron diffraction rings; and finally to the United Kingdom Atomic
Energy Authority for Van der Graff Electrostatic Generator and
Nuclear Research Reactor.
Thanks are due to the following Examining Boards for their kind
permission to translate numerical quantities in past questions to SI
units; the translation is the sole responsibility of the author:
London University School Examinations (L.),
Oxford and Cambridge Schools Examination Board (O. & C),
Joint Matriculation Board (N.),
Cambridge Local Examinations Syndicate (C),
Oxford Delegacy of Local Examinations (O.).
1970 M.N.
Preface to First Edition
This text-book is designed for Advanced Level students of Physics,
and covers Mechanics and Properties of Matter, Heat, Optics, and
Sound. Electricity and Atomic Physics to that standard. It is based on
the experience gained over many years of teaching and lecturing to a
wide variety of students in schools and polytechnics.
In the treatment, an Ordinary Level knowledge of the subject is
assumed. We have aimed at presenting the physical aspect of topics
as much as possible, and then at providing the mathematical arguments
and formulae necessary for a thorough understanding. Historical
details have also been given to provide a balanced perspective of the
subject. As a help to the student, numerous worked examples from past
examination papers have been included in the text.
It is possible here to mention only a few points borne in mind by the
authors. In Mechanics and Properties of Matter, the theory of dimen-
sions has been utilized where the mathematics is difficult, as in the sub-
ject of viscosity, and the "excess pressure" formula has been extensively
used in the treatment of surface tension. In Heat, the kinetic theory of
gases has been fully discussed, and the experiments of Joule and
Andrews have been presented in detail. The constant value of n sin / has
been emphasized in refraction at plane surfaces in Optics, there is a full
treatment of opticaltreatment of optical instruments, and accounts of
interference, diffraction and polarization. In Sound, the physical
principles of stationary waves, and their applications to pipes and
strings, have been given prominence. Finally, in Electricity the electron
and ion have been used extensively to produce explanations of pheno-
mena in electrostatics, electromagnetism, electrolysis and atomic
physics ; the concept of e.m.f. has been linked at the outset with energy ;
and there are accounts of measurements and instruments.
We acknowledge our gratitude to the following for their kindness in
reading sections of the work before the complete volume was compiled :
Mr. J. H. Avery, Stockport Grammar School ; Dr. J. Duffey, formerly
of Watford Technical College; Mr. J. Newton, The City University,
London; Mr. A. W. K. Ingram, Lawrence Sheriff School, Rugby;
Mr. O. C. Gay, College of Technology, Hull ; Mr. T. N. Littledale,
Gunnersbury Grammar School ; Mr. C. R. Ensor, Downside School,
Bath; Mr. L. S. Powell, Garnett College, London; Dr. D. W. Stops,
The City University, London; and Professor H. T. Flint, formerly of
London University.
Preface to Second Edition
In this edition I have added an introduction to Atomic Structure,
which covers the Advanced level syllabus on this topic. I am particu-
larly indebted to Mr. J. Yarwood, M.Sc., F.Inst.P., head of the physics
and mathematics department, Regent Street Polytechnic, London, for
reading this section and for valuable advice, and to Prof. L. Pincherle,
Bedford College, London University, for his kind assistance in parts of
the text.
I am also indebted to G. Ullyott, Charterhouse School and L. G.
Mead, Wellington School, Somerset, for their helpful comments on
dynamics and optics respectively.
Contents
PART ONE: MECHANICS AND PROPERTIES OF MATTER
Page
1 Dynamics 1
2 Circular motion, S.H.M. Gravitation 36
3 Rotation of rigid bodies 75
4 Static bodies, Fluids . 96
5 Surface tension 125
6 Elasticity 153
7 Solid Friction. Viscosity 171
PART TWO: HEAT
8 Introduction 189
9 Calorimetry 199
10 Gases 220
11 Thermal expansion 269
12 Changes of State 294
13 Transfer of heat 330
14 Thermometry and pyrometry 366
PART THREE: OPTICS AND SOUND
15 Optics: Introduction 387
16 Reflection at plane surfaces 391
17 Reflection at curved surfaces 402
1 8 Refraction at plane surfaces 420
19 Refraction through prisms 441
20 Dispersion. Spectra 454
21 Refraction through lenses 471
22 Defects of vision. Defects of lenses 508
23 Optical instruments 525
24 Velocity of light. Photometry . 551
25 Oscillations and waves. Sound waves .... 577
26 Characteristics, properties, velocity of sound ... 606
27 Vibrations in pipes, strings, rods ...... 640
28 Optics: Wave theory of light 675
29 Interference, diffraction, polarization . . 687
PART FOUR: ELECTRICITY AND ATOMIC PHYSICS
Page
30 Electrostatics . 731
31 Capacitors 765
32 Current electricity. Resistance. Power 785
33 Ohm's law applications. Measurement. Networks. 807
34 Chemical effect of current 844
35 Magnetic field. Force on conductor ..... 874
36 Electromagnetic induction 895
37 Magnetic fields due to conductors 931
38 Magnetic properties of material 948
39 A.C. circuits. Electromagnetic waves 967
40 Electrons. Motion in fields 993
41 Valves. C.R.O. Junction diode. Transistors 1009
42 Radioactivity. The nucleus 1040
43 X-rays. Photoelectricity. Energy levels .... 1068
Summary of C.G.S. and SI Units 1095
Answers to exercises 1101
Index 1107
•
PART ONE
Mechanics and Properties of Matter
chapter one
Dynamics
Motion in a Straight Line. Velocity
If a car travels steadily in a constant direction and covers a distance s
in a time t, then its velocity in that direction = s/t. If the car does not
travel steadily, then s/t is its average velocity, and
distance s = average velocity x t.
We are here concerned with motion in a constant direction. The term
'displacement' is given to the distance moved in a constant direction,
for example, from L to C in Fig. 1.1 (i). Velocity may therefore be
defined as the rate of change of displacement.
Velocity can be expressed in centimetres per second (cm/s orcms" *)
or metres per second (m/s or m s _1 ) or kilometres per hour (km/h or
kmh -1 ). By calculations, 36 kmh -1 = 10 ms" 1 . It should be noted
that complete information is provided for a velocity by stating its
direction in addition to its magnitude, as explained shortly.
If an object moving in a straight line travels equal distances in equal
times, no matter how small these distances may be, the object is said
to be moving with uniform velocity. The velocity of a falling stone
increases continuously, and so is a non-uniform velocity.
If, at any point of a journey, As is the small change in displacement in
a small time At, the velocity v is given by v = As/ At. In the limit, using
calculus notation, ,
dt
Vectors
Displacement and velocity are examples of a class of quantities called
vectors which have both magnitude and direction. They may therefore
be represented to scale by a line drawn in a particular direction. Thus
Acceleration Xg
(iii)
Cambridge is 80 km from London in a direction 20° E. of N. We
can therefore represent the displacement between the cities in magnitude
1
2 ADVANCED LEVEL PHYSICS
and direction by a straight line LC 4 cm long 20° E. of N., where
1 cm represents 20 km, Fig. 1.1 (i). Similarly, we can represent the
velocity u of a ball initially thrown at an angle of 30° to the horizontal
by a straight line OD drawn to scale in the direction of the velocity u,
the arrow on the line showing the direction, Fig. 1.1 (ii). The acceleration
due to gravity, g, is always represented by a straight line AO to scale
drawn vertically downwards, since this is the direction of the accelera-
tion, Fig. 1.1 (iii). We shall see later that 'force' and 'momentum' are
other examples of vectors.
Speed and Velocity
A car moving along a winding road or a circular track at 80 km h~ *
is said to have a speed of 80 km h~ 1 . 'Speed' is a quantity which has no
direction but only magnitude, like 'mass' or 'density' or 'temperature'.
These quantities are called scalars.
The distinction between speed and velocity can be made clear by
reference to a car moving round a circular track at 80 km h~ * say. Fig.
1.2. At every point on the track the speed is the same — it is 80 km h _ 1 .
Speed constant
Velocity different
Fig. 1 .2. Velocity and speed
At every point, however, the velocity is different. At A, B or C, for
example, the velocity is in the direction of the particular tangent, AP,
BQ or CR, so that even though the magnitudes are the same, the three
velocities are all different because they point in different directions.
Generally, vector quantities can be represented by a line drawn in the
direction of the vector and whose length represents its magnitude.
Distance-Time Curve
When the displacement, or distance, s of a moving car from some
fixed point is plotted against the time t, a distance-time (s — t) curve of
DYNAMICS
the motion is obtained The velocity of the car at any instant is given
by the change in distance per second at that instant. At E, for example,
if the change in distance s is As and this change is made in a time At,
velocity at E =
As
At
In the limit, then, when At approaches zero, the velocity at E becomes
equal to the gradient of the tangent to the curve at EL Using calculus
notation, As/ At then becomes equal to ds/dt (p. 1).
Uniform
velocity
D
W
Fig. 1.3 Displacement (.s)-time (/) curves
If the distance-time curve is a straight line CD, the gradient is
constant at all points; it therefore follows that the car is moving with
a uniform velocity, Fig. 1.3. If the distance-time curve is a curve CAB,
the gradient varies at different points. The car then moves with non-
uniform velocity. We may deduce that the velocity is zero at the instant
corresponding to A, since the gradient at A to the curve CAB is zero.
When a ball is thrown upwards, the height s reached at any instant
t is given by s = ut—\gt 2 , where u is the initial velocity and g is the
constant equal to the acceleration due to gravity (p. 8). The graph
of s against t is represented by the parabolic curve CXY in Fig. 1.3 ; the
gradient at X is zero, illustrating that the velocity of the ball at its
maximum height is zero.
Velocity-Time Curves
When the velocity of a moving train is plotted against the time, a
'velocity-time (v-t) curve' is obtained. Useful information can be de-
duced from this curve, as we shall see shortly. If the velocity is uniform,
the velocity-time graph is a straight line parallel to the time-axis, as
shown by line (1) in Fig. 1.4. If the train accelerates uniformly from rest,
the velocity-time graph is a straight line, line (2), inclined to the time-
axis. If the acceleration is not uniform, the velocity-time graph is curved.
ADVANCED LEVEL PHYSICS
XY
— t
Fig. 1.4 Velocity (v)-time (t) curves
In Fig. 1.4, the velocity-time graph OAB represents the velocity of a
train starting from rest which reaches a maximum velocity at A, and
then comes to rest at the time corresponding to B; the acceleration and
retardation are both not uniform in this case.
Acceleration is the 'rate of change of velocity', i.e. the change of
velocity per second. The acceleration of the train at any instant is given
by the gradient to the velocity-time graph at that instant, as at E. At the
peak point A of the curve OAB the gradient is zero, i.e., the acceleration
is then zero. At any point, such as G, between A, B the gradient to the
curve is negative, i.e., the train undergoes retardation.
The gradient to the curve at any point such as E is given by :
velocity change
time
Av
At
where Av represents a small change in v in a small time At. In the limit,
the ratio Av/At becomes dv/dt, using calculus notation.
Area Between Velocity-Time Graph and Time-Axis
Consider again the velocity-time graph OAB, and suppose the
velocity increases in a very small time-interval XY from a value rep-
resented by XC to a value represented by YD, Fig. 1.4. Since the small
distance travelled = average velocity x time XY, the distance travelled
is represented by the area between the curve CD and the time-axis,
shown shaded in Fig. 1.4. By considering every small time-interval
between OB in the same way, it follows that the total distance travelled
by the train in the time OB is given by the area between the velocity-time
graph and the time-axis. This result applies to any velocity-time graph,
whatever its shape.
Fig. 1.5 illustrates the velocity-time graph AB of an object moving
with uniform acceleration a from an initial velocity u. From above,
the distance s travelled in a timet or OC is equivalent to the area
OABC. The area OADC = u.t. The area of the triangle ABD =
DYNAMICS
Fig. 1.5 Uniform acceleration
^AD . BD = \t . BD. Now BD = the increase in velocity in a time t
= at. Hence area of triangle ABD = \t.at = \at 2
:. total area OABC = s = ut+^at 2 .
This result is also deduced on p. 6.
Acceleration
The acceleration of a moving object at an instant is the rate of
change of its velocity at that instant. In the case of a train accelerating
steadily from 36 kmh" 1 (10 ms" 1 ) to 54 kmh" 1 (15 ms -1 ) in 10
second, the uniform acceleration
= (54-36) km h~ * -f- 10 seconds = 1-8 km h" 1 per second,
or
(15 — 10) m s * -r- 10 seconds = 0-5 m s l per second.
Since the time element (second) is repeated twice in the latter case, the
acceleration is usually given as 0-5 ms" 2 . Another unit of acceleration
is 'cms -2 '. In terms of the calculus, the acceleration a of a moving
object is given by
a =
dv
dt
where dv/dt is the velocity change per second.
Distance Travelled with Uniform Acceleration. Equations of Motion
If the velocity changes by equal amounts in equal times, no matter
how small the time-intervals may be, the acceleration is said to be
uniform. Suppose that the velocity of an object moving in a straight
6 ADVANCED LEVEL PHYSICS
line with uniform acceleration a increases from a value u to a value v in
a time t. Then, from the definition of acceleration,
v — u
from which v = u+at (1)
Suppose an object with a velocity u accelerates with a uniform
acceleration a for a time t and attains a velocity v. The distance s
travelled by the object in the time t is given by
s = average velocity x t
= %(u + v) X t
But v = u + at
.'. s = %u + u+at)t
.-. s = ut +£at 2 (2)
If we eliminate t by substituting t = (v—u)/a from (1) in (2), we obtain,
on simplifying,
v 2 = u 2 +2as ... (3)
Equations (1), (2), (3) are the equations of motion of an object moving
in a straight line with uniform acceleration. When an object undergoes
a uniform retardation, for example when brakes are applied to a car,
a has a negative value.
EXAMPLES
1. A car moving with a velocity of 54 km h" l accelerates uniformly at the rate
of 2 m s -2 . Calculate the distance travelled from the place where acceleration
began to that where the velocity reaches 72 km h~ 1 , and the time taken to cover
this distance.
(i) 54kmh -1 = 15 ms" 1 , 72 kmh -1 = 20 ms -1 , acceleration a = 2 ms -2 .
Using v 2 = u 2 + 2as,
.. 20 2 = 15 2 +2x2xs
20 2 -15 2 ..3
. . s = — - — - — = 434 m.
2x2 *
(ii) Using v = u + at
.\20= l5 + 2t
DYNAMICS '
2. A train travelling at 72 km h" 1 undergoes a uniform retardation of 2 m s " 2
when brakes are applied. Find the time taken to come to rest and the distance
travelled from the place where the brakes were applied.
(i) 72kmh _1 = 20 ms" 1 , and a = -2ms _2 ,i> = 0.
Using v = u + at
'.-. = 20-2*
;.t=l0s
(ii) The distance, s, = ut+\at 2 .
= 20 x 10-i x 2 x 10 2 = 100 m.
Motion Under Gravity
When an object falls to the ground under the action of gravity,
experiment shows that the object has a constant or uniform acceleration
of about 980 cms" 2 , while it is falling (see p. 49). In SI units this is
9-8 ms -2 or 10 ms -2 approximately. The numerical value of this
acceleration is usually denoted by the symbol g. Suppose that an object
is dropped from a height of 20 m above the ground. Then the initial
velocity u = 0, and the acceleration a = g = 10 m s -2 (approx).
Substituting ins = ut+jat 2 , the distance fallen s in metres is calculated
from
S = \gt 2 = 5t 2 .
When the object reaches the ground, s = 20 m.
.'. 20 = 5t 2 , or t = 2 s
Thus the object takes 2 seconds to reach the ground.
If a cricket-ball is thrown vertically upwards, it slows down owing to
the attraction of the earth. The ball is thus retarded. The magnitude
of the retardation is 9-8 m s -2 , or g. Mathematically, a retardation
can be regarded as a negative acceleration in the direction along which
the object is moving; and hence a = — 9-8 m s~ 2 in this case.
Suppose the ball was thrown straight up with an initial velocity, u,
of 30 ms -1 . The time taken to reach the top of its motion can be
obtained from the equation v = u + at. The velocity, v, at the top is
zero; and since u = 30 m and a = —9-8 or 10 ms" 2 (approx), we have
= 30-10r.
. ♦ 30 -
•< = Io = 3s -
The highest distance reached is thus given by
s = ut+^at 2
= 30x3-5x3 2 = 45 m.
Resultant. Components
If a boy is running along the deck of a ship in a direction OA, and the
8 ADVANCED LEVEL PHYSICS
ship is moving in a different direction OB, the boy will move relatively
to the sea along a direction OC, between OA and OB, Fig. 1.6 (i). Now
in one second the boat moves from O to B, where OB represents the
velocity of the boat, a vector quantity, in magnitude and direction.
The boy moves from O to A in the same time, where OA represents the
velocity of the boy in magnitude and direction. Thus in one second the
net effect relative to the sea is that the boy moves from O to C. It can
now be seen that if lines OA, OB are drawn to represent in magnitude
and direction the respective velocities of the boy and the ship, the
magnitude and direction of the resultant velocity of the boy is repre-
sented by the diagonal OC of the completed parallelogram having OA,
OB as two of its sides; OACB is known as a parallelogram of velocities.
Conversely, a velocity represented completely by OC can be regarded
as having an 'effective part', or component represented by OA, and
another component represented by OB.
(ii)
A
Fig. 1.6. Resultant and component
In practice, we often require to find the component of a vector
quantity in a certain direction. Suppose OR represents the vector F,
and OX is the direction, Fig. 1.6 (ii). If we complete the parallelogram
OQRP by drawing a perpendicular RP from R to OX, and a per-
pendicular RQ from R to OY, where OY is perpendicular to OX, we
can see that OP, OQ represent the components of F along OX, OY
respectively. Now the component OQ has no effect in a perpendicular
direction ; consequently OP represents the total effect of F along the
direction OX. OP is called the 'resolved component' in this direction.
If is the angle ROX, then, since triangle OPR has a right angle at P,
OP = OR cos = F cos
(4)
Components of g
The acceleration due to gravity, g, acts vertically downwards. In
free fall, an object has an acceleration g. An object sliding freely down
an inclined plane, however, has an acceleration due to gravity equal
to the component of g down the plane. If it is inclined at 60° to the
vertical, the acceleration down the plane is then g cos 60° or 9-8 cos 60°
ms -2 , which is 4-9 ms~ 2 .
Consider an object O thrown forward from the top of a cliff OA
DYNAMICS
with a horizontal velocity u of 15 m s _1 . Fig. 1.7. Since u is horizontal,
it has no component in a vertical direction. Similarly, since g acts
vertically, it has no component in a horizontal direction.
i i/ = 15ms"'
Fig. 1.7 Motion under gravity
We may thus treat vertical and horizontal motion independently.
Consider the vertical motion from O. If OA is 20 m, the ball has an
initial vertical velocity of zero and a vertical acceleration of g, which is
9-8 m s~ 2 (10 m s~ 2 approximately). Thus, from s = ut+$at 2 , the time
t to reach the bottom of the cliff is given, using g = 10 m s -2 , by
20 = \ . 10 . t 2 = 5t 2 , or t = 2 s.
So far as the horizontal motion is concerned, the ball continues to
move forward with a constant velocity of 15 m s~ 1 since g has no com-
ponent horizontally. In 2 seconds, therefore,
horizontal distance AB = distance from cliff = 15 x 2 = 30 m.
Generally, in a time t the ball falls a vertical distance, y say, from O
given by y = \gt 2 . In the same time the ball travels a horizontal distance,
x say, from O given by x = ut, where u is the velocity of 15 m s.~ *. If t is
eliminated by using t = x/u in y = \gt 2 , we obtain y = gx 2 /2u. This is
the equation of a parabola. It is the path OB in Fig. 1.7.
Addition of Vectors
Suppose a ship is travelling due east at 30 km h" 1 and a boy runs
across the deck in a north-west direction at 6 kmh -1 , Fig. 1.8 (i). We
k-i
6 km.hT 1 Vector sum
30 km.h:' N.
* 30
(i) (»)
Fig. 1.8 Addition of vectors
10 ADVANCED LEVEL PHYSICS
can find the velocity and direction of the boy relative to the sea by
adding the two velocities. Since velocity is a vector quantity, we draw
a line OA to represent 30 km h~ * in magnitude and direction, and then,
from the end of A, draw a line AC to represent 6 km h~ x in magnitude
and direction, Fig. 1.8 (ii). The sum, or resultant, of the velocities is now
represented by the line OC in magnitude and direction, because a
distance moved in one second by the ship (represented by OA) together
with a distance moved in one second by the boy (represented by AC)
is equivalent to a movement of the boy from O to C relative to the sea.
Fig. 1.9 Subtraction of velocities
In other words, the difference between the vectors?, (} in Fig. 1.9 (i)
is the sum of the vectors? and ( — (^). Now ( — (^) is a vector drawn
exactly equal and opposite to the vector (J. We therefore draw ab to
represent P completely, and then draw be to represent ( — (J) completely,
Fig. 1.9 (ii). Then ? + (-($) = the vector represented by ac =?-Q.
Relative Velocity and Relative Acceleration
If a car A travelling at 50 km h _1 is moving in the same direction as
another car B travelling at 60 km h~ *, the relative velocity of B to A —
60— 50 = 10 kmh -1 . If, however, the cars are travelling in opposite
directions, the relative velocity of B to A = 60— (—50) = 110 kmh -1 .
Suppose that a car X is travelling with a velocity v along a road 30°
east of north, and a car Y is travelling with a velocity u along a road
due east, Fig. 1.10 (i). Since 'velocity' has direction as well as magnitude,
i.e., 'velocity' is a vector quantity (p. 1), we cannot subtract u and v
numerically to find the relative velocity. We must adopt a method
which takes into account the direction as well as the magnitude of the
velocities, i.e., a vector subtraction is required.
Relative \ /v Relative
velocity \ / velocity
Fig. 1.10. Relative velocity.
DYNAMICS
11
The velocity of X relative to Y =~$-u ="u+(-"u). Suppose OA
represents the velocity, v, of X in magnitude and direction, Fig. 1.10 (ii).
Since Y is travelling due east, a velocity AB numerically equal to u
but in the due wsr direction represents the vector (-1J). The vector
sum of OA and AB is OB from p. 0, which therefore represents in
magnitude and direction the velocity of X relative to Y. By drawing an
accurate diagram of the two velocities, OB can be found.
The velocity of Y relative to X = it— if = 1?+ ( -"3), and can be found
by a similar method. In this case, OD represents the velocity, u, of Y
in magnitude and direction, while DE represents the vector (- 1$),
which it is drawn numerically equal to v but in the opposite direction,
Fig. 1.10 (iii). The vector sum of OD and DE is OE, which therefore
represents the velocity of Y relative to X in magnitude and direction.
When two objects P, Q are each accelerating, the acceleration of P
relative to Q = acceleration of P - acceleration of Q. Since 'accelera-
tion' is a vector quantity, the relative acceleration must be found by
vector subtraction, as for the case of relative velocity.
EXAMPLE
Explain the difference between a scalar and a vector quantity.
What is meant by the relative velocity of one body with respect to another?
Two ships are 10 km apart on a line running S. to N. The one farther north
is steaming W. at 20 km h~ *. The other is steaming N. at 20 km h~ *. What is
their distance of closest approach and how long do they take to reach it? (C.)
Suppose the two ships are at X, Y, moving with velocities u, v respectively,
each 20 kmh -1 Fig 1.11 (i). The velocity of Y relative to X =~v-~u=~v+(-^ti).
We therefore draw OA to represent if (20) and add to it AB, which represents
(-"3), Fig 1.11 (ii). The relative velocity is then represented by OB.
u = 20
i/=20
10
u
(')
Since OAB is a right-angled triangle,
Also,
OB = VOA 2 +AB 2 = V20 2 + 20 2 = 28-28 = 28-3 kmh" 1 (i)
AB 20
tan ^oi = 20 =Ue ^ = 45 °
(ii)
12
ADVANCED LEVEL PHYSICS
Thus the ship Y will move along a direction QR relative to the ship X, where
QR is at 45° to PQ, the north-south direction, Fig. 1.11 (iii). If PQ = 10 km,
the distance of closest approach is PN, where PN is the perpendicular from
P to QR.
.'. PN = PQ sin 45° = 10 sin 45° = 707 km.
The distance QN = 10 cos 45° = 707 km. Since, from (i), the relative velocity
is 28-28 km h~\ it follows that
time to reach N = — = £ hour.
28 '28
LAWS OF MOTION. FORCE AND MOMENTUM
Newton's Laws of Motion
In 1686 Sir Isaac Newton published a work called Principia, in
which he expounded the Laws of Mechanics. He formulated in the
book three 'laws of motion' :
Law I. Every body continues in its state of rest or uniform motion in a
straight line, unless impressed forces act on it.
Law IL The change of momentum per unit time is proportional to the
impressed force, and takes place in the direction of the straight line along
which the force acts.
Law HI. Action and reaction are always equal and opposite.
These laws cannot be proved in a formal way; we believe they are
correct because all the theoretical results obtained by assuming their
truth agree with the experimental observations, as for example in
astronomy (p. 58).
Inertia. Mass
Newton's first law expresses the idea of inertia. The inertia of a body
is its reluctance to start moving, and its reluctance to stop once it has
begun moving. Thus an object at rest begins to move only when it is
pushed or pulled, i.e., when a force acts on it. An object O moving in a
Velocity change
Velocity ,
change
O
t
(i) (»)
Fig. 1.12 Velocity changes
DYNAMICS 13
straight line with constant velocity will change its direction or move
faster only if a new force acts on it. Fig. 1.12 (i). This can be demon-
strated by a puck moving on a cushion of gas on a smooth level sheet
of glass. As the puck slides over the glass, photographs taken at succes-
sive equal times by a stroboscopic method show that the motion is
practically that of uniform velocity. Passengers in a bus or car are
jerked forward when the vehicle stops suddenly. They continue in their
state of motion until brought to rest by friction or collision. The use of
safety belts reduces the shock.
Fig. 1.12 (ii) illustrates a velocity change when an object O is whirled
at constant speed by a string. This time the magnitude of the velocity
v is constant but its direction changes.
'Mass' is a measure of the inertia of a body. If an object changes its
direction or its velocity slightly when a large force acts on it, its inertial
mass is high. The mass of an object is constant all over the world ;
it is the same on the earth as on the moon. Mass is measured in kilo-
grammes (kg) or grammes (g) by means of a chemical balance, where
it is compared with standard masses based on the International
Prototype Kilogramme (see also p. 14).
Force. The newton
When an object X is moving it is said to have an amount of momentum
given, by definition, by
momentum = mass ofX x velocity . (1)
Thus an object of mass 20 kg moving with a velocity of 10 m s~ * has a
momentum of 200 kg ms -1 . If another object collides with X its
velocity alters, and thus the momentum of X alters. From Newton's
second law, a. force acts on X which is equal to the change in momentum
per second.
Thus if F is the magnitude of a force acting on a constant mass m,
F ocmx change of velocity per second
.'. F oc ma,
where a is the acceleration produced by the force, by definition of a.
:. F = kma (2)
where k is a constant.
With SI units, the newton (N) is the unit of force. It is defined as
the force which gives a mass of 1 kilogramme an acceleration of
1 metre s -2 . Substituting F = IN, m = 1 kg and a = 1 ms -2 in
the expression for F in (i), we obtain k = 1. Hence, with units as stated,
k = 1.
.". F = ma,
which is a standard equation in dynamics. Thus if a mass of 200 g is
acted upon by a force F which produces an acceleration a of 4 m s~ 2 ,
then, since m = 200 g = 0-2 kg,
F = ma = 0-2(kg) x 4(m s~ 2 ) = 0-8 N.
14 ADVANCED LEVEL PHYSICS
C.g.s. units of force
The dyne is the unit of force in the centimetre-gramme-second
system ; it is defined as the force acting on a mass of 1 gramme which
gives it an acceleration of 1 cms" 2 . The equation F = ma also applies
when m is in grammes, a is in cm s -2 , and F is in dynes. Thus if a force
of 10000 dynes acts on a mass of 200 g, the acceleration a is given by
F = 10000 = 200 x a, or a = 50 cms -2 .
Suppose m = 1 kg = 1000 g, a = 1 m s" 2 = 100 cm s~ 2 . Then,
the force F is given by
F = ma= 1000 x 100 dynes = 10 5 dynes.
But the force acting on a mass of 1 kg which gives it an acceleration of
1 m s -2 is the newton, N. Hence
1 N = 10 5 dynes
Weight. Relation between newton, kgf and dyne, gf
The weight of an object is defined as the force acting on it due to
gravity ; the weight of an object can hence be measured by attaching
it to a spring-balance and noting the extension, as the latter is propor-
tional to the force acting on it (p. 50).
Suppose the weight of an object of mass m is denoted by W. If the
object is released so that it falls to the ground, its acceleration is g.
Now F = ma. Consequently the force acting on it, i.e., its weight, is
given by
W = mg.
If the mass is 1 kg, then, since g = 9-8 m s -2 , the weight W = 1 x 9-8 =
9-8 N (newton). The force due to gravity on a mass of 1 kg where g has
the value 9-80665 m s~ 2 is called a 1 kilogramme force or 1 kgf (this is
roughly equal to 1 kilogramme weight or 1 kg wt, which depends on
the value of g and thus varies from place to place). Hence it follows that
1 kgf = 9-8 N = 10 N approximately.
A weight of 5 kgf is thus about 50 N. Further, 1 N = yq kgf approx =
100 gf. The weight of an apple is about 1 newton.
. The weight of a mass of 1 gramme is called gramme-force (1 gf); it
was formerly called '1 gramme wt'. From F = ma,. it follows that
1 gf = 1 x 980 = 980 dynes.
since g = 980 cms" 2 (approx).
The reader should note carefully the difference between the 'kilo-
gramme' and the 'kilogramme force'; the former is a mass and is
therefore constant all over the universe, whereas the kilogramme force
is a force whose magnitude depends on the value of a. The acceleration
due to gravity, g, depends on the distance of the place considered from
the centre of the earth; it is slightly greater at the poles than at the
DYNAMICS 15
equator, since the earth is not perfectly spherical (see p. 41). It there-
fore follows that the weight of an object differs in different parts of the
world. On the moon, which is smaller than the earth and has a smaller
density, an object would weigh about one-sixth of its weight on the
earth.
The relation F = ma can be verified by using a ticker-tape and
timer to measure the acceleration of a moving trolley. Details are
given in a more basic text, such as Fundamentals of Physics (Chatto and
Windus) by the author.
The following examples illustrate the application of F = ma. It
should be carefully noted that (i) F represents the resultant force on the
object of mass m, (ii) F must be expressed in the appropriate units of a
'force' and m in the corresponding units of a 'mass'.
EXAMPLES
1. A force of 20 kgf pulls a sledge of mass 50 kg and overcomes a constant
frictional force of 4 kgf. What is the acceleration of the sledge?
Resultant force, F, = 20 kgf- 4 kgf = 16 kgf.
To change this to units of newtons, use 1 kgf = 9-8 N = 10 N approx.
.'. 16kgf = 160 N approx.
From F = ma,
.-. 160= 50 x a
.'. a = 3-2 ms~ 2 .
2. An object of mass 200 kg is attached to the hook of a spring-balance, and
the latter is suspended vertically from the roof of a lift. What is the reading on the
spring-balance when the lift is (i) ascending with an acceleration of 20 cms" 2 ,
(ii) descending with an acceleration of 10 cm s" 2 , (iii) ascending with a uniform
velocity of 15 cm s~ *.
Suppose T is the tension (force) in the spring-balance in kgf.
(i) The object is acted upon two forces: (a) The tension T kgf in the spring-
balance, which acts upwards, (b) its weight, 2 kgf, which acts downwards. Since
the object moves upwards, T is greater than 2 kgf. Hence the net force, F, acting
on the object = (T-2) kgf = (T-2) x 10 N, approx. Now
F = ma,
where a is the acceleration inms" 2 .
.-. (T-2)xl0 = 2xa = 2x0-2
.-. T = 204 kgf . . (1)
(ii) When the lift descends with an acceleration of 10 cms -2 or 01 m s~ 2 , the
weight, 2 kgf, is now greater than T t kgf, the tension in the spring-balance.
.-. resultant force = (2 - TJ kgf = (2 - T t ) x 10 N approx.
.'. F = (2 - Tj) x 10 = ma = 2 x 01
.'. T t = 2-002 = 1-98 kgf.
(iii) When the lift moves with constant velocity, the acceleration is zero. In
this case the reading on the spring-balance is exactly equal to the weight, 2 kgf.
16 ADVANCED LEVEL PHYSICS
Linear Momentum
Newton defined the force acting on an object as the rate of change
of its momentum, the momentum being the product of its mass and
velocity (p. 13). Momentum is thus a vector quantity. Suppose that the
mass of an object is m, its initial velocity is u, and its final velocity due
to a force F acting on it for a time t is v. Then
change of momentum = mv — mu,
„ mv — mu
and hence r = —
.'. Ft = mv — mu = momentum change . . (1)
The quantity Ft (force x time) is known as the impulse of the force on
the object, and from (1) it follows that the units of momentum are the
same as those of Pt, i.e., newton second (N s). From 'mass x velocity',
alternative units are 'kgms -1 '.
Force and momentum change
A person of mass 50 kg who is jumping from a heig ht o f 5 metres
will land on the ground with a velocity = yflgh = yjl x 10 x 5
= 10 ms" 1 , assuming g = 980 cm s~ 2 = 10 ms" 2 approx. If he does
not flex his knees on landing, he will be brought to rest very quickly, say
in T^th second. The force F acting is then given by
momentum change
F = -.
time
= 50 ^_ 10 = 5000 N = 500 kgf (approx).
10
This is a force of about 10 times the person's weight and this large
force has a severe effect on the body.
Suppose, however, that the person flexes his knees and is brought to
rest much more slowly on landing, say in 1 second. Then, from above,
the force F now acting is 10 times less than before, or 50 kgf (approx).
Consequently, much less damage is done to the person on landing.
Initial
momentum mu
•
Final <J '
momentum -mu
fe Sand
j.
•j| Horizontal momentum = O
f,_Final mo mentum^ ^Belt Wall
-^S^b D CF=-» ,
5cm s
(i) (»)
Fig .1.13 Linear momentum
DYNAMICS 17
Suppose sand is allowed to fall vertically at a steady rate of 100 g s~ *
on to a horizontal conveyor belt moving at a steady velocity of 5 cm s~ *.
Fig. 1.13 (i). The initial horizontal velocity of the sand is zero. The
final horizontal velocity is 5 cm s~ 1 . Now
mass = 100 g = 01 kg, velocity = 5 cm s -1 = 5 x 10" 2 m s _1
.'.momentum change per second = 01 x 5 x 10 -2 = 5 x 10 -3 newton
= force on belt
Observe that this is a case where the mass changes with time and the
velocity gained is constant. In terms of the calculus, the force is the
rate of -change of momentum mv, which is vx dm/dt, and dm/dt is
100 g s~ * in this numerical example.
Consider a molecule of mass m in a gas, which strikes the wall of
a vessel repeatedly with a velocity u and rebounds with a velocity — u.
Fig. 1.13 (ii). Since momentum is a vector quantity, the momentum
change = final momentum — initial momentum = mu — ( — mu) = 2mu.
If the containing vessel is a cube of side /, the molecule repeatedly
takes a time 2l/u to make an impact with the same side.
.'. average force on wall due to molecule
_ momentum change
time
_ 2mu __ mu 2
~ 2l/u~ i:
The total gas pressure is the average force per unit area on the walls
of the container due to all the numerous gas molecules.
EXAMPLES
1. A hose ejects water at a speed of 20 cm s" 1 through a hole of area 100 cm 2 .
If the water strikes a wall normally, calculate the force on the wall in newton,
assuming the velocity of the water normal to the wall is zero after collision.
The volume of water per second striking the wall = 100 x 20 = 2000 cm 3 .
.'. mass per second striking wall = 2000 g s~ 1 = 2 kg s~ *.
Velocity change of water on striking wall = 20-0 = 20 cm s" 1 = 0-2 m s _1 .
.". momentum change per second = 2 (kg s~ *) x 0-2 (m s~ *) = 0-4 newton.
2. Sand drops vertically at the rate of 2 kg s -1 on to a conveyor belt moving
horizontally with a velocity of 01 m s _1 . Calculate (i) the extra power needed to
keep the belt moving, (ii) the rate of change of kinetic energy of the sand. Why
is the power twice as great as the rate of change of kinetic energy?
(i) Force required to keep belt moving = rate of increase of horizontal
momentum of sand = mass per second (dm/dt) x velocity change = 2x01 =
0-2 newton.
.". power = work done per second = force x rate of displacement
= force x velocity = 0-2x01 = 002 watt (p. 25).
18 ADVANCED LEVEL PHYSICS
(ii) Kinetic energy of sand = \mv 2 .
i 2 ..dm
.'. rate of change of energy = hr x — -, since v is constant,
at
= ±x01 2 x2 = 001 watt.
Thus the power supplied is twice as great as the rate of change of kinetic
energy. The extra power is due to the fact that the sand does not immediately
assume the velocity of the belt, so that the belt at first moves relative to the sand.
The extra power is needed to overcome the friction between the sand and belt.
Conservation of Linear Momentum
We now consider what happens to the linear momentum of objects
which collide with each other.
Experimentally, this can be investigated by several methods :
1. Trolleys in collision, with ticker-tapes attached to measure velocities.
2. Linear Air-track, using perspex models in collision and stroboscopic
photography for measuring velocities.
61 5g
^7
jtfOcms- 1 , A 615g
■ry
. B 620g
K°)
^7
180cm s-
Before collision After collision
Fig. 1.14 Linear momentum experiment
As an illustration of the experimental results, the following measure-
ments were taken in trolley collisions (Fig. 1.14) :
Before collision.
Mass of trolley A = 615 g; initial velocity = 360 cms" 1 .
After collision.
A and B coalesced and both moved with velocity of 180 cm s~ 1 .
Thus the total linear momentum of A and B before collision =
0-615 (kg) x 3-6 (ms _1 ) + = 2-20 kgms -1 (approx). The total
momentum of A and B after collision = 1-235 x 1-8 = 2-20 kgms !
(approx).
Within the limits of experimental accuracy, it follows that the total
moment of A and B before collision = the total momentum after collision.
Similar results are obtained if A and B are moving with different speeds
after collision, or in opposite directions before collision.
Principle of Conservation of Linear Momentum
These experimental results can be shown to follow from Newton's
second and third laws of motion (p. 12).
Suppose that a moving object A, of mass m x and velocity u ly collides
DYNAMICS
19
u 2
Fig. 1.15
Conservation of linear momentum
with another object B, of mass m 2 and velocity u 2 , moving in the same
direction, Fig. 1.15. By Newton's
law of action and reaction, the force
F exerted by A on B is equal and
opposite to that exerted by B on A.
Moreover, the time t during which
the force acted on B is equal to the
time during which the force of reaction acted on A. Thus the magnitude
of the impulse, Ft, on B is equal and opposite to the magnitude of the
impulse on A. From equation (1), p. 16, the impulse is equal to the
change of momentum It therefore follows that the change in the total
momentum of the two objects is zero, i.e., the total momentum of the
two objects is constant although a collision had occurred. Thus if A
moves with a reduced velocity v x after collision, and B then moves
with an increased velocity v 2 ,
m 1 u l +m 2 U2 = m 1 v l +m 2 v 2 .
The principle of the conservation of linear momentum states that,
if no external forces act on a system of colliding objects, the total mo±
mentum of the objects remains constant.
EXAMPLES
1. An object A of mass 2 kg is moving with a velocity of 3 m s~ 1 and collides
head on with an object B of mass 1 kg moving in the opposite direction with a
velocity of 4 m s " l . Fig 1 . 16 (i). After collision both objects coalesce, so that they
move with a common velocity v. Calculate v.
Fig. 1.16 Examples
Total momentum before collision of A and B in the direction of A
= 2x3 — 1x4=2 kg ms~\
Note that momentum is a vector and the momentum of B is of opposite sign to
A.
20 ADVANCED LEVEL PHYSICS
After collision, momentum of A anid B in the direction of A = 2v+ Iv = 3u.
.-. 3v = 2
1 „, e -l
3
2. What is understood by (a) the principle of the conservation of energy, (b) the
principle of the conservation ofmomentwnl
A bullet of mass 20 g travelling horizontally at 100 m s -1 , embeds itself in
the centre of a block of wood of mass 1 kg which is suspended by light vertical
strings 1 m in length. Calculate the maximum inclination of the strings to the
vertical.
Describe in detail how the experiment might be carried out and used to
determine the velocity of the bullet just before the impact of the block. (N.)
Second part. Suppose A is the bullet, B is the block suspended from a point O,
and 6 is the maximum inclination to the vertical, Fig. 1.16(h). If ucms" 1 is the
common velocity of block and bullet when the latter is brought to rest relative
to the block, then, from the principle of the conservation of momentum, since
20 g = 002 kg,
(l + 002)t; = 002x100
2 100 _!
• y = ro2 = ir ms •
The vertical height risen by block and bullet is given by v 2 = Igh, where g =
9-8ms" 2 and/i = /-/cos0 = /(l-cos0).
.-. v 2 = 20/(1- cos 0).
100* 2
, = 2x9-8x1(1- cos 0).
51/
(1 00\ 2 1
ir) x 2^ = 01961
.-. cos 6 = 0-8038, or d = 37° (approx.).
The velocity, v, of the bullet can be determined by applying the conservation
of momentum principle.
Thus mv = (m+M)V, where m is the mass of the bullet, M is the mass of the
block, and V is the common velocity. Then v = (m+M)V/m. The quantities m
and M can be found by weighing. V is calculated from the horizontal displacement
a of the block, since (i) V 2 = 2gh and (ii) h(2l-h) = a 2 from the geometry of the
circle, so that, to a good approximation, 2h = a 2 /l.
Inelastic and elastic collisions
In collisions, the total momentum of the colliding objects is always
conserved. Usually, however, their total kinetic energy is not conserved.
Some of it is changed to heat or sound energy, which is not recoverable.
Such collisions are said to be inelastic. If the total kinetic energy is
conserved, the collision is said to be elastic. The collision between
two smooth billiard balls is approximately elastic. Many atomic
collisions are elastic. Electrons may make elastic or inelastic collisions
DYNAMICS
21
with atoms of a gas. As proved on p. 28, the kinetic energy of a mass m
moving with a velocity v has kinetic energy equal to \mv 2 .
As an illustration of the mechanics associated with elastic collisions,
consider a sphere A of mass m and velocity v incident on a stationary
sphere B of equal mass m. (Fig. 1.17 (i). Suppose the collision is elastic,
and after collision let A move with a velocity v t at an angle of 60° to
its original direction and B move with a velocity v 2 at an angle 6 to
the direction of v.
(i)
'60
Conservation of
momentum
(ii)
Fig. 1.17 Conservation of momentum
Since momentum is a vector (p. 17), we may represent the mo-
mentum my of A by the line PQ drawn in the direction oft;. Fig. 1.17 (ii).
Likewise, PR represents the momentum mv x of A after collision.
Since momentum is conserved, the vector RQ must represent the momentum
mv 2 ofB after collision, that is,
Hence
nW = nW t + rnv* 2 -
-$ =~^ 1 +t 2 ,
or PQ represents v in magnitude, PR represents v x and RQ represents
v 2 . But if the collision is elastic,
\mv 2 = \mv* +\mv 2
:. v 2 = v x 2 +v 2 2 .
Consequently, triangle PRQ is a right-angled triangle with angle R
equal to 90°.
.*. V t = V COS 60° = -r.
Also, 6 = 90° -60° = 30°, and v 2 = v cos 30° =
o __ >/3"»
2 '
22 ADVANCED LEVEL PHYSICS
Coefficient of restitution
In practice, colliding objects do not stick together and kinetic energy is always
lost. If a ball X moving with velocity u x collides head-on with a ball Y moving
with a velocity u 2 in the same direction, then Y will move faster with a velocity
v t say and X may then have a reduced velocity v 2 in the same direction. The co-
efficient of restitution, e, between X and Y is denned as the ratio :
velocity of separation v 2 — v l
velocity of approach
The coefficient of restitution is approximately constant between two given
materials. It varies from e = 0, when objects stick together and the collision is
completely inelastic, to e = 1, when objects are very hard and the collision is
practically elastic Thus, from above, if u t = 4 m s _1 , u 2 = 1ms -1 and e = 0-8,
then velocity of separation, v 2 — v t = 0-8 x (4— 1) = 2-4 m s~ *.
Momentum and Explosive forces
There are numerous cases where momentum changes are produced
by explosive forces. An example is a bullet of mass m = 50 g say, fired
from a rifle of mass M = 2 kg with a velocity v of 100 m s~ 1 . Initially,
the total momentum of the bullet and rifle is zero. From the principle
of the conservation of linear momentum, when the bullet is fired the
total momentum of bullet and rifle is still zero, since no external force
has acted on them. Thus if V is the velocity of the rifle,
mv (bullet) + M Ffrifle) =
.'. MV = —mv, or V = —ri v -
M
The momentum of the rifle is thus equal and opposite to that of the bullet.
Further, V/v = -m/M. Since m/M = 50/2000 = 1/40, it follows that
V = — u/40 = 2-5 m s _1 . This means that the rifle moves back or
recoils with a velocity only about ^th that of the bullet.
If it is preferred, one may also say that the explosive force produces
the same numerical momentum change in the bullet as in the rifle.
Thus mv = MV, where V is the velocity of the rifle in the opposite
direction to that of the bullet. The joule (J) is the unit of energy (p. 24).
The kinetic energy, E u of the bullet = \mv 2 = \ . 005 . 100 2 = 250J
The kinetic energy, E 2 , of the rifle = \MV 2 = \ . 2 . 2-5 2 = 6-25 J
Thus the total kinetic energy produced by the explosion = 256-25 J.
The kinetic energy E t of the bullet is thus 250/256-25, or about 98%, of
the total energy. This is explained by the fact that the kinetic energy
depends on the square of the velocity. The high velocity of the bullet
thus more than compensates for its small mass relative to that of the
rifle. See also p. 26.
Rocket
Consider a rocket moving in outer space where no external forces
act on it. Suppose its mass is M and its velocity is v at a particular
instant. Fig. 1.18 (i). When a mass m of fuel is ejected, the mass of the
rocket becomes (M — m) and its velocity increases to (v + Av). Fig. 1.18
(ii).
DYNAMICS
23
(0
(ii)
Velocity u
relative to rocket
■Ov
v+ A v
Fig. 1.18. Motion of rocket
Suppose the fuel is always ejected at a constant speed u relative to
the rocket Then the velocity of the mass m = v+-= — win the direction
of the rocket, since the initial velocity of the rocket is v and the final
velocity is v + Av, an average of v + Ay/2.
We now apply the principle of the conservation of momentum to the rocket
and fuel. Initially, before m of fuel was ejected, momentum of rocket and fuel
inside rocket = Mv.
After m is ejected, momentum of rocket = (M—m) (v +Av) r
I A« \
and momentum of fuel
= "f + T-7
.'. (M — m)(v+Av)+rnw+— — u\ = Mv.
Neglecting the product of m . Av, then, after simplification,
M.&v—mu = 0,
. m _ Av
"M~ u'
Now m = mass of fuel ejected = —AM,
. _AM = Av
M ~ u'
Integrating between limits of M, M and v, v respectively
K M "Jv
i M
e M n u
/. M = M e- (v - V0)lu .
or v = v - u log e (M/M )
When the mass M decreases to MJ2
v = u +ulog e 2.
(1)
(2)
24 ADVANCED LEVEL PHYSICS
Motion of centre of mass
If two particles, masses m t and m 2 , are distances x t ,x 2 respectively
from a given axis, their centre of mass is at a distance x from the axis
given by m t x t +m 2 x 2 = {m-^+m^x. See p. 104. Since velocity, v =dx/dt
generally, the velocity v of the centre of mass in the particular direction
is given by m 1 v l +m 2 v 2 = {m 1 + m 2 )v, where t? l5 v 2 are the respective
velocities of m 1 ,m 2 . The quantity (m 1 v t + m 2 v 2 ) represents the total
momentum of the two particles. The quantity (m x +m 2 )v = Mv, where
M is the total mass of the particles. Thus we can imagine that the total
mass of the particles is concentrated at the centre of mass while they
move, and that the velocity v of the centre of mass is always given by
total momentum = Mv.
If internal forces act on the particles while moving, then, since action
and reaction are equal and opposite, their resultant on the whole body
is zero. Consequently the total momentum is unchanged and hence
the velocity or motion of their centre of mass if unaffected. If an external
force, however, acts on the particles, the total momentum is changed.
The motion of their centre of mass now follows a path which is due to
the external force.
We can apply this to the case of a shell fired from a gun. The centre
of mass of the shell follows at first a parabolic path. This is due to the
external force of gravity, its weight. If the shell explodes in mid-air,
the fragments fly off in different directions. But the numerous internal
forces which occur in the explosion have zero resultant, since action
and reaction are equal and opposite and the forces can all be paired.
Consequently the centre of mass of all the fragments continues to follow
the same parabolic path. As soon as one fragment reaches the ground,
an external force now acts on the system of particles. A different
parabolic path is then followed by the centre of mass
If a bullet is fired in a horizontal direction from a rifle, where is
their centre of mass while the bullet and rifle are both moving?
Work
When an engine pulls a train with a constant force of 50 units
through a distance of 20 units in its own direction, the engine is said
by definition to do an amount of work equal to 50 x 20 or 1000 units,
the product of the force and the distance. Thus if W is the amount of
work,
W = force x distance moved in direction of force.
Work is a scalar quantity; it has no property of direction but only
magnitude. When the force is one newton and the distance moved is
one metre, then the work done is one joule. Thus a force of 50 N moving
through a distance of 10 m does 50 x 10 or 500 joule of work. Note
this is also a measure of the energy transferred to the object.
The force to raise steadily a mass of 1 kg is 1 kilogram force (1 kgf ),
which is about 10 N (see p. 14). Thus if the mass of 1 kg is raised
vertically through 1 m, then, approximately, work done = 10 (N) x 1 (m)
= 10 joule.
DYNAMICS 25
The c.g.s. unit of work is the erg; it is the work done when a
force of 1 dyne moves through 1 cm. Since 1 N = 10 5 dynes and
1 m = 100 cm, then 1 N moving through 1 m does an amount of work =
10 5 (dyne) x 100 (cm) = 10 7 ergs = 1 joule, by definition of the joule
(p. 24).
P cos 9 Work = P cos 9.s
=£ A
Fig. 1.19 Work
Before leaving the topic of 'work', the reader should note carefully
that we have assumed the force to move an object in its own direction.
Suppose, however, that a force P pulls an object a distance s along a
line OA acting at an angle to it, Fig. 1.19. The component of P along
OA is P cos (p. 8), and this is the effective part of P pulling along the
direction OA. The component of P along a direction perpendicular to
OA has no effect along OA. Consequently
work done = P cos 9 x s.
In general, the work done by a force is equal to the product of the force
and the displacement of its point of application in the direction of the
force.
Power
When an engine does work quickly, it is said to be operating at a
high power; if it does work slowly it is said to be operating at a low
power. 'Power' is defined as the work done per second, i.e.,
_ work done
— time taken'
The practical unit of power, the SI unit, is 'joule per second' or
watt (W); the watt is defined as the rate of working at 1 joule per second.
1 horse-power (hp) = 746 W = f kW (approx),
where 1 kW = 1 kilowatt of 1000 watt. Thus a small motor of £ hp
in a vacuum carpet cleaner has a power of about 125 W.
Kinetic Energy
An object is said to possess energy if it can do work. When an object
possesses energy because it is moving, the energy is said to be kinetic,
e.g., a flying stone can disrupt a window. Suppose that an object of
mass m is moving with a velocity u, and is gradually brought to rest in a
distance s by a constant force F acting against it. The kinetic energy
originally possessed by the object is equal to the work done against F,
and hence
kinetic energy = Fxs.
26 ADVANCED LEVEL PHYSICS
But F = ma, where a is the retardation of the object. Hence Fxs =
mas. From v 2 = u 2 + las (see p. 6), we have, since v = and a is negative
in this case,
= u 2 — 2as, i.e., as = -y.
.". kinetic energy = mas = jmw 2 .
When m is in kg and w is in m s _1 , then \mu 2 is in joule. Thus a
car of mass 1000 kg, moving with a velocity of 36 km h~ * or 10 m s" l ,
has an amount W of kinetic energy given by
W=\mu 2 = ±x 1000x10* = 50000 J
Kinetic Energies due to Explosive Forces
Suppose that, due to an explosion or nuclear reaction, a particle of
mass m breaks away from the total mass concerned and moves with
velocity v, and a mass M is left which moves with velocity V in the
opposite direction. Then
kinetic energy, E t , of mass m _ \mv 2 _ mv 2
kinetic energy, E 2 , of mass M \MV 2 MV 2
Now from the principle of the conservation of linear momentum,
mv = MV. Thus v = MV/m. Substituting for v in (1).
. E t _ mM 2 V 2 _ M _ 1/m
" E 2 ~ m 2 MV 2 ~ m~ 1/M'
Hence the energy is ini;erse/y-proportional to the masses of the
particles, that is, the smaller mass, m say, has the larger energy. Thus if
E is the total energy of the two masses, the energy of the smaller
mass = ME/(M + m). An a-particle has a mass of 4 units and a radium
nucleus a mass of 228 units. If disintegration of a thorium nucleus,
mass 232, produces an a-particle and radium nucleus, and a release of
energy of 4-05 MeV, where 1 MeV = 1-6 x 10 _13 J, then
228
energy of a-particle = — — ^r^ x 405 = 3-98 MeV.
(4 + 228)
The a-particle thus travels a relatively long distance before coming to
rest compared to the radium nucleus.
Potential Energy
A weight held stationary above the ground has energy, because, when
released, it can raise another object attached to it by a rope passing
over a pulley, for example. A coiled spring also has energy, which is
released gradually as the spring uncoils. The energy of the weight or
spring is called potential energy, because it arises from the position or
arrangement of the body and not from its motion. In the case of the
DYNAMICS
27
weight, the energy given to it is equal to the work done by the person or
machine which raises it steadily to that position against the force-of
attraction of the earth. In the case of the spring, the energy is equal to
the work done in displacing the molecules from their normal equilibrium
positions against the forces of attraction of the surrounding molecules.
If the mass of an object is m, and the object is held stationary at a
height h above the ground, the energy released when the object falls to
the ground is equal to the work done
= force x distance = weight of object x h.
Suppose the weight is 5 kgf and h is 4 metre. Then, since 1 kgf =
9-8 N = 10 N approx, then
potential energy P.E. = 50 (N) x 4 (m) = 200 J
(more accurately, P.E. = 192 J).
Generally, at a height of h,
potential energy = mgh,
where m is in kg, h is in metre, g = 9-8.
EXAMPLE
Define work, kinetic energy, potential energy. Give one example of each of the
following : (a) the conversion into kinetic energy of the work done on a body
and (b) the conversion into potential energy of the work done on a body.
A rectangular block of mass 10 g rests on a rough plane which is inclined to
the horizontal at an angle sin - * (005). A force of 003 newton, acting in a direc-
tion parallel to a line of greatest slope, is applied to the block so that it moves up
the plane. When the block has travelled a distance of 110 cm from its initial
position, the applied force is removed. The block moves on and comes to rest
again after travelling a further 25 cm Calculate (i) the work done by the applied
force, (ii) the gain in potential energy of the block and (hi) the value of the coefficient
of sliding friction between the block and the surface of the inclined plane. How
would the coefficient of sliding friction be measured if the angle of the slope
could be altered? (0. and C.)
Fig. 1.20 Example
(i) Force = 003 newton; distance = 110 cm = 11m.
.'. work = 003 x 11 = 0033 J.
28 ADVANCED LEVEL PHYSICS
(ii) Gain in P.E. = wt x height moved = 001 kgf x 1-35 sin 6 m,
= 001 x 9-8 newton x 1-35 x 005 m = 00066 J (approx.).
(iii) Work done against frictional force F = work done by force — gain in P.E.
= 0033 -0-0066 = 00264 J.
.-. F x 1-35 = 0-0264.
Normal reaction,
. r 00264
. F = —r^rr- newton.
1-35
R = mg cos = mg (approx.), since 6 is so small
00264
= F =
M R 1-35 x 0-01x9-8
= 0-2 (approx).
Conservative Forces
If a ball of weight W is raised steadily from the grbund to a point X
at a height h above the ground, the work done is W.h. The potential
energy, P.E., of the ball is thus W. h. Now whatever rc|>ute is taken from
ground level to X, the work done is the same — if a longer path is
chosen, for example, the component of the weight in the particular
direction must then be overcome and so the force Required to move
the ball is correspondingly smaller. The P.E. of the ball at X is thus
independent of the route to X. This implies that if the ball is taken in
a closed path round to X again, the total work done is zero. Work has
been expended on one part of the closed path, ^nd
regained on the remaining part.
When the work done in moving round a closed path
in a field to the original point is zero, the forces in the
field are called conservative forces. The earth's gravita-
tional field is an example of a field containing conserva-
tive forces, as we now show.
Suppose the ball falls from a place Y at a heigh|t h
to another X at a height of x above the ground. Fig. 1|21.
Then, if W is the weight of the ball and m its mass,
P.E. at X = Wx = max
and K.E. at X = \mv 2 = \m . 2g(h — x) = mg(h — x\),
using v 2 = las = 2g(h — x). Hence
r mg
ftng x
B
Fig. 1.21.
Mechanical
energy
P.E. + K.E. = mgx + mg(h — x) =
Thus at any point such as X, the total mechanical energy of the falling
ball is equal to the original energy. The mechanica| energy is hence
constant or conserved. This is the case for a conservative field.
Non-Conservative forces. Principle of Conservation of I Energy
The work done in taking a mass m round a closed! path in the con-
servative earth's gravitational field is zero. Fig. 1.2J> (i). If the work
done in taking an object round a closed path to its original position is
DYNAMICS
29
No work done
returning to A
Earth's
field
Closed
patn
Work done
returning ^->.
7777777777777777777777777777777
Conservative field
(•)
Non-conservative field
(ii)
Fig. 1.22 Non-conservative and conservative fields
not zero, the forces in the field are said to be non-conservative. This is
the case, for example, when a wooden block B is pushed round a closed
path on a rough table to its initial position O. Work is therefore done
against friction, both as A moves away from O and as it returns. In a
conservative field, however, work is done during part of the path and
regained for the remaining part.
When a body falls in the earth's gravitational field, a small part of
the energy is used up in overcoming the resistance of the air. This
energy is dissipated or lost as heat — it is not regained in moving the
body back to its original position. This resistance is another example
of the action of a non-conservative force.
Although energy may be transformed from one form to another,
as in the last example from mechanical energy to heat, the total energy
in a given system is always constant. If an electric motor is supplied
with 1000 joule of energy, 850 joule of mechanical energy, 140 joule
of heat energy and 10 joule of sound energy may be produced. This
is called the Principle of the Conservation of Energy and is one of the
key principles in science.
Mass and Energy
Newton said that the 'mass' of an object was 'a measure of the
quantity of matter' in it. In 1905, Einstein showed from his Special
Theory of Relativity that energy is released from an object when its
mass decreases. His mass-energy relation states that if the mass de-
creases by Am kg, the energy released in joule, AW, is given by
AW = Am . c 2 ,
where c is the numerical value of the speed of light inrns" 1 , which is
3 x 10 8 . Experiments in Radioactivity on nuclear reactions showed that
Einstein's relation was true. Thus mass is a form of energy.
Einstein's relation shows that even if a small change in mass occurs, a
30
ADVANCED LEVEL PHYSICS
relatively large amount of energy is produced. Thu(s if Aw = 1 milli-
gramme = 10" 6 kg, the energy AW released
= Am . c 2 = 1(T 6 x (3 x 10 8 ) 2 = 9 x 10 10 J.
This energy will keep 250000 100-W lamps burning for about an hour.
In practice, significant mass changes occur only in nuclear reactions.
The internal energy of a body of mass m may be considered as £ int =
mc 2 , where m is its rest mass. In nuclear reactions where two particles
collide, a change occurs in their total kinetic energy and in their total
mass. The increase in total kinetic energy is accompanied by an equal
decrease in internal energy, Am . c 2 . Thus the total energy, kinetic plus
internal, remains constant.
Before Einstein's mass-energy relation was known, two independent
laws of science were :
(1) The Principle of the Conservation of Mass (the total mass of a
given system of objects is constant even though collisions or other
actions took place between them) ;
(2) The Principle of the Conservation of Energy (the total energy of a
given system is constant). From Einstein's relation, however, the two
laws can be combined into one, namely, the Principle of the Conserva-
tion of Energy.
The summary below may assist the reader ; it refers to the units of
some of the quantities encountered, and their relations.
Quantity
SI
C.G.S.
Relations
Force
(vector)
newton (N)
dyne
10 5 dyne = 1 N
1 kgf = 9-8 N (approx, 10 N)
1 gf = 00098 N
(approx, 0O1 N)
Mass
(scalar)
kilogramme (kg)
gramme (g)
1000g=lkg
Momentum
(vector)
newton second (Ns)
dyne second
10 5 dyn s = 1 N s
Energy
(scalar)
joule (J)
erg
10 7 erg = 1 J
Power
(scalar)
watt (W)
ergs -1
1W = Us -1
1 h.p. = 746 W
Dimensions
By the dimensions of a physical quantity we mean the way it is
related to the fundamental quantities mass, length and time; these
are usually denoted by M, L, and T respectively. An area, length x
breadth, has dimensions L x L or L 2 ; a volume has dimensions L 3 ;
density, which is mass/volume, has dimensions M/L 3 or ML ~ 3 ; relative
density has no dimensions, since it is the ratio of similar quantities, in
this case two masses (p. 114); an angle has no dimensions, since it is the
ratio of two lengths.
As an area has dimensions L 2 , the unit may be written in terms of the
metre as 'm 2 '. Similarly, the dimensions of a volume are L 3 and hence
DYNAMICS 31
the unit is 'm 3 '. Density has dimensions ML -3 . The density of mercury
is thus written as '13600 kgm -3 '. If some physical quantity has
dimensions ML -1 T -1 , its unit may be written as 'kgm -1 s -1 '.
The following are the dimensions of some quantities in Mechanics :
Velocity. Since velocity = — : , its dimensions are L/T or LT -1 .
Acceleration. The dimensions are those of velocity/time, i.e., L/T 2 or
LT -2 .
Force. Since force = mass x acceleration, its dimensions are MLT -2 .
Work or Energy. Since work = force x distance, its dimensions are
ML 2 T" 2 .
EXAMPLE
In the gas equation (p+j^)(V—b) = RT, what are the dimensions of the
constants a and fe?
p represents pressure, V represents volume. The quantity a/V 2 must represent
a pressure since it is added to p. The dimensions of p = [force]/[area] =
MLT 2 /L 2 = ML _1 T~ 2 ; the dimensions of V = L 3 . Hence
^ = ML-'T" 2 , or [a] = ML 5 T~ 2 .
The constant b must represent a volume since it is subtracted from V. Hence
M = l 3 -
Application of Dimensions. Simple Pendulum
If a small mass is suspended from a long thread so as to form a simple
pendulum, we may reasonably suppose that the period, T, of the oscil-
lations depends only on the mass m, the length / of the thread, and the
acceleration, g, due to gravity at the place concerned. Suppose then that
T = km x l y g z .... (i)
where x, y, z, k are unknown numbers. The dimensions of g are LT -2
from above. Now the dimensions of both sides of (i) must be the same.
.-. T = M*L y (LT- 2 ) z .
Equating the indices of M, L, T on both sides, we have
x = 0,
y + z = 0,
and —2z = 1.
Thus, from (i), the period T is given by
T = kfig-*,
or T = k /i
32 ADVANCED LEVEL PHYSICS
We cannot find the magnitude of A; by the method of dimensions, since
it is a number. A complete mathematical investigation shows that
k = 2n in this case, and hence T = Inyjljg. (See also p. 48).
Velocity of Transverse Wave in a String
As another illustration of the use of dimensions, consider a wave
set up in a stretched string by plucking it. The velocity, V, of the wave
depends on the tension, F, in the string, its length /, and its mass m,
and we can therefore suppose that
V = kF x l y rrf, . . . . ' (i)
where x, y, z are numbers we hope to find by dimensions and k is a
constant.
The dimensions of velocity, V, are LT _1 , the dimensions of tension,
F, are MLT~ 2 , the dimension of length, I, is L, and the dimension of
mass, m, is M. From (i), it follows that
LT 1 = (MLT _2 )*xL/xM z .
Equating powers of M, L, and T on both sides,
.'. = x + z, . . . . (i)
1 = x+ v, . . . • (ii)
and -1 = -2x, .... (iii)
• • x = 2' Z = ~ 2' y = 2-
:. V = k.F i l i m' i ,
olV=k m =k ^ =k r Tension
m yJ m/l yl mass per unit length
A complete mathematical investigation shows that k = 1.
The method of dimensions can thus be used to find the relation
between quantities when the mathematics is too difficult. It has been
extensively used in hydrodynamics, for example. See also pp. 176, 181.
EXERCISES 1
(Assume g = 10 m s~ 2 , unless otherwise given)
What are the missing words in the statements 1-10?
1. The dimensions of velocity are . . .
2. The dimensions of force are . . .
3. Using 'vector' or 'scalar', (i) mass is a . . . (ii) force is a . . . (iii) energy is
a . . . (iv) momentum is a . . .
4. Linear momentum is defined as . . .
5. An 'elastic' collision is one in which the . . . and the . . . are conserved.
6. When two objects collide, their ... is constant provided no . . . forces act.
7. One newton x one metre = . . .
DYNAMICS 33
8. 1 kilogram force = . . . newton, approx.
9. The momentum of two different bodies must be added by a . . . method.
10. Force is the ... of change of momentum.
Which of the following answers, A, B, C, D or E, do you consider is the correct
one in the statements 11-14?
11. When water from a hosepipe is incident horizontally on a wall, the force
on the wall is calculated from A speed of water, B mass x velocity, C mass per
second x velocity, D energy of water, E momentum change.
12. When a ball of mass 2 kg moving with a velocity of 10 m s _1 collides
head-on with a ball of mass 3 kg and both move together after collision, the
common velocity is A 5 m s _1 and energy is lost, B4ms _1 and energy is lost,
C 2 m s -1 and energy is gained, D 6 m s _1 and momentum is gained, £6ms"'
and energy is conserved.
13. An object of mass 2 kg moving with a velocity of 4 m s _1 has a kinetic
energy of A 8 joule, B 16 erg, C 4000 erg, D 16 joule, E 40000 joule.
14. The dimensions of work are A ML 2 T -2 and it is a scalar, B ML 2 T -2 and
it is a vector, C MLT -1 and it is a scalar, D ML 2 T and it is a scalar, E MLT and
it is a vector.
15. A car moving with a velocity of 36 km h " 1 accelerates uniformly at 1 m s _ 2
until it reaches a velocity of 54 kmh -1 . Calculate (i) the time taken, (ii) the
distance travelled during the acceleration, (iii) the velocity reached 100 m from
the place where the acceleration began.
16. A ball of mass 100 g is thrown vertically upwards with an initial speed of
72 km h~ x . Calculate (i) the time taken to return to the thrower, (ii) the maximum
height reached, (iii) the kinetic and potential energies of the ball half-way up.
17. The velocity of a ship A relative to a ship B is 100 km h~ * in a direction
N. 45° E. If the velocity of B is 200 km h _1 in a direction N. 60° W., find the
actual velocity of A in magnitude and direction.
18. Calculate the energy of (i) a 2 kg object moving with a velocity of 10 m s " l ,
(ii) a 10 kg object held stationary 5 m above the ground.
19. A 4 kg ball moving with a velocity of 10O ms" 1 collides with a 16 kg ball
moving with a velocity of 40 m s~ * (i) in the same direction, (ii) in the opposite
direction. Calculate the velocity of the balls in each case if they coalesce on
impact, and the loss of energy resulting from the impact. State the principle
used to calculate the velocity.
20. A ship X moves due north at 300 kmh _1 ;a ship Y moves N. 60° W. at
200 kmh -1 . Find the velocity of Y relative to X in magnitude and direction.
If Y is 10 km due east of X at this instant, find the closest distance of approach
of the two ships.
21. Two buckets of mass 6 kg are each attached to one end of a long inexten-
sible string passing over a fixed pulley. If a 2 kg mass of putty is dropped from a
height of 5 m into one bucket, calculate (i) the initial velocity of the system,
(ii) the acceleration of the system, (iii) the loss of energy of the 2 kg mass due to
the impact.
34 ADVANCED LEVEL PHYSICS
22. A bullet of mass 25 g and travelling horizontally at a speed of 200 ms" 1
imbeds itself in a wooden block of mass 5 kg suspended by cords 3 m long.
How far will the block swing from its position of rest before beginning to return?
Describe a suitable method of suspending the block for this experiment and
explain briefly the principles used in the solution of the problem. (L.)
23. State the principle of the conservation of linear momentum and show
how it follows from Newton's laws of motion.
A stationary radioactive nucleus of mass 210 units disintegrates into an alpha
particle of mass 4 units and a residual nucleus of mass 206 units. If the kinetic
energy of the alpha particle is E, calculate the kinetic energy of the residual
nucleus. (N.)
24. Define linear momentum and state the principle of conservation of linear
momentum. Explain briefly how you would attempt to verify this principle by
experiment.
Sand is deposited at a uniform rate of 20 kilogramme per second and with
negligible kinetic energy on to an empty conveyor belt moving horizontally
at a constant speed of 10 metre per minute. Find (a) the force required to maintain
constant velocity, (b) the power required to maintain constant velocity, and (c) the
-rate of change of kinetic energy of the moving sand. Why are the latter two
quantities unequal? (O. & C.)
25. What do you understand by the conservation of energy? Illustrate your
answer by reference to the energy changes occurring (a) in a body whilst falling
to and on reaching the ground, (b) in an X-ray tube.
The constant force resisting the motion of a car, of mass 1500 kg, is equal to
one-fifteenth of its weight If, when travelling at 48 km per hour, the car is brought
to rest in a distance of 50 m by applying the brakes, find the additional retarding
force due to the brakes (assumed constant) and the heat developed in the brakes.
(N.)
26. Define uniform acceleration. State, for each case, one set of conditions
sufficient for a body to describe (a) a parabola, (b) a circle.
A projectile is fired from ground level, with velocity 500 m s -1 at 30° to the
horizontal. Find its horizontal range, the greatest vertical height to which it rises,
and the time to reach the greatest height. What is the least speed with which it
could be projected in order to achieve the same horizontal range? (The resistance
of the air to the motion of the projectile may be neglected.) (O.)
27. Define momentum and state the law of conservation of linear momentum.
Discuss the conservation of linear momentum in the following cases (a) a
freely falling body strikes the ground without rebounding, (b) during free flight
an explosive charge separates an earth satellite from its propulsion unit, (c) a
billiard ball bounces off the perfectly elastic cushion of a billiard table.
A bullet of mass 10 g travelling horizontally with a velocity of 300 ms" 1 strikes
a block of wood of mass 290 g which rests on a rough horizontal floor. After
impact the block and bullet move together and come to rest when the block has
travelled a distance of 15 m. Calculate the coefficient of sliding friction between
the block and the floor. (0. & C.)
28. Explain the distinction between fundamental and derived units, using two
examples of each.
Derive the dimensions of (a) the moment of a couple and work, and comment
on the results, (b) the constants a and b in van der Waals' equation (p + a/v 2 ) (v - b)
= rT for unit mass of a gas. (N.)
DYNAMICS 35
29. Explain what is meant by the relative velocity of one moving object
with respect to another.
A ship A is moving eastward with a speed of 15 km h - 1 and another ship B, at
a given instant 10 km east of A, is moving southwards with a speed of 20 km h~ *.
How long after this instant will the ships be nearest to each other, how far apart
will they be then, and in what direction will B be sighted from A1 (C.)
30. Define momentum and state the law of conservation of linear momentum.
Outline an experiment to demonstrate momentum conservation and discuss
the accuracy which could be achieved.
Show that in a collision between two moving bodies in which no external
act, the conservation of linear momentum may be deduced directly from Newton's
laws of motion.
A small spherical body slides with velocity v and without rolling on a smooth
horizontal table and collides with an identical sphere which is initially at rest on
the table. After the collision the two spheres slide without rolling away from
the point of impact, the velocity of the first sphere being in a direction at 30° to
its previous velocity. Assuming that energy is conserved, and that there are no
horizontal external forces acting, calculate the speed and direction of travel of
the target sphere away from the point of impact. (O. & C.)
31. Answer the following questions making particular reference to
the physical principles concerned (a) explained why the load on the
back wheels of a motor car increases when the vehicle is accelerating,
(b) the diagram, Fig. 1.23, shows a painter in a crate which hangs
alongside a building. When the painter who weighs 100 kgf pulls
on the rope the force he exerts on the floor of the crate is 45 kgf.
If the crate weighs 25 kgf find the acceleration. (N.)
s
£
32. Derive an expression for the kinetic energy of a moving body. Fig. 1-23
A vehicle of mass 2000 kg travelling at 10 ms" 1 on a horizontal surface is
brought to rest in a distance of 12-5 m by the action of its brakes. Calculate the
average retarding force. What horse-power must the engine develop in order
to take the vehicle up an incline of 1 in 10 at a constant speed of 10 m s~ 1 if the
frictional resistance is equal to 20 kgf? (L.)
33. Explain what is meant by the principle of conservation of energy for a
system of particles not acted upon by any external forces. What modifications
are introduced when external forces are operative?
A bobsleigh is travelling at 10 ms" 1 when it starts ascending an incline of
1 in 100. If it comes to rest after travelling 150 m up the slope, calculate the
proportion of the energy lost in friction and deduce the coefficient of friction
between the runners and the snow. (O. & C.)
34. State Newton's Laws of Motion and deduce from them the relation
between the distance travelled and the time for the case of a body acted upon by
a constant force. Explain the units in which the various quantities are measured.
A fire engine pumps water at such a rate that the velocity of the water leaving
the nozzle is 15 m s _1 . If the jet be directed perpendicularly on to a wall and the
rebound of the water be neglected, calculate the pressure on the wall (1 m 3
water weighs 1000 kg). (O. & C.)
chapter two
Circular motion. S.H.M. Gravitation
Angular Velocity
In the previous chapter we discussed the motion of an object moving
in a straight line. There are numerous cases of objects moving in a
curve about some fixed point. The
earth and the moon revolve continu-
ously round the sun, for example, and
the rim of the balance-wheel of a
watch moves to-.and-fro in a circular
path about the fixed axis of the wheel.
In this chapter we shall study the
motion of an object moving in a circle
with a uniform speed round a fixed
point O as centre, Fig. 2.1.
„,,,', ,. If the object moves from A to B
Fig. 2.1 Circular motion , , J ,. _ . , ,
so that the radius OA moves through
an angle 0, its angular velocity, co, about O is defined as the change of
the angle per second. Thus if t is the time taken by the object to move
from A to B,
co = ?. • • • • CD
Angular velocity is usually expressed in 'radian per second' (rad s _1 ).
From (1),
6 = cot (2)
which is analogous to the formula 'distance = uniform velocity x time'
for motion in a straight line. It will be noted that the time T to describe
the circle once, known as the period of the motion, is given by
T = ^, . . . . (3)
co
since 2n radians = 360° by definition.
If s is the length of the arc AB, then s/r = 0, by definition of an angle
in radians.
.-. s = rd.
Dividing by t, the time taken to move from A to B,
• s - = r-
" t V
But s/t = the velocity, v, of the rotating object, and 0/t is the angular
velocity.
/. v = rco . . • • (4)
36
CIRCULAR MOTION
37
Acceleration in a circle
When a stone is attached to a string and whirled round at constant
speed in a circle, one can feel the force in the string needed to keep the
stone moving. The presence of the force, called a centripetal force,
implies that the stone has an acceleration. And since the force acts
towards the centre of the circle, the direction of the acceleration,
which is a vector quantity, is also towards the centre.
To obtain an expression for the acceleration towards the centre,
consider an object moving with a constant speed v round a circle of
radius r. Fig. 2.2 (i). At A, its velocity v A is in the direction of the tangent
AC; a short time St later at B, its velocity v B is in the direction of the
tangent BD. Since their directions are different, the velocity u B is
different from the velocity v A , although their magnitudes are both
equal to v. Thus a velocity change or acceleration has occurred from
A to B.
Vector diagram
R. ~^
Velocity change
Fig. 2.2 Acceleration in circle
The velocity change from A to B =~v B — t A =~^ B +( — ~^a)- The
arrows denote vector quantities. In Fig. 2.2 (ii), PQ is drawn to represent
v B in magnitude (v) and direction (BD); QR is drawn to represent
(— 1>a) in magnitude (v) and direction (CA). Then, as shown on p. 11,
velocity change = ~v B + ( — ~y a) = P^..
When St is small, the angle AOB or SO is small. Thus angle PQR,
equal to SO, is small. PR then points towards O, the centre of the circle.
The velocity change or acceleration is thus directed towards the centre.
The magnitude of the acceleration, a, is given by
a =
velocity change
time
PR
St'
v.se
St "
38 ADVANCED LEVEL PHYSICS
since PR = v .36. In the limit, when St approaches zero, SO/St =
dO/dt = co, the angular velocity. But v = rco (p. 36). Hence, since
a = vco,
V 2 2
a = — or rco .
r
Thus an object moving in a circle of radius r with a constant speed v
has a constant acceleration towards the centre equal to v 2 /r or rco 2 .
Centripetal forces
The force F required to keep an object of mass m moving in a circle
of radius r = ma = mv 2 /r. It is called a centripetal force and acts
towards the centre of the circle. When a stone A is whirled in a hori-
zontal circle of centre O by means of a string, the tension T provides
the centripetal force. Fig. 2.3 (i). For a racing car moving round a
circular track, the friction at the wheels provides the centripetal force.
Planets such as P, moving in a circular orbit round the sun S, have a
centripetal force due to gravitational attraction between S and P
(p. 59). Fig. 2.3(h).
Tension
'A
(0 («)
Fig. 2.3 Centripetal forces
If some water is placed in a bucket B attached to the end of a string,
the bucket can be whirled in a vertical plane without any water falling
out. When the bucket is vertically above the point of support O, the
weight mg of the water is less than the required force mv 2 /r towards
the centre and so the water stays in. Fig. 2.3 (iii). The reaction R of the
bucket base on the water provides the rest of the force. If the bucket is
whirled slowly and mg > mv 2 /r, part of the weight provides the force
mv 2 /r. The rest of the weight causes the water to accelerate downward
and hence to leave the bucket.
Centrifuges
Centrifuges are used to separate particles in suspension from the
less dense liquid in which they are contained. This mixture is poured
into a tube in the centrifuge, which is then whirled at high speed in a
horizontal circle.
The pressure gradient due to the surrounding liquid at a particular
distance, r say, from the centre provides a centripetal force of mrco 2
for a small volume of liquid of mass m, where co is the angular velocity.
CIRCULAR MOTION
39
If the volume of liquid is replaced by an equal volume of particles of
smaller mass rri than the liquid, the centripetal force acting on the
particles at the same place is then greater than that required by
(m-m')rco 2 . The net force urges the particles towards the centre in
spiral paths, and here they collect. Thus when the centrifuge is stopped,
and the container or tube assumes a vertical position, the suspension
is found at the top of the tube and clear liquid at the bottom. For the
same reason, cream is separated from the denser milk by spinning the
mixture in a vessel. The cream spirals towards the centre and collects
here.
Motion of Bicycle Rider Round Circular Track
When a person on a bicycle rides round a circular racing track, the
frictional force F at the ground provides
the inward force towards the centre or
centripetal force. Fig. 2.4. This produces
a moment about his centre of gravity G
which is counterbalanced, when he leans
inwards, by the moment of the normal
reaction R. Thus provided no skidding
occurs, F .h = R.a = mg .a, since R =
mg for no vertical motion.
Centre
. . t = tan = — ,
h mg
mg
Fig. 2.4 Rider on circular track
2,
where 6 is the angle of inclination to the vertical. Now F = mv 2 /r.
.-. tan = — .
rg
When F is greater than the limiting friction, skidding occurs. In this
case F > /img, or mg tan > fimg. Thus tan > fi is the condition
for skidding.
Motion of Car (or Train) Round Circular Track
Suppose a car (or train) is moving with a velocity i; round a horizontal
circular track of radius r, and let R l9 R 2 be the respective normal re-
Fig. 2.5 Car on circular track
40
ADVANCED LEVEL PHYSICS
actions at the wheels A, B, and F lt F 2 the corresponding frictional
forces, Fig. 2.5. Then, for circular motion we have
^1+^2 =
mv
(i)
(ii)
and vertically R t + R 2 = mg
Also, taking moments about G,
(F 1 +F 2 )h + R l a-R 2 a = O . . (iii)
where la is the distance between the wheels, assuming G is mid-way
between the wheels, and h is the height of G above the ground. From
these three equations, we find
1 i v 2 ti
K 2 = im|0+—
and, vertically,
*i =jm\ff-
ra
R 2 never vanishes since it always has a positive value. But if
v 2 = arg/h, R x = 0, and the car is about to overturn outwards. R x will
be positive if v 2 < arg/h.
Motion of Car (or Train) Round Banked Track
Suppose a car (or train) is moving round a banked track in a circular
path of horizontal radius r, Fig. 2.6. If the only forces at the wheels
Centre ^^_ A w
Fig. 2.6 Car on banked track
A, B are the normal reactions R lt R 2 respectively, that is, there is no
side-slip or strain at the wheels, the force towards the centre of the
track is (/^j + R 2 ) sin 0, where is the angle of inclination of the plane
to the horizontal.
,2
.-. (^ 1 + R 2 )sinf? =
mv
For vertical equilibrium, (R x + R 2 ) cos = mg
,2
Dividing (i) by (ii),
.-. tan0 =
rg
(i)
(")
(iii)
CIRCULAR MOTION 41
Thus for a given velocity v and radius r, the angle of inclination of
the track for no side-slip must be tan^ x {v 2 lrg). As the speed v increases,
the angle increases, from (iii). A racing-track is made saucer-shaped
because at higher speeds the cars can move towards a part of the track
which is steeper and sufficient to prevent side-slip. The outer rail of a
curved railway track is raised about the inner rail so that the force
towards the centre is largely provided by the component of the reaction
at the wheels. It is desirable to bank a road at corners for the same
reason as a racing track is banked.
Thrust at Ground
Suppose now that the car (or train) is moving at such a speed thaUhe frictional
forces at A, B are F„ F 2 respectively, each acting towards the centre of the track.
Resolving horizontally,
2
.-. (R i + R 2 )sin0+(F r +F 2 )cos0 = — . . (i)
Resolving vertically,
Solving, we find
(R t + R 2 )cos0- (F t + F 2 ) sin = mg . . (ii)
= ml— cos 0—
F 1 + F 2 = ml— cos 0-0 sin . . (iii)
If — cos > g sin 0, then (F x + F 2 ) is positive ; and in this case both the thrusts
on the wheels at the ground are towards the centre of the track.
If— cos < g sin 0, then (F t +F 2 ) is negative. In this case the forces F t and
F 2 act outwards away from the centre of the track.
For stability, we have, by moments about G,
{Fi + FJh+R^-R^ =
.. (F 1 + F 2 ^ = R 2 -R 1 .
From (iii), .*. — l— cos 9-g sin d\ = R 2 -R t (iv)
The reactions R v R 2 can be calculated by finding (Ri + RJ from equations
(iX (iiX and combining the result with equation (iv). This is left as an exercise to the
student.
Variation of g with latitude
The acceleration due to gravity, g, varies over the earth's surface.
This is due to two main causes. Firstly, the earth is elliptical, with the
polar radius, b, 6-357 x 10 6 metre and the equatorial radius, a,
6-378 x 10 6 metre, and hence g is greater at the poles than at the equator,
where the body is further away from the centre of the earth. Secondly,
the earth rotates about the polar axis, AB. Fig. 2.7. We shall consider
this effect in more detail, and suppose the earth is a perfect sphere.
In general, an object of mass m suspended by a spring-balance at a
42
ADVANCED LEVEL PHYSICS
point on the earth would be acted on by an upward force T = mg',
where g' is the observed or apparent acceleration due to gravity. There
would also be a downward attractive force mg towards the centre of
the earth, where g is the acceleration in the absence of rotation.
(1) At the poles, A or B, there is no rotation. Hence mg — T = 0, or
mg = T = mg'. Thus g' = g.
(2) At the equator, C or D, there is a resultant force mrco 2 towards the
centre where r is the earth's radius. Since OD is the vertical, we have
mrco = mg
mg—T= mr<x> 2 .
:. T = mg
.'. g' = g-rco-
The radius r of the earth is about
6-37 x 10 6 m, and co =
[(2tc/(24 x 3600)] radian per second.
2 6-37 x 10 6 x {In) 1 nMA
g—g— rar = ^- T J — — 0034.
" (24x3600) 2
Latest figures give g, at the pole,
9-832 ms~ 2 , and g', at the equator,
9-780 m s " 2 , a difference of 0052 ms" 2 .
The earth's rotation accounts for 0034 ms" 2 .
Fig. 2.7 Variation of g
Example
EXAMPLE
Explain the action of a centrifuge when used to hasten
the deposition of a sediment from a liquid.
A pendulum bob of mass 1 kg is attached to a string
1 m long and made to revolve in a horizontal circle of
radius 60 cm. Find the period of the motion and the
tension of the string. (C.)
First part. See text, p. 38.
Second part. Suppose A is the bob, and OA is tht
string, Fig. 2.8. If Tis the tension in newton, and 6 is the
angle of inclination of OA to the horizontal, then, for
motion in the circle of radius r = 60 cm = 0-6 m,
Tcos0 =
0-6
Since the bob A does not move in a vertical direction, then
T sin 6 = mg .
Now cos = -£§q = | ; hence sin 6 = f.
From (ii),
1x9-8
(i)
(ii)
T =
mg
sin 6 4/5
= 12-25 newton.
SIMPLE HARMONIC MOTION
43
From (i)
06T cos 6
0-6 x 12-25 x 3
1x5
= 21ms _1
'. angular velocity, co = - = -p— = - rad s *
r 0-6 2
.-. period, r, = ^ = |£ = ^ second.
^ co 1/2 7
.-. T = 1-8 second.
SIMPLE HARMONIC MOTION
Circular
motion
Y'
i Simple
\ harmonic
motion
When the bob of a pendulum moves
to-and-fro through a small angle, the bob
is said to be moving with simple harmonic
motion. The prongs of a sounding tuning
fork, and the layers of air near it, are moving
with simple harmonic motion, and light
waves can be considered due to simple
harmonic variations.
Simple harmonic motion is closely asso-
ciated with circular motion. An example
is shown in Fig. 2.9. This illustrates an
arrangement used to convert the circular
motion of a disc D into the to-and-fro or
simple harmonic motion of a piston P.
The disc is driven about its axle O by a
peg Q fixed near its rim. The vertical
Fig. 2.9
Simple harmonic motion
motion drives P up and down. Any horizontal component of the
motion merely causes Q to move along the slot S. Thus the simple
harmonic motion of P is the projection on the vertical line YY' of the
circular motion of Q.
An everyday example of an opposite conversion of motion occurs in
car engines. Here the to-and-fro or 'reciprocating' motion of the
piston engine is changed to a regular circular motion by connecting
rods and shafts so that the wheels are turned.
Formulae in Simple Harmonic Motion
Consider an object moving round a circle of radius r and centre Z with
a uniform angular velocity co, Fig. 2.10. If CZF is a fixed diameter, the
foot of the perpendicular from the moving object to this diameter moves
from Z to C, back to Z and across to F, and then returns to Z, while the
object moves once round the circle from O in an anti-clockwise direc-
tion. The to-and-fro motion along CZF of the foot of the perpendicular
is defined as simple harmonic motion.
Suppose the object moving round the circle is at A at some instant,
where angle OZA = 0, and suppose the foot of the perpendicular
from A to CZ is M. The acceleration of the object at A is co 2 r, and this
44
ADVANCED LEVEL PHYSICS
v
C
M
y
B_-_-
N
c
,L
,tf
Sy
z'
p\
/q
r\
7*"
E*
F
H
Fig. 2.10 Simple harmonic curve
acceleration is directed along the radius AZ (see p. 37). Hence the
acceleration of M towards Z
= co 2 r cos AZC = <y 2 r sin 0.
But r sin = MZ = y say.
.'. acceleration of M towards Z = co 2 y.
Now <w 2 is a constant.
.'. acceleration ofM towards Z oc distance of M from Z.
If we wish to express mathematically that the acceleration is always
directed towards Z, we must say
acceleration towards Z =
■co 2 y
(1)
The minus indicates, of course, that the object begins to retard as it
passes the centre, Z, of its motion. If the minus were omitted from
equation (1) the latter would imply that the acceleration increases as
y increases, and the object would then never return to its original
position.
We can now form a definition of simple harmonic motion. It is the
motion of a particle whose acceleration is always (i) directed towards a
fixed point, (ii) directly proportional to its distance from that point.
Period, Amplitude. Sine Curve
The time taken for the foot of the perpendicular to move from C to F
and back to C is known as the period (T) of the simple harmonic motion.
In this time, the object moving round the circle goes exactly once
round the circle from C; and since co is the angular velocity and 2n
radians (360°) is the angle described, the period T is given by
T =
In
co
(1)
The distance ZC, or ZF, is the maximum distance from Z of the foot
of the perpendicular, and is known as the amplitude of the motion. It is
equal to r, the radius of the circle.
We have now to consider the variation with time, t, of the distance,
SIMPLE HARMONIC MOTION 45
y, from Z of the foot of the perpendicular. The distance y = ZM =
r sin 9. But 6 = cot, where co is the angular velocity.
.'. y = r sin cot . . . . (2)
The graph of y v. t is shown in Fig. 2.10, where ON represents the
y-axis and OS the t-axis; since the angular velocity of the object moving
round the circle is constant, 6 is proportional to the time t. Thus as the
foot of the perpendicular along CZF moves from Z to C and back to Z,
the graph OLP is traced out; as the foot moves from Z to F and returns
to Z, the graph PHQ is traced out. The graph is a sine curve. The
complete set of values of v from O to Q is known as a cycle. The
number of cycles per second is called the frequency. The unit '1 cycle
per second' is called '1 hertz (Hz)\ The mains frequency in Great
Britain is 50 Hz or 50 cycles per second.
Velocity during S.H.IVL
Suppose the object moving round the circle is at A at some instant,
Fig. 2.10. The velocity of the object is rco, where r is the radius of the
circle, and it is directed along the tangent at A. Consequently the
velocity parallel to the diameter FC at this instant = rco cos 0, by
resolving.
.'. velocity, v, of M along FC = rco cos 0.
But y = r sin
:. cos = 71 -sin 2 = 71 ~y 2 /r 2 = ± y Jr 2 -y 2
:. v = co^r 2 ~y 2 (1)
This is the expression for the velocity of an object moving with simple
harmonic motion. The maximum velocity, v m , corresponds to y = 0,
and hence
v m = cor. . . . . (2)
Summarising our results:
(1) If the acceleration a of an object = — co 2 y, where v is the distance
or displacement of the object from a fixed point, the motion is simple
harmonic motion.
(2) The period, T, of the motion = 2n/co, where T is the time to make
a complete to-and-fro movement or cycle. The frequency,/, = \/T and
its unit is 'Hz'.
(3) The amplitude, r, of the motion is the maximum distance on
either side of the centre of oscillation.
(4) The velocity at any instant, v, = cOyJr 2 -y 2 ; the maximum
velocity = cor. Fig. 2.11 (i) shows a graph of the variation of v and
acceleration a with displacement y, which are respectively an ellipse
and a straight line.
46
ADVANCED LEVEL PHYSICS
Amplitude
— K
/?=0
XL
Top of
' oscillation
mg =/no>a
(H)
Fig. 2. 1 1 Simple harmonic motion
S.H.M. and g
If a small coin is placed on a horizontal platform connected to a
vibrator, and the amplitude is kept constant as the frequency is in-
creased from zero, the coin will be heard 'chattering' at a particular
frequency f . At this stage the reaction of the table with the coin
becomes zero at some part of every cycle, so that it loses contact
periodically with the surface. Fig. 2.11 (ii).
The maximum acceleration in S.H.M. occurs at the end of the oscilla-
tion because the acceleration is directly proportional to the displace-
ment. Thus maximum acceleration = co 2 a, where a is the amplitude
and co is 2nf .
The coin will lose contact with the table when it is moving down
with acceleration g (Fig. 2. 1 1 (ii) ). Suppose the amplitude is 80 cm. Then
{27if ) 2 a=g
:.4n 2 f 2 xOW = 9-S
•*- /o= >/4^x 8 0-08 =1 " 8Hz -
Damping of S.H.M .
In practice, simple harmonic variations of a pendulum, for example,
will die away as the energy is dissipated by viscous forces due to the air.
The oscillation is then said to be damped. In the absence of any damping
forces the oscillations are said to be free.
A simple experiment to investigate the effect of damping is illustrated
in Fig. 2.12 (i). A suitable weight A is suspended from a helical spring S,
a pointer P is attached to S, and a vertical scale R is set up behind P.
The weight A is then set pulled down and released. The period, and the
time taken for the oscillations to die away, are noted.
As shown in Fig. 2.12 (ii), A is now fully immersed in a damping
medium, such as a light oil, water or glycerine. A is then set oscillating,
SIMPLE HARMONIC MOTION
47
Free I
oscillation 1
Damped
oscillation
(i) ('»)
Fig. 2.12 Experiment on damped oscillations
and the time for oscillations to die away is noted. It is shorter than
before and least for the case of glycerine. The decreasing amplitude in
successive oscillations may also be noted from the upward limit of
travel of P and the results plotted.
Fig. 2.13 (i), (ii) shows how damping produces an exponential fall
in the amplitude with time.
Time
Time
Free oscillation
(0
Damped oscillation
(ii)
Fig. 2. 1 3 Free and damped oscillations
The experiment works best for a period of about ^-second and a
weight which is long and thin so that the damping is produced by
non-turbulent fluid flow over the vertical sides. During the whole
cycle, A must be totally immersed in the fluid.
EXAMPLE
A steel strip, clamped at one end, vibrates with a frequency of 20 Hz and
an amplitude of 5 mm at the free end, where a small mass of 2 g is positioned.
48
ADVANCED LEVEL PHYSICS
Find (a) the velocity of the end when passing through the zero position, (b) the
acceleration at maximum displacement, (c) the maximum kinetic and potential
energy of the mass.
Suppose y = r sin cot represents the vibration of the strip where r is the ampli-
tude.
(a) The velocity, v, = co^J^—y 2 (p. 45). When the end of the strip passes
through the zero position v = 0; and the maximum speed, v m , is given by
v m = cor.
Now co = 2nf=2nx 20, and r = 0-005 m.
.-. v m = 2% x 20 x 0005 = 0628 ms" 1 .
(b) The acceleration = — co 2 y = — co 2 r at the maximum displacement.
.'. acceleration = (2n x 20) 2 x 0O05
= 79ms -2 .
(c) m = 2g = 2xl0~ 3 kg,t) m = 0-628 ms -1 .
.-. maximum K.E. = %mv m 2 = \ x (2 x 10~ 3 ) x 0-628 2 = 3-9 x 10~ 4 J (approx.).
Maximum P.E. (v = 0) = Maximum K.E. = 3-9 x 10 ~ 4 J.
Simple Pendulum
We shall now study some cases of simple har-
monic motion. Consider a simple pendulum, which
consists of a small mass m attached to the end
of a length / of wire, Fig. 2.14. If the other end of
the wire is attached to a fixed point P and the
mass is displaced slightly, it oscillates to-and-fro
along the arc of a circle of centre P. We shall
now show that the motion of the mass about its
original position O is simple harmonic motion.
Suppose that the vibrating mass is at B at
some instant, where OB = y and angle OPB = 6.
At B, the force pulling the mass towards O is
directed along the tangent at B, and is equal to
mgsintf. The tension, T, in the wire has no
component in this direction, since PB is perpen-
dicular to the tangent at B. Thus, since force =
mass x acceleration (p. 13),
— mg sin = ma,
where a is the acceleration along the arc OB; the minus indicates that
the force is towards O, while the displacement, y, is measured along the
arc from O in the opposite direction. When is small, sin 6 == in
radians ; also = y/l. Hence,
y
r-mg0 = — mgj = ma
v m g
Fig. 2.14
Simple pendulum
Q 2
:. a = -yy = -co 2 y,
SIMPLE HARMONIC MOTION
49
where co 2 = g/l. Since the acceleration is proportional to the distance
y from a fixed point, the motion of the vibrating mass is simple harmonic
motion (p. 50). Further, from p. 50, the period T = 2n/co.
T =
4^ = 2* ~
yfg/i Vg
(1)
At a given place on the earth, where g is constant, the formula shows
that the period T depends only on the length, /, of the pendulum.
Moreover, the period remains constant even when the amplitude of the
vibrations diminish owing to the resistance of the air. This result was
first obtained by Galileo, who noticed a swinging lantern one day, and
timed the oscillations by his pulse. He found that the period remained
constant although the swings gradually diminished in amplitude.
Determination of g by Simple Pendulum
The acceleration due to gravity, g, can be found by measuring the
period, T, of a simple pendulum corresponding to a few different lengths,
/, from 80 cm to 180 cm for example. To perform the experiment
accurately : (i) Fifty oscillations
should be timed, (ii) a small
I angle of swing is essential^less
than 10°, (hi) a small sphere
should be tied to the end of a
thread to act as the mass, and
its radius added to the length
of the thread to determine /.
A graph of / againstJ^is now
s ' plotted from the results, and a
straight line AB,jyJiich should
£2 pass through th6 origin, is then
drawn to lie evenly between the
Fig. 2. 1 5 Graph of / v. T 2 points, Fig. 2. 1 5.
+•*•
A + ^
Now
T = 2* J 1 -,
a-2j
T 2 =
AnH
9
g = 4n 2 x— 2 .
(1)
The gradient a/b of the line AB is the magnitude of l/T 2 ; and by sub-
stituting in (1), g can then be calculated.
If the pendulum is suspended from the ceiling of a very tall room and
the string and bob reaches nearly to the floor, then one may proceed
to find g by (i) measuring the period Ti, (ii) cutting off a measured
length a of the string and determining the new period T 2 with the
50 ADVANCED LEVEL PHYSICS
shortened string. Then, if h is the he ight of t he ceiling above the bob
initially, T x = liiyjhfg and T 2 = 2nyJ{h-a)/g. Thus
h = ^\ and h-a = ^X.
4tt An 2
_ An 2 a
■ • 9 ~ T 2 t2'
J l ~ l 2
Thus g can be calculated from a, T x and T 2 . Alternatively, th e period T
can be measured for several lengths a. Then, since T = 2ny](h— a/g,
h - a = J> T 2
Atc
A graph of a v. T 2 is thus a straight line whose gradient is g/4n 2 .
Hence g can be found. The intercept on the axis of a, when T 2 = 0,
is h, the height of the ceiling above the bob initially.
The Spiral Spring or Elastic Thread
When a weight is suspended from the end of a spring or an elastic
thread, experiment shows that the extension of the spring, i.e., the
increase in length, is proportional to the weight, provided that the
elastic limit of the spring is not exceeded (see p. 181).
Generally, then, the tension (force), T, in a spring is
proportional to the extension x produced, i.e., T = kx,
where k is a constant of the spring.
Consider a spring or an elastic thread PA of length /
suspended from a fixed point P, Fig. 2.16. When a mass
m is placed on it, the spring stretches to O by a length
e given by
mg = ke, . . (l)
since the tension in the spring is then mg. If the mass is
pulled down a little and then released, it vibrates up-
and-down above and below O. Suppose at an instant
that B is at a distance x below O. The tension T of the
spring at B is then equal to k(e + x), and hence the force
towards O = k(e+x)—mg. Since force = massx
acceleration, Fig. 2.16
. ' . — [k(a + x) — mg~\ = ma, Spiral spring
the minus indicates that the net force is upward at this instant, whereas
the displacement x is measured from O in the opposite direction at the
same instant. From this equation,
— ke — kx + mg = ma.
But, from (i), mg = ke,
.'. —kx = ma,
k 2
. . a = x = —cox,
m
mg
SIMPLE HARMONIC MOTION 51
where co 2 = k/m. Thus the motion is simple harmonic about O, and
the period T is given by
T-^-2«/=. . . . (1)
co \ k
Also, since mg = ke, it follows that m/k = e/g.
.. . . . (2)
From (1), it follows that T 2 = 4n 2 m/k. Consequently a graph of
T 2 v. m should be a straight line passing through the origin. In
practice, when the load m is varied and the corresponding period T is
measured, a straight line graph is obtained when T 2 is plotted against
m, thus verifying indirectly that the motion of the load was simple
harmonic. The graph does not pass through the origin, however,
owing to the mass and the movement of the various parts of the spring.
This has not been taken into account in the foregoing theory and we
shall now show how g may be found in this case.
Determination of g by Spiral Spring
The mass s of a vibrating spring is taken into account in addition to the mass
m suspended at the end, theory beyond the scope of this book then shows that
the period of vibration, T, is given by
m+ks
where k is approximately | and k is the elastic constant of the spring. Squaring (i)
and re-arranging,
:.^T 2 = m+ks .... (ii)
Thus, since k, k, s are constants, a graph of T 2 v. m should be a straight line
when m is varied and T observed. A straight line graph verifies indirectly that the
motion of the mass at the end of the spring is simple harmonic. Further, the
magnitude of k/4n 2 can be found from the slope of the line, and hence k can be
calculated.
If a mass M is placed on the end of the spring, producing a steady extension e
less than the elastic limit, then Mg — ke.
■■9 = ~xk . . . . . (Hi)
By attaching different masses to the spring, and measuring the corresponding
extension, the magnitude of e/M can be found by plotting ev.M and measuring
the slope of the line. This is called the 'static' experiment on the spring. From
the magnitude of k obtained in the 'dynamic' experiment when the period was
determined for different loads, the value of g can be found by substituting the
magnitudes of e/M and k in (iii).
52
ADVANCED LEVEL PHYSICS
Fig. 2.17
S.H.M. of liquid
Oscillations of a Liquid in a U-Tube
If the liquid on one side of a U-tube T is de-
pressed by blowing gently down that side, the
levels of the liquid will oscillate for a short
time about their respective initial positions O,
C, before finally coming to rest, Fig. 2.17.
The period of oscillation can be found by
supposing that the level of the liquid on the left
side of T is at D at some instant, at a height x
above its original (undisturbed) position O. The
level B of the liquid on the other side is then at
a depth x below its original position C, and hence
the excess pressure on the whole liquid, as shown
on p. 110,
= excess height x density of liquid x g
= 2xpg.
Now pressure = force per unit area.
.'. force on liquid = pressure x area of cross-section of the tube
= 2xpg x A,
where A is the cross-sectional area of the tube.
This force causes the liquid to accelerate. The mass of liquid in the
U-tube = volume x density = 2hAp, where 2h is the total length of
the liquid in T. Now the acceleration, a, towards O or C is given by
force = mass x a.
.'. — IxpgA = lhApa.
The minus indicates that the force towards O is opposite to the dis-
placement measured from O at that instant.
Q i
.'. a = — T x = —cox,
h
where co 2 =. j. The motion of the liquid about O (or C) is thus simple
h
harmonic, and the period T is given by
co \jg
P.E. and K.E. exchanges in oscillating systems
We can now make a general point about oscillations and oscillating
systems. As an illustration, suppose that one end of a spring S of
negligible mass is attached to a smooth object A, and that S and A
are laid on a horizontal smooth table. If the free end of S is attached to
the table and A is pulled slightly to extend the spring and then released,
the system vibrates with simple harmonic motion. This is the case
discussed on p. 50, without taking gravity into account. The centre
of oscillation O is the position of the end of the spring corresponding
SIMPLE HARMONIC MOTION
53
to its natural length, that is, when the spring is neither extended or
compressed. If the spring extension obeys the law force = kx, where k
is a constant, and m is the mass of A, then, as on p. 51, it can easily
be shown that the period T of oscillation is given by :
co y k
The energy of the stretched spring is potential energy, P.E.— its
molecules are continually displaced or compressed relative to their
normal distance apart. The P.E. for an extension x = ]F . dx =
\kx .dx = jkx 2 .
The energy of the mass is kinetic energy, K.E., or jmv 2 , where v is
the velocity. Now from x = a sin cot, v = dx/dt = coa cos cot.
.'. total energy of spring plus mass = \kx 2 +%mv 2
= jka 2 sin 2 cot +%mco 2 a 2 cos 2 cot.
But co 2 = k/m, or k = mco 2 .
.'. total energy = %mco 2 a 2 (sin 2 cot+cos 2 cot) = \rmo 2 a 2 = constant.
Energy
Total energy
End
N— <
Cycle
Fig. 2. 1 8 Energy of S.H.M.
Thus the total energy of the vibrating mass and spring is constant.
When the K.E. of the mass is a maximum (energy = jmco 2 a 2 and
mass passing through the centre of oscillation), the P.E. of the spring
is then zero (x = 0). Conversely, when the P.E. of the spring is a maxi-
mum (energy = \ka 2 = \mco 2 a 2 and mass at end of the oscillation),
the K.E. of the mass is zero (v = 0). Fig. 2.18 shows the variation of
P.E. and K.E. with displacement x ; the force F extending the spring,
also shown, is directly proportional to the displacement from the
centre of oscillation.
The constant interchange of energy between potential and kinetic
energies is essential for producing and maintaining oscillations, what-
ever their nature In the case of the oscillating bob of a simple pendulum,
54 ADVANCED LEVEL PHYSICS
for example, the bob loses kinetic energy after passing through the
middle of the swing, and then stores the energy as potential energy as
it rises to the top of the swing. The reverse occurs as it swings back.
In the case of oscillating layers of air when a sound wave passes,
kinetic energy of the moving air molecules is converted to potential
energy when the air is compressed. In the case of electrical oscillations,
a coil L and a capacitor C in the circuit constantly exchange energy;
this is stored alternately in the magnetic field of L and the electric
field of C.
EXAMPLES
1. Define simple harmonic motion and state the relation between displacement
from its mean position and the restoring force when a body executes simple
harmonic motion.
A body is supported by a spiral spring and causes a stretch of 1-5 cm in the
spring. If the mass is now set in vertical oscillation of small amplitude, what is the
periodic time of the oscillation? (L.)
First part. Simple harmonic motion is the motion of an object whose accelera-
tion is proportional to its distance from a fixed point and is always directed
towards that point. The relation is: Restoring force = — k x distance from fixed
point, where k is a constant.
Second part. Let m be the mass of the body in kg. Then, since 1-5 cm = 0015 m
m0 = fcxO-O15 ..... (i)
where k is a constant of the spring in N m~ l . Suppose the vibrating body is x m
below its original position at some instant and is moving downwards. Then since
the extension is (x + 001 5) m, the net downward force
= mg-k(x + 0-015)
= mg— /cxO-015 — kx = — kx
from (i). Now mass x acceleration = force.
.'. m x acceleration = —kx
.'. acceleration = x.
m
But, from (i), - = ^
acceleration = - x = — co 2 x,
0-015
where co 2 = 0/0-015.
0015
•■■****-* -*jT-*n
= 0-25 second.
2. A small bob of mass 20 g oscillates as a simple pendulum, with amplitude
5 cm and period 2 seconds. Find the velocity of the bob and the tension in the
supporting thread, when the velocity of the bob is a maximum.
First part. See text.
SIMPLE HARMONIC MOTION 55
Second part. The velocity, v, of the bob is a maximum when it passes through
its original position. With the usual notation (see p. 45), the maximum velocity
v m is given by
v m = cor,
where r is the amplitude of 005 m. Since T = 2n/<o,
.'. v m = (oa = 7rx005 = 016ms" 1 .
Suppose P is the tension in the thread. The net force towards the centre of the
circle along which the bob moves is then given by {P-mg). The acceleration
towards the centre of the circle, which is the point of suspension, is v m 2 /l, where
/ is the length of the pendulum.
2
:.P-m t+ m f
^4
. gT 2 gxA
' An 2 An 2
9
K 2
Now
Since m = 002 kg, g = 9-8 m s 2 , it follows from above that
P = 002x9-8 + 0<)2x ^ 2x7r2
= 19-65 x 10" 2 newton
Waves. Wave equation
Waves and their properties can be demonstrated by producing them
on the surface of water, as in a ripple tank. As the wave travels outwards
from the centre of disturbance, it reaches more distant particles of
water at a later time. Thus the particles of water vibrate out of phase
with each other while the wave travels. It should be noted that the
vibrating particles are the origin of the wave. Their mean position
remains the same as the wave travels, but like the simple harmonic
oscillators previously discussed, they store and release energy which is
handed on from one part of the medium to another. The wave shows
the energy travelling through the medium.
If the displacement y of a vibrating particle P is represented by
y = a sin cot, the displacement of a neighbouring particle Q can be
represented by y = a sin (cot + cf>). cf> is called the phase angle between
the two vibrations. If <j> = n/2 or 90°, the vibration of Q is y = a sin
(cot+n/2). In this case, y = when t = for P, but y — a sin n/2 — a
when t = for Q. Comparing the two simple harmonic variations,
it can be seen that Q leads on P by a quarter of a period.
If the wave is 'frozen' at different times, the displacements of the
various particles will vary according to their position or distance x
56 ADVANCED LEVEL PHYSICS
from some chosen origin such as the centre of disturbance. Now the
wavelength, X, of a wave is the distance between successive crests or
troughs. At these points the phase difference is 2n. Consequently the
phase angle for a distance x is (x/X) x In or 2tzx/X. The wave equation,
which takes x into account as well as the time t, can thus be written as :
y = a sin 2^-2^1 = asin27r --| . . (1)
Other forms of the wave equation may be used. The velocity v of
a wave is the distance travelled by the disturbance in 1 second. If the
frequency of the oscillations is f, then / waves travel outwards in 1
second. Each wave occupies a length X. Hence v = fX. Further, the
period Tis the time for 1 oscillation. Thus/= 1/Tand hence v =fX =
X/T. Substituting for T in (1), the wave equation may also be written as :
y = asm-r-(vt—x) ... (2)
A
The wave equation in (1) or (2) is a progressive wave. The energy of
the wave travels outwards through the medium as time goes on.
Longitudinal and transverse waves
Waves can be classified according to the direction of their vibrations.
A longitudinal wave is one produced by vibrations parallel to the direc-
tion of travel of the wave. An example is a sound wave. The layers of
air are always vibrating in a direction parallel to the direction of travel
of the wave. A longitudinal wave can be seen travelling in a 'Slinky'
coil when one end is fixed and the other is pulled to-and-fro in the
direction of the coil.
A transverse wave is one produced by vibrations perpendicular to
the direction of travel of the wave. Light waves are transverse waves.
The wave along a bowed string of a violin is a transverse wave.
Velocity of waves
There are various types of waves. A longitudinal wave such as a
sound wave is a mechanical wave. The speed v with which the energy
travels depends on the restoring stress after particles in the medium are
strained from their original position. Thus v depends on the modulus
of elasticity of the medium It also depends on the inertia of the particles,
of which the mass per unit volume or density p is a measure.
By dimensions, as well as rigorously, it can be shown that
/modulus of elasticity
v =
For a solid, the modulus is Young's modulus, E. Thus v = yJE/p.
For a liqu id or gas, the modulus is the bulk modulus, k. Hence
v = yjk/p. In air, k = yp, where y is the ratio of the principal specific
heats of air and p is the atmospheric pressure. Thus v = yjyp/p (p. 163).
When a taut string is plucke d or b owed, the velocity of the transverse
wave along it is given by v = y/T/m, where T is the tension and m is the
SIMPLE HARMONIC MOTION 57
mass per unit length of the string. In this case T provides the restoring
force acting on the displaced particles of string and m is a measure of
their inertia.
Electromagnetic waves, which are due to electric and magnetic
vibrations, form an important group of waves in nature. Radio waves,
infra-red, visible and ultra-violet light, X-rays, and y-rays are all
electromagnetic waves, ranging from long wavelength such as 1000
metres (radio waves) to short wavelengths such as 10~ 8 m (y-waves).
Unlike the mechanical waves, no material medium is needed to carry
the waves. The speed of all electromagnetic waves in a vacuum is the
same, about 3 x 10 8 metre per second. The speed varies with wave-
length in material media and this explains why dispersion (separation
of colours) of white light is produced by glass.
Stationary waves
The equation y = asm.2n(t/T— x/X) represents a progressive wave
travelling in the x-direction. A wave of the same amplitude and
frequency travelling in the opposite direction is represented by the
same form of equation but with — x in place of x, that is, by y =
a sin 2n(t/T+ x/X).
The principle of superposition states that the combined effect or
resultant of two waves in a medium can be obtained by adding the
displacements at each point due to the respective waves. Thus if the
displacement due to one wave is represented by y t , and that due to the
other wave by y 2 , the resultant displacement y is given by
v = yi+y 2 = asinln Vf~j +asin2^ -+|
t x t
= la sm 2n— . cos 2n— = A sin 2n-=„
where A = 2a cos 2nx/X.
A represents the amplitude at different points in the medium.
When x = 0, y = A; when x = A/4, A = 0; when x = A/2, y = -A;
when x = — 3A/4, y = 0. Thus at some points called antinodes, A,
the amplitude of vibration is a maximum. At points half-way between
the antinodes called nodes, N, the amplitude is zero, that is, there is no
vibration here. Fig. 2.19 (i). This type of wave, which stays in one place
in a medium, is called a stationary or standing wave. Stationary waves
may be produced which are either longitudinal or transverse.
Fig. 2.19 Stationary and progressive waves
58 ADVANCED LEVEL PHYSICS
Unlike the progressive wave, where the energy travels outwards
through the medium, Fig. 2.19 (ii), the energy of the stationary wave
remains stored in one part of the medium. Stationary waves are
produced in musical instruments when they are played. Stationary
radio waves are also produced in receiving aerials. Stationary waves,
due to electron motion, are believed to be present around the nucleus
of atoms.
Interference. Diffraction
A stationary wave is a special case of interference between two waves.
Another example occurs when two tuning forks of nearly equal fre-
quency are sounded together. A periodic variation of loud sounds
called 'beats' is then heard. They are due to the periodic variation of
the amplitude of the resultant wave. If two very close coherent sources
of light are obtained, interference between the two waves may produce
bright and dark bands.
Diffraction is the name given to the interference between waves
coming from coherent sources on the same undivided wavefront. The
effect is pronounced when a wave is incident on a narrow opening
whose width is of comparable order to the wavelength. The wave now
spreads out or is 'diffracted' after passing through the slit. If the width
of the slit, however, is large compared with the wavelength, the wave
passes straight through the opening without any noticeable diffraction.
This is why visible light, which has wavelengths of the order of 6 x
10~ 7 m, passes straight through wide openings and produces sharp
shadows; whereas sound, which has wavelengths over a million times
longer and of the order of say 0-5 m, can be heard round corners.
Further details of wave phenomena are discussed in the Sound and
Optics sections of the book.
GRAVITATION
Kepler's Laws
The motion of the planets in the heavens had excited the interest of
the earliest scientists, and Babylonian and Greek astronomers were
able to predict their movements fairly accurately. It Was considered
for some time that the earth was the centre of the universe, but about
1542 Copernicus suggested that the planets revolved round the sun as
centre. A great advance was made by Kepler about 1609. He had
studied for many years the records of observations on the planets made
by Tycho Brahe, and he enunciated three laws known by his name.
These state :
(1) The planets describe ellipses about the sun as one focus.
(2) The line joining the sun and the planet sweeps out equal areas in
equal times.
GRAVITATION 59
(3) The squares of the periods of revolution of the planets are
proportional to the cubes of their mean distances from the sun.
The third law was announced by Kepler in 1619.
Newton's Law of Gravitation
About 1666, at the early age of 24, Newton discovered a universal
law known as the law of gravitation.
He was led to this discovery by considering the motion of a planet
moving in a circle round the sun S as centre. Fig. 2.20 (i). The force
acting on the planet of mass m is mrco 2 , where r is the radius of the
circle and co is the angular velocity of the motion (p. 38). Since co = 2n/T,
where T is the period of the motion,
. , A (2n\ 2 4n 2 mr
force on planet = w(-^r = t2 •
This is- equal to the force of attraction of the sun on the planet.
Assuming an inverse-square law, then, if k is a constant,
(0 (ii).
Fig. 2.20 Satellites
force
on planet
km
~ r 2 -
. km
•• r 2
An 2 mr
- T 2
.'. T 2
An 2 3
.-. T 2 oc r\
since k, n are constants.
Now Kepler had announced that the squares of the periods of
revolution of the planets are proportional to the cubes of their mean
distances from the sun (see above). Newton thus suspected that the force
between the sun and the planet was inversely proportional to the square of
the distance between them. The great scientist now proceeded to test
the inverse-square law by applying it to the case of the moon's motion
60 ADVANCED LEVEL PHYSICS
round the earth. Fig. 2.20 (ii). The moon has a period of revolution, T,
about the earth of approximately 27-3 days, and the force on it = mRco 2 ,
where jR is the radius of the moon's orbit and m is its mass.
J2n\ 2 An 2 mR
. . force = mR y = T% .
If the planet were at the earth's surface, the force of attraction on it
due to the earth would be mg, where g is the acceleration due to gravity.
Fig. 2.20 (ii). Assuming that the force of attraction varies as the inverse
square of the distance between the earth and the moon,
. An 2 mR 1 1
-^f^- :mg = R- 2: 7 2 '
where r is the radius of the earth.
. An 2 R r 2
T 2 g R 2 '
An 2 R 3
r 2 T
^n is. , 1X
9 = ~^FT .... (1)
Newton substituted the then known values of .R, r, and T, but was
disappointed to find that the answer for g was not near to the observed
value, 9-8 m s~ 2 . Some years later, he heard of a new estimate of the
radius of the moon's orbit, and on substituting its value he found that
the result for g was close to 9-8 m s -2 . Newton saw that a universal law
could be formulated for the attraction between any two particles of
matter. He suggested that : The force of attraction between two given
masses is inversely proportional to the square of their distance apart.
Gravitational Constant, G, and its Determination
From Newton's law, it follows that the force of attraction, F,
between two masses m, M at a distance r apart is given by F oc — j~.
:.r~&£ (2)
where G is a universal constant known as the gravitational constant.
This expression for F is Newton's law of gravitation.
From (2\ it follows that G can be expressed in 'N m 2 kg -2 '. The
dimensions of G are given by
MLT~ 2 x L 2
M 2
-IT 3t-2
[G]= \*2 =M-*L 3 T
Thus the unit of G may also be expressed as m 3 kg -1 s~ 2 .
A celebrated experiment to measure G was carried out by C. V.
Boys in 1895, using a method similar to one of the earliest determina-
tions of G by Cavendish in 1798. Two identical balls, a, b, of gold,
GRAVITATION
61
©
_Torsion
(quartz)
wire
C D
b*
JL
•a
Plan view
Fig. 2.21 Experiment on G
5 mm in diameter, were suspended by a long and a short fine quartz
fibre respectively from the ends, C, D, of a highly-polished bar CD,
Fig. 2.21. Two large identical lead spheres, A, B, 1 15 mm in diameter,
were brought into position near a, b respectively. As a result of the
attraction between the masses, two equal but opposite forces acted on
CD. The bar was thus deflected, and the angle of deflection, 0, was
measured by a lamp and scale method by light reflected from CD. The
high sensitivity of the quartz fibres enabled the small deflection to be
measured accurately, and the small size of the apparatus allowed it to
be screened considerably from air convection currents.
Calculation for G
Suppose d is the distance between a, A, or b, B, when the deflection
is 0. Then if m, M are the respective masses of a, A,
torque of couple on CD = G^f x CD.
a
But torque of couple = cO,
where c is the torque in the torsion wire per unit radian of twist (p. 192).
. . G-^2- x CD = cO.
G =
cdd 2
mMxCD
(1)
The constant c was determined by allowing CD to oscillate through a
small angle and then observing its period of oscillation, T, which was
of the order of 3 minutes. If / is the known moment of inertia of the
system about the torsion wire, then (see p. 75),
T=2n
62 ADVANCED LEVEL PHYSICS
The constant c can now be calculated, and by substitution in (i), G can
be determined. Accurate experiments showed that G = 6-66 xlO -11
N m 2 kg" 2 and Heyl, in 1942, found G to be 6-67 x 10" n N m 2 kg -2 .
Mass and Density of Earth
At the earth's surface the force of attraction on a mass m is mg,
where g is the acceleration due to gravity. Now it can be shown that it
is legitimate in calculations to assume that the mass, M, of the earth
is concentrated at its centre, if it is a sphere. Assuming that the earth is
spherical and of radius r, it then follows that the force of attraction of
the earth on the mass m is GmM/r 2 .
■ ■ G~pr = mg.
GM
Now, g = 9-8 m s~ 2 , r = 6-4 x 10 6 m, G = 6-7 x 1(T X1 N m 2 kg -2 .
a# 9-8x(6-4xl0 6 ) 2 , A in24 ,
'■ M = 6-7x10-" = 60 >< 1Q24k g-
The volume of a sphere is 4nr 3 /3, where r is its radius. Thus the
density, p, of the earth is approximately given by
- — = gr2 = 3flf
p ~ V ~ 47tr 3 G/3 AnrG'
By substituting known values of g, G, and r, the mean density of the
earth is found to be about 5500 kg m -3 . The density may approach a
value of 10000 kg m _ 3 towards the interior.
It is now believed that gravitational force travels with the speed of
light Thus if the gravitational force between the sun and earth were
suddenly to disappear by the vanishing of the sun, it would take about
7 minutes for the effect to be experienced on the earth. The earth would
then fly off along a tangent to its original curved path.
Gravitational and inertial mass
The mass m of an object appearing in the expression F = ma,
force — mass x acceleration, is the inertial mass, as stated on p. 13.
It is a measure of the reluctance of the object to move when forces act
on it. It appears in F = ma from Newton's second law of motion.
The 'mass' of the same object concerned in Newton's theory of
gravitational attraction can be distinguished from the inertial mass.
This is called the gravitational mass. If it is given the symbol m ff then
F g = GMmJr 2 , where F g is the gravitational force, M is the mass of
the earth and r its radius. Now GM/r 2 = g, the acceleration due to
gravity (see above). Thus F g = m g g = W, the weight of the object.
GRAVITATION
63
In the simple pendulum theory on p. 48, we can derive the period
T using W = weight = m % g in place of the symbols adopted there.
Thus
or
-Wg0y = ma,
a = —
m^g
ml
CO V m 8^
Experiments show that to a high degree of accuracy, T = lityflfg no
matter what mass is used, that is, the period depends only on / and g.
Thus m = m g , or the gravitational mass is equal to the inertial mass to
the best of our present knowledge.
Mass of Sun
The mass M s of the sun can be found from the period of a satellite
and its distance from the sun. Consider the case of the earth. Its period
T is about 365 days or 365 x 24 x 3600 seconds. Its distance r s from
the centre of the sun is about 1-5 x 10 11 m. If the mass of the earth is
m, then, for circular motion round the sun,
GM*m , mrAn 2
, 2 - mr SP> = ^2~ »
2, 3
11\3
M c =
47r 2 x(l-5xl0 11 )
4n 2 r,
GT 2 ~ 6-7 x 10" x J x (365 x 24 x 3600) 2
= 2 x 10 30 kg.
Orbits round the earth
Satellites can be launched from the earth's surface to circle the earth.
They are kept in their orbit by the gravitational attraction of the earth.
Orbit 2
Fig. 2.22 Orbits round earth
Consider a satellite of mass m which just circles the earth of mass M
64 ADVANCED LEVEL PHYSICS
close to its surface in an orbit 1. Fig. 2.22 (i). Then, if r is the radius of
the earth,
mv 2 „Mm
= G—r = "W»
r r
where g is the acceleration due to gravity at the earth's surface and
v is the velocity of m in its orbit. Thus v 2 = rg, and hence, using r =
64 x 10 6 m and g = 9-8 m s~ 2 ,
v = Jr~g = ^6-4 x 10* x 9-8 = 8 x 10 3 m s" 1 (approx),
= 8kms _1 .
The velocity v in the orbit is thus about 8 kms -1 . In practice, the
satellite is carried by a rocket to the height of the orbit and then given
an impulse, by firing jets, to deflect it in a direction parallel to the
tangent of the orbit (see p. 66). Its velocity is boosted to 8 km s _1 so
that it stays in the orbit. The period in orbit
circumference of earth _ 2n x 6-4 x 10 6 m
= v 8xl0 3 ms _1
= 5000 seconds (approx) = 83 min.
Parking Orbits
Consider now a satellite of mass m circling the earth in the plane of
the equator in an orbit 2 concentric with the earth. Fig. 2.22 (ii). Suppose
the direction of rotation as the same as the earth and the orbit is at a
distance R from the centre of the earth. Then if v is the velocity in orbit,
mv 2
GMm
R
~ R 2 *
ut
GM
= gr 2 ,
, where r is
i the radius of the earth.
mv 2
~R~
mgr 2
~ R 2
.-. v 2
gr 2
R'
Tis
the period of the satellite in
its orbit, then v
= 2nR/T.
4n 2 R 2
gr 2
• T 2
R
/. T 2
4n 2 R 3
gr 2
(i)
If the period of the satellite in its orbit is exactly equal to the period
of the earth as it turns about its axis, which is 24 hours, the satellite
will stay over the same place on the earth while the earth rotates. This
GRAVITATION
65
is sometimes called a 'parking orbit'. Relay satellites can be placed
in parking orbits, so that television programmes can be transmitted
continuously from one part of the world to another. Syncom was a
satellite used for transmission of the Tokio Olympic Games in 1964.
Since T = 24 hours, the radius R can be found from (i). Thus from
R =
T 2 gr 2
An 2
and g = 9-8 m s 2 , r = 6-4 x 10 6 m,
,«-?
(24 x 3600) 2 x 9-8 x (6-4 x 10 6 )
6\2
An 2
= 42400 km
The height above the earth's surface of the parking orbit
= R-r = 42 400-6 400 = 36 000 km.
In the orbit, the velocity of the satellite
= 2nR = 2n x 42 400
T 24 x 3600 seconds
= 31 km s
Weightlessness
When a rocket is fired to launch a spacecraft and astronaut into
orbit round the earth, the initial acceleration must be very high owing
to the large initial thrust required. This acceleration, a, is of the order
of 15g, where g is the gravitational acceleration at the earth's surface.
Suppose S is the reaction of the couch to which the astronaut is
initially strapped Fig. 2.23 (i). Then, from F = ma, S~mg = ma =
mA5g, where m is the mass of the astronaut. Thus S = 16mg. This
force is 16 times the weight of the astronaut and thus, initially, he
experiences a large force.
5=1 6 mg
Spacecraft
mg
Orbit
a=15gf
/ v \
Rocket' mg ^
(') (H)
Fig. 2.23 Weight and weightlessness
In orbit, however, the state of affairs is different. This time the
acceleration of the spacecraft and astronaut are both g' in magnitude,
66 ADVANCED LEVEL PHYSICS
where g' is the acceleration due to gravity outside the spacecraft at the
particular height of the orbit. Fig. 2.23 (ii). If S' is the reaction of the
surface of the spacecraft in contact with the astronaut, then, for circular
motion, ,
F = mg—S = ma = mg .
Thus S' = 0. Consequently the astronaut becomes 'weightless'; he
experiences no reaction at the floor when he walks about, for example.
At the earth's surface we feel the reaction at the ground and are thus
conscious of our weight. Inside a lift which is falling fast, the reaction
at our feet diminishes. If the lift falls freely, the acceleration of objects
inside is the same as that outside and hence the reaction on them is
zero. This produces the sensation of 'weightlessness'. In orbit, as in
Fig. 2.23 (ii), objects inside a spacecraft are also in 'free fall' because
they have the same acceleration g' as the spacecraft. Consequently the
sensation of weightlessness is experienced.
EXAMPLE
A satellite is to be put into orbit 500 km above the earth's surface. If its vertical
velocity after launching is 2000 m s _1 at this height, calculate the magnitude and
direction of the impulse required to put the satellite directly into orbit, if its mass
is 50 kg. Assume g = 10 m s~ 2 ; radius of earth, R = 6400 km.
Suppose u is the velocity required for orbit, radius r. Then, with usual notation,
mu 2 GmM gR 2 m GM
= — ^— = ■> , as -=-?- = g.
r r 2 r 2 ' R 2 y
r
NowK = 6400 km, r = 6900 km, g = 10ms~ 2 .
2 10x(6400xl0 3 ) 2
• • " _ 6900 xlO 3 '
.. u = 7700 m s _1 (approx.).
At this height, vertical momentum
U Y = mv = 50x 2000 = 100000 kg m s
Fig 2 24 F IG - 2.24 Example
Horizontal momentum required U x = mu = 50 x 7700 = 385 000 kg m s~ *.
-i
.-. impulse needed, U, = S /U Y 2 + U X 2 = ^lOO 000 2 + 385 000 2
= 40xl0 5 kgms _1 . (1)
Direction. The angle 6 made by the total impulse with the horizontal or orbit
tangent is given by tan 6 = U Y /U X = 100000/385 000 = 0-260. Thus 6 = 14-6°.
Magnitudes of acceleration due to gravity
(i) Above the earth's surface. Consider an object of mass m in an
orbit of radius R from the centre, where R > r, the radius of the earth.
Then, if g' is the acceleration due to gravity at this place,
, GmM {i >.
mg =-^2- . . • • (i)
GRAVITATION
67
But, if g is the acceleration due to gravity at the earth's surface,
GmM ,..,
mg = —p- .... (n)
a' r 2 r 2
Dividing (1) by (ii), .\ ^ = -^, or g' = —j . g.
g k k
Thus above the earth's surface, the acceleration due to gravity g' varies
inversely as the square of the distance from the centre. Fig. 2.25.
For a height h above the earth, R ■ = r + h.
.2 !
9 =
(r+h) 2
9 =
-r
9-
since powers of (h/rf and higher can be neglected when h is small
compared with r.
g—g' = reduction in acceleration due to gravity.
2h
-J 9
(1)
k {/(surface )
Assuming uniform /
\ of earth
density inside ^ /
earth ^y
w
NP/
N \ s
7
Inverse-
^^^^^square law
Fig. 2.25 Variation of g
(ii) Below the earth's surface. Consider an object of mass m at a point
below the earth's surface. If its distance from the centre is b, the 'effect
tive' mass M' of the earth which attracts it is that contained in a sphere
of radius b. Assuming a constant density, then, since the mass of a
sphere is proportional to radius 3 ,
b 3
r*
68 ADVANCED LEVEL PHYSICS
where M is the mass of the earth. Suppose g" is the acceleration due to
gravity at the radius b. Then, from above,
„ GmW GmMb
Since GM/r 2 = g, it follows by substitution that
9 =~9-
Thus assuming a uniform density of core, which is not the case in
practice, the acceleration due to gravity g" is directly proportional
to the distance from the centre. Fig. 2.25.
If the depth below the earth's surface is h, then b = r-h.
■ '■ 9" =
■ '■ 9-9" = '^9 ( 2 )
Comparing (1) and (2), it can be seen that the acceleration at a
distance h below the earth's surface is greater than at the same distance
h above the earth's surface.
Potential
The potential, V, at a point due to the gravitational field of the
earth is defined as numerically equal to the work done in taking a unit
mass from infinity to that point. This is analogous to 'electric potential'.
The potential at infinity is conventionally taken as zero.
For a point outside the earth, assumed spherical, we can imagine the whole
mass M of the earth concentrated at its centre. The force of attraction on a unit
mass outside the earth is thus GM/r 2 , where r is the distance from the centre.
The work done by the gravitational force in moving a distance dr towards the
earth = force x distance = GM.dr/r 2 . Hence the potential at a point distant
a from the centre is given by
[ a GM J
if the potential at infinity is taken as zero by convention. The negative sign indi-
cates that the potential at infinity (zero) is higher than the potential close to the
earth.
On the earth's surface, of radius r, we therefore obtain
V=-^ (2)
r
Velocity of Escape. Suppose a rocket of mass m is fired from the earth's surface
Q so that it just escapes from the gravitational influence of the earth. Then
work done = m x potential difference between infinity and Q.
GM
= mx .
r
, , i 2 GM
.". kinetic energy of rocket = \mv J - = mx— — .
2GM
= velocity of escape.
GRAVITATION
69
Now
GM/r 2 = g.
:. v = J2gr.
:. v = 72 x 9-8 x 6-4 x 10 6 = llxl0 3 ms _1 = llkms -1 (approx).
With an initial velocity, then, of about 11 km s _1 , a rocket will
completely escape from the gravitational attraction of the earth. It
can be made to travel towards the moon, for example, so that eventually
it comes under the gravitational attraction of this planet. At present,
'soft' landings on the moon have been made by firing retarding retro
rockets.
Hyperbola (v>\^)
Parabola (v=v £ )
v='/2p?=11 kms
Ellipse (v<v E )
-1
Circle (v=W= 8 km s~ 1 )
Fig. 2.26 Orbits
Summarising, with a velocity of about 8 km s _1 , a satellite can
describe a circular orbit close to the earth's surface (p. 64). With a
velocity greater than 8 km s _1 but less than 11 km s _1 , a satellite
describes an elliptical orbit round the earth. Its maximum and minimum
height in the orbit depends on its particular velocity. Fig. 2.26 illustrates
the possible orbits of a satellite launched from the earth.
The molecules of air at normal temperatures and pressures have an
average velocity of the order of 480 m s _1 or 0-48 kms" 1 which is
much less than the velocity of escape/Many molecules move with
higher velocity than 0-48 km s _1 but gravitational attraction keeps the
atmosphere round the earth. The gravitational attraction of the moon
is much less than that of the earth and this accounts for the lack of
atmosphere round the moon.
EXERCISES 2
(Assume g = 10 ms -2 )
What are the missing words in the statements 1-6?
1. The force towards the centre in circular motion is called the . . . force.
2. In simple harmonic motion, the maximum kinetic energy occurs at the . . .
of the oscillation.
70 ADVANCED LEVEL PHYSICS
3. The constant of gravitation G is related to g by . . .
4. In simple harmonic motion, the maximum potential energy occurs at the
... of the oscillation.
5. Outside the earth, the acceleration due to gravity is proportional to . . .
from the centre.
6. A satellite in orbit in an equatorial plane round the earth will stay at the
same place above the earth if its period is . . . hours.
Which of the following answers, A, B, C, D or E, do you consider is the correct
one in the statements 7-10?
7. The earth retains its atmosphere because A the earth is spherical, B the
velocity of escape is greater than the mean speed of molecules, C the constant of
gravitation is a universal constant, D the velocity of escape is less than the mean
speed of molecules, E gases are lighter than solids.
8. In simple harmonic motion, the moving object has A only kinetic energy,
B mean kinetic energy greater than the mean potential energy, C total energy
equal to the sum of the maximum kinetic energy and maximum potential energy,
D mean kinetic energy equal to the mean potential energy, E minimum potential
energy at the centre of oscillation.
9. If r is the radius of the earth and g is the acceleration at its surface, then the
acceleration g" at an orbit distance R from the centre of the earth is given by
A g'/g = r/R, B g'/g = r 2 /R 2 , C g'/g = R 2 /r 2 , D g'/g = (R-r) 2 /r 2 , E g'/g =
(R-r)/r.
10. When water in a bucket is whirled fast overhead, the water does not fall out
at the top of the motion because A the centripetal force on the water is greater
than the weight of water, B the force on the^water is opposite to gravity, C the
reaction of the bucket on the water is zero, B the centripetal force on the water
is less than the weight of water, E atmospheric pressure counteracts the weight.
Circular Motion
11. An object of mass 4 kg moves round a circle of radius 6 m with a constant
speed of 12 m s _1 . Calculate (i) the angular velocity, (ii) the force towards the
centre.
12. An object of mass 10 kg is whirled round a horizontal circle of radius
4 m by a revolving string inclined to the vertical. If the uniform speed of the
object is 5 m s -1 , calculate (i) the tension in the string in kgf, (ii) the angle of
inclination of the string to the vertical. x
13. A racing-car of 1000 kg moves round a banked track at a constant speed
of 108 km h~ l . Assuming the total reaction at the wheels is normal to the track,
and the horizontal radius of the track is 100 m, calculate the angle of inclination
of the track to the horizontal and the reaction at the wheels.
14. An object of mass 80 kg is whirled round in a vertical circle of radius 2m
with a constant velocity of 6 ms _1 . Calculate the maximum and minimum
tensions in the string.
15. Define the terms (a) acceleration, and (b) force. Show that the acceleration
of a body moving in a circular path of radius r with uniform speed v is v 2 /r, and
draw a diagram to show the direction of the acceleration.
A small body of mass m is attached to one end of a light inelastic string of
GRAVITATION , 71
length /. The other end of the string is fixed. The string is initially held taut and
horizontal, and the body is then released. Find the values of the following quanti-
ties when the string reaches the vertical position : (a) the kinetic energy of the
body, (b) the velocity of the body, (c) the acceleration of the body, and (d) the
tension in the string. (0. & C)
16. Explain what is meant by angular velocity. Derive an expression for the
force required to make a particle of mass m move in a circle of radius r with
uniform angular velocity w.
A stone of mass 500 g is attached to a string of length 50 cm which will break
if the tension in it exceeds 20 kgf. The stone is whirled in a vertical circle, the
axis of rotation being at a height of 100 cm above the ground. The angular speed
is very slowly increased until the string breaks. In what position is this break
most likely to occur, and at what angular speed? Where will the stone hit the
ground? (G.)
Simple Harmonic Motion
17. An object moving with simple harmonic motion has an amplitude of 2 cm
and a frequency of 20 Hz. Calculate (i) the period of oscillation, (ii) the accelera-
tion at the middle and end of an oscillation, (iii) the velocities at the corresponding
instants.
18. Calculate the length in centimetres of a simple pendulum which has a
period of 2 seconds. If the amplitude of swing is 2 cm, calculate the velocity and
acceleration of the bob (i) at the end of a swing, (ii) at the middle, (iii) 1 cm from
the centre of oscillation.
19. Define simple harmonic motion. An elastic string is extended 1 cm when a
small weight is attached at the lower end. If the weight is pulled down £ cm and
then released, show that it moves with simple harmonic motion, and find the
period.
20. A uniform wooden rod floats upright in water with a length of 30 cm
immersed. If the rod is depressed slightly and then released, prove that its motion
is simple harmonic and calculate the period.
21. A simple pendulum, has a period of 4-2 seconds. When the pendulum is
shortened by 1 m, the period is 3-7 seconds. From these measurements, calculate
the acceleration due to gravity and the original length of the pendulum.
22. What is simple harmonic motion"? Show how it is related to the uniform
motion of a particle with velocity v in a circle of radius r.
A steel strip, clamped at one end, vibrates with a frequency of 50 Hz and an
amplitude of 8 mm at the free end. Find (a) the velocity of the end when passing
through the zero position, (b) the acceleration at the maximum displacement.
23. Explain what is meant by simple harmonic motion.
Show that the vertical oscillations of a mass suspended by a light helical spring
are simple harmonic and describe an experiment with the spring to determine the
acceleration due to gravity.
A small mass rests on a horizontal platform which vibrates vertically in simple
harmonic motion with a period of 0-50 second. Find the maximum amplitude of
the motion which will allow the mass to remain in contact with the platform
throughout the motion. (L.)
24. Define simple harmonic motion and state a formula for its period. Show
that under suitable conditions the motion of a simple pendulum is simple har-
monic and hence obtain an expression for its period.
72 ADVANCED LEVEL PHYSICS
If a pendulum bob is suspended from an inaccessible point, by a string whose
length may be varied, describe how to determine (a) the acceleration due to
gravity, (b) the height of the point of suspension above the floor.
How and why does the value of the acceleration due to gravity at the poles
differ from its value at the equator? (L.)
25. Derive an expression for the time period of vertical oscillations of small
amplitude of a mass suspended from the free end of a light helical spring.
What deformation of the wire of the spring occurs when the mass moves? (N.)
26. Give two practical examples of oscillatory motion which approximate to
simple harmonic motion. What conditions must be satisfied if the approximations
are to be good ones.
A point mass moves with simple harmonic motion. Draw on the same axes
sketch graphs to show the variation with position of (a) the potential energy,
(b) the kinetic energy, and (c) the total energy of the particle.
A particle rests on a horizontal platform which is moving vertically in simple
harmonic motion with an amplitude of 10 cm. Above a certain frequency, the
thrust between the particle and the platform would become zero at some point
in the motion. What is this frequency, and at what point in the motion does the
thrust become zero at this frequency? (C.)
27. In what circumstances will a particle execute simple harmonic motion?
Show how simple harmonic motion can be considered to be the projection on
the diameter of a circle of the motion of a particle describing the circle with uniform
speed.
The balance wheel of a watch vibrates with an angular amplitude of n radians
and a period of 0-5 second. Calculate (a) the maximum angular speed, (b) the
angular speed when the displacement is n/2, and (c) the angular acceleration
when the displacement is n/4. If the radius of the wheel is r, calculate the maximum
radial force acting on a small dust particle of mass m situated on the rim of the
wheel. (O. & C.)
28. Prove that the bob of a simple pendulum may move with simple harmonic
motion, and find an expression for its period.
Describe with full details how you would perform an experiment, based on the
expression derived, to measure the value of the acceleration due to gravity.
What factors would influence your choice of (a) the length of the pendulum,
(b) the material of the bob, and (c) the number of swings to be timed? (O. & C.)
29. Define simple harmonic motion and show that the free oscillations of a
simple pendulum are simple harmonic for small amplitudes.
Explain what is meant by damping of oscillations and describe an experiment
to illustrate the effects of damping on the motion of a simple pendulum. Briefly
discuss the difficulties you would encounter and indicate qualitatively the results
you would expect to observe. (O. & C.)
30. What is meant by simple harmonic motion? Obtain an expression for the
kinetic energy of a body of mass m, which is performing S.H.M. of amplitude a
and period 2n/co, when its displacement from the origin is x.
Describe an experiment, or experiments, to verify that a mass oscillating at
the end of a helical spring moves with simple harmonic motion. (C.)
31. State the dynamical condition under which a particle will describe simple
harmonic motion. Show that it is approximately fulfilled in the case of the bob of
a simple pendulum, and derive, from first principles, an expression for the period
of the pendulum.
GRAVITATION 73
Explain how it can be demonstrated from observations on simple pendulums,
that the weight of a body at a given place is proportional to its mass. (O. & C.)
32. Define simple harmonic motion. Show that a heavy body supported by a
light spiral spring executes simple harmonic motion when displaced vertically
from its equilibrium position by an amount which does not exceed a certain value
and then released How would you determine experimentally the maximum
amplitude for simple harmonic motion?
A spiral spring gives a displacement of 5 cm for a load of 500 g. Find the
maximum displacement produced when a mass of 80 g is dropped from a height
of 10 cm on to a light pan attached to the spring. (N.)
Gravitation
33. Calculate the force of attraction between two small objects of mass 5 and
8 kg respectively which are 10 cm apart. (G = 6-7 x 10 -11 N m 2 kg -2 .)
34. If the acceleration due to gravity is 9-8 m s~ 2 and the radius of the earth is
6400 km, calculate a value for the mass of the earth. (G = 6-7 x 10 ~ "Nm 2 kg -2 .}
Give the theory.
35. Assuming that the mean density of the earth is 5500 kg m~ 3 , that the
constant of gravitation is 6-7 x 10" u Nm 2 kg - 2 , and that the radius of the earth
is 6400 km, find a value for the acceleration due to gravity at the earth's surface.
Derive the formula used.
36. How do you account for the sensation of 'weightlessness' experienced by
the occupant of a space capsule (a) in a circular orbit round the earth, {b) in outer
space? Give one other instance in which an object would be 'weightless'. (N.)
37. State Newton's law of universal gravitation. Distinguish between the
gravitational constant (G) and the acceleration due to gravity (g) and show the
relation between them.
Describe an experiment by which the value of g may be determined. Indicate
the measurements taken and how to calculate the result. Derive any formula
used. (L.)
38. State Newton's law of gravitation. What experimental evidence is there
for the validity of this law?
A binary star consists of two dense spherical masses of 10 30 kg and 2 x 10 30 kg
whose centres are 10 7 km apart and which rotate together with a uniform angular
velocity co about an axis which intersects the line joining their centres. Assuming
that the only forces acting on the stars arise from their mutual gravitational
attraction and that each mass may be taken to act at its centre, show that the
axis of rotation passes through the centre of mass of the system and find the
value of co. (G = 6-7 x 10" u m 3 kg" 1 s" 2 .) (O. & C.)
39. Assuming that the planets are moving in circular orbits, apply Kepler's
laws to show that the acceleration of a planet is inversely proportional to the
square of its distance from the sun. Explain the significance of this and show
clearly how it leads to Newton's law of universal gravitation.
Obtain the value of g from the motion of the moon, assuming that its period of
rotation round the earth is 27 days 8 hours and that the radius of its orbit is 60-1
times the radius of the earth (Radius of earth = 6-36 x 10 6 m.) (N.)
40. Explain what is meant by the gravitation constant (G), and describe an
accurate laboratory method of measuring it. Give an outline of the theory of
your method.
74 ADVANCED LEVEL PHYSICS
Assuming that the earth is a sphere of radius 6370 km and that G = 6-66 x 10" 1 *
N m 2 kg -2 , calculate the mean density of the earth. (O. & C.)
41. Assuming the earth to be perfectly spherical, give sketch graphs to show
how (a) the acceleration due to gravity, {b) the gravitational potential due to the
earth's mass, vary with distance from the surface of the earth for points external
to it. If any other assumption has been made, state what it is.
Explain why, even if the earth were a perfect sphere, the period of oscillation of
a simple pendulum at the poles would not be the same as at the equator. Still
assuming the earth to be perfectly spherical, discuss whether the velocity required
to project a body vertically upwards, so that it rises to a given height, depends on
the position on the earth from which it is projected. (C.)
42. Explain what is meant by the constant oj gravitation. Describe a laboratory
experiment to determine it, showing how the result is obtained from the observa-
tions.
A proposed communication satellite would revolve round the earth in a
circular orbit in the equatorial plane, at a height of 35880 km above the earth's
surface. Find the period of revolution of the satellite in hours, and comment
on the result. (Radius of earth = 6370 km, mass of earth = 5-98 x 10 24 kg
constant of gravitation = 6-66 x 10 -11 N m 2 kg -2 .) (N.)
chapter three
Rotation of Rigid Bodies
So far in this book we have considered the equations of motion and
other dynamical formulae associated with a particle. In practice,
however, an object is made of millions of particles, each at different
places, and we need now to consider the motion of moving objects.
Moment of Inertia, I
Suppose a rigid object is rotating about a fixed
axis O, and a particle A of the object makes an
angle with a fixed line OY in space at some in-
stant, Fig. 3.1. The angular velocity, d0/dt or co,
of every particle about O is the same, since we
are dealing with a rigid body, and the velocity v t
of A at this instant is given by r x co, where r t =
OA. Thus the kinetic energy of A = ^m^ 2 =
\m x r 2 co 2 . Similarly, the kinetic energy of another
particle of the body = \m 2 r 2 2 co 2 , where r 2 is
its distance from O and m 2 is its mass. In this
way we see that the kinetic energy, K.E., of the
whole object is given by
Fig. 3.1
Rotating rigid body
K.E. = \m 1 r 2 co 2 +\m 2
r 2 co 2 +jm 3 r 3 2 co 2 + ,
= %co 2 (m 1 r 1 2 + m 2 r 2 2 + m 3 r 3 2 + ...)
= io> 2 (2mr 2 ),
where Ymr 2 represents the sum of the magnitudes of ''mr 2 ' for all the
particles of the object. We shall see shortly how the quantity 2,mr 2
can be calculated for a particular object. The magnitude of l.mr 2 is
known as the moment of inertia of the object about the axis concerned,
and we shall denote it by the symbol J. Thus
Kinetic energy, K.E., = \lco 2 . . . (1)
The units of / are kg metre 2 (kgm 2 ). The unit of co is 'radian s -1 '
(rad s" 1 ). Thus if J = 2 kg m 2 and co = 3 rad s~ *, then
K.E. = \lco 2 = | x 2 x 3 2 joule = 9 J.
The kinetic energy of a particle of mass m moving with a velocity v
is \mv 2 . It will thus be noted that the formula for the kinetic energy of a
rotating object is similar to that of a moving particle, the mass m being
replaced by the moment of inertia J and the velocity uj>eifig replaced
by the angular velocity co. As we shall require values of J, the moment
of inertia of several objects about a particular axis will first be calculated.
75
76
ADVANCED LEVEL PHYSICS
Moment of Inertia of Uniform Rod
(1) About axis through middle. The moment of inertia of a small
element Sx about an axis PQ through its centre O perpendicular to the
length = 1-^Mbc 2 , where / is the length of the rod, M is its mass, and x
is the distance of the small element from O, Fig. 3.2.
~M-
I IT
u £ Sx .
Fig. 3.2 Moment of inertia — uniform rod
moment of inertia, /, =
2MC 112 2j Ml
= —r- 1 x z dx =
1 Jo
12
(1)
Thus if the mass of the rod is 60 g and its length is 20 cm, M = 6 x 10" 2
kg, / = 0-2 m, and / = 6 x 10" 2 x 0-2 2 /12 = 2 x 10~ 4 kg m 2 .
(2) About the axis through one end, A. In this case, measuring dis-
tances x from A instead of O,
moment of inertia, /, =
r Mxx 2 = ^-
(2)
Moment of Inertia of Ring
Every element of the ring is the same distance from the centre.
Hence the moment of inertia about an axis through the centre per-
pendicular to the plane of the ring = Ma 2 , where M is the mass of the
ring and a is its radius.
Moment of Inertia of Circular Disc
Consider the moment of inertia of a circular
disc about an axis through its centre perpen-
dicular to its plane, Fig. 3.3. If we take a small
ring of the disc enclosed between radii x and
2,7LX$X
x + Sx, its mass = 5— M , where a is the
7t<r
radius of the disc and M is its mass. Each
element of the ring is distant x from the Centre,
and hence the moment of inertia of the ring
about the axis through O = I — Ml x x 2
Fig. 3.3
Moment of inertia-
disc
na
ROTATION OF RIGID BODIES 77
.'. moment of inertia of whole disc = I — ^AjMxx 2
■r
Thus if the disc weighs 60 g and has a radius of 10 cm, M = 60 g =
6x10 2 kg, a = 01 m, so that / = 6x KT 2 x0-l 2 /2 = 3xl0 _4 kgm 2 .
Moment of Inertia of Cylinder
If a cylinder is solid, its moment of inertia about the axis of symmetry
is the sum of the moments of inertia of discs into which we may imagine
the cylinder cut. The moment of inertia of each disc = \ mass x a 2 ,"
where a is the radius; and hence, if M is the mass of the cylinder,
moment of inertia of solid cylinder = \Ma 2 (i)
If a cylinder is hollow, its moment of inertia about the axis of
symmetry is the sum of the moments of inertia of the curved surface
and that of the two ends, assuming the cylinder is closed at both ends.
Suppose a is the radius, h is the height of the cylinder, and a is the mass
per unit area of the surface. Then
mass of curved surface = Inaha,
and moment of inertia about axis = mass x a 2 = 2na 3 ha,
since we can imagine the surface cut into rings.
The moment of inertia of one end of the cylinder = mass x a 2 /2 =
na 2 a x a 2 /2 = na 4 a/2. Hence the moment of inertia of both ends =
ncfio.
:. moment inertia of cylinder, /, = 27ta 3 ha+na*o:
The mass of the cylinder, M, = 2naha + 2na 2 a
■ r — 2na 3 ha + na A o M
2naha + 2na 2 a
= 2a 2 h + a 3 M
2h + 2a
If a hollow and a solid cylinder have the same mass M and the
same radius and height, it can be seen from (i) and (ii) that the moment
of inertia of the hollow cylinder is greater than that of the solid cylinder
about the- axis of symmetry. This is because the mass is distributed
on the average at a greater distance from the axis in the former case.
78
ADVANCED LEVEL PHYSICS
Moment of Inertia of Sphere
The moment of inertia of a sphere about
an axis PQ through its centre can be found
by cutting thin discs such as S perpendicular
to the axis, Fig. 3.4. The volume of the disc,
of thickness Sy and distance j; from the centre,
= nr 2 Sy = n{a 2 -y 2 )dy.
'. mass M' of disc =
n(a 2 -y 2 )Sy
W/3
M
Fig. 3.4
Moment of inertia — sphere
3M, 2 2\s
= 4? <fl y)Sy '
where M is the mass of the sphere and a is its radius, since the volume
of the sphere = 4na 3 /3. Now the moment of inertia of the disc about
PQ
„ radius 2 3M. 2 ,, s (a 2 -y 2 )
" - — = x->(« -y )*y x ~^—
= M'x
.'. moment of inertia of sphere =
4a ;
3M
8a 3
\y-
2a 2 y 2 + y*)dy
= \Ma 2
(1)
Thus if the sphere has mass 4 kg and a radius of 0-2 m, the moment
of inertia = f x 4 x 0-2 2 = 0064 kg m 2 .
Radius of Gyration
The moment of inertia of an object about an axis, Hmr 2 , is sometimes
written as Mk 2 , where M is the mass of the object and k is a quantity
called the radius of gyration about the axis.
For example, the moment of inertia of a rod
about an axis through one end = M/ 2 /3
(p. 76) = M(/A/3) 2 . Thus the radius of gyration,
k, = 1/^/3 = 0-58/. The moment of inertia of a
sphere about its centre = \Ma 2 = M x (^/f a) 2 .
Thus the radius of gyration, k, = J^a = 0-63a
in this case.
Relation Between Moment of Inertia About C.G.
and Parallel Axis.
Suppose / is the moment of inertia of a body
about an axis CD and I G is the moment of
inertia about a parallel axis PQ through the
centre of gravity, G, distant h from the axis
CD, Fig. 3.5. If A is a particle of mass m whose
distance from PQ is x, its moment of inertia
about CD = m(h-x) 2
;. I = Xm(h-x) 2 = Xmh 2 + £mx 2 -Ylmhx.
Fig. 3.5
Theorem of parallel axes
ROTATION OF RIGID BODIES 79
Now Em/i 2 = /i 2 x2m = Mh 2 , where M is the total mass of the object,
and Zmx 2 = I G , the moment of inertia through the centre of gravity.
Also, Z2m/ix = IHZmx = 0,
since Emx, the sum of the moments about the centre of gravity, is
zero; this follows because the moment of the resultant (the weight)
about G is zero.
1 = I G + Mh 2
(1)
From this result, it follows that the moment of inertia, /, of a disc of
radius a and mass M about an axis through a point on its circumference
= I G + Ma 2 , since h = a = radius of disc in this case. But I G =
moment of inertia about the centre = Ma 2 /2 (p. 77).
.-. moment of inertia, 7, = ^-+Ma 2 = ^^-.
Similarly the moment of inertia of a sphere of radius a and mass M
about an axis through a point on its circumference = I G + Ma 2 =
2Ma 2 /5 + Ma 2 = !Ma 2 /5, since I G , the moment of inertia about an
axis through its centre, is 2Ma 2 /5.
Relation Between Moments of Inertia about Perpendicular Axes
Suppose OX, OY are any two perpendicular axes and OZ is an axis
perpendicular to OX and OY, Fig. 3.6 (i). The moment of inertia, J, of
(') (ii)
Fig. 3.6 Theorem of perpendicular axes
a body about the axis OZ = I.mr 2 , where r is the distance of a particle
A from OZ and m is its mass. But r 2 = x 2 + y 2 , where x, y are the
distances of A from the axis OY, OX respectively.
'. I = Zm(x 2 + y 2 ) = 2mx 2 + Emy 2 .
I = Iy + I,
(1)
where I r l x are the moments of inertia about OX, OY respectively.
80
ADVANCED LEVEL PHYSICS
As a simple application, consider a ring R and two perpendicular
axes OX, OY in its plane, Fig. 3.6 (ii). Then from the above result,
I y +I x = I = moment of inertia through O perpendicular to ring.
But I y = !# by symmetry.
.-. /„ +I X = Ma 2 .
.'. I x +I x = Ma 2 ,
■ I = Mq2
This is the moment of inertia of the ring about any diameter in its plane.
In the same way, the moment of inertia, /, of a disc about a diameter
in its plane is given by "
i + i - %
since the moments of inertia, J, about the two perpendicular diameters
are the same and Ma 2 /2 is the moment of inertia of the disc about an
axis perpendicular to its plane.
• I _ Mfl2
Couple on a Rigid Body
Consider a rigid body rotating about a fixed axis O with an angular
velocity to at some instant. Fig. 3.7.
Couple about
= I x angular
acceleration
Fig. 3.7 Couple on rigid body
The force acting on the particle A = m 1 x acceleration = m 1 x
iir.co) = m,x r A = m.r^-, since co = ^-. The moment of this
dV * ' x 1 d( dt 2 dt
ROTATION OF RIGID BODIES 81
force about the axis O = force x perpendicular distance from O =
d 2 6
m i r i~n2 x r i» smce tne force acts perpendicularly to the line OA.
d 2
.'. moment or torque = rn 1 r 1 -^ r .
.'. total moment of all forces on body about O, or torque,
2 d 2 e 2 d 2 e 2 d 2 e
= m 1 r l -^ + m 2 r 2liI + m 3 r 3 2 1F+ ...
/v 2, d 2
since the angular acceleration, d 2 B/dt 2 , about O is the same for all
particles.
d 2
.'. total torque about O = J-j-j, . (1)
where J = I,mr 2 = moment of inertia about O. The moment about O
is produced by external forces which together act as a couple of torque
C say. Thus, for any rotating rigid body,
Couple, C = I-Tz-
This result is analogous to the case of a particle of mass m which
undergoes an acceleration a when a force F acts on it. Here F = ma.
In place of F we have a couple C for a rotating rigid object ; in place of
m we have the moment of inertia I; and in place of linear acceleration a,
we have angular acceleration d 2 0/dt 2 {dca/dt).
EXAMPLES
1. A heavy flywheel of mass 15 kg and radius 20 cm is mounted on a horizontal
axle of radius 1 cm and negligible mass compared with the flywheel. Neglecting
friction, find (i) the angular acceleration if a force of 4 kgf is applied tangentially
to the axle, (ii) the angular velocity of the flywheel after 10 seconds.
(i) Moment of inertia = ^ = 15 x °" 22 = 0-3 kg m 2 .
Couple C = 4 x 9-8 (N) x 001 (m) = 0-4 N m approx.
0-4
.". angular acceleration = — = 1-3 rads -2 .
(ii) After 10 seconds, angular velocity = angular acceleration x time.
= 1-3x10= 13 rads -1 .
2. The moment of inertia of a solid flywheel about its axis is 01 kg m 2 . It
is set in rotation by applying a tangential force of 2 kgf with a rope wound
round the circumference, the radius of the wheel being 10 cm. Calculate the
82
ADVANCED LEVEL PHYSICS
angular acceleration of the flywheel. What would be
the acceleration if a mass of 2 kg were hung from the
end of the rope? (0. 61 C.)
d 2
Couple C = Vj2 =f moment of inertia x angular
acceleration.
NowC= 2x9-8x01 Nm.
.'. angular acceleration =
2x9-8x01
01
= 19-6 rad s~ 2 .
If a mass of 2kg is hung from the end of the
rope, it moves down with an acceleration a. Fig.
3.8. In this case, if T is the tension in the rope,
2kfg
Fig. 3.8 Example
mg — T = ma
For the flywheel,
T.r
couple = ^
,<P8
dt 1
(1)
(2)
where r is the radius of the flywheel. Now the mass of 2 kg descends a distance
given by rd, where is the angle the flywheel has turned. Hence the acceleration
a = rd 2 6/dt 2 . Substituting in (1),
(3)
. - rr d 2 6
.. mg -T = mr w .
2 d 2
mgr-T.r = mr 2 -^2
Adding (2) and (3),
n^ 2,d 2
. . mgr = (I + mr)-^.
d 2 mgr 2x10x01
•• dt 2 ~ I + mr 2 ~ 0-1 + 2 x 0-1 2 '
= 16-7 rad s -2 .
using g = 10 ms -2
Angular Momentum and Conservation
In linear or straight-line motion, an important property of a moving
object is its linear momentum (p. 18). When an object spins or rotates
about an axis, its angular momentum plays an important part in its
motion.
Consider a particle A of a rigid object rotating about an axis O.
Fig. 3.9. The momentum of A = mass x velocity = m x v = m^r^co.
The 'angular momentum' of A about O is denned as the moment of the
momentum about O. Its magnitude is thus m x vxp, where p is the
perpendicular distance from O to the direction of v. Thus angular
momentum of A = m x vp = m^co x r t = m x r*co.
ROTATION OF RIGID BODIES
83
.". total angular momentum of whole body = "Lm^^co = coLm^^
= let),
where J is the moment of inertia of the body about O.
Angular momentum is analogous to 'linear momentum', mv, in the
dynamics of a moving particle. In place of m we have J, the moment of
inertia; in place of v we have co, the angular velocity.
Further, the conservation of angular momentum, which corresponds
to the conservation of linear momentum, states that the angular
momentum about an axis of a given rotating body or system of bodies is
constant, if no external couple acts about that axis. Thus when a high
diver jumps from a diving board, his moment of inertia, J, can be
decreased by curling his body more, in which case his angular velocity
co is increased. Fig. 3.9 (ii). He may then be able to turn more somer-
saults before striking the water. Similarly, a dancer on skates can spin
faster by folding her arms.
Angular
momentum
r
3
Fig. 3.9 Angular momentum
High /
Low to
o
Low /
High co
o
\
(")
The earth is an object which rotates about an axis passing through
its geographic north and south poles with a period of 1 day. If it is
struck by meteorites, then, since action and reaction are equal, no
external couple acts on the earth and meteorites. Their total angular
momentum is thus conserved. Neglecting the angular momentum of the
meteorites about the earth's axis before collision compared with that
of the earth, then
angular momentum of earth plus meteorites after collision =
angular momentum of earth before collision.
Since the effective mass of the earth has increased after collision the
moment of inertia has increased. Hence the earth will slow up slightly.
84 ADVANCED LEVEL PHYSICS
Similarly, if a mass is dropped gently on to a turntable rotating freely
at a steady speed, the conservation of angular momentum leads to a
reduction in the speed of the table.
Angular momentum, and the principle of the conservation of angular
momentum, have wide applications in physics.- They are used in con-
nection with enormous rotating masses such as the earth, as well as
minute spinning particles such as electrons, neutrons and protons
found inside atoms.
Experiment on Conservation of Angular Momentum
A simple experiment on the principle of the conservation of angular
momentum is illustrated below.
m TT
Fig. 3.10 Conservation of angular momentum
Briefly, in Fig. 3.10 (i) a bicycle wheel A without a tyre is set rotating
in a horizontal plane and the time for three complete revolutions is
obtained with the aid of a white tape marker M on the rim. A ring D of
known moment of inertia, J, is then gently placed on the wheel con-
centric with it, by 'dropping' it from a small height. The time for the
next three revolutions is then determined. This is repeated with several
more rings of greater known moment of inertia.
If the principle of conservation of angular momentum is true, then
I (o = (J + / 1 )(o 1 , where I is the moment of inertia of the wheel alone,
co is the angular frequency of the wheel alone, and co 1 is the angular
frequency with a ring. Thus is t , t x are the respective times for three
revolutions,
-ip + M _ *_0
Thus a graph of tjt v. I t should be a straight line. Within the limits
of experimental error, this is found to be the case.
EXAMPLE
Consider a disc of mass 100 g and radius 10 cm is rotating freely about axis O
through its centre at 40 r.p.m. Fig. 3.11. Then, about O,
moment of inertia / = ~- = \ x 01 (kg) x 01 2 (m 2 ) = 5 x 10~ 4 kg m 2 ,
and angular momentum = Ico = 5 x 10 _4 co,
where co is the angular velocity corresponding to 40 r.p.m.
ROTATION OF RIGID BODIES
85
Suppose some wax W of mass m 20 g is dropped gently on to the disc at a
distance r of 8 cm from the centre O. The disc then slows down to another speed,
corresponding to an angular velocity co^ say. The total angular momentum
about O of disc plus wax
<0 fc>
= Ia) 1 +mr 2 (D l = SxKT^+OK^xO-OS 2 .^
= 6-28 xlO -4 ^.
From the conservation of angular momentum
for the disc and wax about O
Fig. 3. 1 1 Example
6-28 xKTV = 5xl0 _4 a>.
• coi_'500_n_
' " ~~co ~ 628 ~ 40'
where n is the r.p.m. of the disc.
n = — x 40 = 32 (approx).
Kepler's law and angular momentum
Consider a planet moving in an orbit round the sun S. Fig. 3.12.
Orbit
Fig. 3.12 Angular momentum and planets
At an instant when the planet is at O, its velocity v is along the tangent
to the orbit at O. Suppose the planet moves a very small distance Ss
from O to B in a small time St, so that the velocity v = Ss/St and its
direction is practically along OB. Then, if the conservation of angular
momentum is obeyed,
mvxp = constant,
where m is the mass of the planet and p is the perpendicular from S to
OB produced.
m.ds .p
St
= constant.
86 ADVANCED LEVEL PHYSICS
But the area SA of the triangle SBO = \ base x height = Ss x p/2.
.. m. 2-r- = constant
ot
SA
.'. -=- = constant,
St
since 2m is constant. Thus if the conservation of angular momentum
is true, the area swept out per second by the radius SO is constant while
the planet O moves in its orbit. In other words, equal areas are swept
out in equal times. But this is Kepler's second law, which has been
observed to be true for centuries (see p. 58). Consequently, the principle
of the conservation of angular momentum has stood the test of time.
From the equality of the angular 'momentum values at O and C, where
p is less than p x , it follows that v is greater than v v Thus the planet
speeds up on approaching S.
The force on O is always one of attraction towards S. It is described
as a central force. Thus the force has no moment about O and hence
the angular momentum of the planet about S is conserved.
Kinetic Energy of a Rolling Object
When an object such as a cylinder or ball rolls on a plane, the object
is rotating as well as moving bodily along the plane; therefore it has
rotational energy as well as translational energy.
/
Fig. 3.13 Rolling object
Consider a cylinder C rolling along a plane without slipping, Fig. 3.13.
At any instant the line of contact, PQ, with the plane is at rest, and we
can consider the whole of the cylinder to be rotating about this axis.
Hence the energy of the cylinder = \l x (o 2 , where I t is the moment of
inertia about PQ and co is the angular velocity.
But if J is the moment of inertia about a parallel axis through the
centre of gravity of the cylinder, M is the mass of the cylinder and a its
radius, then ■ , , . , ,
I t = I + Ma 2 ,
from the result on p. 79.
.*. energy of cylinder = %(I + Ma 2 )co 2
= \l(x> 2 +%Ma 2 co 2
:. Energy = ^Ico 2 +\Mv 2 . . (1)
ROTATION OF RIGID BODIES 87
since, by considering the distance rolled and the angle then turned,
v = aco = velocity of centre of gravity. This energy formula is true
for any moving object.
As an application of the energy formula, suppose a ring rolls along a
plane. The moment of inertia about the centre of gravity, its centre,
= Ma 2 (p. 76); also, the angular velocity, co, about its centre = v/a,
where v is the velocity of the centre of gravity.
.". kinetic energy of ring = \Mv 2 +^Ico 2
= Mv 2 .
By similar reasoning, the kinetic energy of a sphere rolling down a plane
= \Mv 2 +\lco 2
= \Mv 2 +\x\Ma 2 x\ V ^
since J = 2Ma 2 /5 (p. 78).
Acceleration of Rolling Object
We can now deduce the acceleration of a rolling object down an
inclined plane.
As an illustration, suppose a solid cylinder rolls down a plane. Then
kinetic energy = ^Mv 2 +%Ico 2 .
But moment of inertia, J, about an axis through the centre of gravity
parallel to the plane = \Ma 2 , and co = v/a, where a is the radius.
.". kinetic energy = \Mv 2 +|Md 2 = \Mv 2 .
If the cylinder rolls from rest through a distance s, the loss of potential
energy = Mgs sin a, where a is the inclination of the plane to the
horizontal. ~ , ,
.'. %Mv = Mgs sin a
2 % •
.'. v = -|-s sin a
But v 2 = las, where a is the linear acceleration.
Ag .
.". las = -r-s sin a
la . ...
.". a = jsina . . . . (i)
The acceleration if sliding, and no rolling, took place down the plane
is g sin a. The cylinder has thus a smaller acceleration when rolling.
The time t taken to move through a distance s from rest is given by
s = \at 2 . Thus, from (i),
s = \gt 2 sin a,
3s
or t =
g sin a
88
ADVANCED LEVEL PHYSICS
If the cylinder is hollow, instead of solid as assumed, the moment of
inertia about an axis through the centre of gravity parallel to the plane
is greater than that for a solid cylinder, assuming the same mass and
dimensions (p. 88). The time taken for a hollow cylinder to roll a
given distance from rest on the plane is then greater than that taken by
the solid cylinder, from reasoning similar to that above; and thus if
no other means were available, a time test on an inclined plane will
distinguish between a solid and a hollow cylinder of the same dimen-
sions and mass. If a torsion wire is available, however, the cylinders
can be suspended in turn, and the period of torsional oscillations
determined. The cylinder of larger moment of inertia, the hollow
cylinder, will have a greater period, as explained on p. 89.
Measurement of Moment of Inertia of Flywheel
The moment of inertia of a flywheel W about a horizontal axle A
can be determined by tying one end of some string to a pin on the axle,
winding the string round the axle, and attaching a mass M to the other
end of the string, Fig. 3.14. The length of string is such that M reaches
the floor, when released, at the same instant as the string is completely
unwound from the axle.
Fig. 3.14 Moment of inertia of flywheel
M is released, and the number of revolutions, n, made by the wheel
W up to the occasion when M strikes the ground is noted. The further
number of revolutions n t made by W until it comes finally to rest, and
the time t taken, are also observed by means of a chalk-mark on W.
Now the loss in potential energy of M = gain in kinetic energy of
M+gain in kinetic energy of fly wheel + work done against friction.
Mgh = \Mr 2 a> % +^Ieo 2 + nf,
(i)
where h is the^hstance M has fallen, r is the radius of the axle, co is the
angular velocity, J is the moment of inertia, and/is the energy per turn
expended against friction. Since the energy of rotation of the flywheel
ROTATION OF RIGID BODIES
89
when the mass M reaches the ground = work done against friction
in n x revolutions, then
hlu 2 = nj.
Substituting for /in (i),
Mgh = \Mr 2 co 2 +^Ico 2 1 +
(ii)
Since the angular velocity of the wheel when M reaches the ground is
co, and the final angular velocity of the wheel is zero after a time t, the
average angular velocity = co/2 = Innjt. Thusw =
Annjt. Knowing co and the magnitude of the other
quantities in (ii), the moment of inertia J of the fly-
wheel can be calculated.
Period of Oscillation of Rigid Body
On p. 81 we showed that the moment of the forces
acting on rotating objects = Idco/dt = ld 2 0/dt 2 ,
where / is the moment of inertia about the axis
concerned and d 2 0/dt 2 is the angular acceleration
about the axis. Consider a rigid body oscillating
about a fixed axis O, Fig. 3.15. The moment of the
weight mg (the only external force) about O is
mgh sin 0, or mgh0 if is small, where h is the
distance of the centre of gravity from O.
.:I^=-mgh0,
the minus indicating that the moment due to the weight always opposes
the growth of the angle 0.
d 2 -mgh L
" dt 2
where co 2 = mgh/I.
mg
Fig. 3.15
Compound
pendulum
uu = -™^ = _ a>2ft
.'. the motion is simple harmonic motion (p. 44),
and
period, T, = — = —, — —
co yjmghl
= 2n
I
mgh
(1)
T = 2n
(2)
If / = mk 2 , where k x is the radius of gyration about O,
\ mk x 2 = fky 1
mgh \j gh
A ring of mass m and radius a will thus oscillate about an axis
through a point O on its circumference normal to the plane of the
ring with a period T given by
T = 2n
mga
90 ADVANCED LEVEL PHYSICS
But I = I G + ma 2 (theorem of parallel axes) = ma 2 + ma 2 = 2ma 2 .
{2a
T = In .
Thus if a = 05 m, g = 9-8 m s
■2
T = In \ = 2-0 seconds (approx).
\j 9-8
Measurement of Moment of Inertia of Plate
The moment of inertia of a circular disc or other plate about an axis
perpendicular to its plane, for example, can be measured by means of
torsional oscillations. The plate is suspended horizontally from a
vertical torsion wire, and the period Tj of torsional oscillations is
measured. Then, from (1),
7; = 2* /5 . . . . (i)
where I x is the moment of inertia and c is the constant (opposing couple
per unit radian) of the wire (p. 164). A ring or annulus of known
moment of inertia l 2 is now placed on the plate concentric with the
axis, and the new period T 2 is observed. Then
T 2 = 2nlh+±.
■ (")
V c
By squaring (i) and (ii), and then eliminating c, we obtain
j - J i T
l l — t 2 T 2' J 2'
l 2 ~ J l
Thus knowing T v T 2 , and I 2 , the moment of inertia I x can be calculated.
Compound Pendulum. Since I = I G + mh 2 = mk 2 + mh 2 , where I G is the moment
of inertia about the centre of gravity, h is the distance of the axis O from the
centre of gravity, and k is the radius of gyration about the centre of gravity, then,
from previous, .
T = 2n / l = In l™ k +mh
mgh
Hence
where / = — : — ..... (i)
h
Thus (k 2 + h 2 )/h is the length, /, of the equivalent simple pendulum.
From(i), h 2 -hl+k 2 = 0.
.'. h 1 + h 2 = /, and h x h 2 = k 2 ,
where h t and h 2 are the roots of the equation.
ROTATION OF RIGID BODIES
91
(i)
S ^
( =
all
4n2
"x-x * p
I—
■" — ^ ^-^*v.
Q V x x** S
/7—
Fig. 3.16 Compound pendulum experiment
+h
By timing the period of vibration, T, of a long rod about a series of axes at
varying distances h on either side of the centre of gravity, and then plotting a
graph of T v. h, two different values of h giving the same period can be obtained,
Fig. 3.16 (i), (ii). Suppose h x , h 2 are the two values. Then from the result just
obtained , h t + h 2 = 4 the length of the equivalent simple pendulum. Thus, since
_4n 2 l _ 4n 2 {h i + h 2 )
In Fig. 3.16 (ii), PQ + QS = h t + h 2 = /.
Kater's Pendulum. The acceleration due to gravity was first measured by the
simple pendulum method, and calculated from the relation g = 4n 2 l/T 2 , with
the usual notation. The length /, the distance from the point of suspension to the
centre of gravity of the bob, however, cannot be determined with very great
accuracy.
In 1817 Captain Kater designed a reversible pendulum, with knife-edges for
the suspension; it was a compound pendulum. Now it has just been shown
that the same period is obtained between two non-symmetrical points on a
compound pendulum when their distance apart is /, the length of the equivalent
simple pendulum. Thus if T is the period about either knife-edge when this
occurs, g = 4n 2 l/T 2 , where / is now the distance between the knife-edges. The
pendulum is made geometrically symmetrical about the mid-point, with a brass
bob at one end and a wooden bob of the same size at the other. A movable large
and small weight are placed between the knife-edges, which are about one metre
apart. The period is then slightly greater than 2 seconds.
To find g, the pendulum is set up in front of an accurate seconds clock, with the
bob of the clock and that of the Kater pendulum in line with each other, and
both sighted through a telescope. The large weight on the pendulum is moved
until the period is nearly the same about either knife-edge, and the small weight is
used as a fine adjustment When the periods are the same, the distance / between
the knife-edges is measured very accurately by a comparator method with a
microscope and standard metre. Thus knowing T and /, g can be calculated from
g = 4n 2 l/T 2 .
For details of the experiment the reader should consult Advanced Practical
Physics for Students by Worsnop and Flint (Methuen).
92 ADVANCED LEVEL PHYSICS
Summary
The following table compares the translational (linear) motion of a
small mass m with the rotational motion of a large object of moment of
inertia /.
Linear Motion Rotational Motion
1. Velocity, v Velocity v = rco
2. Momentum = mv Angular momentum = Ico
3. Energy = \mv 2 Rotational energy = ^Ico 2
4. Force, F, = ma Torque, C, = Ix ang. accn. (d 2 0/dt 2 )
5. Simple pendulum: T = 2n - Compound pendulum : T= 2n \——r
yjg yjmgh
6. Motion down inclined plane — Rotating without slipping down in-
energy equation : clined plane — energy equation :
\mv 2 = mgh sin jMv 2 +jIco 2 = Mgh sin 6
7. Conservation of linear momentum Conservation of angular momentum
on collision, if no external forces on collision, if no external couple
EXAMPLE
What is meant by the moment of inertia of an object about an axis?
Describe and give the theory of an experiment to determine the moment of
inertia of a flywheel mounted on a horizontal axle.
A uniform circular disc of mass 20 kg and radius 15 cm is mounted on a hori-
zontal cylindrical axle of radius 1-5 cm and negligible mass. Neglecting frictional
losses in the bearings, calculate (a) the angular velocity acquired from rest by
the application for 12 seconds of a force of 20 kgf tangential to the axle, (b) the
kinetic energy of the disc at the end of this period, (c) the time required to bring
the disc to rest if a braking force of 01 kgf were applied tangentially to its rim. (L.)
Moment of inertia of disc,/, = \Ma 2 = \ x 20 (kg) x 015 2 (m 2 ) = 0-225 kg m 2 .
(a) Torque due to 2 kgf tangential to axle
= 2x9-8 (N) x 0015 (m) = 0-294 N m.
, , .. torque 0-294 , _ 2
. . angular acceleration = — ? — = rad s .
.". after 12 seconds, angular velocity = — — =15-7 rad s -1 .
(b) K.E. of disc after 12 seconds = jlco 2
= £x 0-225 xl5-7 2 = 27-8 J.
(c) Decelerating torque = 01 x 9-8 (N) x 015 (m).
, , , torque 01x9-8x0-15 , _,
.'. angular deceleration = — - — = — rad s .
initial angular velocity
.'. time to bring disc to rest = ; — - — ; :
angular deceleration
15-7x0-225
01x9-8x015
= 24 seconds.
ROTATION OF RIGID BODIES 93
EXERCISES 3
(Assume g = 10 m s~ 2 unless otherwise stated)
What are the missing words in the statements 1-4?
1. The kinetic energy of an object rotating about an axis is calculated from . . .
2. The angular momentum of the object is calculated from ...
3. 7 x angular acceleration' is equal to the ... on the object.
4. The period of oscillation of an object about an axis is calculated from . . .
Which of the following answers, A, B, C, D or E, do you consider is the correct
one in the statements 5-8?
5. When a sphere of moment of inertia / about its centre of gravity, and mass
m, rolls from rest down an inclined plane without slipping, its kinetic energy is
calculated from A %Ico 2 , B \mv 2 , C Ico + mv, D jIco 2 +%mv 2 , E lea.
6. If a hoop of radius a oscillates about an a xis th rough its circumference
perpe ndicul ar to its plane, the period is A 2nja/g, B InJla/g, C InJgJa,
D 2n^gj2a,E all.
7. Planets moving in orbit round the sun A increase in velocity at points near
the sun because their angular momentum is constant, B increase in velocity
near the sun because their energy is constant, C decrease in velocity near the
sun owing to the increased attraction, D sweep out equal mass in equal times
because their energy is constant, E always have circular orbits.
8. If a constant couple of 500 newton metre turns a wheel of moment of inertia
100 kg m 2 about an axis through its centre, the angular velocity gained in two
seconds is A 5 rad s~\ B 100 m s~\ C 200 m s"\ D 2 m s~\ E 10 rad s _1 .
9. A uniform rod has a mass of 60 g and a length 20 cm. Calculate the moment
of inertia about an axis perpendicular to its length (i) through its centre, (ii)
through one end. Prove the formulae used.
10. What is the Theorem of Parallel Axes'! A uniform disc has a mass of 4 kg
and a radius of 2 m. Calculate the moment of inertia about an axis perpendicular
to its plane (i) through its centre, (ii) through a point of its circumference.
11. What is the Theorem of Perpendicular Axes! A ring has a radius of 20 cm
and a mass of 1 00 g. Calculate the moment of inertia about an axis (i) perpendicular
to its plane through its centre, (ii) perpendicular to its plane passing through a
point on its circumference, (iii) in its plane passing through the centre.
12. What is the formula for the kinetic energy of (i) a particle, (ii) a rigid body
rotating about an axis through its centre of gravity, (iii) a rigid body rotating
about an axis through/4ny point? Calculate the kinetic energy of a disc of mass
5 kg and radius 1 m rolling along a plane with a uniform velocity of 2 m s~ *.
13. A sphere rolls down a plane inclined at 30° to the horizontal. Find the
acceleration and velocity of the sphere after it has moved 50 m from rest along
the plane, assuming the moment of inertia of a sphere about a diameter is
2Ma 2 /5, where M is the mass and a is the radius.
14. A uniform rod of length 3-0 m is suspended at one end so that it can move
about an axis perpendicular to its length, and is held inclined at 60° to the vertical
and then released. Calculate the angular velocity of the rod when (i) it is inclined
at 30° to the vertical, (ii) reaches the vertical.
94 ADVANCED LEVEL PHYSICS
15. Define the moment of inertia of a rigid object about an axis.
A ring of radius 2-0 m oscillates about an axis on its circumference which is
perpendicular to the plane of the ring. Calculate the period of oscillation. Give
an explanation of any formula used.
16. A flywheel with an axle 1-0 cm in diameter is mounted in frictionless
bearings and set in motion by applying a steady tension of 200 gf to a thin
thread wound tightly round the axle. The moment of inertia of the system about
its axis of rotation is 50 x 10" 4 kg m 2 . Calculate (a) the angular acceleration of
the flywheel when 1 m of thread has been pulled off the axle, .(b) the constant
retarding couple which must then be applied to bring the flywheel to rest in one
complete turn, the tension in the thread having been completely removed. (N.)
17. Define the moment of inertia of a body about a given axis. Describe how
the moment of inertia of a flywheel can be determined experimentally.
A horizontal disc rotating freely about a vertical axis makes 100 r.p.m. A small
piece of wax of mass 10 g falls vertically on to the disc and adheres to it at a
distance of 9 cm from the axis. If the number of revolutions per minute is thereby
reduced to 90, calculate the moment of inertia of the disc. (N.)
18. Describe an experiment using a bar pendulum to determine the accelera-
tion due to gravity. Show how the result is calculated from the observations.
A uniform disc of diameter 120 cm and mass 810 g is suspended with its plane
horizontal by a torsion wire and allowed to perform small torsional oscillations
about a vertical axis through its centre. The disc is then replaced by a uniform
sphere which is allowed to oscillate similarly about a diameter. If the period of
oscillation of the sphere is 1-66 times that of the disc, determine the moment of
inertia of the sphere about a diameter. (L.)
19. A plane sheet of metal of uniform thickness and of irregular shape is
pierced with a number of small holes, irregularly distributed, so that it can be
pivoted about an axis through any one of them to swing in its own plane. Describe
how you would proceed in order to find the value of the moment of inertia of the
sheet about an axis through its centre of gravity normal to its plane.
A rigid bar pendulum is pivoted at a distance h from its centre of gravity.
When a piece of lead of negligible size and of mass M equal to that of the pendulum,
is attached at the centre of gravity the periodic time of the pendulum is reduced
to 0-80 of its former value. Find an expression for the moment of inertia of the
pendulum about the pivot. (L.)
20. Define moment of inertia and derive an expression for the kinetic energy of
a rigid body of moment of inertia / about a given axis when it is rotating about
that axis with a uniform angular velocity co.
Give two examples of physical phenomena in which moment of inertia is a
necessary concept for a theoretical description, in each case showing how the
concept is applied.
A uniform spherical ball starts from rest and rolls freely without slipping down
an inclined plane at 10° to the horizontal along a line of greatest slope. Calculate
its velocity after it has travelled 5 m. (M.L of sphere about a diameter = 2Mr 2 /5.)
(O. & C.)
21. Explain the meaning of the term moment of inertia. Describe in detail how
you would find experimentally the moment of inertia of a bicycle wheel about
the central line of its hub.
A uniform cylinder 20 cm long, suspended by a steel wire attached to its mid-
point so that its long axis is horizontal, is found to oscillate with a period of
ROTATION OF RIGID BODIES 95
2 seconds when the wire is twisted and released. When a small thin disc, of mass
10 g, is attached to each end the period is found to be 2-3 seconds. Calculate the
moment of inertia of the cylinder about the axis of oscillation. (AT.)
22. What is meant by 'moment of inertia'? Explain the importance of this
concept in dealing with problems concerning rotating bodies.
Describe, with practical details, how you would determine whether a given
cylindrical body were hollow or not without damaging it. (C.)
23. Define moment of inertia, and find an expression for the kinetic energy of a
rigid body rotating about a fixed axis.
A sphere, starting from rest, rolls (without slipping) down a rough plane
inclined to the horizontal at an angle of 30°, and it is found to travel a distance of
13-5 m in the first 3 seconds of its motion. Assuming that F, the frictional
resistance to the motion, is independent of the speed, calculate the ratio of F
to the weight of the sphere. (For a sphere of mass m and radius r, the moment
of inertia about a diameter is f mr 2 .) (O. & C.)
chapter four
Static Bodies. Fluids
STATIC BODIES
Statics
1. Statics is a subject which concerns the equilibrium of forces, such
as the forces which act on a bridge. In Fig. 4.1 (i), for example, the
joint O of a light bridge is in equilibrium under the action of the two
forces P, Q acting in the girders meeting at O and the reaction S of the
masonry at O.
h^-v c
0) 0»
Fig. 4. 1 Equilibrium of forces
(iii)
Parallelogram of Forces
A force is a vector quantity, i.e., it can be represented in magnitude
and direction by a straight line (p. 1). If AB, AC represent the forces
P, Q respectively at the joint O, their resultant, R, is represented in
magnitude and direction by the diagonal AD of the parallelogram
ABDC which has AB, AC as two of its adjacent sides, Fig. 4.1 (ii).
This is known as the parallelogram of forces, and is exactly analogous
to the parallelogram of velocities discussed on p. 8. Alternatively, a
line ab may be drawn to represent the vector P and bd to represent Q,
in which case ad represents the resultant R.
By trigonometry for triangle ABD, we have
AD 2 = BA 2 + BD 2 - 2BA . BD cos ABD.
.-. R 2 = P 2 + Q 2 + 2PQcos0,
where = angle BAC; the angle between the forces P, Q, =
180° -angle ABD. This formula enables R to be calculated when P, Q
96
STATIC BODIES, FLUIDS 97
and the angle between them are known. The angle BAD, or a say,
between the resultant R and the force P can then be found from the
relation
R Q
sin sin a
applying the sine rule to triangle ABD and noting that angle ABD =
180° -0.
Resolved component. On p. 8 we saw that the effective part, or
resolved component, of a vector quantity X in a direction inclined to
it is given by X cos 0. Thus the resolved component of a force P in a
direction making an angle of 30° with it is P cos 30° ; in a perpendicular
direction to the latter the resolved component is P cos 60°, or P sin 30°.
In Fig. 4.1 (i), the downward component of the force P on the joint of
O is given by P cos BOS.
Forces in Equilibrium. Triangle of Forces
Since the joint O is in equilibrium, Fig. 4.1 (i), the resultant of the
forces P, Q in the rods meeting at this joint is equal and opposite to
the reaction S at O. Now the diagonal AD of the parallelogram ABDC
in Fig. 4.1 (ii) represents the resultant R of P, Q since ABDC is the
parallelogram of forces for P, Q ; and hence DA represents the force S.
Consequently the sides of the triangle ABD represent the three forces
at O in magnitude and direction:
This result can be generalised as
follows. If three forces are in equi-
librium, they can be represented by the
three sides of a triangle taken in order.
r ~ > y This theorem in Statics is known as
the triangle of forces. In Fig. 4.1 (ii),
AB, BD, DA, in this order, represent,
P, Q, S respectively in Fig. 4.1 (i)
We can derive another relation be-
tween forces in equilibrium. Suppose
X, Y are the respective algebraic sums
•Fig. 4.2 Resolution of forces of the resolved components in two
perpendicular directions of three
forces P, Q, T in equilibrium, Fig. 4.2. Then, since X, Y can each be
represented by the sides of a rectangle drawn to scale, their resultant R
is given by
R 2 = X 2 +Y 2 (i)
Now if the forces are in equilibrium, R is zero. It then follows from (i)
that X must be zero and Y must be zero. Thus if forces are in equili-
brium the algebraic sum of their resolved components in any two perpen-
dicular directions is respectively zero. This result applies to any number
of forces in equilibrium.
98
ADVANCED LEVEL PHYSICS
7"=0-1N
W=0049N
EXAMPLE
State what is meant by scalar and vector quantities, giving examples of each.
Explain how a flat kite can be flown in a
wind that is blowing horizontally. The line
makes an angle of 30° with the vertical and
is under a tension of 01 newton ; the mass
of the kite is 5 g. What angle will the plane
of the kite make with the vertical, and what
force will the wind exert on it? (O. & C.)
Second part. When the kite AB is inclined
to the horizontal, the wind blowing hori-
zontally exerts an upward force F normal to
AB, Fig. 4.3. For equilibrium of the kite,
F must be equal and opposite to the resultant
of the tension T, 01 newton, and the weight
W, 0005 x 9-8 or 0049 N. By drawing the
parallelogram of forces for the resultant of
T and W, F and the angle 6 between F and
T can be found. 6 is nearly 10°, and F is
about 01 48 N. The angle between AB and
the vertical = 60° + = 70° (approx.). Also,
since F is the component of the horizontal force P of the wind.
Pcos(6O° + 0) = F.
• p= F = 0148
cos (60° + 0) cos 70°
= 0-43 N.
Moments
When the steering-wheel of a car is turned, the applied force is said
to exert a moment, or turning-effect, about the axle attached to the
wheel. The magnitude of the moment of a force P about a point O is
defined as the product of the force P and the perpendicular distance OA
for all the forces in Fig. 4.4 (ii), we have
moment = P x AO.
The magnitude of the moment is expressed in newton metre (N m) when
Fig. 4.3 Example
90°~X
O
A I B
06m
<7
10
03rr|
I
I
v
20
0-4m
1 30
I
V
15
/?=100
(ii)
8m
V
25
Fig. 4.4 Parallel forces
STATIC BODIES, FLUIDS
99
P is in newtons and AO is in metres. We shall take an anticlockwise
moment as positive in sign and a clockwise moment as negative in
sign.
Parallel Forces
If a rod carries loads of 10, 20, 30, 15, and 25 N at O, A, B, C, D
respectively, the resultant R of the weights, which are parallel forces,
for all the forces in Fig. 4.4 (ii), we have
resultant, R, = 10 + 20 + 30 + 1 5 + 25 = 100 N.
Experiment and theory show that the moment of the resultant of a
number of forces about any point is equal to the algebraic sum of the
moments of the individual forces about the same point. This result
enables us to find where the resultant R acts. Taking moments about O
for all the forces in Fig. 4.4 (ii), we have
(20x0-6) + (30x0-9) + (15xl-3)+(25x21),
because the distances between the forces are 0-6 m, 0-3 m, 0-4 m, 0-8 m,
as shown. If x m is the distance of the line of action of R from O, the
moment of R about O = R x x = 100 x x.
:. KXbc = (20x0-6) + (30x0-9) + (15xl-3)+(25x21),
from which x = Mm.
Equilibrium of Parallel Forces
The resultant of a number of forces in equilibrium is zero ; and the
moment of the resultant about any point is hence zero. It therefore
follows that the algebraic sum of the moments of all the forces about
any point is zero when those forces are in equilibrium. This means
that the total clockwise moment of the forces about any point = the
total anticlockwise moment of the remaining forces about the same
point.
B
2m
3m
1m'
1*0
4m
V
20
Fig. 4.5 Example
V
4
As a simple example of the equilibrium of parallel forces, suppose a
light beam XY rests on supports, A, B, and has loads of 10, 20, and 4 N
concentrated at X, O, Y respectively, Fig. 4.5. Let R, S be the reactions
at A, B respectively. Then, for equilibrium in a vertical direction,
R + S= 10 + 20 + 4 = 34 N
(i)
100
ADVANCED LEVEL PHYSICS
To find R, we take moments about a suitable point such as B, in
which case the moment of S is zero. Then, for the remaining four forces,
+ 10.6 + 20. 1-R. 4-4. 4 = 0,
from which R = 16 N. From (i), it follows that S = 34-16 = 18 N.
Equilibrium of Three Coplanar Forces
If any object is in equilibrium under the action of three forces, the
resultant of two of the forces must be equal and opposite to the third
force. Thus the line of action of the third force must pass through the
point of intersection of the lines of action of the other two forces.
As an example of calculating unknown forces in this case, suppose
that a 12 m ladder of 20O kgf is placed at an angle of 60° to the hori-
zontal, with one end B leaning against a smooth wall and the other end
/?«
777777777777777777777,777777777777/
vw=20kgi
Fig. 4.6 Triangle of forces
A on the ground, Fig. 4.6. The force R at B on the ladder is called the
reaction of the wall, and if the latter is smooth, R acts perpendicularly
to the wall. Assuming the weight, W, of the ladder acts at its mid-point
G, the forces W and R meet at O, as shown. Consequently the frictional
force F at A passes through O.
The triangle of forces can be used to find the unknown forces R, F.
Since DA is parallel to R, AO is parallel to F, and OD is parallel to W,
the triangle of forces is represented by AOD. By means of a scale draw-
ing R and F can be found, since
W(20) F R
OD ~ AO - DA
A quicker method is to take moments about A for all the forces. The
algebraic sum of the moments is zero about any point since the object
is in equilibrium, and hence
R.a-W. AD = 0,
STATIC BODIES, FLUIDS
101
where a is the perpendicular from A to R. (F has zero moment about
A.) But a = 12 sin 60°, and AD = 6 cos 60°.
.'. R x 12 sin 60° - 20 x 6 cos 60° = 0.
cos 60°
R = 10
sin 60°
= 5-8 kgf.
Suppose 6 is the angle F makes with the vertical.
Resolving the forces vertically, F cos 6 = W = 20 kgf.
Resolving horizontally, F sin = R = 5-8 kgf.
.-. F 2 cos 2 0+F 2 sin 2 = F 2 = 20 2 + 5-8 2 .
.-. F = V20 2 + 5-8 2 = 20-8 kgf.
Couples and Torque
There are many examples in practice where two forces, acting
together, exert a moment or turning-effect on some object. As a very
pimple case, suppose two strings are tied to a wheel at X, Y, and two
'e^ual and opposite forces, F, are exerted tangentially to the wheel,
Fig. 4.7 (i). If the wheel is pivoted at its centre, O, it begins to rotate
about O in an anticlockwise direction.
Two equal and opposite forces whose lines of action do not coincide
are said to constitute a couple in Mechanics. The two forces always
have a turning-effect, or moment, called a torque, which is defined by
torque = one force x perpendicular distance between forces (1)
Since XY is perpendicular to each of the forces F in Fig. 46 (i), the
Af=10N
F= BUN
F=BILN
F= 10N
(i) ('•)
Fig. 4.7 Couple and torque
moment of the couple acting on the wheel = Fx XY = F x diameter
of wheel. Thus if F = 10 newt on and the diameter is 2 metre, the
moment of the couple or torque = 20 newton metre (N m).
In the theory of the moving-coil electrical instrument, we meet a case
where a coil rotates when a current J is passed into it and comes to rest
102 ADVANCED LEVEL PHYSICS
after deflection through an angle 0. Fig. 4.7 (ii). The forces F on the
two sides X and Y of the coil are both equal to BUN, where B is the
strength of the magnetic field, / is the length of the coil and N is the
number of turns (see Electricity section, chapter 35). Thus the coil is
deflected by a couple. The moment or torque of the deflecting couple
= Fxb, where b = XY = breadth of coil. Hence
torque = BUN x b = BAN I,
where A = lb = area of coil. The opposing couple, due to the spring S,
is c0, where c is its elastic constant (p. 164). Thus, for equilibrium,
BAN I = cO.
Work Done by a Couple
Suppose two equal and opposite forces F act tangentially to a wheel
W, and rotate it through an angle while the forces keep tangentially
to the wheel, Fig. 4.8. The moment of the couple is then constant.
Fig. 4.8 Work done by couple
The work done by each force = F x distance = F xr0, since rd is the
distance moved by a point on the rim if is in radians.
.*. total work done by couple = Fr0+Fr0 = 2Fr0.
But moment of couple = F x 2r = 2Fr
.'. work done by couple = torque or moment of couple x
Although we have chosen a simple case, the result for the work done
by a couple is always given by torque x angle of rotation. In the formula,
it should be carefully noted that is in radians. Thus suppose
F = 100 gf = 01 kgf = 01 x 9-8 newton, r = 4 cm = 004 metre, and
the wheel makes 5 revolutions while the moment of the couple is kept
constant. Then
torque or moment of couple = 01 x 9-8 x 008 newton metre,
and angle of rotation = 2n x 5 radian.
.-. work done = 01 x 9-8 x 008 x 2n x 5 = 2-5 J
STATIC BODIES, FLUIDS
103
Centre of Gravity
Every particle is attracted towards the centre of the earth by the
force of gravity, and the centre of gravity of a body is the point where the
resultant force of attraction or weight of the body acts. In the simple
case of a ruler, the centre of gravity is the point of support when the
ruler is balanced. A similar method can be used to find roughly the
centre of gravity of a flat plate. A more accurate method consists of
suspending the object in turn from two points on it, so that it hangs
freely in each case, and finding the point of intersection of a plumb-line,
suspended in turn from each point of suspension. This experiment is
described in elementary books.
An object can be considered to consist of many small particles. The
forces on the particles due to the attraction of the earth are all parallel
since they act vertically, and hence their resultant is the sum of all the
forces. The resultant is the weight of the whole object, of course. In
the case of a rod of uniform cross-sectional area, the weight of a particle
A at one end, and that of a corresponding particle A' at the other end,
have a resultant which acts at the mid-point O of the rod, Fig. 4.9 (i).
A B
O
B' A'
n
Resultant
(')
Fig. 4.9 Centre of gravity and mass
Similarly, the resultant of the weight of a particle B, and that of a
corresponding particle at B', have a resultant acting at O. In this way,
i.e., by symmetry, it follows that the resultant of the weights of all the
particles of the rod acts at O. Hence the centre of gravity of a uniform
rod is at its mid-point.
The centre of gravity, C.G., of the curved surface of a hollow cylinder
acts at the midpoint of the cylinder axis. This is also the position of the
C.G. of a uniform solid cylinder. The C.G. of a triangular plate or
lamina is two-thirds of the distance along a median from corresponding
point of the triangle. The C.G. of a uniform right solid cone is three-
quarters along the axis from the apex.
Centre of Mass
The 'centre of mass' of an object is the point where its total mass
acts or appears to act Fig. 4.9 (ii) illustrates how the position of the
centre of mass of an object may be calculated, using axes Ox, Oy.
If m x is the mass of a small part of the object and x t is the perpendi-
cular distance to Oy, then m l x 1 represents a product similar to the
104
ADVANCED LEVEL PHYSICS
moment of a weight at m l about Oy. Likewise, m 2 x 2 is a 'moment'
about Oy, where m 2 is another small part of the object. The sum of the
total 'moments' about Oy of all the parts of the object can be written
Zmx. The total mass = Em = M say. The distance x of the centre of
mass C from Oy is then given by
x =
Lmx
~M~'
Similarly, the distance y of the centre of mass C from Ox is given by
y =
Lmy
If the earth's field is uniform at all parts of the body, then the
weight of a small mass m of it is typically mg. Thus, by moments, the
distance of the centre of gravity from Oy is given by
Lmg x x 'Lmx Lmx
Lmg
Em
M
The acceleration due to gravity, g, cancels in numerator and denomi-
nator. It therefore follows that the centre of mass coincides with the
centre of gravity. However, if the earth's field is not uniform at all parts
of the object, the weight of a small mass m x is then m l g 1 say and the
weight of another small mass m 2 is m 2 g 2 . Clearly, the centre of gravity
does not now coincide with the centre of mass. A very long or large
object has different values of g at various parts of it.
EXAMPLE
What is meant by (a) the centre of mass of a body, (b) the centre of gravity of a
body?
A cylindrical can is made of a material of mass 10
gem -2 and has no lid. The diameter of the can is 25 cm
and its height 50 cm. Find the position of the centre of
mass when the can is half full of water. (C)
The area of the base = nr 2 = n x (25/2) 2 cm 2 ; hence
the mass is n x (25/2) 2 x 10 g, and acts at A, the centre of
the base, Fig. 4.10.
The mass of the curved surface of the centre = 2nrh x
10 g = 27i x (25/2) x 50 x 10 g, and acts at B, half-way
along the axis.
The mass of water = nr 2 h g = n x (25/2) 2 x 25 g and
acts^at C, the mid-point of AB.
Thus the resultant mass in gramme
^?Ch™
^HHH^A^e
Fig. 4.10 Example
n x 625 x 10
In x 25 x 50 x 10 , n x 625 x 25
= n x 625 x 284.
STATIC BODIES, FLUIDS
Taking moments about A,
105
7rx625x28f xx = (nx 12500) xAB+lrcx £ — |xAC
where x is the distance of the centre of mass from A.
.-. 28|x = 20x25+^xl2±
.'. x = 20 (approx).
.'. centre of mass is 20 cm from the base.
Types of Equilibrium
If a marble A is placed on the curved surface of a bowl S, it rolls
Unstable
(i) (ii)
Fig. 4. 1 1 Stable and unstable equilibrium
down and settles in equilibrium at the lowest point. Fig. 4.11 (i). Its
potential energy is then a minimum. This is the case for objects in any
field, gravitational, magnetic or electrical. The equilibrium position
corresponds to minimum potential energy.
If the marble A is disturbed and displaced to B, its energy increases.
When it is released, the marble rolls back to A. Thus the marble at A
is said to be in stable equilibrium. Note that the centre of gravity of A
is raised on displacement to B. On this account the forces in the field
return the marble from B to A, where its potential energy is lower.
Suppose now that the bowl S is inverted and the marble is placed at its
top point at A. Fig. 4.11 (ii). If A is displaced slightly to C, its potential
energy and centre of gravity are then lowered. A now continues to
move further away from B under the action of the forces in the field.
Thus in Fig. 4.10 (ii), A is said to be in unstable equilibrium.
Stable
Unstable
Neutral
(ii)
Fig. 4. 12 Movement of C.G.
(i»)
106
ADVANCED LEVEL PHYSICS
Fig. 4.12 (i) shows a cone C with its base on a horizontal surface. If it is
slightly displaced to D, its centre of gravity G rises to G v As previously
explained, D returns to C when the cone is released, so that the equili-
brium is stable. In Fig. 4.12 (ii), the cone is balanced on its apex. When
it is slightly displaced, the centre of gravity, G x is lowered to G 2 . This
is unstable equilibrium. Fig. 4.12 (iii) illustrates the case of the cone
resting on its curved surface. If it is slightly displaced, the centre of
gravity G remains at the same height G 3 . The cone hence remains in
its displaced position. This is called neutral equilibrium.
EXAMPLE
A rectangular beam of thickness a is balanced on the curved surface of a rough
cylinder of radius r. Show that the beam is stable if r is greater than a/2.
Fig. 4.13 Example
Suppose the beam is tilted through a small angle 0. The point of contact C
then moves to A, the radius of the cylinder moves through an angle 0, and the
vertical GB through the centre of gravity G of the beam makes an angle 6 with
CG. (Fig 4.13 (\\ As shown in the exaggerated sketch in Fig 4.13 (ii), AC = r6.
The beam is in stable equilibrium if the vertical through G lies to the left of A,
since a restoring moment is then exerted. Thus for stable equilibrium, AD must
be greater than DB, where CD is the vertical through C.
Now
AD = r0cos 6, DB = CL = ^sin 0.
.'. rO cos > -sin 9.
When is very small, cos #-►!, sin 0-»0.
:.re>- 2 e.
r >
STATIC BODIES, FLUIDS 107
Common Balance
The common balance is basically a lever whose two arms are equal,
Fig. 414. The fulcrum, about which the beam and pointer tilt, is an
agate wedge resting on an agate plate; agate wedges, B, at the ends of
the beam, support the scale-pans. The centre of gravity of the beam and
pointer is vertically below the fulcrum, to make the arrangement
stable. The weights placed on the two scale-pans are equal when there
is a 'balance'.
Fig. 4.14 Common balance
On rare occasions the arms of the balance are slightly unequal. The
mass Wof an object is then determined by finding the respective masses
W u W 2 required to balance it on each scale-pan. Suppose a, b are the
lengths of the respective arms. Then, taking moments,
.-. W t . a = W. b, and W. a = W 2 .b.
• • W x ~ b~ W
.-. w 2 = w x w 2
:. w = y Jw 1 w 2 .
Thus Wean be found from the two masses W u W 2 .
Sensitivity of a Balance
A balance is said to be very sensitive if a small difference in weights
on the scale-pans causes a large deflection of the beam. To investigate
the factors which affect the sensitivity of a balance, suppose a weight
W x is placed on the left scale-pan and a slightly smaller weight W 2 is
placed on the right scale-pan, Fig. 4.15. The beam AOB will then be
inclined at some angle 6 to the horizontal, where O is the fulcrum.
108
ADVANCED LEVEL PHYSICS
B
Fig. 4. 1 5 Theory of balance
The weight W of the beam and pointer acts at G, at a distance h
below O. Suppose AO = OB = a. Then, taking moments about O,
W x a cos = Wh sin + W 2 a cos
:. (W t - W 2 )a cos = Wh sin
tan# =
(W,-W 2 )a
Wh
Thus for a given value of (W 1 — W 2 ), the difference of the weights
on the scale-pans, will increase when a increases and W, h both
decrease. In theory, then, a sensitive balance must be light and have
long arms, and the centre of gravity of its beam and pointer must be
very close to the fulcrum. Now a light beam will not be rigid. Further,
a beam with long arrns will take a long time to settle down when it is
deflected. A compromise must therefore be made between the require-
ments of sensitivity and those of design.
If the knife-edges of the scale-pan and beam are in the same plane,
corresponding to A, B and O in Fig. 4.15, then the weights W l9 W 2 on
them always have the same perpendicular distance from O, irrespective
of the inclination of the beam. In this case the net moment about O is
(W 1 — W 2 )a cos 0. Thus the moment depends on the difference, W x — W 2 ,
of the weights and not on their actual values. Hence the sensitivity is
independent of the actual load value over a considerable range.
When the knife-edge of the beam is below the knife-edges of the two
scale-pans, the sensitivity increases with the load ; the reverse is the
case if the knife-edge of the beam is above those of the scale-pans.
Buoyancy Correction in Weighing
In very accurate weighing, a correction must be made for the
buoyancy of the air. Suppose the body weighed has a density p and
a mass m. From Archimedes principle (p. 114), the upthrust due to the
air of density a is equal to the weight of air displaced by the body, and
hence the net downward force
m
m . a \g, since the volume of the
P
STATIC BODIES, FLUIDS
109
body is m/p. Similarly, if the weights restoring a balance have a total
mass m v and a density p v the net downward force = \m 1 ~-a\g.
Since there is a balance, \ Pi-
ma m,a
m = m, —
P l Pi
m = m x
1 —
Pi
1-°
Thus knowing the density of air, a, and the densities p, p lt the true
mass m can be found in terms of m v The pressure and temperature of
air, which may vary from day to day, affects the magnitude of its
density a, from the gas laws ; the humidity of the air is also taken into
account in very accurate weighing, as the density of moist air differs
from that of dry air.
FLUIDS
Pressure
Liquids and gases are called fluids. Unlike solid objects, fluids can
flow.
If a piece of cork is pushed below the surface of a pool of water and
then released, the cone rises to the surface again. The liquid thus
exerts an upward force on the cork and this is due to the pressure
exerted on the cork by the surrounding liquid. Gases also exert pres-
sures. For example, when a thin closed metal can is evacuated, it
Fig. 4.16 Pressure in liquid
usually collapses with a loud explosion. The surrounding air now
exerts a pressure on the outside which is no longer counter-balanced
by the pressure inside, and hence there is a resultant force.
Pressure is defined as the average force per unit area at the particular
region of liquid or gas. In Fig. 4.16, for example, X represents a small
110
ADVANCED LEVEL PHYSICS
horizontal area, Y a small vertical area and Z a small inclined area,
all inside a vessel containing a liquid. The pressure p acts normally
to the planes of X, Y or Z. In each case
F
average pressure, p, =
A'
where F is the normal force due to the liquid on an area A of X, Y or Z.
Similarly, the pressure p on the sides L or M of the curved vessel act
normally to L and M have magnitude F/A. In the limit, when the area
is very small, p = dF/dA.
At a given point in a liquid, the pressure can act in any direction.
Thus pressure is a scalar, not a vector. The direction of the force
on a particular surface is normal to the surface.
Formula for Pressure
Observation shows that the pressure increases with the depth,
h, below the liquid surface and with its
density p.
To obtain a formula for the pressure,
p, suppose that a horizontal plate X of
area A is placed at a depth h below the
liquid surface, Fig. 4.17. By drawing
Liquid
| weight
Si IS :§- iiffilffif vertical lines from points on the perimeter
§§§§§§§ of X, we can see that the force on X due
EzfgBSg to the liquid is equal to the weight of
Xjjjlfjl liquid of height h and uniform cross-
|E=E^B8;1 section A. Since the volume of this liquid
is Ah, the mass of the liquid = Ahxp.
Fig. 4.17 Pressure and depth
weight = Ahpg newton,
where g is 9-8, h is in m, A is in m 2 , and p is in kg m 3 .
force Ahpg
'. pressure, p, on X =
area
. . p = hpg
(1)
When h, p, g have the units already mentioned, the pressure p is in
newton m~ 2 (N m -2 ).
1 bar = 10 6 dyne cm -2 . To change 10 6 dyne cm -2 to Nm~ 2 , we
may proceed as follows :
N m ■ cnr l(r
cm
4.io 4 = io 5 4
nr 4 m z
1 bar = 10 5 N m
-2
(2)
Pressure is often expressed in terms of that due to a height of mercury
(Hg). One unit is the torr (after Torricelli) :
1 torr = 1 mmHg = 133-3 N m 2 (approx).
STATIC BODIES, FLUIDS
111
A BCD
Fig. 4.18 Pressure and cross-section
From p = hpg it follows that the pressure in a liquid is the same at all
points on the same horizontal level in it. Experiment also gives the
same result. Thus a liquid filling the vessel shown in Fig. 4.18 rises to
the same height in each section if ABCD is horizontal. The cross-
sectional area of B is greater than that of D; but the force on B is the
sum of the weight of water above it together with the downward
component of reaction R of the sides of
the vessel, whereas the force on D is the
weight of water above it minus the upward
component of the reaction S of the sides
of the vessel. It will thus be noted that the
pressure in a vessel is independent of the
cross-sectional area of the vessel.
Water
Fig. 4.19
Comparison of densities
Liquids in U-tube
Suppose a U-tube is partly filled with
water, and oil is then poured into the left
side of the tube. The oil will then reach
some level B at a height h t above the
surface of separation, A, of the water and oil, while the water on the
right side of the tube will then reach some level D at a height h 2 above
the level of A, Fig. 4.19.
Since the pressure in the water at A is equal to the pressure at C on
the same horizontal level, it follows that
H+hiptf = H+h 2 p 2 g,
where H is the atmospheric pressure, and p v p 2 are the respective
densities of oil and water. Simplifying,
h t Pi = h 2 p 2
hi
Since p 2 (water) = 1000 kg m -3 , and
density p x of the oil can be found.
h 2 , h x can be measured, the
Atmospheric Pressure
The pressure of the atmosphere was first measured by Galileo, who
observed the height of a water column in a tube placed in a deep well.
112
ADVANCED LEVEL PHYSICS
About 1640 Torricelli thought of the idea of using mercury instead
of water, to obtain a much shorter column. He completely filled a
glass tube about a metre long with mercury, and then inverted it in a
vessel D containing the liquid, taking care that no air entered the tube.
He observed that the mercury in the tube fell to a level A about 76 cm
or 0-76 m above the level of the mercury in D, Fig. 4.20. Since there was
r\ ^Vacuum
Atmospheric
pressure
Fig. 4.20 Atmospheric pressure
no air originally in the tube, there must be a vacuum above the mercury
at A, and it is called a Torricellian vacuum. This was the first occasion
in the history of science that a vacuum had been created.
If the tube in Fig. 4.20 is inclined to the vertical, the mercury ascends
the tube to a level B at the same vertical height H above the level of the
mercury in D as A.
The pressure on the surface of the mercury in D is atmospheric
pressure; and since the pressure is transmitted through the liquid, the
atmospheric pressure supports the column of mercury in the tube.
Suppose A is at a height H above the level of the mercury in D. Now
the pressure, p, at the bottom of a column of liquid of height H and
density p is given by p = Hpg (p. 1 10). Thus if H = 760 mm = 0-76 m
and p = 13600 kg m" 3 ,
p = Hpg = 0-76 x 13600 x 9-8 = 1013 x 10 5 newton metre -2 .
The pressure at the bottom of a column of mercury 76 cm high for a
particular mercury density and value of g is known as standard pressure
or one atmosphere. By definition, 1 atmosphere = 1 01 325 x 10 5 N m -2 .
Standard temperature and pressure (S.T.P.) is 0°C and 76 cm Hg pressure.
A bar is the name given to a pressure of one million (10 6 ) dyne cm -2 ,
and is thus very nearly equal to one atmosphere. 1 bar = 10 5 newton
m~ 2 (p. 110).
STATIC BODIES, FLUIDS
113
Fortin's Barometer
A barometer is an instrument for measuring the
pressure of the atmosphere, which is required in
weather-forecasting, for example. The most accurate
form of barometer is due to Fortin, and like the
simple arrangement already described, it consists
basically of a barometer tube containing mercury,
with a vacuum at the top, Fig. 4.21. One end of
the tube dips into a pool of mercury contained in a
washleather bag B. A brass scale C graduated in
centimetres and millimetres is fixed at the top of the
barometer. The zero of the scale correspondings to
the tip of an ivory tooth P, and hence, before the
level of the top of the mercury is read from the scales,
the screw S is adjusted until the level of the mercury
in B just reaches the tip of P. A vernier scale V
can be moved by a screw D until the bottom of it
just reaches the top of the mercury in the tube, and
the reading of the height of the mercury is taken
from C and V. Torricelli was the first person to
observe the variation of the barometric height as
the weather changed.
'Correction' to the Barometric Height
For comparison purposes, the pressure read on a
barometer is often 'reduced' or 'corrected' to the
magnitude the pressure would have at 0°C and at
sea-level, latitude 45°. Suppose the 'reduced' pres-
sure is H cm of mercury, and the observed pressure
is H t cm of mercury, corresponding to a temperature of t°C. Then,
since pressure = hpg (p. 1 10),
iS
Fig. 4.21
Fortin barometer
H Po9 = H t Pt9'>
where g is the acceleration due to gravity at sea-level, latitude 45°, and
g' is the acceleration at the latitude of the place where the barometer
was read.
:.H=H t x^x^.
f P 9
The magnitude of g'/g can be obtained from standard tables. The
ratio pjp of the densities = 1/(1 + yt). where y is the absolute or true
cubic expansivity of mercury. Further, the observed height H t , on
the brass scale requires correction for the expansion of brass from
the temperature at which it was correctly calibrated. If the latter is
0°C, then the corrected height is H t (l + at), where a is the mean linear
114 ADVANCED LEVEL PHYSICS
expansivity of brass. Thus, finally, the 'corrected' height H is given by
* 1 + yt g
For further accuracy, a correction must be made for the surface tension
of mercury (p. 132).
Variation of atmospheric pressure with height
The density of a liquid varies very slightly with pressure. The density
of a gas, however, varies appreciably with pressure. Thus at sea-level
the density of the atmosphere is about 1-2 kg m -3 ; at 1000 m above
sea-level the density is about 11 kgm -3 ; and at 5000 m above sea-
level it is about 0-7 kgm -3 . Normal atmospheric pressure is the
pressure at the base of a column of mercury 760 mm high, a liquid
which has a density of about 13600 kg m -3 . Suppose air has a constant
density of about 1-2 kg m~ 3 . Then the height of an air column of this
density which has a pressure equal to normal atmospheric pressure
760 13600 QAV
In fact, the air 'thins' the higher one goes, as explained above. The
height of the air is thus much greater than 8-4 km.
Density, Relative Density
As we have seen, the pressure in a fluid depends on the density of the
fluid.
The density of a substance is defined as its mass per unit volume. Thus
, . _ mass of substance ,.-,
volume of substance
The density of copper is about 90 g cm -3 or 9x 10 3 kg m -3 ; the
density of aluminium is 2-7 g cm -3 or 2-7 x 10 3 kg m -3 ; the density of
water at 4°C is 1 g cm -3 or 1000 kg m -3 .
Substances which float on water have a density less than 1000 kg m - 3
(p. 117). For example, ice has a density of about 900 kg m - 3 ; cork has a
density of about 250 kgm -3 . Steel, of density 8500 kgm -3 , will float
on mercury, whose density is about 13600 kg m -3 at 0°C.
The density of a substance is often expressed relative to the density
of water. This is called the relative density or specific gravity of the
substance. It is a ratio or number, and has no units. The relative
density of mercury is 13-6. Thus the density of mercury is 13-6 times
the density of water, 1000 kg m - 3 , and is hence 13600 kg m - 3 . Copper
has a relative density of 90 and hence a density of 9000 kg m -3 .
Archimedes' Principle
An object immersed in a fluid experiences a resultant upward force
owing to the pressure of fluid on it. This upward force is called the
upthrust of the fluid on the object. Archimedes stated that the upthrust
is equal to the weight of fluid displaced by the object, and this is known as
STATIC BODIES, FLUIDS
115
Archimedes' Principle. Thus if an iron cube of volume 400 cm 3 is
totally immersed in water of density 1 g cm -3 , the upthrust on the
cube = 400 x 1 = 400 gf. If the same cube is totally immersed in oil of
density 0-8 g cm -3 , the upthrust on it = 400 x 0-8 = 320 gf.
Fig. 4.22 Archimedes' Principle
Fig. 4.22 shows why Archimedes' Principle is true. If S is a solid
immersed in a liquid, the pressure on the lower surface C is greater
than on the upper surface B, since the pressure at the greater depth h 2
is more than that at h v The pressure on the remaining surfaces D and
E act as shown. The force on each of the four surfaces is calculated by
summing the values of pressure x area over every part, remembering
that vector addition is needed to sum forces. With a simple rectangular-
shaped solid and the sides, D, E vertical, it can be seen that (i) the
resultant horizontal force is zero, (ii) the upward force on C =
pressure x area A = h 2 pgA, where p is the liquid density and the
downward force on B = pressure x area A = h^pgA. Thus
resultant force on solid = upward force (upthrust) = {h 2 — h^)pgA.
But (h 2 — h 1 )A = volume of solid, V,
.'. upthrust = Vpg = mg, where m = Vp.
.'. upthrust = weight of liquid displaced.
With a solid of irregular shape, taking into account horizontal and
vertical components of forces, the same result is obtained. The upthrust
is the weight of liquid displaced whatever the nature of the object
immersed, or whether it is hollow or not. This is due primarily to the
fact that the pressure on the object depends on the liquid in which it is
placed.
Density or Relative Density measurement by Archimedes' Principle
The upthrust on an object immersed in water, for example, is the
difference between (i) its weight in air when attached to a spring-
balance and (ii) the reduced reading on the spring-balance or 'weight'
116 ADVANCED LEVEL PHYSICS
when it is totally immersed in the liquid. Suppose the upthrust is
found to be 100 gf. Then, from Archimedes' Principle, the object dis-
places 100 gf of water. But the density of water is 1 g cm -3 . Hence the
volume of the object = 100 cm 3 , which is numerically equal to the
difference in weighings in (i) and (ii).
The density or relative density of a solid such as brass or iron can
thus be determined by (1) weighing it in air, m gf say, (2) weighing
it when it is totally immersed in water, m^^ gf say. Then
upthrust = m — m 1 = wt. of water displaced.
.*. relative density of solid =
m
m n -m,
and density of solid, p, = — X density of water.
The density or relative density of a liquid can be found by weighing a
solid in air (m ), then weighing it totally immersed in the liquid (m t ),
and finally weighing it totally immersed in water (m 2 ).
Now m -m 2 = upthrust in water = weight of water displaced,
and m -»ii = upthrust in liquid = weight of liquid displaced.
.'. — = relative density of liquid,
m -m 2
or — X density of water= density of liquid.
m -m 2
Density of Copper Sulphate crystals
If a solid dissolves in water, such as a copper sulphate crystal for example,
its density can be found by totally immersing it in a liquid in which it is insoluble.
Copper sulphate can be weighed in paraffin oil, for example. Suppose the apparent
weight is m v and the weight in air is m . Then
m — wii = upthrust in liquid = Vp,
where V is the volume of the solid and p is the density of the liquid.
. v= m -m l
P
■ A : f VA maSS m ° m <>
. . density of solid = — : = — = .p.
volume V m — m^
The density, p, of the liquid can be found by means of a density bottle,
for example. Thus knowing m^ and m lf the density of the solid can be calculated.
Density of Cork
If a solid floats in water, cork for example, its density can be found by attaching
a brass weight or 'sinker' to it so that both solids become totally immersed in
water. The apparent weight (m^ of the sinker and cork together is then obtained.
Suppose m 2 is the weight of the sinker in air, m 3 is*the weight of the sinker alone
in water, and m is the weight of the cork in air.
STATIC BODIES, FLUIDS 117
Then m 2 — m 3 = upthrust on sinker in water.
'. m + m 2 — m 1 — (m 2 — m 3 ) = upthrust on cork in water
= mo — mi+mj
relative density of cork =
'o
m — m 1 +m 3
Flotation
When an object floats in a liquid, the upthrust on the object must be
equal to its weight for equilibrium. Cork has a density of about 0-25 g
em" 3 , so that 100 cm 3 of cork has a mass of 25 g. In water, then, cork
sinks until the upthrust is 25 gf. Now from Archimedes' Principle,
25 gf is the weight of water displaced. Thus the cork sinks until 25 cm 3
of its 100 cm 3 volume is immersed. The fraction of the volume im-
mersed is hence equal to the relative density.
Ice has a density of about 0-9 g cm -3 . A block of ice therefore
floats in water with about ^ths of it immersed.
Hydrometer
Hydrometers use the principle of flotation to measure density or
relative density. Fig. 4.23. Since they have a constant weight, the
upthrust when they float in a liquid is always the
same. Thus in a liquid of density 1-0 g cm -3 , a
hydrometer of 20 gf will sink until 20 cm 3 is im-
mersed. In a liquid of 20 g cm -3 , it will sink until
only 10 cm 3 is immersed. The density or relative ;==^;
density readings hence increase in a downward ;§=§:
direction, as shown in Fig. 4.23. |||=Eh
Practical hydrometers have a weighted end M l?E=E=E=ql §c=]
for stability, a wide bulb to produce sufficient up- 11BB88 IS
thrust to counterbalance the weight, and a narrow tlEgEg^rjRr
stem BL for sensitivity. If V is the whole volume of fE-Ef€?<=BE==H
the hydrometer in Fig. 4.23, a is the area of the stem FlG 4 23
and y is the length not immersed in a liquid of Hydrometer
density p, then
upthrust = wt. of liquid displaced = (V— ay)p = w,
where w is the weight of the hydrometer.
EXAMPLE
An ice cube of mass 500 g floats on the surface of a strong brine solution of
volume 200-0 cm 3 inside a measuring cylinder. Calculate the level of the liquid
in the measuring cylinder (i) before and (ii) after all the ice is melted, (iii) What
happens to the level if the brine is replaced by 2000 cm 3 water and 50-0 g of ice is
again added? (Assume density of ice, brine = 900, 1 100 kg m~ 3 or 0-9, 11 gem -3 .)
(i) Floating ice displaces 50 g of brine since upthrust is 50 gf.
level on measuring cylinder = 245-5 cm 3
volume displaced = — = — - = 45-5 cm 3 .
density 11
118
ADVANCED LEVEL PHYSICS
(ii) 50 g of ice forms 50 g of water when all of it is melted.
.'. level on measuring cylinder falls to 2500 cm 3 .
(iii) Water. Initially, volume of water displaced = 50cm 3 ,sinceupthrust = 50 g.
.'. level on cylinder = 2500 cm 3 .
If 1 g of ice melts, volume displaced is 1 cm 3 less. But volume of water formed
is 1 cm 3 . Thus the net change in water level is zero. Hence the water level remains
unchanged as the ice melts.
Fluids in Motion. Streamlines and velocity
A stream or river flows slowly when it runs through open country
and faster through narrow openings or constrictions. As shown shortly,
this is due to the fact that water is practically an incompressible fluid,
that is, changes of pressure cause practically no change in fluid density
at various parts.
High velocity v
Low velocity vj
Fig. 4.24 Bernoulli's theorem
Fig. 4.24 shows a tube of water flowing steadily between X and Y,
where X has a bigger cross-sectional area A x than the part Y, of cross-
sectional area A 2 . The streamlines of the flow represent the directions
of the velocities of the particles of the fluid and the flow is uniform or
laminar (p. 204). Assuming the liquid is incompressible, then, if it moves
from PQ to RS, the volume of liquid between P and R is equal to the
volume between Q and S. Thus AJ X = A 2 l 2 , where l t is PR and l 2 is QS,
or l 2 /l x = AJA 2 . Hence l 2 is greater than l v Consequently the velocity
of the liquid at the narrow part of the tube, where, it should be noted,
the streamlines are closer together, is greater than at the wider part Y,
where the streamlines are further apart. For the same reason, slow-
running water from a tap can be made into a fast jet by placing a finger
over the tap to narrow the exit.
Pressure and velocity. Bernoulli's Principle
About 1740, Bernoulli obtained a relation between the pressure and
velocity at different parts of a moving incompressible fluid. If the
viscosity is negligibly small, there are no frictional forces to overcome
(p. 174). In this case the work done by the pressure difference per unit
STATIC BODIES, FLUIDS 119
volume of a fluid flowing along a pipe steadily is equal to the gain of
kinetic energy per unit volume plus the gain in potential energy per unit
volume.
Now the work done by a pressure in moving a fluid through a distance
= force x distance moved = (pressure x area) x distance moved =
pressure x volume moved, assuming the area is constant at a particular
place for a short time of flow. At the beginning of the pipe where the
pressure is p lt the work done per unit volume on the fluid is thus p x ; at
the other end, the work done per unit volume by the fluid is likewise p 2 .
Hence the net work done on the fluid per unit volume ■= p t —p 2 . The
kinetic energy per unit volume = \ mass per unit volume x velocity 2 .
= jpx velocity 2 , where p is the density of the fluid. Thus if v 2 and v t
are the final and initial velocities respectively at the end and the
beginning of the pipe, the kinetic energy gained per unit volume =
jp(v 2 — t>i) 2 . Further, if h 2 and h t are the respective heights measured
from a fixed level at the end and beginning of the pipe, the potential
energy gained per unit volume = mass per unit volume xgx(h 2 ~ h^
Thus, from the conservation of energy,
Pi~P2=jPiv 2 2 -v 1 2 )+P9(h 2 -h 1 )
•'• Pi+if»>i 2 + pghi = P2+lPV 2 2 +pgh 2
.'. p+%pv 2 +pgh = constant,
where p is the pressure at any part and v is the velocity there. Hence it
can be said that, for streamline motion of an incompressible non-viscous
fluid,
the sum of the pressure at any part plus the kinetic energy per unit
volume plus the potential energy per unit volume there is always
constant.
This is known as Bernoulli's principle.
Bernoulli's principle shows that at points in a moving fluid where
the potential energy change pgh is very small, or zero as in flow through
a horizontal pipe, the pressure is low where the velocity is high; con-
versely, the pressure is high where the velocity is low. The principle
has wide applications.
EXAMPLE
As a numerical illustration of the previous analysis, suppose the area of
cross-section A x of X in Fig. 4.25 is 4 cm 2 , the area A 2 of Y is 1 cm 2 , and water
flows past each section in laminar flow at the rate of 400 cm 3 s _ 1 . Then
at X, speedy of water = voLpersecond = 100cms -i = lms -i.
area
at Y, speed v 2 of water = 400 cms" 1 = 4 m s" 1 .
The density of water, p = 1000 kg m~ 3 .
•'• P = $Piv 2 2 -Vi 2 ) = ix 1000x(4 2 -l 2 ) = 7-5 x 10 3 newton m~ 2 .
120
ADVANCED LEVEL PHYSICS
If h is in metres, p = 1000 kg m 3 for water, g = 9-8 m s 2 , then, from hpg,
7-5 x 10 3
h =
1000 x 9-8
= 077 m (approx.).
The pressure head h is thus equivalent to 0-77 m of water.
Applications of Bernoulli's Principle
1. A suction effect is experienced by a person standing close to the
platform at a station when a fast train passes. The fast-moving air
between the person and train produces a decrease in pressure and the
excess air pressure on the other side pushes the person towards the
train.
High velocity
low pressure
Constriction
Air flow
■^_ Air
A — flow
(■)
(ii)
Low velocity
high pressure
Fig. 4.25 Fluid velocity and pressure
2. Filter pump. A filter pump has a narrow section in the middle, so
that a jet of water from the tap flows faster here. Fig. 4.25 (i). This causes
a drop hi pressure near it and air therefore flows in from the side tube
to which a vessel is connected. The air and water together are expelled
through the bottom of the filter pump.
3. Aerofoil lift. The curved shape of an aerofoil creates a fast flow of
air over its top surface than the lower one. Fig. 4.25 (ii). This is shown
by the closeness of the streamlines above the aerofoil compared with
those below. From Bernoulli's principle, the pressure of the air below
is greater than that above, and this produces the lift on the aerofoil.
4. Flow of liquid from wide tank. Suppose a liquid flows through a
hole H at the bottom of a wide tank, as shown in Fig. 4.26. Assuming
B
-\:-
h~
B
■v
Fig. 4.26 Torricelli's theorem
negligible viscosity and streamline flow at
a small distance from the hole, which is an
approximation, Bernoulli's theorem can be
applied. At the top X of the liquid in the
tank, the pressure is atmospheric, say B, the
height measured from a fixed level such as
the hole H is h, and the kinetic energy is
negligible if the tank is wide so that the
level falls very slowly. At the bottom, Y,
near H, the pressure is again B, the height
above H is now zero, and the kinetic energy
is ?pv 2 , where p is the density and v is the
STATIC BODIES, FLUIDS 121
velocity of emergence of the liquid. Thus, from Bernoulli's Principle,
B+phg = B+^pv 2
.-. v 2 = 2gh
Thus the velocity of the emerging liquid is the same as that which
would be obtained if it fell freely through a height h, and this is known
as TorricelWs theorem. In practice the velocity is less than that given by
yjlgh owing to viscous forces, and the lack of streamline flow must also
be taken into account.
EXERCISES 4
What are the missing words in the statements 1-6?
1. In SI units, the moment or torque of a couple is measured in . . .
2. In stable equilibrium, when an object is slightly displaced its centre of
gravity . . .
3. When an object is in equilibrium under the action of three non-parallel
forces, the three forces must . . . one point.
4. The component of a force F in a direction inclined to it at an angle 6 is . . .
5. The sensitivity of a beam balance depends on the depth of the . . . below the
fulcrum.
6. When an object floats, the weight of fluid displaced is equal to the . . .
7. In laminar flow of non- viscous fluid along a pipe, at regions of high pressure
the... is low.
Which of the following answers, A, B, C, D or E, do you consider is the correct
one in the statements 8-10?
8. If a cone is balanced on its apex on a horizontal table and then slightly
displaced, the potential energy of the cone is then A increased, B decreased,
G constant,,/) a minimum, E a maximum.
9. If a hydrometer of mass 20 g and volume 30 cm 3 has a graduated stem of
1 cm 2 , and floats in water, the exposed length of stem is A 30 cm, B 25 cm, C 20 cm,
D 10 cm, E 1 cm.
10. In laminar flow of a non-viscous fluid along a horizontal pipe, the work
per second done by the pressure at any section is equal to A the pressure, B the
volume per second there, C pressure x volume per second there, D pressure x
volume, E pressure x area of cross-section.
11. A flat plate is cut in the shape of a square of side 20-0 cm, with an equi-
lateral triangle of side 200 cm adjacent to the square. Calculate the distance of
the centre of mass from the apex of the triangle.
12. The foot of a uniform ladder is on a rough horizontal ground, and the top
rests against a smooth vertical wall. The weight of the ladder is 40 kgf, and a man
weighing 80 kgf stands on the ladder one-quarter of its length from the bottom.
If the inclination of the ladder to the horizontal is 30°, find the reaction at the
wall and the total force at the ground.
122 ADVANCED LEVEL PHYSICS
13. A rectangular plate ABCD has two forces of 10 kgf acting along AB and
DC in opposite directions. IfAB = 3m,BC = 5m, what is the moment of the
couple acting on the plate? What forces acting along BC and AD respectively
are required to keep the plate in equilibrium?
14. A hollow metal cylinder 2 m tall has a base of diameter 35 cm and is
filled with water to a height of (i) 1 m, (ii) 50 cm. Calculate the distance of the
centre of gravity in metre from the base in each case if the cylinder has no top.
(Metal weighs 20 kg m~ 2 of surface. Assume n = 22/7.)
15. A trap-door 120 cm by 120cmis kept horizontal by a string attached to
the mid-point of the side opposite to that containing the hinge. The other end of
the string is tied to a point 90 cm vertically above the hinge. If the trap-door
weight is 5 kgf, calculate the tension in the string and the reaction at the hinge.
16. Two smooth inclined planes are arranged with their lower edges in contact ;
the angles of inclination of the plane to the horizontal are 30°, 60° respectively,
and the surfaces of the planes are perpendicular to each other. If a uniform rod
rests in the principal section of the planes with one end on each plane, find the
angle of inclination of the rod to the horizontal.
17. Describe and give the theory of an accurate beam balance. Point out the
factors which influence the sensitivity of the balance. Why is it necessary, in very
accurate weighing, to take into account the pressure, temperature, and humidity
of the atmosphere? (0. & C.)
18. Summarise the various conditions which are being satisfied when a body
remains in equilibrium under the action of three non-parallel forces.
A wireless aerial attached to the top of a mast 20 m high exerts a horizontal
force upon it of 60 kgf. The mast is supported by a stay-wire running to the
ground from a point 6 m below the top of the mast, and inclined at 60° to the
horizontal. Assuming that the action of the ground on the mast can be regarded
as a single force, draw a diagram of the forces acting on the mast, and determine
by measurement or by calculation the force in the stay-wire. (C.)
19. The beam of a balance has mass 150 g and its moment of inertia is 5 x 10~ 4
kg m 2 . Each arm of the balance is 10 cm long. When set swinging the beam
makes one complete oscillation in 6 seconds. How far is the centre of gravity of
the beam below its point of support, and through what angle would the beam be
deflected by a weight of 1 milligram placed in one of the scale pans? (C.)
20. Under what conditions is a body said to be in equilibrium? What is meant
by (a) stable equilibrium and (b) unstable equilibrium! Give one example of each.
A pair of railway carriage wheels, each of radius r, are joined by a thin axle;
the mass of the whole ism. A light arm of length /(< r) is attached perpendicularly
to the axle and the free end of the arm carries a point mass M. The wheels rest,
with the axle horizontal, on rails which are laid down a slope inclined at an angle
<j> to the horizontal Show that provided that <j> is not too large and that the
wheels do not slip on the rails, there are two values of the angle d that the arm
makes with the horizontal when the system is in equilibrium, and find these
values of 0. Discuss whether, in each case, the equilibrium is stable or unstable.
(O.&C.)
21. Give a labelled diagram to show the structure of a beam balance. Show
if the knife-edges are collinear the sensitivity is independent of the load. Discuss
other factors which then determine the sensitivity.
A body is weighed at a place on the equator, both with a beam balance and a
very sensitive spring balance, with identical results. If the observations are
STATIC BODIES. FLUIDS 123'
repeated at a place near one of the poles, using the same two instruments, discuss
whether identical results will again be obtained. (L.)
22. Three forces in one plane act on a rigid body. What are the conditions for
equilibrium?
The plane of a kite of mass 6 kg is inclined to the horizon at 60°. The resultant
thrust of the air on the kite acts at a point 25 cm above its centre of gravity, and
the string is attached at a point 30 cm above the centre of gravity. Find the thrust
of the air on the kite, and the tension in the string (C.)
23. In what circumstances is a physical system in equilibrium? Distinguish
between stable, unstable and neutral equilibria.
Discuss the stability of the equilibrium of a uniform rough plank of thickness t,
balanced horizontally on a rough cylindrical-fixed log of radius r, it being assumed
that the axes of plank and log lie in perpendicular directions. (N.)
24. State the conditions of equilibrium for a body subjected to a system of
coplanar parallel forces and briefly describe an experiment which you could
carry out to verify these conditions.
Show how the equilibrium of a beam balance is achieved and discuss the
factors which determine its sensitivity. Explain how the sensitivity of a given
balance may be altered and why, for a particular adjustment, the sensitivity
may be practically independent of the mass in the balance pans. Why is it in-
convenient in practice to attempt to increase the sensitivity of a given balance
beyond a certain limit? (O. & C.)
Fluids
25. An alloy of mass 588 g and volume 100 cm 3 is made of iron of relative
density 80 and aluminium of relative density 2-7. Calculate the proportion
(i) by volume, (ii) by mass of the constituents of the alloy.
26. A string supports a solid iron object of mass 180 g totally immersed in a
liquid of density 800 kg m~ 3 . Calculate the tension in the string if the density of
iron is 8000 kg m -3 .
27. A hydrometer floats in water with 6-0 cm of its graduated stem unimmersed,
and in oil of relative density 0-8 with 40 cm of the stem unimmersed. What is
the length of stem unimmersed when the hydrometer is placed in a liquid of
relative density 0-9?
28L An alloy of mass 170 g has an apparent weight of 95 gf in a liquid of density
1-5 g cm~ 3 . If the two constituents of the alloy have relative densities of 4-0 and
30 respectively, calculate the proportion by volume of the constituents in the
alloy.
29. State the principle of Archimedes and use it to derive an expression for
the resultant force experienced by a body of weight W and density a when it is
totally immersed in a fluid of density p.
A solid weighs 237-5 g in air and 12-5 g when totally immersed in a liquid of
relative density 0-9. Calculate (a) the specific gravity of the solid, (b) the
relative density of a liquid in which the solid would float with one-fifth of its
volume exposed above the liquid surface. (£.)
30. Distinguish between mass and weight. Define density.
Describe and explain how you would proceed to find an accurate value for
the density of gold, the specimen available being a wedding ring of pure gold.
124 ADVANCED LEVEL PHYSICS
What will be the reading of (a) a mercury barometer, (b) a water barometer,
when the atmospheric pressure is 10 5 N m -2 ? The density of mercury may be
taken as 13600 kg m -3 and the pressure of saturated water vapour at room
temperature as 13 mm of mercury. (L.)
31. Describe an experiment which demonstrates the difference between laminar
and turbulent flow in a fluid.
A straight pipe of uniform radius R is joined, in the same straight line, to a
narrower pipe of uniform radius r. Water (which may be assumed to be incom-
pressible) flows from the wider into the narrower pipe. The velocity of flow in
the wider pipe is V and in the narrower pipe is v. By equating work done against
fluid pressures with change of kinetic energy of the water, show that the hydro-
static pressure is lower where the velocity of flow is higher.
Describe and explain one practical consequence or application of this dif-
ference in pressures. (O. &C.)
32. Describe some form of barometer used for the accurate measurement of
atmospheric pressure, and point out the corrections to be applied to the observa-
tion.
Obtain an expression for the correction to be applied to the reading of a
mercurial barometer when the reading is made at a temperature other than
0°C. (L.)
33. State the principle of Archimedes, and discuss its application to the deter-
mination of specific gravities by means of a common hydrometer. Why is this
method essentially less accurate than the specific gravity bottle?
A common hydrometer is graduated to read specific gravities from 0-8 to 10.
In order to extend its range a small weight is attached to the stem, above the
liquid, so that the instrument reads 0-8 when floating in water. What will be the
specific gravity of the liquid corresponding to the graduation 10? {O. & C.)
34. A hydrometer consists of a bulb of volume Vand a uniform stem of volume
v per cm of its length. It floats upright in water so that the bulb is just completely
immersed. Explain for what density range this hydrometer may be used and
how you would determine the density of such liquids. Describe the graph which
would be obtained by plotting the reciprocal of the density against the length of
the stem immersed.
A hydrometer such as that described sinks to the mark 3 on the stem, which is
graduated in cm, when it is placed in a liquid of density 0*95 gem -3 . If the
volume per cm of the stem is 01 cm 3 , find the volume of the bulb. (L.)
35. A straight rod of length I, small cross sectional area a and of material
density p is supported by a thread attached to its upper end. Initially the rod
hangs in a vertical position over a liquid of density o and then is lowered until it is
partially submerged Derive and discuss the equilibrium conditions of the rod
neglecting surface tension. (N.)
chapter five
Surface Tension
Intermolecular Forces
The forces which exist between molecules can explain many of the
bulk properties of solids, liquids and gases. These intermolecular
forces arise from two main causes :
(1) The potential energy of the molecules, which is due to interactions
with surrounding molecules (this is principally electrical, not gravita-
tional, in origin).
(2) The thermal energy of the molecules — this is the kinetic energy
of the molecules and depends on the temperature of the substance
concerned.
We shall see later that the particular state or phase in which matter
appears — that is, solid, liquid or gas — and the properties it then has,
are determined by the relative magnitudes of these two energies.
Potential energy and Force
In bulk, matter consists of numerous molecules. To simplify the
situation, Fig. 5.1 shows the variation of the potential energy V between
two molecules at a distance r apart.
Along the part BCD of the curve, the potential energy V is negative.
Along the part AB, the potential energy Vis positive. The force between
the molecules is always given by F = —dV/dr = —potential gradient.
Along CD the force is attractive and it decreases with distance r
according to an inverse-power of r. Along ABC, the force is repulsive.
Fig. 5.1 shows the variation of F with r.
At C, the minimum potential energy point of the curve, the molecules
would be at their normal distance apart in the absence of thermal
energy. The equilibrium distance OM, r , is of the order 2 or 3 x
10~ 10 m (2 or 3 A) for a solid. At this distance apart, the attractive and
repulsive forces balance each other. If the molecules are closer, (r < r ),
they would repel each other. If they are further apart, (r > r \ they
attract each other.
Phases or States of Matter
The molecules in a solid are said to be in a 'condensed' phase or state.
Their thermal energy is then relatively low compared with their
potential energy V and the molecules are 'bound' to each other.
They may now vibrate about C, the minimum of the curve in Fig. 5.1.
When the thermal energy increases by an amount corresponding to
CC in Fig. 5.1, the molecule can then oscillate between the limits
125
126
ADVANCED LEVEL PHYSICS
Fig. 5.1 Molecular potential energy and force
corresponding to X and Y. From the graph of force, F, it can be seen
that when the molecule is on the left of C it experiences a greater
force towards it than when on the right. Consequently the molecule
returns quicker to C Thus the mean position G is on the right of C.
This corresponds to a mean separation of molecules which is greater
than r . Thus the solid expands when its thermal energy is increased.
As the thermal energy increases further, at some particular tem-
perature the molecules are able to move comparatively freely relative
to neighbouring molecules. The solid then loses its rigid form and
becomes a. liquid. The molecules in the liquid constantly exchange
places with other molecules, whereas in a solid the neighbours of a
particular molecule remain unchanged. Further, the molecules of a
liquid have translational as well as vibrational energy, that is, they
move about constantly through the liquid, whereas molecules of a
solid have vibrational energy only.
As the temperature of the liquid rises, the thermal energy of the
molecules further increases. The average distance between the molecules
then also increases and so their mean potential energy approaches
zero, as can be seen from Fig. 5.1. At some stage the increased thermal
SURFACE TENSION 127
energy enables the molecules to completely break the bonds of attrac-
tion which keep them in a liquid state. The molecules then have little
or no interaction and now form a gas. At normal pressures the forces
of attraction between the gas molecules are comparatively very small
and the molecules move about freely inside the volume they occupy.
Gas molecules which are monatomic such as helium have translational
energy only. Gas molecules such as oxygen or carbon dioxide, with
two or more atoms, have rotational and vibrational energies in addition
to translational energy.
Gases
At normal pressure, permanent gases such as air or oxygen obey
Boyle's law, pV = constant, to a very good approximation. Now in
the absence of attractive forces between the molecules, and assuming
their actual volume is negligibly small, the kinetic theory of gases
shows that Boyle's law is obeyed by this ideal gas. Consequently, the
attractive forces between the gas molecules at normal pressure are
unimportant. They increase appreciably when the gas is at high pres-
sure as the molecules are then on the average very much closer.
In the bulk of the gas, the resultant force of attraction between a
particular molecule and those all round it is zero when averaged over a
period. Molecules which strike the wall of the containing vessel,
however, are retarded by an unbalanced force due to molecules behind
them. The observed pressure p of a gas is thus less than the pressure in
the ideal case, when the attractive forces due to molecules is zero.
Van der Waals derived an expression for this pressure 'defect'. He
considered that it was proportional to the product of the number of
molecules per second striking unit area of the wall and the number per
unit volume behind them, since this is a measure of the force of attrac-
tion. For a given volume of gas, both these numbers are proportional
to the density of the gas. Consequently the pressure defect, p x say, is
proportional to p x p or p 2 . For a fixed mass of gas, p oc 1/V, where V
is the volume. Thus p x = a/V 2 , where a is a constant for the particular
gas. Taking into account the attractive forces between the molecules,
it follows that, if p is the observed pressure, the gas pressure in the bulk
of the gas = p+a/V 2 .
The attraction of the walls on the molecules arriving there is to
increase their velocity from v say to v + Av. Immediately after rebound-
ing from the walls, however, the force of attraction decreases the velocity
to v again. Thus the attraction of the walls has no net effect on the
momentum change due to collision. Likewise, the increase in momen-
tum of the walls due to their attraction by the molecules arriving is lost
after the molecules rebound.
The effect of the volume actually occupied by all the molecules is
represented by a constant b, so that the volume of the space in which
they move is not V but (V—b). The magnitude of b is not the actual
volume of the molecules, as if they were swept into one corner of the
space, since they are in constant motion, b has been estimated to be
about four times the actual volume.
128
ADVANCED LEVEL PHYSICS
Surf ace Tension.
We now consider in detail a phenomenon of a liquid surface called
surface tension. As we shall soon show, surface tension is due to inter-
molecular attraction.
It is a well-known fact that some insects, for example a water-carrier,
are able to walk across a water surface ; that a drop of water may remain
suspended for some time from a tap before falling, as if the water
particles were held together in a bag; that mercury gathers into small
droplets when spilt; and that a dry steel needle may be made, with
Fig. 5.2 Needle floating on water
care, to float on water, Fig. 5.2. These observations suggest that the
surface of a liquid acts like an elastic skin covering the liquid or is in a
state of tension. Thus forces 5 in the liquid support the weight W of
the needle, as shown in Fig. 5.2.
Energy of Liquid Surface. Molecular theory
The feet that a liquid surface is in a state of tension can be explained
by the intermolecular forces discussed on p. 125. In the bulk of the
liquid, which begins only a few molecular diameters downwards from
the surface, a particular molecule such as A is surrounded by an equal
number of molecules on all sides. This can be seen by drawing a sphere
round A. Fig. 5.3. The average distance apart of the molecules is such
that the attractive forces balance the repulsive forces (p. 145), Thus the
average intermolecular force between A and the surrounding molecules
is zero. Fig. 5.3.
,'' C \ Liquki surface / B ^
PBBlll
Resultant
.. . , , inward force
No inward force nn R
on A on a
Fig. 5.3 Molecular forces in liquid
SURFACE TENSION 129
Consider now a molecule such as C or B in the surface of the liquid.
There are very few molecules on the vapour side aboveC or B compared
with the liquid below, as shown by drawing a sphere round C or B.
Thus if C is displaced very slightly upward, a resultant attractive force
F on C, due to the large number of molecules below C, now has to be
overcome. It follows that if all the molecules in the surface were removed
to infinity, a definite amount of work is needed. Consequently molecules
in the surface have potential energy. A molecule in the bulk of the
liquid forms bonds with more neighbours than one in the surface.
Thus bonds must be broken, i.e. work must be done, to bring a molecule
into the surface. Molecules in the surface of the liquid hence have more
potential energy than those in the bulk.
Surface area. Shape of drop
The potential energy of any system in stable equilibrium is a mini-
mum. Thus under surface tension forces, the area of a liquid surface will
have the least number of molecules in it, that is, the surface area of a
given volume of liquid is a minimum. Mathematically, it can be shown
that the shape of a given volume of liquid with a rninimum surface area
is a sphere.
Raindrops
° fe=>^^zBKK>>^_ Aniline
° F>=-=->^ l ^^^^" s P r,enca '
Mercury I
.Warm
water
(i) (>')
Fig. 5.4 Liquid drops
This is why raindrops, and small droplets of mercury, are approxi-
mately spherical in shape. Fig. 5.4 (i). To eliminate completely the
effect of gravitational forces, Plateau placed a drop of oil in a mixture
of alcohol and water of the same density. In this case the weight of the
drop is counterbalanced by the upthrust of the surrounding liquid.
He then observed that the drop was a perfect sphere. Plateau's 'spherule'
experiment can be carried out by warming water in a beaker and then
carefully introducing aniline with the aid of a pipette. Fig. 5.4 (ii).
At room temperature the density of aniline is slightly greater than water.
At a higher temperature the densities of the two liquids are roughly the
same and the aniline is then seen to form spheres, which rise and fall in
the liquid.
A soap bubble is spherical because its weight is extremely small and
the liquid shape is then mainly due to surface tension forces. Although
the density of mercury is high, small drops of mercury are spherical.
The ratio of surface area (4nr 2 ) to weight (or volume, 4nr 3 /3) of a
sphere is proportional to the ratio r 2 /r 3 , or to 1/r. Thus the smaller
130
ADVANCED LEVEL PHYSICS
the radius, the greater is the influence of surface tension forces compared
to the weight Large mercury drops, however, are flattened on top.
This time the effect of gravity is relatively greater. The shape of the
drop conforms to the principle that the sum of the gravitational poten-
tial energy and the surface energy must be a minimum, and so the
centre of gravity moves down as much as possible.
Lead shot is manufactured by spraying lead from the top of a tall
tower. As they fall, the small drops form spheres under the action of
surface tension forces.
Surface tension definition. Units, dimensions
Since the surface of a liquid acts like an elastic skin, the surface is
in a state of tension. A blown-up football bladder has a surface in a
state of tension. This is a very rough analogy because the surface
tension of a bladder increases as the surface area increases, whereas
the surface tension of a liquid is independent of surface area. Any
line in the bladder surface is then acted on by two equal and opposite
forces, and if the bladder is cut with a knife the rubber is drawn away
from the incision by the two forces present.
R. C. Brown and others have pointed out that molecules in the sur-
face of a liquid have probably a less dense packing than those in the
bulk of the liquid, as there are fewer molecules in the surface when its
area is a minimum. The average separation between molecules in the
surface are then slightly greater than those inside. On average, then,
the force between neighbouring molecules in the surface are attractive
(see p. 125). This would explain the existence of surface tension.
The surface tension, y, of a liquid, sometimes called the coefficient
of surface tension, is defined as the force per unit length acting in the
surface at right angles to one side of a line drawn in the surface. In Fig.
5.5 AB represents a line 1 m long. The unit of y is newton metre' 1
(Nm -1 ).
-iy^-z^bi-z-z-z- z-e: *: - - - -
Fig. 5.5 Surface tension
The 'magnitude' of y depends on the temperature of the liquid and
on the medium on the other side of the surface. For water at 20°C in
contact with air, y = 7-26 x 10 ~ 2 newton metre -1 . For mercury at
20°C in contact with air, y = 46-5 x 10" 2 Nm -1 . The surface tension
of a water-oil (olive-oil) boundary is 2-06 x 10 ~ 2 Nm -1 , and for a
mercury- water boundary it is 42-7 xlO _2 Nm _1 .
SURFACE TENSION 131
Since surface tension y is a 'force per unit length', the dimensions of
dimensions of force MLT " 2
surface tension =
dimensions of length
= MT~ 2 .
We shall see later that surface tension can be defined also in terms
of surface energy (p. 146).
Some surface tension phenomena
The effect of surface tension forces in a soap film can be demonstrated
by placing a thread B carefully on a soap film formed in a metal ring
A, Fig. 5.6 (i). The surface tension forces on both sides of the thread
counterbalance, as shown in Fig. 5.6 (i). If the film enclosed by the
thread is pierced, however, the thread is pulled out into a circle by the
surface tension forces F at the junction of the air and soap-film, Fig. 5.6
(ii). Observe that the film has contracted to a minimum area.
Another demonstration of surface tension forces can be made by
sprinkling light dust or lycopodium powder over the surface of water
contained in a dish. If the middle of the water is touched with the end of
a glass rod which had previously been dipped into soap solution, the
powder is carried away to the sides by the water. The explanation lies
in the fact that the surface tension of water is greater than that of a
r A
(i) (H)
Fig. 5.6 Contraction of surface
soap-film (p. 136). The resultant force at the place where the rod
touched the water is hence away from the rod, and thus the powder
moves^away from the centre towards the sides of the vessel.
A toy duck moves by itself across the surface of water when it has a
small bag of camphor attached to its base. The camphor lowers the
surface tension of the water in contact with it, and the duck is urged
across the water by the resultant force on it.
Capillarity
When a capillary tube is immersed in water, and then placed verti-
cally with one end in the liquid, observation shows that the water
rises in the tube to a height above the surface. The narrower the tube,
132
ADVANCED LEVEL PHYSICS
the greater is the height to which the water rises, Fig. 5.7 (i). See also
p. 140). This phenomenon is known as capillarity, and it occurs when
blotting-paper is used to dry ink. The liquid rises up the pores of the
paper when it is pressed on the ink.
„' -T^. > * .r » < : »
(')
Fig. 5.7
rciiryJ
(ii)
Capillary rise and fall
When a capillary tube is placed inside mercury, however, the liquid
is depressed below the outside level, Fig. 5.7 (ii). The depression in-
creases as the diameter of the capillary tube decreases. See also p. 141.
Angle of Contact
In the case of water in a glass capillary tube, observation of the
meniscus shows that it is hemispherical if the glass is clean, that is, the
glass surface is tangential to the meniscus where the water touches it.
In other cases where liquids rise in a capillary tube, the tangent BN to
the liquid surface where it touches the glass may make an acute angle
with the glass, Fig. 5.8 (i). The angle is known as the angle of contact
between the liquid and the glass, and is always measured through the
liquid. The angle of contact between two given surfaces varies largely
with their freshness and cleanliness. The angle of contact between
water and very clean glass is zero, but when the glass is not clean the
angle of contact may be about 8° for example. The angle of contact
between alcohol and very clean glass is zero.
B
-3d
EH
:-c|
N
(>) (ii)
Fig. 5.8 Angle of contact
SURFACE TENSION
133
When a capillary tube is placed inside mercury, observation shows
that the surface of the liquid is depressed in the tube and is convex
upwards. Fig. 5.8 (ii). The tangent BN to the mercury at the point B
where the liquid touches the glass thus makes an obtuse angle, 0, with
the glass when measured through the liquid. We shall see later (p. 160)
that a liquid will rise in a capillary tube if the angle of contact is acute,
and that a liquid will be depressed in the tube if the angle of contact is
obtuse. For the same reason, clean water spreads over, or 'wets', a
clean glass surface when spilt on it, Fig. 5.9 (i); the angle of contact is
zero. On the other hand, mercury gathers itself into small pools or
globules when spilt on glass, and does not 'wet' glass, Fig. 5.9 (ii). The
angle of contact is obtuse.
Water
77777777777777777777
Glass
(i)
Mercury
Glass
(ii)
Fig. 5.9 Water and mercury on glass
The difference in behaviour of water and mercury on clean glass can
be explained in terms of the attraction between the molecules of these
substances. It appears that the force of cohesion between two molecules
of water is less than the force of adhesion between a molecule of water
and a molecule of glass; and thus water spreads over glass. On the
other hand, the force of cohesion between two molecules of mercury
is greater than the force of adhesion between a molecule of mercury
and a molecule of glass; and thus mercury gathers in pools when
spilt on glass.
Angle of Contact measurement
The angle of contact can be found by means of the method outlined
inFig.5.10(i),(ii).
Acute _ ^
Obtuse
0) ('0
Fig. 5.10 Angle of contact measurement
A plate X of the solid is placed at varying angles to liquid until the
surface S appears to be plane at X. The angle made with the liquid
surface is then the angle of contact. For an obtuse angle of contact,
a similar method can be adopted. In the case of mercury and glass, a
thin plane mirror enables the liquid surface to be seen by reflection.
For a freshly-formed mercury drop in contact with a clean glass plate,
the angle of contact is 137°.
134
ADVANCED LEVEL PHYSICS
Measurement of Surface Tension by Capillary Tube Method
Theory. Suppose y is the magnitude of the surface tension of a
liquid such as water, which rises up a clean glass capillary tube and has
an angle of contact zero. Fig. 5.11 shows a section of the meniscus
M at B, which is a hemisphere. Since the glass AB is a tangent to the
liquid, the surface tension forces, which act along the boundary of the
liquid with the air, act vertically downwards on the glass. By the law
of action and reaction, the glass exerts an equal force in an upward
direction on the liquid. Now surface tension, y, is the force per unit
length acting in the surface of the liquid, and the length of liquid in
contact with the glass is 2nr, where r is the radius of the capillary tube.
. (D
1
2nr x y = upward force on liquid
A Y*2nr
B-^
Meniscus
M
I Weight =
v nr 2 hpg
Fig. 5.1 1 Rise in capillary tube— theory
If y is in newton metre _1 and r is in metres, then the upward force is in
newtons.
This force supports the weight of a column of height h above the
outside level of liquid. The volume of the liquid = nr 2 h, and thus the
mass, m, of the liquid column = volume x density = nr 2 hp, where p
is the density. The weight of the liquid = mg = nr 2 hpg.
If p is in kg m~ 3 ,r and h in metres, and g = 9-8 m s -2 , then nr 2 hpg
is in newtons.
From (1), it now follows that
.'. 2nry = nr 2 hpg
rhpg
7 =
(2)
If r = 0-2 mm = 0-2 x 10" 3 m, h = 6-6 cm for water = 6-6 x 10" 2 m,
and p = 1 g cm 3 = 1000 kg m , then
y =
0-2 x 10 _ 3 x 6-6 x 10 -2 x 1000 x 9-8
= 6-5xlO _2 Nm
-l
In deriving this formula for y it should be noted that we have (i)
assumed the glass to be a tangent to the liquid surface meeting it,
(ii) neglected the weight of the small amount of liquid above the
bottom of the meniscus at B, Fig. 5.11.
SURFACE TENSION
135
Experiment In the experiment, the capillary tube C is supported in
a beaker Y, and a pin P, bent at right angles at two places, is attached
to C by a rubber band, Fig. 5.12. P is adjusted until its point just touches
the horizontal level of the liquid in the beaker. A travelling microscope
is now focussed on to the meniscus M in C, and then it is focussed on to
the point of P, the beaker being removed for this observation. In this
way the height h of M above the level in the beaker is determined. The
<>
h M — d
Fig. 5.12 Surface tension by capillary rise
radius of the capillary at M can be found by cutting the tube at this
place and measuring the diameter by the travelling microscope ; or by
measuring the length, /, and mass, m, of a mercury thread draw n into
the tube, and calculating the radius, r, from the relation r = y/m/nlp,
where p is the density of mercury. The surface tension y is then calcu-
lated from the formula y = rhpg/2. Its magnitude for water at 15°C
is 7-33 x 10 ~ 2 newton metre -1 .
Measurement of Surface Tension by Microscope Slide
Besides the capillary tube method, the surface tension of water can
be measured by weighing a microscope slide in air, and then lowering it
until it just meets the surface of water, Fig. 5.13. The surface tension
force acts vertically downward round the boundary of the slide, and
pulls the slide down. If a and b are the length and thickness of the slide,
Fig. 5.13 Surface tension by microscope slide
136
ADVANCED LEVEL PHYSICS
then, since y is the force per unit length in the liquid surface and
(2a + 26) is the length of the boundary of the slide, the downward
force = y(2a+2b). If the mass required to counterbalance the force is
m, then
y(2a + 2b) = mg,
mg
' " ' 2a + 2b'
If m = 0-88 gramme, a = 60 cm, b = 0-2 cm, then :
0'88xl(T 3 (kg)x9-8(ms- 2 )
2x(6 + O2)xl0- 2 (m)
= 7-0xl(T 2 Nm-
Surface Tension of a Soap Solution
The surface tension of a soap solution can be found by a similar
method. A soap-film is formed in a three-sided metal frame ABCD, and
the apparent weight is found, Fig. 5.14. When the film is broken by
piercing it, the decrease in the apparent weight, mg, is equal to the
Fig. 5.14 Surface tension of soap film
surface tension force acting downwards when the film existed. This is
equal to 2yb, where b = BC, since the film has two sides.
.'. 2yb = mg,
y =
mg
2b'
It will be noted that the surface tension forces on the sides AB, CD
of the frame act horizontally, and their resultant is zero.
A soap film can be supported in a vertical rectangular frame but a
film of water can not. This is due to the fact that the soap drains down-
ward in a vertical film, so that the top of the film has a lower concentra-
tion of soap than the bottom. The surface tension at the top is thus
greater than at the bottom (soap diminishes the surface tension of
pure water). The upward pull on the film by the top bar is hence
greater than the downward pull on the film by the lower bar. The net
upward pull supports the weight of the film. In the case of pure water,
however, the surface tension would be the same at the top and bottom,
and hence there is no net force in this case to support a water film in a
rectangular frame.
SURFACE TENSION
137
Pressure Difference in a Bubble or Curved Liquid Surface
As we shall see presently, the magnitude of the curvature of a liquid,
or of a bubble formed in a liquid, is related to the surface tension of
Consider a bubble formed inside a
liquid, Fig. 5.15. If we consider the
equilibrium of one half, B, of the
bubble, we can see that the surface
tension force on B plus the force on
B due to the external pressure p A =
the force on B due to the internal
pressure p 2 inside the bubble. The
force on B due to the pressure p t is
given by nr 2 x p x , since nr 2 is the
area of the circular face of B and
pressure is 'force per unit area'; the
force on B due to the pressure p 2
is given similarly by nr 2 x p 2 . The surface tension force acts round
the circumference of the bubble, which has a length 2nr; thus the force
is 2nry. It follows that
2nry+nr 2 p l = nr 2 p 2 .
Fig. 5.15 Excess pressure in bubble
Simplifying,
or
.'. 2y = r{p 2 -p x ),
2y
Now (p 2 —Pi) is the excess pressure, p, in the bubble over the outside
pressure. 2
.". excess pressure, p, — — . . . (1)
Although we considered a bubble, the same formula for the excess
pressure holds for any curved liquid surface or meniscus, where r is
its radius of curvature and y is its surface tension, provided the angle
of contact is zero. If the angle of contact is 6, the formula is modified
by replacing y by ycos 6. Thus, in general,
2ycosd
excess pressure, p, = — . . . (2)
Excess Pressure in Soap Bubble
A soap bubble has two liquid surfaces in contact with air, one inside
the bubble and the other outside the bubble. The force on one half, B,
of the bubble due to surface tension forces is thus y x 2nr X 2, i.e.,
y x Anr, Fig. 5.16. For the equilibrium of B, it follows that
4nry + nr 2 p 1 = nr 2 p 2 ,
where p 2 , p x are the pressures inside and outside the bubble respectively.
Simplifying,
4y
Ay
excess pressure p = -*-
(3)
138
ADVANCED LEVEL PHYSICS
■
Hi
Fig. 5.16 Bubble in water
This result for excess pressure should be compared with the result
obtained for a bubble formed inside a liquid, equation (1).
If y for a soap solution is 25xl0 -3 Nm" 1 , the excess pressure
inside a bubble of radius 0-5 cm or 0-5 x 10 -2 m is hence given by:
P =
4 x 25 x 10
0-5 x 10 _
-3
= 20Nm
Two soap-bubbles of unequal size can be blown on the ends of a tube,
communication between them being prevented by a closed tap in the
middle. If the tap is opened, the smaller bubble is observed to collapse
gradually and the size of the larger bubble increases. This can be
explained from our formula p = 4y/r, which shows that the pressure
of air inside the smaller bubble is greater than that inside the larger
bubble. Consequently air flows from the smaller to the larger bubble
when communication is made between the bubbles, and the smaller
bubble thus gradually collapses.
Since the excess pressure in a bubble is inversely-proportional to the
radius, the pressure needed to form a very small bubble is high. This
explains why one needs to blow hard to start a balloon growing. Once
the balloon has grown, less air pressure is needed to make it expand
more.
Surface Tension of Soap-Bubble
A
K
O
Surface
Fig. 5.17
tension of soap-bubble
The surface tension of a soap solution
can be measured by blowing a small soap-
bubble at the end B of a tube connected
to a manometer M, Fig. 5.17. The tap T
is then closed, the diameter d of the bubble
is measured by a travelling microscope,
and the difference in levels h of the liquid
in the manometer is observed with the
same instrument. The excess pressure, p,
in the bubble = hpg, where p is the
density of the liquid in M.
7 =
hpgd
SURFACE TENSION
139
Rise or Fall of Liquids in Capillary Tubes
From our knowledge of the angle of contact and the excess pressure
on one side of a curved liquid surface, we can deduce that some liquids
will rise in a capillary tube, whereas others will be depressed.
Suppose the tube A is placed in water, for example, Fig. 5.18 (i). At
first the liquid surface becomes concave upwards in the tube, because
the angle of contact with the glass is zero. Consequently the pressure
on the air side, X, of the curved surface is greater than the pressure on
the liquid side Y by 2y/r, where y is the surface tension and. r is the
radius of curvature of the tube. But the pressure at X is atmospheric,
H. Hence the pressure at Y must be less than atmospheric by 2y/r.
Fig. 5.18 (i) is therefore impossible because it shows the pressure at Y
equal to the atmospheric pressure. Thus, as shown in Fig. 5.18 (ii), the
liquid ascends the tube to a height h such that the pressure at N is less
than at M by 2y/r, Fig. 5.18 (ii). A similar
argument shows that a liquid rises in a
capillary tube when the angle of contact
is acute.
The angle of contact between mercury
and glass is obtuse (p. 133). Thus when
a capillary tube is placed in mercury the
liquid first curves downwards. The pres-
sure inside the liquid just below the
curved surface is now greater than the
pressure on the other side, which is
atmospheric, and the mercury therefore
moves down the tube until the excess
pressure = 2y cos 0/r, with the usual no-
tation. A liquid thus falls in a capillary
tube if the angle of contact is obtuse.
H.
H
,*K 2
II
gyjp;|
H
(0
(ii)
Fig. 5.18
Capillary rise by excess pressure
Capillary Rise and Fall by Pressure Method
We shall now calculate the capillary rise of water by the excess
pressure formula p = 2y/r, or p = 2y cos 0/r.
In the case of a capillary tube dipping into water, the angle of contact
is practically zero, Fig. 5.19 (i). Thus if p 2 is the pressure of the atmos-
phere, and p t is the pressure in the liquid, we have
2y
Pi- Pi =y-
I r
(0
Fig. 5.19
Excess
pressure
application
(ii)
140 ADVANCED LEVEL PHYSICS
Now if if is the atmospheric pressure, h is the height of the liquid in
the tube and p its density,
p 2 = H and p t = H — hpg,
.;H-(H-hpg) = Q
.'• hpg = -£
.•.* = -& . . . (i)
rpg y '
The formula shows that h increases as r decreases, i.e., the narrower the
tube, the greater is the height to which the water rises (see Fig. 5.7 (i),
p. 132).
If the height / of the tube above the water is less than the calculated
value of h in the above formula, the water surface at the top of the tube
now meets it at an acute angle of contact 0. The radius of the meniscus
is therefore r/cos 0, and Ipg = 2y/(r/cos 0), or
Z = ?*~? .... (ii)
rpg w
Dividing (ii) by (i), it follows that
COS = -r.
h
Thus suppose water rises to a height of 10 cm in a capillary tube
when it is placed in a beaker of water. If the tube is pushed down until
the top is only 5 cm above the outside water surface, then cos =
jq = 0-5. Thus = 60°. The meniscus now makes an angle of contact
of 60° with the glass. As the tube is pushed down further, the angle of
contact increases beyond 60°. When the top of the tube is level with the
water in the beaker, the^meniscus in the tube becames plane. (See
Example 2, p. 141.)
With Mercury in Glass
Suppose that the depression of the mercury inside a tube of radius r
is h, Fig. 5.19 (ii). The pressure p 2 below the curved surface of the
mercury is then greater than the (atmospheric) pressure p 1 outside the
curved surface; and, from our general result,
2y cos
P2~Pl = y ,
where is the supplement of the obtuse angle of contact of mercury
with glass, that is, is an acute angle and its cosine is positive. But
p x —H and p 2 = H + hpg, where H is the atmospheric pressure.
/rr , x rx 2ycos0
.-. (H + hpg)-H = ^— .
SURFACE TENSION
'• hpg =
h =
2y cos
2y cos
rpg
141
(1)
The height of depression, h, thus increases as the radius r of the tube
decreases. See Fig. 5.7 (ii), p. 132.
EXAMPLES
1. Define surface tension of a liquid and describe
a method of finding this quantity for alcohol.
It water rises in a capillary tube 5-8 cm above the
free surface of the outer liquid, what will happen to
the mercury level in the same tube when it is placed
in a dish of mercury? Illustrate this by the aid of a
diagram. Calculate the difference in level between the
mercury surfaces inside the tube and outside. (S.T. of
water = 75 x 10" 3 N m~ 1 . S.T. of mercury = 547 x 10" 3
N m~ l . Angle of contact of mercury with clean glass =
130°. Density of mercury = 13600 kg m -3 .) (L.)
Second part. The mercury is depressed a distance
h below the outside level, and is convex upward, Fig. 5.20.
Suppose r is the capillary tube radius.
For water, h ~ 5-8 cm = 5-8x 10~ 2 m, y = 75 x 10" 3 newton m -1 , p = 1000
kgm -3 , g = 9-8 ms~ 2 .
Fig. 5.20 Example
From y = rhpg/2,
.-. 75 x 10 -3 = r x 5-8 x 10" 2 x 1000 x 9-8/2 (r in metre).
For mercury, p = 13-6 x 10 3 kg m -3 , y~ 547 x 10~ 3 newton m _1 .
. . ^ 2ycos50°
rpg
__ 2 x 547 x 10~ 3 cos 50° x 5-8 x 10~ 2 x 1000 x 9-8
13-6xl0 3 x9'8x75xl0 _3 x2
= 002 m = 2 cm.
2. On what grounds would you anticipate some connection between the
surface tension of a liquid and its latent heat of vaporization?
A vertical capillary tube 10 cm long tapers uniformly from an internal diameter
of 1 mm at the lower end to 0-5 mm at the upper end. The lower end is just touching
the surface of a pool of liquid of surface tension 6 x 10~ 2 Nm" 1 , density 1200 kg
m " 3 and zero angle of contact with the tube. Calculate the capillary rise, justifying
your method Explain what will happen to the meniscus if the tube is slowly
lowered vertically until the upper end is level with the surface of the pool (0. <£ C.)
Suppose S is the meniscus at a height h cm above the liquid surface. The tube
tapers uniformly and the change in radius for a height of 10 cm is (005— 0025)
or 0025 cm, so that the change in radius per cm height is 00025 cm. Thus at a
height h cm, radius of meniscus S is given by
r = (0-05 -0-0025 h) x 10 ~ 2 m
142
ADVANCED LEVEL PHYSICS
0025
005
Fig. 5.21 Example
The pressure above S is atmospheric, A. The pressure below S is (A — hpg).
.: pressure difference = (ftx 10" >, = ^ = ^^^
■ 00»-0«B»> = *»? = ,^ X ^°"' = 0102.
pg 10 2 x 1200x9-8
.. h 2 -20h= -40(approx.).
.-. {h- 10) 2 = 100-40=60.
.'. fc = 10-760= 2-2 cm.
If the tube is slowly lowered the meniscus reaches the top at some stage. On
further lowering the tube the angle of contact changes from zero to an acute angle.
When the upper end is level with liquid surface the meniscus becomes plane.
3. The surface tension of water is 7-5 x 10~ 2 newton m _1 and the angle of
contact of water with glass is zero.' Explain what these statements mean.
Describe an experiment to determine either (a) the surface tension of water, or
(b) the angle of contact between paraffin wax and water.
A glass U-tube is inverted with the open ends of the straight limbs, of diameters
respectively 0-500 mm and 100 mm, below the surface of water in a beaker. The
air pressure in the upper part is increased until the meniscus in one limb is level
with the water outside. Find the height of water in the other limb. (The density
of water may be taken as 1000 kg m~ 3 .) (L.)
r^
.Q-.
r^= 0025 |g^ = 005;
Fig. 5.22 Example (radii in cm)
SURFACE TENSION
143
Suppose p is the air pressure inside the U-table when the meniscus Q is level
with the water outside and P is the other meniscus at a height h. Let A be the
atmospheric pressure. Then, if r x is the radius at P,
2y
p-(A-hpg)
' i
since the pressure in the liquid below P is (A — hpg).
The pressure in the liquid below Q = A. Hence, for Q,
2y
p-A =
(i)
(«)
where r 2 is the radius.
From (i) and (ii), it follows that
»P9 = ?-?
r l r 2
•*-L&!-an
P9\J\ r 2j
1 | " 2x0075
9800[p-25xl(r
2x0-075 |
0-5 xKT 3 ]
= 31 x 10~ 2 m (approx).
Effects of surface tension in measurements
When a hydrometer is used to measure relative density or density,
the surface tension produces a downward force F on the hydrometer.
If r is the radius of the stem and the angle of contact is~ zero, then
F = 2nry. For a narrow stem, the error produced in reading the
relative density from the graduations is small.
Another case of an undesirable surface tension effect occurs in
measurements of the height of liquid columns in glass tubes. The height
of mercury in a barometer, for example, is depressed by surface ten-
sion (p. 114). If the tubes are wide, surface tension forces can be neg-
lected. If they are narrow, the forces must be taken into account. As
an illustration, consider an inverted U-tube dipping into two liquids
B and C. Fig. 5.23. These can be drawn up into the tubes to heights
Fig. 5.23 Comparison of densities
144 ADVANCED LEVEL PHYSICS
h lt h 2 respectively above the outside level. In the absence of surface
tension forces, p + h^^g = atmospheric pressure H, where p is the air
pressure at the top of the tubes = p+h 2 p 2 g. Thus h x p t = h 2 p 2 , or
^i/h 2 = PilPv Thus the liquid densities may be compared from the
ratio of the heights of the liquid columns.
To take account of surface tension, we proceed as follows. Using the notation
on p. 137.
n n - 2yi
P-Pi = -zr,
where p is the air pressure at the top of the tubes, p t is the pressure in the liquid
near the meniscus of the tube in B, y t is the surface tension of the liquid, and
r 1 is the radius. But, from hydrostatics, p t = H—h l p 1 g.
;.p-(H-h lPl g) = ^
'i
:.H-p = h 1 p 1 g-^ . (i)
If y 2 is the surface tension of the liquid in C, and r 2 is the radius of the tube in
the liquid, then, by similar reasoning,
2y 2
H-p = h 2 p2g-—l ... (ii)
• ■>
From (i) and (ii),
Re-arranging,
2y 2 2vj
thPiff-— = KPi9— — ■
~2 '1
Pi Pl9Vl
which is an equation of the form y = mx+c, where c is a constant, h 2 = y,
h l = x, and pjp 2 = m. Thus by taking different values of h 2 and h u and plotting
h 2 against h it a straight-line graph is obtained whose slope is equal to pjp 2 , the
ratio of the densities^ Irrthis way the effect of the surface tension can be eliminated.
Variation of Surface Tension with Temperature. Jaeger's Method
By forming a bubble inside a liquid, and measuring the excess
pressure, Jaeger was able to determine the variation of the surface
tension of a liquid with temperature. One form of the apparatus is
shown in Fig. 5.24 (i). A capillary or drawn-out tubing A is connected
to a vessel W containing a funnel C, so that air is driven slowly through
A when water enters W through C, so that air is driven slowly through
A when water enters W through C. The capillary A is placed inside a
beaker containing the liquid L, and a bubble forms slowly at the end of
A when air is passed through it at a slow rate.
Fig. 5.24. (ii) shows the bubble at three possible stages of growth.
The radius grows from that at a to a hemispherical shape at b. Here
its pressure is larger since the radius is smaller. If we consider the
bubble growing to c, the radius of c would be greater than that of b
and hence it cannot contain the increasing pressure. The downward
force on the bubble due to the pressure, in fact, would be greater than
SURFACE TENSION
145
\l/c
A
lH
II*
M
Air
W
c=:a
(i)
Fig. 5.24 Jaeger's method
C
(i«)
the upward force due to surface tension. Hence the bubble becomes
unstable and breaks away from A when its radius is the same as that of A.
Thus as the bubble grows the pressure in it increases to a maximum,
and then decreases as the bubble breaks away. The maximum pressure
is observed from a manometer M containing a light oil of density p, and
a series of observations are taken as several bubbles grow.
The maximum pressure inside the bubble = H + hpg where h is the
maximum difference in levels in the manometer M, and H is the
atmospheric pressure. The pressure outside the bubble = H+h^p^
where h x is the depth of the orifice of A below the level of the liquid L,
and p x is the latter 's density.
But
.*. excess pressure = (H + hpg)— (H + h^p^g) = hpg — h x p x g.
2y
excess pressure = — ,
where r is the radius of the orifice of A (p. 158).
:.^ = hpg-h x p x g,
:.y = T j{hp-h 1 p 1 ).
By adding warm liquid to the vessel containing L, the variation of
the surface tension with temperature can be determined. Experiment
shows that the surface tension of liquids, and water in particular,
decreases with increasing temperature along a fairly smooth curve.
Various formulae relating the surface tension to temperature have been
proposed, but none has been found to be completely satisfactory. The
decrease of surface tension with temperature may be attributed to the
greater average separation of the molecules at higher temperature.
The force of attraction between molecules is then reduced, and hence
the surface energy is reduced, as can be seen from the potential energy
curve on p. 126.
146
ADVANCED LEVEL PHYSICS
Surface Tension and Surface Energy
We now consider the surface energy of a liquid and its relation to
its surface tension y. Consider a film of liquid stretched across a hori-
zontal frame ABCD, Fig. 5.25. Since y is the force per unit length, the
force on the rod BC of length I = yx2l, because there are two surfaces
to the film.
Suppose the rod is now moved a
distance b from BC to B'C against
the surface tension forces, so that the
surface area of the film increases. The
temperature of the film then usually
decreases, in which case the surface
tension alters (p. 145). If the surface
area increases under isothermal (con-
stant temperature) conditions, how-
ever, the surface tension is constant;
and we can then say that, if y is the
surface tension at that temperature,
Fig. 5.25
Surface energy and work
work done in enlarging surface area = force x distance,
= 2ylxb = yxllb.
But lib is the total increase in surface area of the film.
.". work done per unit area in enlarging area = y.
Thus the surface tension, y, can be defined as the work done per unit
area in increasing the surface area of a liquid under isothermal conditions.
This is also called the free surface energy.
Surface energy and Latent heat
Inside a liquid molecules move about in all directions, continually
breaking and reforming bonds with neighbours. If a molecule in the
surface passes into the vapour outside, a definite amount of energy is
needed to permanently break the bonds with molecules in the liquid.
This amount of energy is the work done in overcoming the inward
force on a molecule in the surface, discussed on p. 129. Thus the energy
needed to evaporate a liquid is related to its surface energy or surface
tension. The latent heat of vaporisation, which is the energy needed to
change liquid to vapour at the boiling point, is therefore related to
surface energy.
Surface energy
As we have seen, when the surface area of a liquid is increased, the surface
energy is increased The molecules which then reach the surface are slowed
up by the inward force, so the average translational kinetic energy of all the liquid
molecules is reduced. On this account the liquid cools while the surface is in-
creased, and heat flows in from the surroundings to restore the temperature.
The increase in the total surface energy per unit area E is thus given by
E = y+H . . . (1)
SURFACE TENSION
147
where if is the heat per unit area from the surroundings. Advanced theory shows that
dy\
H = — 0\-jq\, where 8 is the absolute temperature and dyjdB is the corresponding
gradient of the y \. graph, the variation of surface tension with temperature.
Thus
B-. r -$
(2)
In practice, since y decreases with rising temperature, dy/d6 is negative, and E
is thus greater than y. At 15°C, for example, y = 74x lQ~ 3 Nm~ 1 ,dy/d0 = —015
x 10- 3 N nT 1 K" 1 , 9 = 288 K. Thus, from (2),
E = (74+288 x0-15) x 10 -3 = 0117 N m _1 = 0117 J m -2 .
The variation of E with temperature is shown in Fig. 5.26, together with the
similar variation of L, the latent heat of vaporisation (see p. 146). Both vanish
at the critical temperature, since no liquid exists above the critical temperature
whatever the pressure.
Critical
temperature
8 (temp)
Fig. 5.26 Variation of E and L with temperature
EXAMPLES
1. A soap bubble in a vacuum has a radius of 3 cm and another soap bubble
in the vacuum has a radius of 6 cm If the two bubbles coalesce under isothermal
conditions, calculate the radius of the bubble formed.
Since the bubbles coalesce under isothermal conditions, the surface tension y
is constant. Suppose R is the radius in cm, R x 10 ~ 2 m, of the bubble formed.
Then work done = y x surface area = y x SnR 2 x 10 ~ 4
But original work done = (y x 8n: . 3 2 x y x Sn . 6 2 ) x 10 -4
.-. yxSnR 2 = yxSn.3 2 + y.Sn.6 2 .
.-. R 2 = 3 2 + 6 2 .
.-. R = V3 2 + 6 2 = 6-7cm.
2. (i) Calculate the work done against surface tension forces in blowing a
soap bubble of 1 cm diameter if the surface tension of soap solution is 2-5 x 10" 2
Nm" 1 , (ii) Find the work required to break up a drop of water of radius 0-5 cm
into drops of water each of radii 1 mm. (Surface tension of water = 7x 10" 2 N
m _1 .)
148 ADVANCED LEVEL PHYSICS
(i) The original surface area of the bubble is zero, and the final surface area =
2 x 4nr 2 (two surfaces of bubble) = (2 x 4tt x 0-5 2 ) x 10 -4 = 2% x 10~ 4 m 2 .
.'. work done = yx increase in surface area.
= 2-5xl(T 2 x27txl(r 4 = 1-57 x NT 5 J.
(ii) Since volume of a drop = f nr 3 ,
-nxO-5 3
number of drops formed = 4— — — -=• = 125.
f^xOl 3
.'. final total surface area of drops
= 125 x4w 2 = 125x4ti x01 2 xl0 -4 ,
= 57rxl0- 4 m 2 .
But original surface area of drop = An x 0-5 2 x 10 -4 = n x 10~ 4 m 2 .
.'. work done = yx change in surface area,
= 7x 10 -2 x(57i-7t)x 10 -4 = 8-8 x 1(T 5 J.
EXERCISES 5
What are the missing words in the statements 1-8?
1. The units of surface tension are ...
2. The dimensions of surface tension are . . .
3. Small drops of mercury are spherical because the surface area is a . . .
4. The excess pressure in a soap-bubble is given by . . .
5. The excess pressure at the meniscus of water in a capillary tube is . . .
6. A liquid will not 'wet' the surface of a solid if the angle of contact is . . .
7. Surface tension may be defined as the 'force . . .'
8. Surface tension may also be defined as the '. . . per unit area'.
Which of the following answers, A, B, C, D or E, do you consider is the correct
one in the statements 9-12?
9. A molecule of a liquid which reaches the surface from the interior gains
energy because A it reaches the surface with higher speed than when inside the
liquid, B it overcomes a force of repulsion on molecules at the surface, C it
overcomes a force of attraction on molecules at the surface, D its temperature
increases, E the gravitational potential energy due to the earth is then higher.
10. If a section of a soap bubble through its centre is considered, the force on
one half due to surface tension is A 2nry, B 4nry, C nr 2 y, D 2y/r, E 2nr 2 y.
11. If water has a surface tension of 7 x 10" 2 N m" l and an angle of contact
with water of zero, it rises in a capillary of diameter 0-5 mm to a height of A 70 cm,
B 70 cm, C 6-2 cm, D 5-7 cm, E 0-5 cm.
12. In an experiment to measure the surface tension of a liquid by rise in a
capillary tube which tapers, the necessary radius r would be best obtained
A by cutting the tube at the position of the meniscus and measuring the diameter
here directly, B by drawing up a thread of mercury of length I and using 'mass =
SURFACE TENSION 149
nr 2 lp\ C by measuring the diameter of the lower end of the tube with a travelling
microscope, D by measuring the upper end of the tube with a travelling micro-
scope, E by finding the average of the two measurements in C and D.
13. Define surface tension. A rectangular plate of dimensions 6 cm by 4 cm
and thickness 2 mm is placed with its largest face flat on the surface of water.
Calculate the force due to surface tension on the plate. What is the downward
force due to surface tension if the plate is placed vertical and its longest side just
touches the water? (Surface tension of water = 70 x 10~ 2 N m" 1 .)
14. What are the dimensions of surface tension? A capillary tube of 0-4 mm
diameter is placed vertically inside (i) water of surface tension 6-5 x 10" 2 N m" x
and zero angle of contact, (ii) a liquid of density 800 kgm -3 , surface tension
5-0 x 10" 2 N m" 1 and angle of contact 30°. Calculate the height to which the
liquid rises in the capillary in each case.
15. Define the angle of contact. What do you know about the angle of contact
of a liquid which (i) wets glass, (ii) does not wet glass?
A capillary tube is immersed in water of surface tension 70 x 10" 2 Nn." 1 and
rises 6-2 cm. By what depth will mercury be depressed if the same capillary is
immersed in it? (Surface tension of mercury = 054Nm _1 ; angle of contact
between mercury and glass = 140°; density of mercury = 13600 kgm" 3 .)
16. (i) A soap-bubble has a diameter of 4 mm. Calculate the pressure inside
it if the atmospheric pressure is 10 5 N m" 2 . (Surface tension of soap solution
= 2-8 x 10" 2 N m" l .) (ii) Calculate the radius of a bubble formed in water if the
pressure outside it is 1-000 x 10 5 Nm~ 2 and the pressure inside it is 1001 x 10 5
Nm -2 . (Surface tension of water = 7-0x 10~ 2 Nm" 1 .)
17. Define surface tension of a liquid. State the units in which it is usually
expressed and give its dimensions in mass, length, and time.
Derive an expression for the difference between the pressure inside and outside
a spherical soap bubble. Describe a method of determining surface tension,
based on the difference of pressure on the two sides of a curved liquid surface
or film. (L.)
18. Explain briefly (a) the approximately spherical shape of a rain drop,
(b) the movement of tiny particles of camphor on water, (c) the possibility of
floating a needle on water, (d) why a column of water will remain in an open
vertical capillary tube after the lower end has been dipped in water and with-
drawn. (N.)
19. Define the terms surface tension, angle of contact Describe a method
for measuring the surface tension of a liquid which wets glass. List the principal
sources of error and state what steps you would take to minimize them.
A glass tube whose inside diameter is 1 mm is dipped vertically into a vessel
containing mercury with its lower end 1 cm below the surface. To what height
will the mercury rise in the tube if the air pressure inside it is 3 x 10 3 N m~ 2
below atmospheric pressure? Describe the effect of allowing the pressure in the
tube to increase gradually to atmospheric pressure. (Surface tension of mercury =
0-5 N m" 1 , angle of contact with glass = 180°, density of mercury = 13600 kg
m" 3 ,0 = 9-81ms" 2 .)(O.<£C.)
20. Explain how to measure the surface tension of a soap film.
The diameters of the arms of a U-tube are respectively 1 cm and 1 mm A
liquid of surface tension 70 x 10" 2 N m ~ * is poured into the tube which is placed
vertically. Find the difference in levels in the two arms. The density may be
taken as 1000 kg m _3 and the contact angle zero. (L.)
1 50 ADVANCED LEVEL PHYSICS
21. Explain what is meant by surface tension, and show how its existence is
accounted for by molecular theory.
Find an expression for the excess pressure inside a soap-bubble of radius R
and surface tension T. Hence find the work done by the pressure in increasing
the radius of the bubble from a to b. Find also the increase in surface area of the
bubble, and in the light of this discuss the significance of your result. (C.)
22. A clean glass capillary tube, of internal diameter 004 cm, is held vertically
with its lower end below the surface of clean water in a beaker, and with 10 cm
of the tube above the surface. To what height will the water rise in the tube?
What will happen if the tube is now depressed until only 5 cm of its length is
above the surface? The surface tension of water is 7-2 x 10" 2 N m~ 1 .
Describe, and give the theory of some method, other than that of the rise in a
capillary tube, of measuring surface tension. (O. & C.)
23. Explain (a) in terms of molecular forces why the water is drawn up above
the horizontal liquid level round a steel needle which is held vertically and
partly immersed in water, (b) why, in certain circumstances, a steel needle will
rest on a water surface. In each case show the relevant forces on a diagram. (N.)
24. The force between two molecules may be regarded as an attractive force
which increases as their separation decreases and a repulsive force which is only
important at small separations and which there varies very rapidly. Draw sketch
graphs (a) for force-separation, (b) for potential-energy separation. On each
graph mark the equilibrium distance and on (b) indicate the energy which would
be needed to separate two molecules initially at the equilibrium distance.
With the help of your graphs discuss briefly the resulting motion if the molecules
are displaced from the equilibrium position. (AT.)
25. Explain briefly the meaning of surface tension and angle of contact.
Account for the following: (a) A small needle may be placed on the surface of
water in a beaker so that it 'floats', and (b) if a small quantity of detergent is
added to the water the needle sinks.
A solid glass cylinder of length i, radius r and density a is suspended with its
axis vertical from one arm of a balance so that it is partly immersed in a liquid
of density p. The surface tension of the liquid is y and its angle of contact with the
glass is a. If W l is the weight required to achieve a balance when the cylinder is
in air and W 2 is the weight required to balance the cylinder when it is partly
immersed with a length h(< I) below the free surface of the liquid, derive an
expression for the value of W x — W 2 . If this method were used to measure the
surface tension of a liquid, why would the result probably be less accurate than
that obtained from a similar experiment using a thin glass plate? (O. & C.)
26. Explain in terms of molecular forces why some liquids spread over a solid
surface whilst others do not.
A glass capillary tube of uniform bore of diameter 0-050 cm is held vertically
with its lower end in water. Calculate the capillary rise. Describe and explain
what happens if the tube is lowered so that 40 cm protrudes above the water
surface. Assume that the surface tension of water is 7 Ox 10" 2 N m -1 . (JV.)
27. Define surface tension. Describe how the surface tension of water at room
temperature may be determined by using a capillary tube. Derive the formula
used to calculate the result.
A hydrometer has a cylindrical glass stem of diameter 0-50 cm. It floats in
water of density 1000 kg m -3 and surface tension 7-2 x 10 -2 N m~ l . A drop of
SURFACE TENSION 151
liquid detergent added to the water reduces the surface tension to 50 x 10 -2
N m _1 . What will be the change in length of the exposed portion of the glass
stem? Assume that the relevant angle of contact is always zero. (N.)
28. The lower end of a vertical clean glass capillary tube is just immersed in
water. Why does water rise up the tube?
A vertical capillary tube of internal radius r m has its lower end dipping in
water of surface tension Tnewton m _1 . Assuming the angle of contact between
water and glass to be zero, obtain from first principles an expression for the
pressure excess which must be applied to the upper end of the tube in order
just to keep the water levels inside and outside the tube the same.
A capillary of internal diameter 0-7 mm is set upright in a beaker of water
with one end below the surface; air is forced slowly through the tube from the
upper end, which is also connected to a U-tube manometer containing a liquid
of density 800 kg m~ 3 . The difference in levels on the manometer is found to build
up to 91 cm, drop to 40 cm, build up to 91 cm again, and so on Estimate (a) the
depth of the open end of the capillary below the free surface of the water in the
beaker, (b) the surface tension of water. [State clearly any assumptions you have
made in arriving at these estimates.] (O.)
29. It is sometimes stated that, in virtue of its surface tension, the surface of a
liquid behaves as if it were a stretched rubber membrane. To what extent do
you think this analogy is justified?
Explain why the pressure inside a spherical soap bubble is greater than that
outside. How would you investigate experimentally the relation between the
excess pressure and the radius of the bubble? Show on a sketch graph the form
of the variation you would expect to obtain.
If olive oil is sprayed on to the surface of a beaker of hot water, it remains as
separated droplets on the water surface; as the water cools, the oil forms a
continuous thin film on the surface. Suggest a reason for this phenomenon. (C.)
30. Describe the capillary tube method of measuring the surface tension of a
liquid.
An inverted U-tube (Hare's apparatus) for measuring the specific gravity of a
liquid was constructed of glass tubing of internal diameter about 2 mm The
following observations of the heights of balanced columns of water and another
liquid were obtained :
Height of water (cm) 2-8 4-2 5-4 6-9 8-5 9-8 11-6
Height of liquid (cm) 20 3-8 53 7 91 10-7 13
Plot the above results, explain why the graph does not pass through the origin,
and deduce from the graph an accurate value for the specific gravity of the
liquid. (N.)
31. How does simple molecular theory account for surface tension? Illustrate
your account by explaining the rise of water up a glass capillary.
A light wire frame in the form of a square of side 5 cm hangs vertically in
water with one side in the water-surface. What additional force is necessary to
pull the frame clear of the water? Explain why, if the experiment is performed
with soap-solution, as the force is increased a vertical film is formed, whereas
with pure water no such effect occurs. (Surface tension of water is 7-4 x 10 ~ 2
Nm-K)(0.&C.)
152 ADVANCED LEVEL PHYSICS
32. Define surface tension and state the effect on the surface tension of water
of raising its temperature.
Describe an experiment to measure the surface tension of water over the range
of temperatures from 20°C to 70°C. Why is the usual capillary rise method
unsuitable for this purpose?
Two unequal soap bubbles are formed one on each end of a tube closed in the
middle by a tap. State and explain what happens when the tap is opened to put
the two bubbles into connection Give a diagram showing the bubbles when
equilibrium has been reached. (L.)
chapter six
Elasticity
Elasticity
A bridge, when used by traffic during the day, is subjected to loads
of varying magnitude. Before a steel bridge is
erected, therefore, samples of the steel are sent to a
research laboratory, where they undergo tests to find
out whether the steel can withstand the loads to
which it is likely to be subjected.
Fig. 6.1 illustrates a simple laboratory method of
discovering useful information about the property
of steel we are discussing. Two long thin steel wires,
P, Q, are suspended beside each other from a rigid
support B, such as a girder at the top of the ceiling.
The wire P is kept taut by a weight A attached to its
end and carries a scale M graduated in centimetres.
The wire Q carries a vernier scale V which is along-
side the scale M.
When a load W such as 1 kgf is attached to
the end of Q, the wire increases in length by an
amount which can be read from the change in the
reading on the vernier V. If the load is taken off and
the reading on V returns to its original value, the
wire is said to be elastic for loads from zero to 1
kgf, a term adopted by analogy with an elastic
thread. When the load W is increased to 2 kgf the
extension (increase in length) is obtained from V
again; and if the reading on V returns to origin
value when the load is removed the wire is said to be elastic at least
for loads from zero to 2 kgf.
Tensile
force
on Q
Fig. 6.1
Tensile force
The extension of a thin wire such as Q for increasing loads may
be found by experiments to be as follows :
W(kgf)
1
2
3
4
5
6
7
8
Extension (mm.)
014
0-28
0-42
0-56
0-70
085
101
119
Elastic Limit
When the extension, e, is plotted against the load, W, a graph is obtained
which is a straight line OA, followed by a curve ABY rising slowly
153
154
ADVANCED LEVEL PHYSICS
at first and then very sharply, Fig. 6.2. Up to about 5 kgf, then, the
results in the table show that the extension increased by 014 mm for
every kgf which is added to the wire. Further, the wire returned to its
Extension (e)
Stress
D
A = Elastic limit
B = Yield point
C = Breaking (max) stress
D = Wire breaks
BC = Plastic deformation
Strain
Fig. 6.2 Extension v. Load
original length when the load was removed. For loads greater than
about 5 kgf, however, the extension increases relatively more and
more, and the wire now no longer returns to its original length when
it is unloaded. The wire is thus permanently strained, and A corresponds
to its elastic limit.
Hooke's Law
From the straight line graph OA, we deduce that the extension is
proportional to the load or tension in the wire when the elastic limit is not
exceeded. This is known as Hooke's law, after Robert Hooke, founder
of the Royal Society, who discovered the relation in 1676. The law
shows that when a molecule of a solid is displaced farther from its
mean position, the restoring force is proportional to its displacement
(see p. 126). One may therefore conclude that the molecules of a solid
are undergoing simple harmonic motion (p. 44).
The measurements also show that it would be dangerous to load the
wire with weights greater than 5 kilogrammes, the elastic limit, because
the wire then suffers a permanent strain. Similar experiments in the
research laboratory enable scientists to find the maximum load which
a steel bridge, for example, should carry for safety. Rubber samples
are also subjected to similar experiments, to find the maximum safe
tension in rubber belts used in machinery.
Yield Point. Ductile and Brittle Substances. Breaking Stress
Careful experiments show that, for mild steel and iron for example,
the molecules of the wire begin to 'slide' across each other soon after
the load exceeds the elastic limit, that is, the material becomes plastic.
This is indicated by the slight 'kink' at B beyond A in Fig. 6.2 (i), and it
is called the yield point of the Wire. The change from an elastic to
ELASTICITY 155
a plastic stage is shown by a sudden increase in the extension, and as
the load is increased further the extension increases rapidly along the
curve YN and the wire then snaps. The breaking stress of the wire is the
corresponding force per unit area of cross-section of the wire. Sub-
stances such as those just described, which elongate considerably and
undergo plastic deformation until they break, are known as ductile
substances. Lead, copper and wrought iron are ductile. Other sub-
stances, however, break just after the elastic limit is reached ; they are
known as brittle substances. Glass and high carbon steels are brittle.
Brass, bronze, and many alloys appear to have no yield point These
materials increase in length beyond the elastic limit as the load is
increased without the appearance of a plastic stage.
The strength and ductility of a metal, its ability to flow, are dependent
on defects in the metal crystal lattice. Such defects may consist of a
missing atom at a site or a dislocation at a plane of atoms. Plastic
deformation is the result of the 'slip' of atomic planes. The latter is
due to the movement of dislocations, which spreads across the crystal.
Tensile Stress and Tensile Strain. Young's Modulus
We have now to consider the technical terms used in the subject of
elasticity of wires. When a force or tension F is applied to the end of a
wire of cross-sectional area A, Fig. 6.3,
-HH
A
Fig. 6.3 Tensile stress and tensile strain
the tensile stress = force per unit area = — . (1)
A
If the extension of the wire is e, and its original length is /,
the tensile strain = extension per unit length = | . (2)
Suppose 2 kg is attached to the end of a wire of length 2 metres of
diameter 0-64 mm, and the extension is 0-60 mm. Then
F = .2 kgf = 2 x 9-8 N, A = n x 0-032 2 cm 2 = n x O032 2 x 10" 4 m 2 .
■•. tensile stress %x ^ lfl . 4 Nr»,
and tensile strain = Q^IQ- 3 metre = ^ x 1Q _ 3
2 metre
It will be noted that 'stress' has units such as 'newton m~ 2 '; 'strain'
has no units because it is the ratio of two lengths.
1 56 ADVANCED LEVEL PHYSICS
A modulus of elasticity of the wire, called Young's modulus (E), is
defined as the ratio
r tensile stress ,~,
E = -. — . . . (3)
tensile strain
Thus E = ^4.
e/l
Using the above figures,
2x9-8/(ttx(M)32 2 x10~ 4 )
E =
0-3 xlO -3
2x9-8
n x 0032 2 x 10~ 4 x 0-3 x 1(T 3 '
= 2-0xl0 11 N
m
-2
It should be noted that Young's modulus, E, is calculated from the
ratio stress .strain only when the wire is under 'elastic' conditions, that
is, the load does not then exceed the elastic limit (p. 154). Fig. 62 (ii)
shows the general stress-strain diagram for a ductile material.
Dimensions of Young's Modulus
As stated before, the 'strain' of a wire has no dimensions of mass,
length, or time, since, by definition, it is the ratio of two lengths. Now
dimensions of force
dimensions of stress =
dimensions of area
MLT -2
~ L 2
= ML" 1 !" 2 .
.'. dimensions of Young's modulus, E,
_ dimensions of stress
dimensions of strain
= ML- 1 T~ 2 .
Determination of Young's Modulus
The magnitude of Young's modulus for a material in the form of a
wire can be found with the apparatus illustrated in Fig. 6.1, p. 153, to
which the reader should now refer. The following practical points
should be specially noted :
(1) The wire is made thin so that a moderate load of several kilo-
grams produces a large tensile stress. The wire is also made long so
that a measurable extension is produced.
(2) The use of two wires, P, Q, of the same material and length,
eliminates the correction for (i) the yielding of the support when loads
are added to Q, (ii) changes of temperature.
(3) Both wires should be free of kinks, otherwise the increase in
length cannot be accurately measured. The wires are straightened by
attaching weights to their ends, as shown in Fig. 6.1.
ELASTICITY
157
(4) A vernier scale is necessary to measure the extension of the wire
since this is always small. The 'original length' of the wire is measured
from the top B to the vernier V by a ruler, since an error of 1 millimetre
is negligible compared with an original length of several metres. For
very accurate work, the extension can be measured by using a spirit
level between the two wires, and adjusting a vernier screw to restore the
.spirit level to its original reading after a load is added.
(5) The diameter of the wire must be found by a micrometer screw
gauge at several places, and the average value then calculated. The
area of cross-section, A, = nr 2 , where r is the radius.
(6) The readings on the vernier are also taken when the load is
gradually removed in steps of 1 kilogramme; they should be very
nearly the same as the readings on the vernier when the weights were
added, showing that the elastic limit was not exceeded. Suppose the
reading on V for loads, W, of 1 to 6 kilogramme are a, b, c, d, e, f, as
follows :
W(kgf)
1
2
3
4
5
6
Reading on V
a
b
c
d
e
/
The average extension for 3 kilogramme is found by taking the average
of (d— a), (e — b), and (f—c). Young's modulus can then be calculated
from the relation stress/strain, where the stress = 3x9-S/nr 2 , and the
strain = average extension/original length of wire (p. 155).
Magnitude of Young's Modulus
Mild steel (0-2% carbon) has a Young's modulus value of about
20 x 10 11 newton m~ 2 , copper has a value about 1-2 x 10 11 newton
m~ 2 ; and brass a value about 1-0 x 10 11 newton m -2 .
The breaking stress (tenacity) of cast-iron metal is about 1-5 x 10 8
newton m~ 2 ; the breaking stress of mild steel metal is about 4-5 x 10 8
newton m~ 2 .
At Royal Ordnance and other Ministry of Supply factories, tensile
testing is carried out by placing a sample of the material in a machine
known as an extensometer, which applies stresses of increasing value
along the length of the sample and automatically measures the slight
increase in length. When the elastic limit is reached, the pointer on the
dial of the machine flickers, and soon after the yield point is reached the
sample becomes thin at some point and then breaks. A graph showing
the load v. extension is recorded automatically by a moving pen while
the sample is undergoing test.
EXAMPLE
Find the maximum load in kgf which may be placed on a steel wire of diameter
010 cm if the permitted strain must not exceed ^o an d Young's modulus for
steel is 2-0 xlO 11 Nm -2 .
We have
max. stress
max. strain
max. stress =
= 2xlO xl .
158 ADVANCED LEVEL PHYSICS
Now area of cross-section in m 2 = — = rc x °l 2 *10~ 4
4 4
and stress = ^^
area
.-. F = stress x area = 2 x 10 8 x nx012 x 10 ~ 4 new ton
4
= 157 newton = 15-7 kgf (approx.).
since 10 newtons = 1 kgf (approx.).
Force in Bar Due to Contraction or Expansion
When a bar is heated, and then prevented from contracting as it
cools, a considerable force is exerted at the ends of the bar. We can
derive a formula for the force if we consider a bar of Young's modulus
E, a cross-sectional area A, a linear expansivity of magnitude a, and a
decrease in temperature of t °C Then, if the original length of the bar
is /, the decrease in length e if the bar were free to contract = alt.
Now
E =
F/A
ell'
.-. F =
EAe
I
EAalt
I '
.'. F =
EAat.
As an illustration, suppose a steel rod of cross-sectional area 20 cm 2
is heated to 100°C, and then prevented from contracting when it is
cooled to 10°C. The linear expansivity of steel = 12 x 10" 6 K" 1 and
Young's modulus = 20 x 10 11 newton m~ 2 . Then
A = 2 cm 2 = 2 x 10" 4 m 2 , t = 90 deg C.
.". F = EAat = 2 x 10 11 x 2 x 10" 4 x 12 x 10" 6 x 90 newton (N),
= 43200N = ^^kgf = 4400 kgf.
Energy Stored in a Wire
Suppose that a wire has an original length I and is stretched by a
length e when a force F is applied at one end. If the elastic limit is not
exceeded, the extension is directly proportional to the applied load
(p. 154). Consequently the force in the wire has increased in magnitude
from zero to F, and hence the average force in the wire while stretching
was F/2. Now
work done = force x distance.
.'. work = average force x extension
=\Fe (1)
ELASTICITY
159
This is the amount of energy stored in the wire. The formula \Fe gives
the energy in joule when F is in newton and e is in metre.
Further, since F = EAe/l,
energy = \EA-
As an illustration, suppose £ = 2-0x10" newton m~ 2 , A =
3 x 10 -2 cm 2 = 3 x 10 -6 m 2 , e = 1 mm = 1 x 10 -3 m, / = 400 cm =
4 m. Then
. i^e 2 i 2x 10" x3xl0 _6 x(lxKT 3 ) 2 joule,
energy stored = \FA-r = \ x
= 0075 J.
The volume of the wire = AX. Thus, from (1),
i Fe \ F e
energy per unit volume = \—. = j-t x -.
But F/A = stress, e/l = strain,
.'. energy per unit volume = \ stress x strain
(2)
Graph of F v . e and energy
The energy stored in the wire when it is stretched can also be found
from the graph of F v. e. Fig. 6.4. Suppose the wire extension is e x
when a load F x is applied, and the extension increases to e 2 when the
load increases to F 2 . If F is the load between F t and F 2 at some stage,
and Ax is the small extension which then occurs, then
energy stored = work done = FAx.
O
Area = /: Ax
= energy stored
e 1 lyJ e 2
Ax
Fig. 6.4 Energy in stretched wire
Now F.Ax is represented by the small area between the axis of e and
the graph, shown shaded in Fig. 6.4. Thus the total work done between
e r and e 2 is represented by the area CBDH.
160 ADVANCED LEVEL PHYSICS
If the extension occurs on the straight part of the curve, when Hooke's
law is obeyed, then CBDH is a trapezium. The area of a trapezium =
half the sum of the parallel sides x perpendicular distance between them
= -KBC + DH)xCH = %F 1+ F 2 ){e 2 - ei ).
.'. energy stored = average force x increase in length.
If the extension occurs beyond the elastic limit, for example, along
the curved part of the graph in Fig. 6.4, the energy expended can be
obtained from the area between the curve and the axis of e.
EXAMPLES
1. A 20 kg weight is suspended from a length of copper wire 1 mm in radius.
If the wire breaks suddenly, does its temperature increase or decrease? Calculate
the change in temperature; Young's modulus for copper = 12 x 10 10 N m~ 2 ;
density of copper = 9000 kg m" 3 ; specific heat capacity of copper = 0-42 J g~ 1
K- l .)(C.S.)
When the wire is stretched, it gains potential energy equal to the work done
on it. When the wire is suddenly broken, this potential energy is released as the
molecules return to their original position. The energy is converted into heat
and thus the temperature rises.
Gain in potential energy of molecules = work done in stretching wire
= \ force (F) x extension (e).
With the usual notation, F = EAj
e = -fL. = ^ w- * y °'* ' m = 5-2x 10- 4 /m,
Fl = (20x9-8)xf
E.A 12xl0 10 X7tx(KT 3 )
.'. potential energy gained= \ x 20 x 9-8 x 5-2 x 10 -4 / = 51 x 10~ 2 / J
Heat capacity of wire = mass x specific heat capacity
= it x (10" 3 ) 2 x 9000/ x (0-42x1000)= 11-9/JK" 1
potential energy 5-1 xlO -2 /
.'. temperature rise = —_ r- 2 ^- = — tt-=t,
heat capacity 11-9/
= 4-3xlO~ 3 degC.
2. Define stress and strain. Describe the behaviour of a copper wire when it is
subjected to an increasing longitudinal stress. Draw a stress-strain diagram
and mark on it the elastic region, yield point and breaking stress.
A wire of length 5 m, of uniform circular cross-section of radius 1 mm is
extended by 1-5 mm when subjected to a uniform tension of 100 newt on. Calculate
from first principles the strain energy per unit volume assuming that deformation
obeys Hooke's law.
Show how the stress-strain diagram may be used to calculate the work done in
producing a given strain, when the material is stretched beyond the Hooke's
law region. (O. & C.)
Strain energy = \ tension x extension
Tension = 100 newton. Extension = 1-5 x 10 -3 m.
.-. energy = \ x 100 x 1-5 x 10 -3 = 0075 J.
ELASTICITY 161
Volume of wire = length x area = 5x7rxlxl0 -6 m 3 .
.'. energy per unit volume = ; — ——* = 4-7 x 10 3 Jm" 3 (approx.).
5x7rxlxlO b
Bulk Modulus
When a gas or a liquid is subjected to an increased pressure the
substance contracts. A change in bulk thus occurs, and the bulk strain
is defined by :
change in volume
strain = — r-? — ^ — -. .
original volume
The bulk stress on the substance is the increased force per unit area,
by definition, and the bulk modulus, K, is given by:
bulk stress
K =
bulk strain
increase in force per unit area
change in volume/original volume'
At- *>
bulk stress
=bulk strain
Fig. 6.5 Bulk stress and bulk strain
If the original volume of the substance is v, the change in volume
may be denoted by — Ay when the pressure increases by a small amount
Ap; the minus indicates that the volume decreases. Thus (Fig. 6.5)
Ap
K= -
Av/v
When dp and Sv become very small, then, in the limit,
dp
dv
K =
(1)
The bulk modulus of water is about 2 x 10 9 N m 2 for pressures in
the range 1 — 25 atmospheres ; the bulk modulus of mercury is about
162 ADVANCED LEVEL PHYSICS
27 x 10 9 Nm~ 2 . The bulk modulus of gases depends on the pressure,
as now explained. Generally, since the volume change is relatively
large, the bulk modulus of a gas is low compared with that of a liquid.
Bulk Modulus of a Gas
If the pressure, p, and volume, v, of a gas change under conditions
such that . .
pv = constant,
which is Boyle's law, the changes are said to be isothermal ones. In
this case, by differentiating the product pv with respect to v, we have
p+4- = o.
dv
dp
■"=-%■
dp
But the bulk modulus, K, of the gas is equal to — v-j- by definition
(seep. 161). dv
:.K = p . . . (2)
Thus the isothermal bulk modulus is equal to the pressure.
When the pressure, p, and volume, v, of a gas change under conditions
such that
pv 7 = constant,
where y = c p /c v = the ratio of the specific heat capacities of the gas, the
changes are said to be adiabatic ones. This equation is the one obeyed
by local values of pressure and volume in air when a sound wave travels
through it. Differentiating both sides with respect to v,
dp
■■■yp=-»T v >
.". adiabatic bulk modulus = yp . . (3)
For air at normal pressure, K = 10 5 newton m~ 2 isothermally and
1-4 x 10 5 newton m -2 adiabatically. The values of K are of the order
10 5 times smaller than liquids as gases are much more compressible.
Velocity of Sound
The velocity of sound waves through any material depends on (i) its
density p, (ii) its modulus of elasticity, E. Thus if V is the velocity, we
may say that
V = kE x p y .... (i),
where k is a constant and x, y are indices we can find by the theory of
dimensions (p. 34).
The units of velocity, V, are LT~ J ; the units of density p are ML - 3 ;
and the units of modulus of elasticity, E, are ML -1 T~ 2 (see p. 156).
Equating the dimensions on both sides of (i),
.-. LT _1 = (ML- 1 T- 2 ) x x(ML- 3 ) y .
ELASTICITY 163
Equating the indices of M, L, T on both sides, we have
= x+y,
1 = — x — 3y,
-1 = -2x.
Solving, we find x = \, y = -^. Thus V = kE i p~ i . A rigid investiga-
tion shows k = 1, and thus
V = E±p-± =
In the case of a solid, E is Young's modulus. In the case of air and
other gases, and of liquids, E is replaced by the bulk modulus K.
Laplace showed that the adiabatic bulk modulus must be used in the
case of a gas, and since this is yp, the velocity of sound in a gas is given
by the expression
Modulus of Rigidity
So far we have considered the strain in one direction, or tensile
strain, to which Young's modulus is applicable and the strain in bulk
or volume, to which the bulk modulus is applicable.
Shear stress=F/area
F / B B'
Shear strain
D 7 1 ~ C
Fig. 6.6 Shear stress and shear strain
Consider a block of material ABCD, such as pitch or plastic for
convenience. Fig. 6.6. Suppose the lower plane CD is fixed, and a
stress parallel to CD is applied by a force F to the upper side AB.
The block then changes its shape and takes up a position A'B'CD.
It can now be seen that planes in the material parallel to DC are dis-
placed relative to each other. The plane AB, for example, which was
originally directly opposite the plane PQ, is displaced to A'B' and
PQ is displaced to P'Q'. The angular displacement a is defined as
the shear strain, a is the angular displacement between any two planes,
for example, between CD and P'Q'.
No volume change occurs in Fig. 6.6. Further, since the force along
CD is F in magnitude, it forms a couple with the force F applied to the
upper side AB. The shear stress is defined as the 'shear force per unit
area' on the face AB (or CD), as in Young's modulus or the bulk
164
ADVANCED LEVEL PHYSICS
modulus. Unlike the case for these modulii, however, the shear stress
has a turning or 'displacement' effect owing to the couple present.
The solid does not collapse because in a strained equilibrium position
such as A'B'CD in Fig. 6.6, the external couple acting on the solid
due to the forces F is balanced by an opposing couple due to stresses
inside the material.
If the elastic limit is not exceeded when a shear stress is applied,
that is, the solid recovers its original shape when the stress is removed,
the modulus of rigidity, G, is defined by:
shear stress (force per unit area)
G =
shear strain (angular displacement, a)'
Shear strain has no units; shear stress has units of newton m~ 2 . The
modulus of rigidity of copper is 4-8 x 10 10 Nm" 2 ; for phosphor-
bronze it is 4-4 x 10 10 Nm" 2 , and for quartz fibre it is 3-0 x 10 10 Nm" 2 .
If a spiral spring is stretched, all parts of the spiral become twisted.
The applied force has thus developed a 'torsional' or shear strain. The
extension of the spring hence depends on its modulus of rigidity, in
addition to its dimensions.
Torsion wire
In sensitive current-measuring instruments, a very weak control is
needed for the rotation of the instrument coil. This may be provided
by using a long elastic or torsion wire of phosphor bronze in place of a
spring. The coil is suspended from the lower end of the wire and when
it rotates through an angle 0, the wire sets up a weak opposing couple
equal to cO, where c is the elastic constant of the wire. Quartz fibres
are very fine but comparatively strong, and have elastic properties.
They are also used for sensitive control (see p. 61).
The magnitude of c, the elastic constant, can be derived as follows. Consider
a wire of radius a, length /, modulus of rigidity G,
fixed at the upper end and twisted by a couple
of moment C at the other end. If we take a section
of the cylindrical wire between radii r and r+Sr,
then a 'slice 'of the material ODBX has been sheared
through an angle a to a position ODB x X where X
is the centre of the lower end of the wire. Fig. 6.7.
From the definition of modulus of rigidity, G =
torsional stress 4- torsional strain = F/A + ol, where
F is the tangential force applied over an area A.
Now A = area of circular annulus at lower
end = Inr.dr.
.'. F = GAa = G.lnr.dr.a.
From Fig 6.7, it follows that BBi = fa, and BB X =
rO.
.'. la. = rd, or a. = rd/l.
. ^ G.lnr dr.rO 2nG0r 2 .Sr
Fig. 6.7
Shear (Torsion) in wire
/
/
ELASTICITY
165
'. moment of F about axis OX of wire = F.r
2%GQ 3 c
= — - — .r 3 .or.
.'. total moment, or couple torque C,
■t
a 2nG0 , , 2nG0 a*
-.r or = . —
/ 4
C =
o
nGa*d
2/
(i)
If the wire is a hollow cylinder of radii a, b respectively, the limits of integration
are altered accordingly, and
moment of couple
1
b 2nGd 3j nG(b*-a*)9
r 3 dr =
I
21
Determinations of modulus of rigidity. Dynamical method. One method oi
measuring the modulus of rigidity of a wire E is to clamp it vertically at one end,
attach a horizontal disc D of known moment of inertia, /, at the other end, and
then time the horizontal torsional oscillations of D. Fig. 6.8 (i). On p. 90, it was
shown that the period of oscillation, T, = 2ny/I/c, where c is the opposing couple
per unit angle of twist. Thus, with our previous notation, as 6 = 1,
TiGa 4
21
or
T=2n
G =
2H
nGa 4
Snll
a*T 2
Hence G can be evaluated from measurements of /, a, I, T.
rf
JZL
a
®
7
IT
M%7
^Z
W
M.
7
M
Fig. 6.8 Modulus of rigidity measurement
Statical method. The modulus of rigidity, G, of the wire E can also be found
by measuring the steady deflection 6 at the lower end on a scale S graduated in
degrees when a couple is applied round a wheel W. Fig. 6.8 (ii). If M is the mass
166 ADVANCED LEVEL PHYSICS
in each scale-pan, and d is the diameter of W, the moment of the couple on the
wire = Mgd = nGa*6/2l. The angle 8 in radians, and a, I, are known, and hence
G can be evaluated.
Poisson's Ratio
When a rubber cord is extended its diameter usually decreases at the same
time. Poisson's ratio, a, is the name given to the ratio
lateral contraction/original diameter
longitudinal extension/original length
and is a constant for a given material. If the original length of a rubber strip is
100 cm and it is stretched to 102 cm, the fractional longitudinal extension =
2/100. If the original diameter of the cord is 0-5 cm and it decreases to 0-495 cm,
the fractional lateral contraction = 0005/0-5 = 1/100. Thus, from the definition
of Poisson's ratio,
2/100 2
When the volume of a strip of material remains constant while an extension
and a lateral contraction takes place, it can easily be shown that Poisson's ratio
is 0-5 in this case. Thus suppose that the length of the strip is / and the radius is r.
Then volume, V, = nr 2 l.
By differentiating both sides, noting that V is a constant and that we have a
product of variables on the right side,
.-. = nr 2 xdl+lx2nrdr.
.-. rSl= -2ldr.
■ 6r l r = i
' ■ Sl/l 2 '
But — Sr/r is the lateral contraction in radius/original radius, and Sl/l is the
longitudinal extension/original length.
.". Poisson's ratio, a, = \.
Experiments show that a is 0-48 for rubber, 0-29 for steel 0-27 for iron, and
0-26 for copper. Thus the three metals increase in volume when stretched, whereas
rubber remains almost unchanged in volume.
Summary
The three modulii of elasticity are compared in the table below :
Young's modulus, E
Modulus of Rigidity, G
Bulk modulus, K
1.
Definition :
tensile stress
tensile strain
shear stress
shear strain
Definition :
pressure change
— Av/v
2.
Relates to change in
length ('tensile')
Relates to change in
shape ('shear')
Relates to change in
volume ('bulk')
ELASTICITY
167
E =
F/A
ell
G =
F/A
a
Fig. 6.9
K =
Ap
— Av/v
4.
Applies only to solids
Applies to solids and
Applies to all materials.
liquids
Low value for gases
5.
Used in stretching wires,
Used in torsion wires,
Used in velocity of
bending beams, linear
helical springs
sound formula for all
expansion and contrac-
materials. In gas, K =
tion with temperature
p (isothermal) or yp
change
(adiabatic)
EXERCISES 6
(Assume g = 9-8 m s~ 2 unless otherwise stated)
What are the missing words in the statements 1-6?
1. When a weight is attached to a suspended long wire, it produces a . . . strain.
2. The units of Young's modulus are . . .
3. In measuring Young's modulus, the . . . must not be exceeded.
4. The energy gained by a wire when stretched = . . . x extension.
5. Bulk stress is denned as the . . . change.
6. When a wire is twisted, a . . . strain is produced.
Which of the following answers, A, B, C, D or E, do you consider is the correct
one in the statements 9-12?
7. If a metal bar, coefficient of linear expansion a, Young's modulus E, area
of cross-section A and length /, is heated through t°C when clamped at both
ends, the force in the bar is calculated from A EAlt, B EAt/a, C EAoct, D E 2 Accl,
E Acxtl.
8. When a spiral spring is stretched by a weight attached to it, the strain is
A tensile, B shear, C bulk, D elastic, E plastic.
9. The energy in a stretched wire is A j load x extension, B load x extension,
C stress x extension, D load x strain, E \ load x strain.
10. In an experiment to measure Young's modulus, the wire is thin and long
so that A very heavy weights can be attached, B the wire can be suspended from
the ceiling, C another identical wire can be arranged parallel to it, D the stress
is large and the extension is measurable for laboratory loads, E a micrometer
gauge can be used for accurate measurement.
168 ADVANCED LEVEL PHYSICS
11. Define tensile stress, tensile strain, Young's modulus. What are the units
and dimensions of each?
A load of 2 kgf is applied to the ends of a wire 4 m long, and produces an ex-
tension of 0-24 mm. If the diameter of the wire is 2 mm, calculate the stress on
the wire, its strain, and the value of Young's modulus.
12. What load in kilogrammes must be applied to a steel wire 6 m long and
diameter 1-6 mm to produce an extension of 1 mm? (Young's modulus for
steel = 20 xlO u Nm- 2 .)
13. Find the extension produced in a copper wire of length 2 m and diameter
3 mm when a load of 3 kgf is applied. (Young's modulus for copper = 11 x 10 11
Nm -2 .)
14. What is meant by (i) elastic limit, (ii) Hooke's law, (iii) yield point, (iv)
perfectly elastic? Draw sketches of stress v. strain to illustrate your answers.
15. 'In an experiment to determine Young's modulus, the strain should not
exceed 1 in 1000.' Explain why this limitation is necessary and describe an
experiment to determine Young's modulus for the material of a metal wire.
In such an experiment, a brass wire of diameter 00950 cm is used. If Young's
modulus for brass is .9-86 x 10 10 newton m -2 , find in kilogram force the greatest
permissible load. (L.)
16. Define stress and strain, and explain why these quantities are useful in
studying the elastic behaviour of a material.
State one advantage and one disadvantage in using a long wire rather than a
short stout bar when measuring Young's modulus by direct stretching.
Calculate the minimum tension with which platinum wire of diameter 01 mm
must be mounted between two points in a stout invar frame if the wire is to
remain taut when the temperature rises 100K. Platinum has coefficient of
linear expansion 9 x 10" 6 K _1 and Young's modulus 17 x 10 10 Nm -2 . The
thermal expansion of invar may be neglected. (0. & C.)
17. Explain the terms stress, strain, modulus of elasticity and elastic limit.
Derive an expression in terms of the tensile force and extension for the energy
stored in a stretched rubber cord which obeys Hooke's law.
The rubber cord of a catapult has a cross-sectional area 1-0 mm 2 and a total
unstretched length 10-0 cm. It is stretched to 120 cm and then released to project
a missile of mass 50 g From energy considerations, or otherwise, calculate the
velocity of projection, taking Young's modulus for the rubber as 50 x 10 8 Nm"" 2 .
State the assumptions made in your calculation.
18. State Hooke's law, and describe in detail how it may be verified experi-
mentally for copper wire. A copper wire, 200 cm long and 1 -22 mm diameter, is
fixed horizontally to two rigid supports 200 cm long. Find the mass in grams of
the load which, when suspended at the mid-point of the wire, produces a sag of
2 cm at that point. Young's modulus for copper = 12-3 x 10 10 N m~ 2 . (L.)
19. Distinguish between Young's modulus, the bulk modulus and the shear
modulus of a material. Describe a method for measuring Young's modulus.
Discuss the probable sources of error and assess the magnitude of the contri-
bution from each.
A piece of copper wire has twice the radius of a piece of steel wire. Young's
modulus for steel is twice that for the copper. One end of the copper wire is
joined to one end of the steel wire so that both can be subjected to the same
longitudinal force. By what fraction of its length will the steel have stretched
when the length of the copper has increased by 1%? (O. & C.)
ELASTICITY 169
20. In an experiment to measure Young's modulus for steel a wire is suspended
vertically and loaded at the free end. In such an experiment, (a) why is the wire
long and thin, (b) why is a second steel wire suspended adjacent to the first?
Sketch the graph you would expect to obtain in such an experiment showing
the relation between the applied load and the extension of the wire. Show how it
is possible to use the graph to determine (a) Young's modulus for the wire, (b) the
work done in stretching the wire.
If Young's modulus for steel is 200 x 10 11 N m~ 2 , calculate the work done
in stretching a steel wire 100 cm in length and of cross-sectional area 0030 cm 2
when a load of 10 kgf is slowly applied without the elastic limit being reached.
(N.)
21. Describe the changes which take place when a wire is subjected to a steadily
increasing tension Include in your description a sketch graph of tension against
extension for (a) a ductile material such as drawn copper and (b) a brittle one
such as cast iron.
Show that the energy stored in a rod of length L when it is extended by a length
/ is jEl 2 /L 2 per unit volume where E is Young's modulus of the material.
A railway track uses long welded steel rails which are prevented from expanding
by friction in the clamps. If the cross-sectional area of each rail is 75 cm 2 what is
the elastic energy stored per kilometre of track when its temperature is raised
by 10°C? (Coefficient of thermal expansion of steel =1-2 x 10~ 5 K" 1 ;
Young's modulus for steel = 2 x 10 11 N m~ 2 .) (O. & C.)
22. What is meant by saying that a substance is 'elastic'?
A vertical brass rod of circular section is loaded by placing a 5 kg weight on
top of it If its length is 50 cm, its radius of cross-section 1 cm, and the Young's
modulus of the material 3-5 x 10 10 N m -2 find (a) the contraction of the rod,
(b) the energy stored in it. (C.)
23. Give a short account of what happens when a copper wire is stretched
under a gradually increasing load. What is meant by modulus of elasticity, elastic
limit, perfectly elastic!
When a rubber cord is stretched the change in volume is very small compared
with the change in shape. What will be the numerical value of Poisson's ratio for
rubber, i.e., the ratio of the fractional decrease in diameter of the stretched cord
to its fractional increase in length? (L.)
24. Describe an accurate method of determining Young's modulus for a wire.
Detail the precuations necessary to avoid error, and estimate the accuracy
attainable.
A steel tyre is heated and slipped on to a wheel of radius 40 cm which it fits
exactly at a temperature t°C. What is the maximum value oft if the tyre is not to
be stretched beyond its elastic limit when it has cooled to air temperature (17°C)?
What will then be the tension in the tyre, assuming it to be 4 cm wide and 3 mm
thick? The value of Young's modulus for steel is 1-96 x 10 11 N m~ 2 , its coefficient
of linear expansion is 11 x 10" 5 K _1 , and its elastic limit occurs for a tension
of 2-75 x 10 8 N m~ 2 . The wheel may be assumed to be at air temperature
throughout, and to be incompressible. (O. & C.)
25. State Hooke's law and describe, with the help of a rough graph, the be-
haviour of a copper wire which hangs vertically and is loaded with a gradually
increasing load until it finally breaks. Describe the effect of gradually reducing
the load to zero (a) before, (b) after the elastic limit has been reached.
170 ADVANCED LEVEL PHYSICS
A uniform steel wire of density 7800 kg m~ 3 weighs 16 g and is 250 cm long. It
lengthens by 1-2 mm when stretched by a force of 8 kgf. Calculate (a) the value
of Young's modulus for the steel, (b) the energy stored in the wire. (N.)
26. Describe an experimental method for the determination of (a) Young's
modulus, (b) the elastic limit, of a metal in the form of a thin wire.
A steel rod of mass 97-5 g and of length 50 cm is heated to 200°C and its ends
securely clamped. Calculate the tension in the rod when its temperature is
reduced to 0°C, explaining how the calculation is made. (Young's modulus for
steel = 20 x 10 1 1 Nm" 2 ; linear expansivity = 11 x 10" 5 K -1 ; density of
steel = 7800 kg m~ 3 .)(L.)
27. What do you understand by Hooke's law of elasticity? Describe how you
would verify it in any particular case.
A wire of radius 0-2 mm is extended by 01% of its length when it supports a
load of 1 kg; calculate Young's modulus for the material of the wire. (L.)
28. Define Young's modulus of elasticity. Describe an accurate method of
determining it. The rubber cord of a catapult is pulled back until its original
length has been doubled. Assuming that the cross-section of the cord is 2 mm
square, and that Young's modulus for rubber is 10 7 N m -2 calculate the tension
in the cord. If the two arms of the catapult are 6 cm apart, and the unstretched
length of the cord is 8 cm what is the stretching force? (O. & C.)
29. Define Young's modulus of elasticity and coefficient of linear expansion.
State units in which each may be expressed and describe an experimental deter-
mination of Young's modulus.
For steel. Young's modulus is 1-8 x 10 11 N m~ 2 and the coefficient of expan-
sion llxlO -5 K _1 . A steel wife 1 mm in diameter is stretched between two
supports when its temperature is 200°C. By how much will the force the wire
exerts on the supports increase when it cools to 20°C, if they do not yield?
Express the answer in terms of the weight of a kilogramme. (L.)
30. Define elastic limit and Young's modulus and describe how you would
find the values for a copper wire.
What stress would cause a wire to increase in length by one-tenth of one
per cent if Young's modulus for the wire is 12 x 10 10 N m~ 2 ? What load in
kg wt. would produce this stress if the diameter of the wire is 0-56 mm? (L.)
chapter seven
Solid Friction. Viscosity
SOLID FRICTION
Static Friction
When a person walks along a road, he or she is prevented from
slipping by the force of friction at the ground. In the absence of friction,
for example on an icy surface, the person's shoe would slip when placed
on the ground. The frictional force always opposes the motion of the
shoe.
The frictional force between the surface of a table and a block of
wood A can be investigated by attaching one end of a string to A and
the other to a scale-pan S, Fig. 7.1. The string passes over a fixed
Fig. 7.1 Solid friction
grooved wheel B. When small weights are added to S, the block does
not move. The frictional force between the block and table is thus
equal to the total weight on S together with the weight of S. When
more weights are added, A does not move, showing that the frictional
force has increased, but as the weight is increased further, A suddenly
begins to slip. The frictional force now present between the surfaces is
called the limiting frictional force, and we are said to have reached
limiting friction. The limiting frictional force is the maximum frictional
force between the surfaces.
Coefficient of Static Friction
The normal reaction, R, of the table on A is equal to the weight of A.
By placing various weights on A to alter the magnitude of R, we can
find how the limiting frictional force F varies with R by the experiment
just described. The results show that, approximately,
limiting frictional force (F)
normal reaction (R)
171
= //, a constant,
172
ADVANCED LEVEL PHYSICS
and n is known as the coefficient of static friction between the two sur-
faces. The magnitude of // depends on the nature of the two surfaces ;
for example, it is about 0-2 to 0-5 for wood on wood, and about 0-2 to
0-6 for wood on metals. Experiment also shows that the limiting
frictional force is the same if the block A in Fig. 7.1 is turned on one
side so that its surface area of contact with the table decreases, and thus
the limiting frictional force is independent of the area of contact when
the normal reaction is the same.
mg sin
Fig. 7.2 Coefficient by inclined plane
The coefficient of static friction, fi, can also be found by placing the
block A on the surface S, and then gently tilting S until A is on the
point of slipping down the plane, Fig. 7.2. The static frictional force F
is then equal to mg sin 0, where is the angle of inclination of the plane
to the horizontal ; the normal reaction R is equal to mg cos 0.
F ma sin
.'. n = - = -^ = tan 0,
R mg cos
and hence n can be found by measuring 0.
Kinetic Friction. Coefficient of Kinetic (Dynamic) Friction
When brakes are applied to a bicycle, a frictional force is exerted
between the moving wheels and brake blocks. In contrast to the case
of static friction, when one of the objects is just on the point of slipping,
the frictional force between the moving wheel and brake blocks is
called a kinetic (or dynamic) frictional force. Kinetic friction thus
occurs between two surfaces which have relative motion.
The coefficient of kinetic (dynamic) friction, p!, between two surfaces
is defined by the relation
F'
V
R'
where F' is the frictional force when the object moves with a uniform
velocity and R is the normal reaction between the surfaces. The
coefficient of kinetic friction between a block A and a table can be
found by the apparatus shown in Fig. 7.1. Weights are added to the
scale-pan, and each time A is given a slight push. At one stage A
continues to move with a constant velocity, and the kinetic frictional
SOUD FRICTION, VISCOSITY 173
force F is then equal to the total weight in the scale-pan together with
the latter 's weight. On dividing F by the weight of A, the coefficient
can be calculated. Experiment shows that, when weights are placed
on A to vary the normal reaction R, the magnitude of the ratio F'/R is
approximately constant. Results also show that the coefficient of
kinetic friction between two given surfaces is less than the coefficient
of static friction between the same surfaces, and than the coefficient of
kinetic friction between two given surfaces is approximately indepen-
dent of their relative velocity.
Laws of Solid Friction ,
Experimental results on solid friction are summarised in the laws
of friction, which state :
(1) The frictional force between two surfaces opposes their relative
motion.
(2) The frictional force is independent of the area of contact of the
given surfaces when the normal reaction is constant.
(3) The limiting frictional force is proportional to the normal reaction
for the case of static friction. The frictional force is proportional to the
normal reaction for the case of kinetic (dynamic) friction, and is
independent of the relative velocity of the surfaces.
Theory of Solid Friction
The laws of solid friction were known hundreds of years ago, but
they have been explained only in comparatively recent years, mainly
by F. P. Bowden and collaborators. Sensitive methods, based on
electrical conductivity measurements, reveal that the true area of
contact between two surfaces is extremely small, perhaps one ten-
thousandth of the area actually placed together for steel surfaces. This
is explained by photographs which show that some of the atoms of a
metal project slightly above the surface, making a number of crests or
'humps'. As Bowden has stated : 'The finest mirror, which is flat to a
millionth of a centimetre, would to anyone of atomic size look rather
like the South Downs — valley and rolling hills a hundred or more
atoms high.' Two metal surfaces thus rest on each others projections
when placed one on the other.
Since the area of actual contact is extremely small, the pressures at
the points of contact are very high, perhaps 1000 million kgf per m 2
for steel surfaces. The projections merge a little under the high pressure,
producing adhesion or 'welding' at the points, and a force which
opposes motion is therefore obtained. This explains Law 1 of the laws
of solid friction. When one of the objects is turned over, so that a
smaller or larger surface is presented to the other object, measurements
show that the small area of actual contact remains constant. Thus the
frictional force is independent of the area of the surfaces, which explains
Law 2. When the load increases the tiny projections are further squeezed
174 ADVANCED LEVEL PHYSICS
by the enormous pressures until the new area of contact becomes big
enough to support the load. The greater the load, the greater is the
area of actual contact, and the frictional force is thus approximately
proportional to the load, which explains Law 3.
VISCOSITY
If we move through a pool of water we experience a resistance to our
motion. This shows that there is a frictional force in liquids. We say
this is due to the viscosity of the liquid. If the frictional force is com-
paratively low, as in water, the viscosity of the liquid is low ; if the
frictional force is large, as in glue or glycerine, the viscosity of the
liquid is high. We can compare roughly the viscosity of two liquids by
filling two measuring cylinders with each of them, and allowing identical
small steel ball-bearings to fall through each liquid. The sphere falls
more slowly through the liquid of higher yiscosity.
As we shall see later, the viscosity of a lubricating oil is one of the
factors which decide whether it is suitable for use in an engine. The
Ministry of Aircraft Production, for example, listed viscosity values
to which lubricating oils for aero-engines must conform. The subject
of viscosity has thus considerable practical importance.
Newton's Formula. Coefficient of Viscosity
When water flows slowly and steadily through a pipe, the layer A of
the liquid in contact with the pipe is practically stationary, but the
central part C of the water is moving relatively fast, Fig. 7.3. At other
layers between A and C, such as B, the water has a velocity less than at
C, the magnitude of the velocities being represented by the length of the
arrowed lines in Fig. 7.3. Now as in the case of two solid surfaces
Pipe^
Fig. 7.3 Laminar (uniform) flow through pipe
moving over each other, a frictional force is exerted between two liquid
layers when they move over each other. Thus because the velocities of
neighbouring layers are different, as shown in Fig. 7.3, a frictional
force occurs between the various layers of a liquid when flowing
through a pipe.
The basic formula for the frictional force, F, in a liquid was first
suggested by Newton. He saw that the larger the area of the surface
of liquid considered, the greater was the frictional force F. He also
stated that F was directly proportional to the velocity gradient at the
p^rt of the liquid considered. This is the case for most common liquids,
SOLID FRICTION. VISCOSITY 175
called Newtonian liquids. If v v v 2 are the velocities of C, B respectively
in Fig. 7.3, and h is their distance apart, the velocity gradient between
the liquids is defined as (v l — v 2 )/h. The velocity gradient can thus be
expressed in (m/s)/m, or as 's~ 1 \
Thus if A is the area of the liquid surface considered, the frictional
force F on the surface is given by
F oc A x velocity gradient,
or F = r/Ax velocity gradient, . . (1)
where n is a constant of the liquid known as the coefficient of viscosity.
This expression for the frictional force in a liquid should be contrasted
with the case of solid friction, in which the frictional force is independent
of the area of contact and of the relative velocity between the solid
surfaces concerned (p. 173).
Definition, Units, and Dimensions of Coefficient of Viscosity
The magnitude of n is given by
= F
A x velocity gradient"
The unit of F is a newton, the unit of A is m 2 , and the unit of velocity
gradient is 1 m/s per m. Thus n may be defined as the frictional force
per unit area of a liquid when it is in a region of unit velocity gradient.
The 'unit velocity gradient' = 1 ms -1 change per m. Since the 'm'
cancels, the 'unit velocity gradient' = 1 per second. From n = F/(A x
velocity gradient), it follows that r\ may be expressed in units of newton
s m~ 2 (N s m -2 ), or 'dekapoise'.
The coefficient of viscosity of water at 10°C is 1-3 x 10" 3 Nsm -2 .
Since F = nA x velocity gradient, the frictional force on an area of
10 cm 2 in water at 10°C between two layers of water 01 cm apart
which move with a relative velocity of 2 cms -1 is found as follows:
Coefficient of viscosity a/=1-3x10 -3 newton m~ 2 , 4 = 10xl0~ 4
m 2 , velocity gradient = 2 x 10~ 2 m s" 1 -r 01 x 10" 2 m = 2/01 s~ l .
:. F = 1-3 xl0 -3 x 10 xl0 -4 x 2/0-1 = 2-6 x 10" 5 newton.
Dimensions. The dimensions of a force, F, (= mass x acceleration =
mass x velocity change/time) are MLT -2 . See p. 31. The dimensions
of an area, A, are L 2 . The dimensions of velocity gradient
_ velocity change _ L _ 1
distance T ' T
Now // =
dimensions of r\ =
A x velocity gradient'
MLT 2
L 2 x 1/T'
= ML _1 T-
Thus n may be expressed in units 'kg m~ 1 s~
176 ADVANCED LEVEL PHYSICS
Steady Flow of Liquid Through Pipe. Poiseuille's Formula
The steady flow of liquid through a pipe was first investigated
thoroughly by Poiseuille in 1844, who derived an expression for the
volume of liquid issuing per second from the pipe. The proof of the
formula is given on p. 208, but we can derive most of the formula by
the method of dimensions (p. 31).
The volume of liquid issuing per second from the pipe depends on
(i) the coefficient of viscosity, rj, (ii) the radius, a, of the pipe, (hi), the
pressure gradient, g, set up along the pipe. The pressure gradient = p/l,
where p is the pressure difference between the ends of the pipe and
/ is its length. Thus x, y, z being indices which require to be found,
suppose
volume per second = hfa y g z . . (1)
Now the dimensions of volume per second are L 3 T _1 ; the dimensions
of n are ML ~ *T ~ *, see p. 205 ; the dimension of a is L ; and the dimen-
sions of g are
[pressure] [force] MLT~ 2 ML^T 2
[length] ' ° r [area] [length]' ° r L 2 x L ' Wmcn 1S ML l ■
Thus from (i), equating dimensions on both sides,
L 3 T _1 = (ML- 1 T- 1 YL y (ML- 2 T- 2 y.
Equating the respective indices of M, L, T on both sides, we have
x + z = 0,
— x+y — 2z = 3,
x+2z = 1.
Solving, we obtain x = — 1, z = 1, y = 4. Hence, from (1),
4 4
j ,<*g ,P<*
volume per second = k — - = kh—.
n In
We cannot obtain the numerical factor k from the method of dimen-
sions. As shown on p. 209, the factor of n/% enters into the formula,
which is :
Volume per second = „ . . . (2)
8^1
EXAMPLE
Explain as fully as you can the phenomenon of viscosity, using the viscosity
of a gas as the basis of discussion. Show by the method of dimensions how the
volume of liquid flowing in unit time along a uniform tube depends on the radius
of the tube, the coefficient of viscosity of the liquid, and the pressure gradient
along the tube.
The water supply to a certain house consists of a horizontal water main 20 cm
in diameter and 5 km long to which is joined a horizontal pipe 15 mm in diameter
and 10 m long leading into the house. When water is being drawn by this house
SOLID FRICTION, VISCOSITY
177
only, what fraction of the total pressure drop along the pipe appears between the
ends of the narrow pipe? Assume that the rate of flow of the water is very small.
(O. & C.)
Volume per second = -5-7-, with usual notation.
8;//
Thus volume per second =
Trpi.01 4 7cp 2 .0-0075 4
8^ . 5 x 10 3
8»/.10
where p x , p 2 are the respective pressures in the two pipes, since the volume per
second is the same.
. p x 0-0075 4 5xl0 3 1 , ,
• p7 = ~0l- X -10- = 63 (apprOx) -
.'. Px = — x total pressure = 0-016 x total pressure.
64
Turbulent Motion
Poiseuille's formula holds as long as the velocity of each layer of the
liquid is parallel to the axis of the pipe and the flow pattern has been
developed. As the pressure difference between the ends of the pipe is
increased, a critical velocity is reached at some stage, and the motion
of the liquid changes from an orderly to a turbulent one. Poiseuille's
formula does not apply to turbulent motion.
(i) W
Uniform M-E=>r=>E=:
iHiS ^^^s^issgf ^i^Cifi" ' 1? n^i'pwJ^JL
(io =P i
*
Turbulence
Fig. 7.4 Laminar and turbulent flow
The onset of turbulence was first demonstrated by O. Reynolds in
1883, and was shown by placing a horizontal tube T, about 0-5 cm in
diameter, at the bottom of a tank W of water, Fig. 7.4 (i). The flow of
water along T is controlled by a clip C on rubber tubing connected to T.
A drawn-out glass jet B, attached to a reservoir A containing coloured
water, is placed at one end of T, and at low velocities of flow a thin
coloured stream of water is observed flowing along the middle of T. As
the rate of flow of the water along T is increased, a stage is reached
178 ADVANCED LEVEL PHYSICS
when the colouring in T begins to spread out and fill the whole of the
tube, Fig. 7.4 (ii). The critical velocity has now been exceeded, and
turbulence has begun.
Fig. 7.4 shows diagrammatically in inset: (i) laminar or uniform
flow — here particles of liquid at the same distance from the axis always
have equal velocities directed parallel to the axis, (ii) turbulence — here
particles at the same distance from the axis have different velocities,
and these vary in magnitude and direction with time.
Analogy with Ohm's Law
For orderly flow along a pipe, Poiseuille's formula in equation (2)
states :
Volume per second flowing = -§—r,
otjl
p x na 2
Snnx — j
na z
Now p x na 2 = excess pressure x area of cross-section of liquid =
excess force F on liquid, and l/na 2 = l/A, where A is the area of cross-
section.
f
.'. volume per second flowing = . . (i)
Snrj x —
The volume of liquid per second is analogous to electric current (J) if
we compare the case of electricity flowing along a conductor, and the
excess force F is analogous to the potential difference (V) along the
conductor. Also, the resistance R of the conductor = pi/ A, where p is
its resistivity, / is its length, and A is the cross-sectional area. Since,
from Ohm's law, J = V/R; it follows from (i) that
Snrj is analogous to p, the resistivity ;
that is, the coefficient of viscosity r\ is a measure of the 'resistivity' of
a liquid in orderly flow.
Proof of Poiseuille's Formula. Suppose a pipe of radius a has a liquid flowing
steadily along it. Consider a cylinder of the liquid of radius r having the same
axis as the pipe, where r is less than a. Then the force on this cylinder due to the
excess pressure p = px nr 2 . We can imagine the cylinder to be made up of
cylindrical shells; the force on the cylinder due to viscosity is the algebraic sum
of the viscous forces on these shells. The force on one shell is given by r/Adv/dr,
where dv/dr is the corresponding velocity gradient and A is the surface area of
the shell. And although dv/dr changes as we proceed from the narrowest shell
outwards, the forces on the neighbouring shells cancel each other out, by the law
of action and reaction, leaving a net force of rjAdv/dr, where dv/dr is the velocity
gradient at the surface of the cylinder. The viscous force on the cylinder, and the
force on it due to the excess pressure p, are together zero since there is no accelera-
tion of the liquid, i.e., we have orderly or laminar flow.
SOLID FRICTION, VISCOSITY
179
. . i]A— + nr'p = 0.
.'. tj.2nr1~ + Kr 2 p = 0, since A = 2nd.
dv
dr
Irji
~^-r 2 + c
4rjl r +C '
where c is a constant. Since v = when r = a, at the surface of the tube, c =
pa 2 /4ril.
n
■ (i)
— (a 2 -r 2 )
4t]l K }
Consider a cylindrical shell of the liquid between radii r and (r+Sr). The liquid
in this shell has a velocity t; given by the expression in (i), and the volume per
second of liquid flowing along this shell = v x cross-sectional area of shell,
since v is the distance travelled in one second, = vx Inr.dr.
.'. total volume of liquid per second along tube
Jo
Inr.dr
_ npa*
Determination of Viscosity by Poiseuille's Formula
The viscosity of a liquid such as water can be measured by connecting
one end of a capillary tube T to a constant pressure apparatus A, which
provides a steady flow of liquid, Fig. 7.5. By means of a beaker B
Fig. 7.5 Absolute measurement of viscosity
and a stop-clock, the volume of water per second flowing through the
tube can be measured. The pressure difference between the ends of T
is hpg, where h is the pressure head, p is the density of the liquid, and
g is 9-8 ms~ 2 .
.". volume per second =
npa 4 _ nhpga 4
180
ADVANCED LEVEL PHYSICS
where / is the length of T and a is its radius. The radius of the tube can
be measured by means of a mercury thread or by a microscope. The
coefficient of viscosity t] can then be calculated, since all the other
quantities in the above equation are known.
Comparison of Viscosities. Ostwald Viscometer
An Ostwald viscometer, which contains a
vertical capillary tube T, is widely used for
comparing the viscosities of two liquids, Fig. 7.6.
The liquid is introduced at S, drawn by suction
above P, and the time t x taken for the liquid
level to fall between the fixed marks P, Q is
observed. The experiment is then repeated with
the same volume of a second liquid, and the time
t 2 for the liquid level to fall from P to Q is
noted.
Suppose the liquids have respective densities
p t , p 2 . Then, since the average head h of liquid
forcing it through T is the same in each case,
the pressure excess between the ends of T =
hp x g, hp 2 g respectively. If the volume between
the marks P, Q is V, then, from Poiseuille's
formula, we have
V = nihp^a*
h
Fig. 7.6
Ostwald viscometer
StjJ
(i),
where a is the radius of T, rj x is the coefficient of
viscosity of the liquid, and / is the length of T. Similarly, for the second
liquid,
V n(hp 2 g)a A
Srj 2 l
(ii)
Dividing (ii) by (i),
Hi
Vi
1iP
ViPx
h El
h' Pi
(iii)
Thus knowing t lt t 2 and the densities p lt p 2 , the coefficients of viscosity
can be compared. Further, if a pure liquid of a known viscosity is used,
the viscometer can be used to measure the coefficient of viscosity of a
liquid. Since the viscosity varies with temperature, the viscometer
should be used in a cylinder C and surrounded by water at a constant
temperature, Fig. 7.6. The arrangement can then also be used to
investigate the variation of viscosity with temperature. In very accurate
work a small correction is required in equation (iii). Barr, an authority
on viscosity, estimates that nearly 90% of petroleum oil is tested by
an Ostwald viscometer.
SOLID FRICTION, VISCOSITY
181
Experiment shows that the viscosity coefficient of a liquid diminishes
as its temperature rises. Thus for water, n at 15°C is 11 x 10" 3 N s m -2 ,
at 30°C it is 0-8 x KT 3 N s m~ 2 and at 50°C it is 0-6 x 10~ 3 N s m~ 2 .
Lubricating oils for motor engines which have the same coefficient of
viscosity in summer and winter are known as 'viscostatic' oils.
Stokes' Law. Terminal Velocity
When a small object, such as a steel ball-bearing, is dropped into a
viscous liquid like glycerine it accelerates at first, but its velocity soon
reaches a steady value known as the terminal velocity. In this case the
viscous force acting upwards, and the upthrust due to the liquid on the
object, are together equal to its weight acting downwards, so that the
resultant force on the object is zero. An object dropped from an
aeroplane at first increases its speed v, but soon reaches its terminal
speed. Fig. 7.7 shows that variation of v with time as the terminal
velocity v is reached.
Y
Terminal (constant)
velocity
Acceleration
O Time
Fig. 7.7 Motion of falling sphere
Suppose a sphere of radius a is dropped into a viscous liquid of
coefficient of viscosity rj, and its velocity at an instant is v. The frictional
force, F, can be partly found by the method of dimensions. Thus
suppose F = ka x tj y v z , where k is a constant. The dimensions of F are
MLT" 2 ; the dimension of a is L; the dimensions of rj are ML -1 T _1 ;
and the dimensions of v are LT~ 1 .
.'. MLT -2 = L^xrML^T-^xfLT" 1 ) 2 .
Equating indices of M, L, T on both sides,
.-. v = 1,
x-y + z = 1,
—y — z = —2.
Hence z = 1, x = 1, v = 1. Consequently F = krjav. In 1850 Stokes
showed mathematically that the constant k was 6k, and he arrived at
the formula
F = 6natjv .... (1)
182
ADVANCED LEVEL PHYSICS
Comparison of Viscosities of Viscous Liquids
Stokes' formula can be used to compare the coefficients of viscosity
of very viscous liquids such as gly-
cerine or treacle. A tall glass vessel G
is filled with the liquid, and a small
ball-bearing P is dropped gently into
~x Ijplf^lP^ll tlie liquid so that it falls along the axis
<* Agm^^Eii of G, Fig. 7.8. Towards the middle
of the liquid P reaches its terminal
velocity v , which is measured by
timing its fall through a distance AB
orBC.
The upthrust, U, on P due to the
liquid = 47ta*<rg/3, where a is the
radius of P and a is the density of the
liquid. The weight, W, of P is 4na 3 pg/3,
where p is density of the bearing's
material. The net downward force is
thus 4na 3 g(p— <r)/3. When the opposing frictional force grows to this
magnitude, the resultant force on the bearing is zero. Thus for the
terminal velocity v , we have
A
B
O
G
F 1
Fig. 7.8 Stokes' law
6nrjav = %na 3 g{p — a),
2ga 2 (p-a)
n =
9v f
(i)
When the experiment is repeated with a liquid of coefficient of
viscosity tj t and density <j v using the same ball-bearing, then
>7i =
_ 2gfa 2 (p-g 1 )
9v t
where v t is the new terminal velocity. Dividing (i) by (ii),
(ii)
(iii)
Thus knowing v v v, p, a v a, the coefficients of viscosity can be com-
pared. In very accurate work a correction to (iii) is required for the
effect of the walls of the vessel containing the liquid.
Molecular theory of viscosity
Viscous forces are detected in. gases as well as in liquids. Thus if a
disc is spun round in a gas close to a suspended stationary disc, the
latter rotates in the same direction. The gas hence transmits frictional
forces. The flow of gas through pipes, particularly in long pipes as in
transmission of natural gas from the North Sea area, is affected by the
viscosity of the gas.
The viscosity of gases is explained by the transfer of momentum which
SOLID FRICTION, VISCOSITY 183
takes place between neighbouring layers of the gas as it flows in a
particular direction. Fast-moving molecules in a layer X cross with
their own velocity to a layer Y say where molecules are moving with a
slower velocity. Fig. 7.9. Molecules in Y likewise move to X. The net
Pipe
Y Sl fw
X Fast
Gas flow
Fig. 7.9 Viscosity of gas-momentum effect
effect is an increase in momentum in Y and a corresponding decrease
in X, although on the average the total number of molecules in the two
layers is unchanged. Thus the layer Y speeds up and the layer X slows
down, that is, a force acts on the layers of the gas while they move.
This is the viscous force. We consider the movement of molecules in
more detail shortly.
Although there is transfer of momentum as in the gas, the viscosity
of a liquid is mainly due to the molecular attraction between molecules
in neighbouring layers. Energy is needed to drag one layer over the
other against the force of attraction. Thus a shear stress is required to
make the liquid move in laminar flow.
EXERCISE 7
What are the missing words in the statements 1-6?
1. The coefficient of dynamic (kinetic) friction is the ratio . . .
2. The coefficient of friction between two given surfaces is ... of the area in
contact.
3. In orderly or laminar flow of liquids in a pipe, the volume per second
flowing past any section is given by the formula . . .
4. The dimensions of coefficient of viscosity are . . .
5. When a small sphere of radius a falls through a liquid with a constant
velocity v, the frictional force is given by the formula . . .
6. In comparing the viscosities of water and alcohol by an Ostwald viscometer,
the same liquid . . . must used.
Which of the following answers, A, B, C, D or E, do you consider is the correct
one in the statements 7-10?
7. In orderly or laminar flow of a liquid through a pipe, A tensile forces act
on the layers and the volume per second V is proportional to the pressure at one
end, B shear forces act on the layers and V is proportional to the pressure at one
end, C shear forces act on the layers and V is proportional to the pressure dif-
ference between the ends, D bulk forces act throughout the liquid, E V is directly
proportional to a 4 and to the coefficient of viscosity.
184 ADVANCED LEVEL PHYSICS
8. When a small steel sphere is dropped gently down the axis of a wide jar of
glycerine, the sphere A travels with constant velocity throughout its motion,
B accelerates at first and then reaches a constant velocity, C decelerates at first
and then reaches a constant velocity, D accelerates throughout its motion,
E slowly comes to rest.
9. When a gas flows steadily along a pipe, the viscous forces in it are due to
A transfer of energy from one layer to another, B the uniform speed of the mole-
cules, C the varying density along the pipe, D the transfer of momentum from one
layer to another, E the varying pressure at a given section of the pipe.
10. A pipe P has twice the diameter of a pipe Q, and P has a liquid X flowing
along it which has twice the viscosity of a liquid Y flowing through Q. If the flow
is orderly or laminar in each, and the volume per second in P and Q is the same,
the pressure difference at the ends of P compared to that of Q is A l:8,Bl-4
C8:1,D4:1,£1:1.
Solid Friction
11. State the laws of solid friction. Describe an experiment to determine the
coefficient of dynamic (or sliding) friction between two surfaces.
A horizontal circular turntable rotates about its centre at the uniform rate of
120 revolutions per minute: Find the greatest distance from the centre at which
a small body will remain stationary relative to the turntable, if the coefficient of
static friction between the turntable and the body is 0-80. (L.)
12. State (a) the laws of solid friction, (b) the triangle law for forces in equili-
brium. Describe an experiment to determine the coefficient of sliding (dynamic)
friction between two wooden surfaces.
A block of wood of mass 150 g rests on an inclined plane. If the coefficient of
static friction between the surfaces in contact is 030, find (a) the greatest angle
to which the plane may be tilted without the block slipping, (b) the force parallel
to the plane necessary to prevent slipping when the angle of the plane with the
horizontal is 30°, showing that this direction of the force is the one for which the
force required to prevent slipping is a minimum. (L.)
13. Distinguish between static and sliding (kinetic) friction and define the
coefficient of sliding friction.
How would you investigate the laws of sliding friction between wood and
iron?
An iron block, of mass 10 kg, rests on a wooden plane inclined at 30° to the
horizontal. It is found that the least force parallel to the plane which causes the
block to slide up the plane is 10 kgf. Calculate the coefficient of sliding friction
between wood and iron. (N.)
14. Give an account of the factors which determine the force of friction
(i) between solids, (ii) in liquids.
A block of mass 12 kg is drawn along a horizontal surface by a steadily
applied force of 4 kg weight acting in the direction of motion. Find the kinetic
energy acquired by the block at the end of 10 seconds and compare it with the
total work done on the block in the same time. (Coefficient of friction = 0-28.) (L.)
15. State the laws of solid friction.
Describe experiments to verify these laws, and to determine the coefficient of
static friction, for two wooden surfaces.
A small coin is placed on a gramophone turntable at a distance of 7-0 cm from
the axis of rotation. When the rate of rotation is gradually increased from zero
SOLID FRICTION, VISCOSITY 185
the coin begins to slide outwards when the rate reaches 60 revolutions per
minute. Calculate the rate of rotation for which sliding would commence if
(a) the coin were placed 12-0 cm from the axis, (b) the coin were placed in the
original position with another similar coin stuck on top of it. (L
16. Define coefficient of sliding friction, coefficient of viscosity. Contrast the
laws of solid friction with those which govern the flow of liquids through tubes.
Sketch the apparatus you would employ to determine the coefficient of sliding
friction between a wood block and a board and show how you would deduce
the coefficient from a suitable graph. (L.)
Viscosity
17. Define coefficient of viscosity of a fluid.
When the flow is orderly the volume V of liquid which flows in time t through a
tube of radius r and length / when a pressure difference p is maintained between
V Tzpr*
its ends is given by the equation — = — — where n is the coefficient of viscosity
t Olt]
of the liquid. Describe an experiment based on this equation either (a) to deter-
mine the value of n for a liquid, or (b) to compare the values of rj for two liquids,
pointing out the precautions which must be taken in the experiment chosen to
obtain an accurate result.
Water flows steadily through a horizontal tube which consists of two parts
joined end to end; one part is 21 cm long and has a diameter of 0-225 cm and
the other is 70 cm long and has a diameter of 0075 cm. If the pressure difference
between the ends of the tube is 14 cm of water find the pressure difference between
the ends of each part. (L.)
18. The dimensions of energy, and also those of moment of a force are found
to be 1 in mass, 2 in length and —2 in time. Explain and justify this statement.
, (a) A sphere of radius a moving through a fluid of density p with high velocity
V experiences a retarding force F given by F = k.a x .p y . V z , where k is a non-
dimensional coefficient. Use the method of dimensions to find the values of
x, y and z.
(b) A sphere of radius 2 cm and mass 100 g, falling vertically through air of
density 1-2 kg m~ 3 , at a place where the acceleration due to gravity is 9-81 m
s -2 , attains a steady velocity of 30 m s -1 . Explain why a constant velocity is
reached and use the data to find the value of k in this case. (O. & C.)
19. Mass, length and time are fundamental units, whereas acceleration, force
and energy are derived units. Explain the distinction between these two types of
unit. Define each of the three derived units and apply your definition in each case
to deduce its dimensions.
An incompressible fluid of viscosity rj flows along a straight tube of length /
and uniform circular cross-section of radius r. Provided the pressure difference p
between the ends of the tube is not too great the velocity u of fluid flow along the
axis of the tube is found to be directly proportional to p. Apply the method of
dimensions to deduce this result assuming u depends only on r, I, n and p.
How may the viscosity of an ideal gas be accounted for by elementary molecular
theory? (O. & C.)
20. Define coefficient of viscosity. Describe an experiment to compare the
coefficients of viscosity of water and benzene at room temperature.
A small metal sphere is released from rest in a tall wide vessel of liquid. Discuss
the forces acting on the sphere (a) at the moment of release, (fo) soon after release,
(c) after the terminal velocity has been attained.
186 ADVANCED LEVEL PHYSICS
Castor oil at 20°C has a coefficient of viscosity 2-42 N s m~ 2 and a density
940 kg m 3 . Calculate the terminal velocity of a steel ball of radius 20 mm
falling under gravity in the oil, taking the density of steel as 7800 kg m~ 3 . (L.)
21. Define coefficient of viscosity. What are its dimensions?
By the method of dimensions, deduce how the rate of flow of a viscous liquid
through a narrow tube depends upon the viscosity, the radius of the tube, and
the pressure difference per unit length. Explain how you would use your results
to compare the coefficients of viscosity of alcohol and water. (C.)
2 . Define coefficient of viscosity. For orderly flow of a given liquid through
a capillary tube of length /, radius r, the volume of liquid issuing per second is
proportional to pr 4 // where p is the pressure difference between the ends of the
tube. How would you verify this relation experimentally for water at room
temperature? How would you detect the onset of turbulence? (N.)
23. The viscous force acting on a small sphere of radius a moving slowly
through a liquid of viscosity rj with velocity v is given by the expression 6ntjav,
Sketch the general shape of the velocity-time graph for a particle falling from rest
through a viscous fluid, and explain the form of the graph. List the observations
you would make to determine the coefficient of viscosity of the fluid from the
motion of the particle.
Some particles of sand are sprinkled on to the surface of the water in a beaker
filled to a depth of 10 cm. Estimate the least time for which grains of diameter
010 mm remain in suspension in the water, stating any assumptions made.
[Viscosity of water = 11 x 10" 3 N s m~ 2 ; density of sand = 2200 kg m -3 .]
(C.)
24. Define coefficient of friction and coefficient of viscosity.
Describe how you would (a) measure the coefficient of sliding friction between
iron and wood, and (b) compare the viscosities of water and paraffin oil. (L.)
PART TWO
Heat
chapter eight
Introduction
Temperature
We are interested in heat because it is the commonest form of energy,
and because changes of temperature have great effects on our personal
comfort, and on the properties of substances, such as water, which we
use every day. Temperature is a scientific quantity which corresponds
to primary sensations— hotness and coldness. These sensations are
not reliable enough for scientific work, because they depend on
Steam point
Fine bore
stem
Mercury-
Ice point
■100°C
•0°C
Triple point _ 273-16 K
of water
Thin walled,
bulb
I
L
Celsius
Thermodynamic
(Kelvin)
Fig. 8.1. Mercury-in-glass thermometer (left); °C and K scales.
contrast — the air in a thick-walled barn or church feels cool on a
summer's day, but warm on a winter's day, although a thermometer
may show that it has a lower temperature in the winter* A thermometer,
such as the familiar mercury-in-glass instrument (Fig. 8.1), is a device
whose readings depend on hotness or coldness, and which we choose
to consider more reliable than our senses. We are justified in considering
it more reliable because different thermometers of the same type agree
with one another better than different people do.
The temperature of a body, then, is its degree of hotness, as measured
on a thermometer. The thermometer was invented in Italy about 1630:
it consisted of an open-ended tube, with a bulb full of water at its lower
end. The water rose in the tube when the bulb was warmed, and fell
when it was cooled.
189
190 ADVANCED LEVEL PHYSICS
As a liquid for use in thermometers, water soon gave way to linseed
oil or alcohol, and by about 1 660 thermometer-makers had begun to seal
the top of the tube. Early thermometers had no definite scale, like that of
a modern thermometer; some of them were used for showing the tem-
peratures of greenhouses, and were mounted on wooden backboards,
which were carved with grapes and peaches for example, to indicate the
correct temperatures for growing the different fruits. The thermometer
as we know it to-day, containing pure mercury and graduated according
to a universal scale, was developed by Fahrenheit in 1724.
Temperature Scales. Celsius
When a mercury thermometer is to be graduated, it is placed first
in melting ice, and then in steam from boiling water (Fig. 8.2). The
temperature of the steam depends on the atmospheric pressure, as we
Thermometer —
Stirrer
Thermos
flask
f
1
Double- walled
copper vessel
/ to prevent cooling
of steam in 'A'
Water manometer to
/ show when steam
1 / SIIUW Wll«3
irT^ II in 'A' is at
pi U y atmospheric
i " \d? pressure
-Water
(a) Ice point (b) Steam point
Fig. 8.2. Determination of fixed points.
shall see in Chapter 12; for calibrating thermometers, an atmospheric
pressure of 76 cm mercury is chosen. In both steam and ice, when the
level of the mercury has become steady, it is marked on the glass : the
level in ice is called the lower fixed point, or ice-point, and the level in
steam is called the upper fixed point, or steam-point. The distance
between the fixed points is called the fundamental interval of the ther-
mometer. For scientific work, the fundamental interval is divided into
100 equal parts (Fig. 8.1). This division was first proposed by Celsius
in 1742, and the graduations are called degrees Centigrade or, in
modern nomenclature, degrees Celsius (°C); the ice-point is 0°C and the
steam-point 100°C.
Thermodynamic Scale
The thermodynamic scale of temperature is adopted as the SI
temperature scale. On this scale, the kelvin is the unit of temperature.
It is defined as 1/27316 of the thermodynamic temperature of the triple
point of water (p. 319).
The symbol for temperature is 'K' without a degree sign. Thus the
triple point of water, T tT , is 273- 16 K exactly. The absolute zero, K,
INTRODUCTION 191
is -27315°C. To a good approximation, 0°C = 273 K and 100°C
= 373 K. (Fig. 8.1.)
The temperature change or interval of one degree Celsius, 1 degC
or 1°C, is exactly the same as the temperature interval of one degree
on the thermodynamic scale. On this account the interval 'degC or °C
is written 'K' in SI units. Similarly, 'per degC or per °C' is written
*K -1 ' in SI units. For example, the linear expansivity (formerly,
linear coefficient of expansion) of steel is written '12 x 10 -6 K *' in
SI units, in place of '12 x 10~ 6 per degC\ The use of 'K 1 ' occurs
frequently in units throughout the subject and should be noted by the
reader.
Types of Thermometer
The mercury-in-glass thermometer depends on the change in volume
of the mercury with hotness; it is cheap and simple, but is not reliable
enough for accurate work (Chapter 14). Other types of thermometer
depend on the change, with hotness, of the pressure of a gas at constant
volume or the electrical resistance of a metal (Fig. 8.3). Another type of
To electrical
imMMMj — '* ^5
Coil of fine wire instrument
Fig. 8.3. A resistance thermometer; the wire is usually of platinum.
thermometer depends on the electromotive force change with tem-
perature of two metals joined together. Fig. 8.4 (a) shows two wires,
one of copper and one of iron, soldered together at A. The ends of the
wires are joined to a galvanometer G. When the junction A is heated,
a current flows which deflects the galvanometer. The current usually
increases with the temperature difference between the hot and cold
ends of the wires. For temperature measurement two junctions are
used, as in Fig. 8.4 (b); the second one, called the cold junction, is
maintained at 0°C by ice-water.
Each of these quantities — e.m.f., resistance, pressure — gives its own
temperature scale, and the different scales agree only at the fixed points,
where their readings are defined as 0°C and 100°C. (When we speak of
(a) (b)
Fig. 8.4. Thermojunctions or thermocouples.
192
ADVANCED LEVEL PHYSICS
a temperature scale, we refer to the quantity used to define it; the dif-
ference between °C and K is only a difference in the graduation of a
given scale.) If, for example, a given platinum wire has resistances R
and R l00 at the ice- and steam-point respectively, then its fundamental
interval is R l00 -R . And if it has a resistance R at an unknown tem-
perature, the value of that temperature, t P , on the platinum resistance
Celsius scale, is given by
h= R R ~ R ° R xlOO( C).
^100 — ^0
The platinum-resistance scale differs appreciably from the mercury-in-
glass scale, as the following table shows :
Mercury-in-glass
50
100
200
300
°C
Platinum-resistance
50-25
100
197
291
°c
We shall discuss temperature scales again later (p. 366) ; here we wish
only to point out that they differ from one another, that no one of
them is any more 'true' than any other, and that our choice of which
to adopt is arbitrary, though it may be decided by convenience.
Effects of Temperature
Most bodies, when they are made hotter, become larger; their in-
crease in size is called thermal expansion. Thermal expansion may be
useful, as in a thermometer, or it may be a nuisance, as in bridges and
railway lines. If the thermal expansion of a solid or liquid is resisted,
great forces are set up : that is why gaps are left between railway lines,
and why beer-bottles are never filled quite full. If the thermal expansion
of a gas is resisted, however, the forces set up are not so great ; the
pressure of the gas increases, but not catastrophically. The increase of
pressure is, in fact, made use of in most forms of engine ; it is also made
use of in accurate thermometry.
Besides causing a change in size or pressure, a change of temperature
may cause a change of
state — from solid to
liquid, liquid to gas, or
vice versa. If we heat
some crystals of lead
acetate in a crucible, and
measure their tempera-
ture, with a thermometer
reading to 300°C, we
find that the crystals
warm steadily up to
75°C and then start to
melt. Their temperature
does not rise further until
they have all melted
(Fig. 8.5). After it has
melted, the lead acetate
warms up to 280°C, and
280
o
Time
Fig. 8.5. Warming curve of lead acetate.
INTRODUCTION 193
then keeps a constant temperature until it has all boiled away. We call
280°C the boiling-point of lead acetate; likewise we call 75°C the
melting- (or freezing-) point of lead acetate.
HEAT AND ENERGY
Heat and Temperature
If we run hot water into a lukewarm bath, we make it hotter; if we
run cold water in, we make it cooler. The hot water, we say, gives heat
to the cooler bathrwater; but the cold water takes heat from the
warmer bath-water. The quantity of heat which we can get from hot
water depends on both the mass of water and on its temperature: a
bucket-full at 80°C will warm the bath more than a cup-full at 100°C.
Roughly speaking, temperature is analogous to electrical potential,
and heat is analogous to quantity of electricity. We can perceive tem-
perature changes, and whenever the temperature of a body rises, that
body has gained heat. The converse is not always true ; when a body is
melting or boiling, it is absorbing heat from the flame beneath it, but
its temperature is not rising.
Latent Heat and Specific Heat
The heat which a body absorbs, in melting or boiling, it gives out
again in freezing or condensing; such heat is called latent, or hidden,
heat, because it does not show itself by a change in temperature. When
a body absorbs heat without changing its state, its temperature rises,
and the heat absorbed was first called 'sensible heat'.
The term 'latent heat' was used by Black (1728-99); he and a Swede,
Wilcke, discovered latent heats independently at about the same date-
Black by hanging a bucket of ice in a warm room, Wilcke by pouring
boiling water on to snow.
Also independently, Black and Wilcke studied what we now call
specific heats; the name is due to Wilcke. In his experiments Wilcke
dropped various hot bodies into cold water, and measured the tem-
perature rises which they caused. In this way he showeq* that a given
mass of glass, for example, gave out only one-fifth as much heat as an
equal mass of water, in cooling through the same temperature range.
He therefore said that the specific heat of glass was 0-2.
In the seventeenth and eighteenth centuries the nature of heat was
disputed; some thought of heat as the motion of the particles of a
body, others thought of it as a fluid, filling the body's pores. Measure-
ments of heat were all relative, and no unit of the quantity of heat was
defined. In the nineteenth century, however, the increasing technical
importance of heat made a unit of it essential. The units of heat chosen
were :
(i) the calorie (cal): this is the amount of heat required to warm
1 gramme (g) of water through 1 deg C (see also p. 194);
(ii) the British Thermal Unit (Btu) : this is the amount of heat required
to warm 1 lb of water through 1 deg F.
"■M ADVANCED LEVEL PHYSICS
Heat and Energy
Steam-engines became common in the early part of the eighteenth
century ; but they were not thought of as heat-engines until the latter
part of that century. Consequently the early engines were wasteful of
fuel, squandering useful heat in warming and cooling the cylinder at
every stroke of the piston. Watt reduced this waste of heat by his
invention of the separate condenser in 1769. Trevithick, about 1800,
devised an engine which was driven by steam which entered the cylinder
at a pressure above atmospheric, and therefore at a temperature above
100°C (p. 304). In this engine, the steam came out of the exhaust at a
temperature no higher than in earlier engines, so that a greater fraction
of the heat which it carried from the boiler was used in the engine.
The idea of heat as a form of energy was developed particularly by
Benjamin Thompson (1753-1814); he was an American who, after
adventures in Europe, became a Count of the Holy Roman Empire,
and war minister of Bavaria. He is now generally known as Count
Rumford. While supervising his arsenal, he noticed the great amount
of heat which was liberated in the boring of cannon. The idea common
at the time was that this heat was a fluid, pressed out of the chips of
metal as they were bored out of the barrel. To measure the heat pro-
duced, Rumford used a blunt borer, and surrounded it and the end of
the cannon with a wooden box, filled with water (Fig. 8.6). From the
weight of water, and the rate at which its temperature rose, he concluded
that the boring operation liberated heat at the same rate as 'nine wax
candles, burning with a clear flame'. He showed that the amount of heat
liberated was in no way connected with the mass of metal bored away,
and concluded that it depended only on the work done against friction.
It followed that heat was a form of energy.
Rumford published the results of his experiments in 1798. No similar
experiments were made until 1840, when Joule began his study of
heat and other forms of energy. Joule measured the work done, and the
heat produced, when water was churned, in an apparatus which we shall
describe on p. 197. He also measured the work done and heat produced
when oil was churned, when air was compressed, when water was
forced through fine tubes, and when cast iron bevel wheels were rotated
one against the other. Always, within the limits of experimental error,
he found that the heat liberated was proportional to the mechanical
work done, and that the ratio of the two was the same in all types of
experiment. His last experiments, made in 1878, showed that about
772 ft-lbf of work were equivalent to one British thermal unit of heat.
This ratio Joule called the mechanical equivalent of heat. The metric
unit of work or energy is the joule, J. Since experiment shows that heat
is a form of energy, the joule is now the scientific unit of heat. 'Heat per
second' is expressed in 'joules per second' or watts, W.
Today, from definition, about 4-2 J = 1 calorie (more accurately,
4-187 J = 1 calorie). Calories are units still used by chemists, for ex-
ample. Values in calories met can be converted to joules approximately
by multiplying them by 4-2. See p. 199.
In other experiments, Joule measured the heat liberated by an electric
INTRODUCTION
195
current in flowing through a resistance; at the same time he measured
the work done in driving the dynamo which generated the current.
He obtained about the same ratio for work done to heat liberated as
in his direct experiments. This work linked the ideas of heat, mechanical,
and electrical energy. He also showed that the heat produced by a
current is related to the chemical energy used up.
Fig. 8.6. Rumford's apparatus for converting work into heat.
Fig. 1 shows the cannon. Fig. 2 shows the complete apparatus; w is connected to
machinery driven round by horses, m is joined to a blunt borer n in a cylinder shown
enlarged in Fig. 3 and Fig. 4. Figs. 5, 6, 7, 8 show further details of m and n.
The Conservation of Energy
As a result of all his experiments, Joule developed the idea that
energy in any one form could be converted into any other. There might
be a loss of useful energy in the process — for example, some of the heat
from the furnace of a steam-engine is lost up the chimney, and some
more down the exhaust — but no energy is destroyed. The work done
by the engine added to the heat lost as described and the heat developed
as friction, it is equal to the heat provided by the fuel burnt. The
idea underlying this statement is called the Principle of the Conserva-
tion of Energy. It implies that, if we start with a given amount of energy
in any one form, we can convert it in turn into all other forms; we may
not always be able to convert it completely, but if we keep an accurate
balance-sheet we shall find that the total amount of energy, expressed
in any one form — say heat or work— is always the same, and is equal
to the original amount.
The conservation of energy applies to living organisms — plants and
animals — as well as to inanimate systems. For example, we may put a
man or a mouse into a box or a room, give him a treadmill to work,
and feed him. His food is his fuel; if we burn a sample of it, we can
measure its chemical energy, in heat units. And if we now add up the
196
ADVANCED LEVEL PHYSICS
heat value of the work which the man does, and the heat which his body
gives off, we find that their total is equal to the chemical energy of the
food which the man eats. Because food is the cource of man's energy,
food values are commonly expressed in kilocalories, which is the heat
required to warm 1 kilogramme of water through 1 deg C. A man
needs about 3000 kilocalories per day.
Muscles are unique in their capacity to turn chemical energy directly
into mechanical energy. When a muscle is stimulated, complex phos-
phates in its tissues break down; in doing so, they cause the muscle
fibres to swell and shorten. Thus, via the bones and joints, the muscle
does external work. When the muscle is recovering after contraction,
the phosphates are built up again by a series of reactions, involving the
oxidation of sugars. The sugars and oxygen are brought to the muscle
in the arterial blood; the waste products of the reactions, water and
carbon dioxide, are carried away in the venous blood. 1 Recently
physiologists have found evidence that muscles may also convert
mechanical energy into chemical. 2 For example, when we walk down-
stairs, gravity does work on our leg-muscles ; some of this appears as
heat, but some, it now seems, is used in reversing the chemical actions
of muscle activity.
All the energy by which we live comes from the sun. The sun's
ultra-violet rays are absorbed in the green matter of plants, and make
them grow ; the animals eat the plants, and we eat them — we are all
vegetarians at one remove. The plants and trees of an earlier age de-
cayed, were buried, and turned into coal. Even water-power comes
Engines,
expansion of gases in explosions
cell-charging electrolysis
Fig. 8.7. Forms of energy, and their interconversions.
1 Davson, General Physiology, Chap. XVII (Churchill).
2 Abbott, Aubert and Hill, Jour. Physiology, Vol. III.
INTRODUCTION
197
from the sun — we would have no lakes if the sun did not evaporate the
sea and provide the rainfall which fills the lakes. The relationship
between all the principal forms of energy are summarized in Fig. 8.7.
Joule's Historic Experiments
About 1 847 Joule measured the mechanical equivalent of heat by an apparatus
of the form shown in Fig. 8.8. C is a copper cylinder, about 30 cm in diameter,
containing water. The water is churned by paddles P, and prevented from
whirling round en masse by baffles B. The paddles are connected by a coupling
a
II 1
1 1 s
1
1 —
Fig. 8.8. Joule's apparatus for mechanical equivalent.
K to a drum D, which is rotated by strings S attached to lead weights M. A
thermometer T shows the temperature of the water.
Joule would start an experiment by allowing the weights M to fall to the
ground and turn the paddles. He would then break the coupling K, and re-wind
the weights without disturbing the paddles. In this way he would make the
weights fall twenty times or more, in a single experiment, and so increase the
work done on the water and consequently the temperature rise.
Suppose n is the number of falls ; then the work W done on the water = 2nMgh,
where M is the mass of one weight, g is the acceleration of gravity, and h is the
height of the fall. The heat Q gained by the cylinder and the water is (mc w + C)0,
where m is the mass of water, c w the heat capacity of the cylinder and paddles,
and 6 is their rise in temperature. The rise 6 includes the correction for heat losses.
The mechanical equivalent is given by W'/Q
InMgh
(mc w + C)0 ' ' ' • " K)
An experiment of this kind takes a long time — about half an hour — because
a great deal of work, by everyday standards, must be done to produce a measurable
amount of heat. The cooling correction is therefore relatively great.
Many people refused to accept Joule's work at first, because of the very small
temperature differences on which it rested. Nevertheless, Joule's final result
differs only by about one part in 400 from the value given by the best modern
198
ADVANCED LEVEL PHYSICS
///f/,////
experiments. In calculating his final result, Joule made corrections for the kinetic
energy of the weights as they struck the floor, the work done against friction in the
pulleys and the bearings of the paddle wheel, and the energy stored by the
stretching of the strings; he even estimated the energy in the hum which the
strings emitted, but found it was negligible.
Joule's Large-scale Experiments
In his last experiments, about 1878, Joule rotated the paddles with an engine,
thereby eliminating many
of the corrections just
mentioned. He suspended
the cylinder on a wire,
and kept it in equilibrium
by an opposing couple,
applied by means of the
wheel D (Fig. 8.9). This
method was repeated in
1880 by Rowland, who
had holes drilled in the
paddles and baffles, to
make their churning ac-
tion more thorough. The
moment, T, of the couple
applied by the engine is
equal and opposite to that
of the couple applied by
the masses M. Its value
Engine
Fig. 8.9. Joule's final apparatus, also Rowland's,
is therefore
T = Mgd,
where d is the diameter of the wheel. Now the work done by a couple is equal to
the product of its moment and the angle 6 in radians through which it turns.
Hence if the paddles make n revolutions, the work done on the water, since
= 2nn, is
W = 2nnT
= 2%nMgd.
The number of revolutions was measured on a revolution counter attached to
the paddle spindle. If 6 is the rise in temperature measured by the thermometer T,
corrected for cooling, the heat developed is
Q = {mc w + C)B,
in our previous notation. Hence the mechanical equivalent
Q
InnMgd
(mc w + C)9
The term 'mechanical equivalent of heat', used in the past, has no
meaning nowadays because heat is measured in joules in SI units
(p. 194). Experiments in which mechanical energy is converted to heat
energy are now regarded as experiments which measure the specific
heat capacity of the heated substance in 'joule per kilogramme (or
gramme) per deg K' (J kg -1 K _1 or J g _1 K _1 ). The specific heat
capacity of water may be found by the energy conversion method
described on p. 206.
chapter nine
Calorimetry
Calorimetry is the measurement of heat; here we shall be concerned
with the measurement of specific heat capacities and specific latent
heats.
Heat (Thermal) Capacity, Specific Heat Capacity
The heat capacity of a body, such as a lump of metal, is the quantity
of heat required to raise its temperature by 1 degree. It is expressed in
joules per deg K (J K - 1 ).
The specific heat capacity of a substance is the heat required to warm
unit mass of it through 1 degree ; it is the heat capacity per unit mass of
the substance. Specific heat capacities are expressed in joule per kilo-
gramme per deg K (J kg -1 K _1 ) or in joule per gramme per deg K
(J g~ x K" 1 ). The specific heat of water, c w , is about 4-2 J g~ * K~ \ or
4200 J kg -1 K~\ or 4-2 kJ kg' 1 K~\ where 1 kJ = 1 kilojoule =
1000 J. Formerly, specific heat capacities were expressed in calories per
g per deg C — the values in joule are about 4-2 times as great.
From the definition of specific heat capacity, it follows that
heat capacity, C = mass x specific heat capacity.
The specific heat capacity of copper, for example, is about 0-4 J g~ 1 K~ 1
or 400 J kg" 1 K~ *. Hence the heat capacity of 5 kg of copper = 5 x 400
= 2000JK- 1 = 2kJK" 1 .
SPECIFIC HEAT CAPACITIES
Substance
Sp.Ht.:
Jkg^K -1
Substance
Sp.Ht.:
Jkg-'K- 1
Aluminium
0-91 xlO 3
Ice
2-1 x 10 3
Brass
0-38
Paraffin wax
2-9
Copper .
0-39
Quartz
0-7
Iron
0-47
Rubber
1-7
Lead
013
Stone .
0-9
Mercury
014
Wood.
1-7
Nickel .
0-46
Alcohol
2-5
Platinum
013
Brine (25% by wt.)
3-4
Silver .
0-24
Carbon tetra-
chloride
084
Solder .
018
Ether .
2-4
Steel .
0-45
Glycerine .
2-5
Ebonite .
1-7
Paraffin oil .
2-1
Glass .
07
Turpentine .
1-76
199
200
ADVANCED LEVEL PHYSICS
MEASUREMENT OF SPECIFIC HEAT CAPACITY
Method of Mixtures
A common way of measuring specific heat capacities is the method
of mixtures, used by Wilcke (p. 193). Fig. 9.1 shows how we may
/*
6,
Steam ■
Stirrer
m,c
™c-c,
Fig. 9.1. Specific heat capacity by mixtures.
apply it to a solid, such as a metal. We weigh the specimen (m g) and
hang it on a thread in a steam jacket, J, fitted with a thermometer.
The jacket is plugged with cotton wool to prevent loss of heat by
convection. While the solid is warming, we weigh a thin- walled copper
vessel A called a calorimeter (m c g), then run about 50 cm 3 of water
into it, and by subtraction find the mass m r of this water. We put the
calorimeter into a draught-shield S, and take the temperature, X , of
the water in it. After we have given the specimen time to warm up —
say half an hour — we read its temperature, 3 ; then we slide the calori-
meter under the jacket, and drop the specimen into it. After stirring
the mixture, we measure its final temperature, 2 . If no heat leaves the
calorimeter by radiation, conduction, or convection, after the hot
specimen has been dropped into the calorimeter, we have :
heat lost by solid in cooling from 3 to 2 = heat gained by water
and calorimeter in warming from X to 2 .
(The heat gained by the thermometer and stirrer may be neglected
if high accuracy is not required.)
Therefore, if c is the specific heat of the solid, c w that of water and c c
that of the calorimeter :
mc(0 3 -0 2 ) = m x c w (0 2 -0 x ) + m c c c (0 2 -0 x )
= (m 1 c w + m c c c )(0 2 -0 1 ),
{m x c w + m c c c ){9 2 -0 x )
whence c =
(1)
m{0 3 -0 2 )
Liquids
The specific heat capacity of a liquid can be found by putting some
in a calorimeter and dropping a hot solid, of known specific heat
capacity c, into it. If m„ c, are the mass and specific heat capacity of the
CALORIMETRY 201
liquid, then the product m,c, replaces m in equation (1), from which c,
can be calculated.
Calculations
As an illustration of a specific heat capacity determination, suppose a
metal of mass 200 g at 100°C is dropped into 80 g of water at 15°C
contained in a calorimeter of mass 120 g and specific heat capacity
0-4 kJ kg" * K~ 1 . The final temperature reached is 35°C. Then :
heat capacity of calorimeter = 120 x 0-4 = 48 J K~ 1
heat capacity of water = 80 x 4-2 = 336 J K~ l
.'. heat gained by water + cal. = (336 + 48) x (35 — 15) J
and heat lost by hot metal = 0-2 x c x (100— 35)J
.'. 0-2xcx65 = 384x20
•'• c = M^5 = 590J kg_1 K_1 ( a PP rox -)-
Heat Losses
In a calorimetric experiment, some heat is always lost by leakage.
Leakage of heat cannot be prevented, as leakage of electricity can, by
insulation, because even the best insulator of heat still has appreciable
conductivity (p. 333).
When convection is prevented, gases are the best thermal insulators.
Hence calorimeters are often surrounded with a shield S, as in Fig. 9.1,
and the heat loss due to conduction is made small by packing S with
insulating material or by supporting the calorimeter on an insulating
ring, or on threads. The loss by radiation is small at small excess
temperatures over the surroundings. In some simple calorimetric ex-
periments the final temperature of the mixture is reached quickly, so
that the time for leakage is small. The total loss of heat is therefore
negligible in laboratory experiments on the specific heats of metals,
but not on the specific heat capacities of bad conductors, such as rubber,
which give up their heat slowly. When great accuracy is required, the
loss of heat by leakage is always taken into account.
Newton's Law of Cooling
Newton was the first person to investigate the heat lost by a body in
air. He found that the rate of loss of heat is proportional to the excess
temperature over the surroundings. This result, called Newton's law of
cooling, is approximately true in still air only for a temperature excess
of about 20°C or 30°C; but it is true for all excess temperatures in
conditions of forced convection of the air, i.e. in a draught. With
natural convection Dulong and Petit found that the rate of loss of
heat was proportional to 5 ' 4 , where is the excess temperature, and
this appears to be true for higher excess temperatures, such as from
50°C to 300°C. At low excess temperatures, however, less than 1°C,
202
ADVANCED LEVEL PHYSICS
G. T. P. Tarrant has pointed out that radiation, not convection, is
the major contributing factor to the rate of cooling of an object.
To demonstrate Newton's law of cooling, we plot a temperature (re-
time (t) cooling curve for hot water in a calorimeter placed in a draught
(Fig. 9.2 (a)). If R is the room temperature, then the excess temperature
of the water is (6-0 R ). At various temperatures, such as in Fig. 9.2 (b),
we drew tangents such as APC to the curve. The slope of the tangent,
in degrees per second, gives us the rate of fall of temperature, when the
water is at the temperature :
♦
(a) Apparatus
Excess temp., °C — ■
(c) Treatment of results
o
Time, sec
(b) Results
Fig. 9.2. Newton's law of cooling.
rate of fall = ^ = °1^°1.
We then plot these rates against the excess temperature, — 9 R , as in
Fig. 9.2 (c), and find a straight line passing through the origin. Since
the heat lost per second by the water and calorimeterls proportional
to the rate of fall of the temperature, Newton's law is thus verified.
Heat Loss and Temperature Fall
Besides the excess temperature, the rate of heat loss depends on the
exposed area of the calorimeter, and on the nature of its surface : a dull
surface loses heat a little faster than a shiny one, because it is a better
radiator (p. 343). This can be shown by doing a cooling experiment
twice, with equal masses of water, but once with the calorimeter
polished, and once after it has been blackened in a candle-flame. In
CALORIMETRY
203
general, for any body with a uniform surface at a uniform temperature 0,
we may write, if Newton's law is true,
heat lost/second = ~r = kS(0-0 R )
(2)
where S is the area of the body's surface, 6 R is the temperature of its
surroundings, k is a constant depending on the nature of the surface,
and Q denotes the heat lost from the body.
When a body loses heat Q, its temperature falls; if m is its mass,
and c its specific heat capacity, then its rate of fall of temperature,
d0/dt, is given by
dQ d0
dt dt
Now the mass of a body is proportional to its volume. The rate of heat
loss, however, is proportional to the surface area of the body. The
rate of fall of temperature is therefore proportional to the ratio of
surface to volume of the body. For bodies of similar shape, the ratio
of surface to volume is inversely proportional to any linear dimension.
If the bodies have surfaces of similar nature, therefore, the rate of fall
of temperature is inversely proportional to the linear dimension: a
small body cools faster than a large one. This is a fact of daily experience :
a small coal which falls out of the fire can be picked up sooner than a
large one; a tiny baby should be more thoroughly wrapped up than a
grown man. In calorimetry by the method of mixtures, the fact that a
small body cools faster than a large one means that, the larger the
specimen, the less serious is the heat loss in transferring it from its
heating place to the calorimeter. It also means that the larger the scale
of the whole apparatus, the less serious are the errors due to loss of
heat from the calorimeter.
Correction for Heat Losses in Calorimetry
Newton's law of cooling enables us to estimate the heat lost in an
experiment on the method of mixtures.
In doing the experiment, we take the temperature of the mixture at
half-minute intervals, and plot it against time, as in Fig. 9.3. The broken
line shows how We would
expect the temperature to rise
if no heat were lost ; we have
therefore to estimate the dif-
ference, p, between the plateau
of this imaginary curve, and
the crest of the experimental
curve, C p is known as the
'cooling correction'.
We start by drawing an
ordinate CN through the
crest, and another LM
through any convenient point
' 1 * 2 — *~Time l further along the curve;
Fig. 9.3. Cooling correction. OM should be not less than
204 ADVANCED LEVEL PHYSICS
twice ON— the greater it is, the more accurate the correction. We next
draw an abscissa O'PQ through the room temperature, R ; and by
counting the squares of the graph paper, we measure the areas
O'CP (A^ PCLQ (A 2 ). Then, if q is the fall in temperature from C to L:
p = O'CP = A t
q PCLQ A 2 ' (3)
Before establishing this equation let us see how to use it. Suppose
m lt c, are the mass and specific heat capacity of the specimen; m, c w
are the mass and specific heat capacity of water ; and C the heat capacity
of the calorimeter. Then the heat which these lose to their surroundings
is the heat which would have raised their temperature by p. Thus
heat lost = (m 1 c + mc w + C)p.
Let X denote the initial temperature of the specimen, 6 C the highest
temperature of the mixture; and 2 the original temperature of the
water and calorimeter. Then we have :
heat given out = heat taken in + heat lost.
•'• mic(0i-0c) = (mc w + C)(0 c -0 2 ) + (m 1 c + mc w + C)p,
from which m 1 c{B x -0 c + p) = (mc w + C)(0 c + p- d 2 ).
To correct for the heat losses we must therefore add the correction p to
the crest temperature 6 on each side of the heat balance equation.
In equation (1), p. 200, p must be added to 9 2 in both numerator and
denominator.
Theory of the Correction. To establish equation (3), we write down the expres-
sion for the heat lost per second from the calorimeter, assuming Newton's law
of cooling:
^ = kS(9-9 R ) (4)
where k is a constant, and S the exposed area of the calorimeter. Between times
t = and t = t u the total heat lost is
Q t = { U kSiO-e^ddt
Jo
= ks\ l
Jo
(d-9 R )dt
= kSx area O'CP = kSA v
This is the heat; which, if it had not been lost, would have warmed the calorimeter
and contents by p degrees. Therefore
{m l c+mc w + C)p = kSA l .... (5)
CALORIMETRY
205
Similarly the heat lost between t x and t 2 is given by Q 2 = kSA 2 , and since this
loss caused a fall in temperature of q, we have, by the argument above
{miC+mCn + Qq = kSA 2 .... (6)
On dividing equation (5) by equation (6), we find
fix
a:
or p = q-
A
Specific Heat Capacity of Liquid by Cooling
Specific heat capacities of liquids which react with water are often
measured by the so-called method of cooling. The cooling curve of a
calorimeter is plotted, first when it contained a known volume of hot
water, and then when it contains an equal volume of hot liquid (Fig. 9.4).
The volumes are made equal so as to make the temperature distribution,
Water
Time *"
Fig. 9.4. Specific heat capacity by cooling.
over the surface of the calorimeter, the same in each experiment. From
the curves, the respective times t, and t w are found which the calorimeter
and contents take to cool from 6 X to 2 - Whatever the contents of the
calorimeter, it gives off heat at a rate which depends only on its excess
temperature, since the area and nature, of its surface are constant.
Therefore, at each temperature between X and 2 , the calorimeter
gives off heat at the same rate whatever its contents. Thus the average
rate at which it loses heat, over the whole range, is the same with water
and with liquid. Consequently
Kc+c)(9 r e 2 ) = (mc w +c){0 i -e 2 )
t t ' t w
where m x ,c, are the mass and specific heat capacity of the liquid, m, c w
that of water, and C is the heat capacity of the calorimeter. Thus
m x c + C mc w + C
from which c can be calculated.
206
ADVANCED LEVEL PHYSICS
Specific Heat Capacity by Electrical Method
The simplest way to measure the specific heat capacity of a liquid in
the laboratory is by electrical heating, as illustrated in Fig. 9.5. In
this case, the energy supplied = IVt joules, where I is the current
in amperes the coil R of resistance wire, V is the potential difference
across it in volts and t is the time
in seconds for which current flows.
The coil may be immersed in a
suitable oil of mass m in a calori-
meter of heat capacity C.
We pass a steady current /
through the coil, and measure the
potential difference V across it.
Stirring continuously, we plot the
temperature of the oil against the
time. After a time t long enough to
give several degrees rise, we switch
off the current and plot the cooling
curve. If 9 is the corrected rise in
temperature, we have
IVt = {mc + C)0,
Fig. 9.5. Laboratory electrical
method for specific heat capacity.
whence we can calculate c in J kg
Specific Heat Capacity of Water by Continuous Flow Method
In 1899, Callendar and Barnes devised a method for specific heat
capacity in which only steady temperatures are measured. They used
platinum resistance thermometers, which are more accurate than
mercury ones but take more time to read. In the measurement of
steady temperatures, however, this is no drawback. As we shall see
shortly the heat capacity of the apparatus is not required, which is a
great advantage of the method.
Fig. 9.6 shows Callendar and Barnes' apparatus. Water from the
constant-head tank K flows through the glass tube U, and can be
collected as it flows out It is heated by the spiral resistance wire R,
which carries a steady electric current J. Its temperature, as it enters
Water
Fig. 9.6. Callendar and Barnes' apparatus (contracted several times in
length relative to diameters).
CALORIMETRY 207
and leaves, is measured by the thermometers T x and T 2 . (In a simplified
laboratory experiment, these may be mercury thermometers.) Sur-
rounding the apparatus is a glass jacket G, which is evacuated, so that
heat cannot escape from the water by conduction or convection.
When the apparatus is running, it settles down eventually to a steady
state, in which the heat supplied by the current is all carried away by
the water. None is then taken in warming the apparatus, because every
part of it is at a constant temperature. The mass of water m, which flows
out of the tube in t seconds, is then measured. If the water enters at a
temperature t and leaves at 2 , then if c w is its mean specific heat
capacity,
heat gained by water = Q = mc w {0 2 -0 l ) joules.
The energy which liberates this heat is electrical. To find it, the
current /, and the potential difference across the wire V, are measured
with a potentiometer. If/ and Fare in amperes and volts respectively,
then, in t seconds :
energy supplied to wire = IVt joules.
.'.mcJLOt-ej^IVt
IVt
•• Cw m{0 2 -0 x )
To get the highest accuracy from this experiment, the small heat
losses due to radiation, and conduction along the glass, must be allowed
for. These are determined by the temperatures t and 9 2 . For a given
pair of values of t and 9 2 , and constant-temperature surroundings
(not shown), let the heat lost per second be h Then, in t seconds,
heat supplied by heating coil - mc w (0 2 -fljHM
.\IVt = mc w (0 2 -0 1 ) + ht . (1)
To allow for the loss K the rate of flow of water is changed, to about
half or twice its previous value. The current and voltage are then
adjusted to bring 2 back to its original value. The inflow temperature,
V is fixed by the temperature of the water in the tank. If/', V, are the
new values of/, V, and rri is the new mass of water flowing in the same
t seconds, then :
rV't = m'c w (8 2 -0 1 ) + ht . . . (2)
On subtracting equation (2) from equation (1), we find
(IV-I'V')t = (m-m')c w (0 2 -Oi),
When the temperature rise, 2 — V is made small, for example,
t = 200°C, 2 = 220°C, then c w may be considered as the specific
heat at 210°C, the mean temperature. If the inlet water temperature is
now raised to say t = 40-0°C and 2 is then 42-0°C, c w is now the
208
ADVANCED LEVEL PHYSICS
specific heat at 410°C In this way it was found that the specific heat
capacity of water varied with temperature. The continuous flow method
can be used to find the variation in specific heat capacity of any liquid
in the same way.
The '15°C-calorie' was defined as the heat required to raise the tem-
perature of 1 gramme of water from 14-5°C to 15-5°C. The table shows
the relative variation of the specific heat capacity of water, taking the
15°C-calorie as 10000 in magnitude.
SPECIFIC HEAT CAPACITY OF WATER
Temperature (°C)
15
25
40
70
100
1-0047 10000 0-9980 0-9973 10000 10057
Nernst's Method
Modern methods of measuring specific heat capacities use electrical
heating. Nernst's method for the specific heat capacity of a metal is
shown in Fig. 9.7. The metal S has a heating coil R of insulated platinum
wire wound round the outside, and is covered
with silver foil F to minimize heat loss by
radiation. It is suspended by the leads to the
coil in a glass vessel, which is then evacuated,
to prevent losses by convection and conduc-
tion. The resistance of the coil is measured,
and from it the temperature is calculated.
A steady current /, at a known potential
difference V, is then passed through the coil
for t seconds. After the current has been
switched off, the resistance of the coil is
again measured, to find the rise in tempera-
ture of the specimen. Resistance measurements
are made at intervals, and enable the cooling
curve to be plotted. If m is the mass of the
specimen, C the heat capacity of the coil and
foil, and 6 the corrected rise in temperature,
then the specific heat capacity c of the specimen
is given by
Vac. pump
Fig. 9.7. Nernst's
calorimeter.
IVt = (mc + C)0.
SPECIFIC LATENT HEAT
Fusion
The specific latent heat effusion of a solid is the heat required to
convert unit mass of it, at its melting-point, into liquid at the same
temperature. It is expressed in joules per kilogramme (J kg -1 ). High
values can be more conveniently expressed in kJ kg -1 .
CALORIMETRY
209
MELTING-POINTS AND SPECIFIC LATENT HEATS OF FUSION
Substance
MP
°C
S.L.H.
J kg' 1
Substance
MP.
°C-
S.L.H.
J kg- 1
Aluminium
658
393 x 10 3
Acetic acid
17-5
184 xlO 3
Antimony
630
163
Beeswax .
62
176
Bismuth
271
58
Brass
900
—
Copper
1083
180
Naphthalene .
80
146
Gold .
1063
67
Paradichlorbenzene
53
—
Iron
1530
205
(non-flam.)
Lead .
327
25
Ice .
333
Mercury
—38-9
12-5
Paraffin wax
50-60
—
Nickel .
1452
272
Solder (soft) .
c. 180
50-85
Platinum
1773
113
Solder (hard) .
c.900
—
Sulphur
113-119
38
Steel
1400
—
Silver .
960-8
109
Glass
300-400
—
Sodium
97-5
113
Quartz (fused) .
1,700
—
Tin .
232
58
Hypo
48-2
c. 170
Tungsten
3380
—
Zinc .
419
109
Ice is one of the substances whose specific latent heat of fusion we
are likely to have to measure. To do so, place warm water, at a tempera-
ture 6 x a few degrees above room temperature, inside a calorimeter. Then
add small lumps of ice, dried by blotting paper, until the temperature
reaches a value 2 as much below room temperature as X was above.
In this case a 'cooling correction' is not necessary. Weigh the mixture,
to find the mass m of ice which has been added. Then the specific latent
heat 7 is given by:
heat given by calori-
meter ;
cooling
j heat used in warming
meter and water in } = Jj^tingk^ | + j melted ice from 0°C to
.-. (m i c w + C)(e i -e 2 ) = ml+mc w (0 2 -O),
where m 1 = mass of water and c w = specific heat capacity, C = ther-
mal capacity of calorimeter, and 0- x = initial temperature.
, _ {m x c w +Q{0 1 - e 2 )
Hence
m
"w l/ 2-
A modern electrical method, similar to Nernst's for specific heat
capacities, gives
/ = 334kJkg- 1 or 334 Jg" 1 .
Bunsen's Ice Calorimeter
Bunsen's ice calorimeter is a device for measuring a quantity of heat
by using it to melt ice.
When ice turns to water, it shrinks ; the volume of 1 g of ice at 0°C
is 10908 cm 3 , whereas that of 1 g of water at 0°C is 10001 cm 3 (p. 296).
Thus the melting of 1 g of ice causes a contraction of 00907 cm 3 .
210
ADVANCED LEVEL PHYSICS
In the Bunsen calorimeter, the contraction due to the melting is
measured, and from it
the mass of ice melted
is calculated. The
apparatus is shown in
Fig. 9.8. It consists of a
test-tube T fused into
a wider tube Y. The
wider tube leads to a
capillary C, and is filled
with mercury from X
to Y. The space above
Y is filled with water
from which all dissolved
Fig. 9.8. Bunsen's ice calorimeter. air has been boiled
Except for the capillary, the whole apparatus is placed in ice-water in
the vessel V, and, after some time, it all settles down to 0°C. A little
ether is then poured into T, and air is blown through it via a thin tube ;
the ether evaporates and cools the tube T, so that ice forms on the
outside of it. A pad of cotton wool is then dropped to the bottom of T
and the apparatus is left for some more time, to allow the newly formed
ice to settle down to 0°C.
When the apparatus is ready for use, the end of the mercury thread
in C is observed by a travelling microscope. If the specific heat capacity
of a solid is to be measured, the specimen is weighed (m g) and left to
come to room temperature 6. The solid is then gently dropped into the
tube T. As it cools, it melts ice, and causes the mercury thread to run
back along the capillary. When the thread has ceased to move, its end is
again observed. If it has moved through / cm, and the cross-section of
the capillary is a cm 2 , then the contraction is al cm 3 . The mass of ice
melted is therefore a//0-0907 g, and the heat absorbed is 334 a//0-0907
joules. This heat is given out by m g of solid cooling from 9 to 0°C ; the
specific heat capacity c of the solid is therefore given by
n 334 al
mcl) - 0W07-
In practice, the cross-section is not measured, and the instrument is
calibrated by dropping into it a solid of known mass, m t , and specific
heat capacity, c v If the room temperature is constant, then
mc _ j_
m l c 1 1 1
where l t is the displacement of the mercury in the calibration experi-
ment. Thus c can be found.
Advantages of the Ice Calorimeter
The advantages of the ice calorimeter are :
(i) no correction for heat capacity of the container: the speci-
men tube starts at 0°C and finishes at 0°C — all the heat from
the specimen is used to melt ice, at constant temperature ;
CALORIMETRY
211
(ii) no heat losses from the apparatus — it is surrounded by a bath
at the same temperature as itself, and therefore neither loses
heat to the outside, nor gains any from it ;
(iii) no loss of heat from the specimen before it enters the calori-
meter — the specimen starts at room temperature, and therefore
gives up no heat until it enters the specimen tube (contrast the
method of mixtures, in which the specimen is heated to 100°C
or so) : this is a great advantage when the specimen is small ;
(iv) easy,- and therefore accurate, thermometry — the only tempera-
ture to be measured is the room temperature, which is constant
and can be determined at leisure.
An advantage sometimes asserted is that specimens can be added
one after another, without having to re-set the apparatus. That is true,
because each specimen comes to 0°C in turn, and then behaves simply
like part of the apparatus, taking no heat from any following specimen.
But it does not mean that the calorimeter has the advantage of speed—
the time taken to set it up would be enough for half a dozen measure-
ments by the method of mixtures. A disadvantage of this calorimeter is
that it never settles down completely — the mercury is always slowly
creeping along the capillary, and the creep during an experiment must
be estimated and allowed for.
The calorimeter was devised in 1871; it is rarely used nowadays,
because electrical methods of calorimetry are more convenient and
accurate. However, it has been used for measuring the specific heat
capacities of rare earths of which only small specimens were available.
Evaporation
The specific latent heat of evaporation of a liquid is the heat required
to convert unit mass of it, at its boiling-point, into vapour at the same
temperature. It is expressed in joule per kilogramme (J kg" *), or, with
high values, in kJ kg - *.
BOILING-POINTS AND SPECIFIC LATENT HEATS OF EVAPORATION
Substance
B.P.
°C
S.L.H.
J kg" 1
Aluminium
1800
—
Acetone ....
56-7
—
Alcohol (ethyl) .
78-3
867 x 10 3
Alcohol (methyl)
64-7
1120
Benzene ....
80-2
389
Carbon disulphide
46-2
351
Carbon tetrachloride (non-flam.)
767
193
Ether ....
34-6
370
Glycerine
290
—
Turpentine
161
—
Mercury .
357
272
Platinum .
3910
^-
Sodium .
877
—
Sulphur .
444-6
—
212
ADVANCED LEVEL PHYSICS
To find the specific latent heat of evaporation of water, we pass steam
into a calorimeter with water (Fig. 9.9). On its way the steam passes
through a vessel, T in the figure, which traps any water carried over
by the steam and is called a steam-trap. The mass m of condensed
steam is found by weighing. If 1 and 6 2 are the initial and final tem-
peratures of the water, the specific latent heat / is given by :
EZ
Z
Heat shield
Fig. 9.9. Latent heat of evaporation
of water.
Small
gasring
Fig. 9.10. Berthelot's apparatus for latent
heat of evaporation.
heat ' en b t ) ( ^ eat 8* ven ^ con " ) ( neat taken by
, • )+{ densed water cooling } = ( calorimeter and
condensing J | from 100 °c to 2 J (water
ml + mc w (100-6 2 ) ={m 1 c w + Q(0 2 -O i )
where m 1 c w and C have their usual meanings.
Hence
/ =
(m 1 c w +Q(e 2 -e 1 )
c w (lOO-0 2 ).
The accepted value of the specific latent heat of evaporation of water
is about / = 2260 kJ kg - 1 or 2260 J g" K
Berthelot's Apparatus. An apparatus suitable for use with liquids
other than water was devised by Berthelot in 1877 (see Fig. 9.10).
The liquid is boiled in the flask F, and its vapour passes out through the
tube T. This fits with a ground joint G into the glass spiral S, which is
surrounded by water in a calorimeter. The vapour condenses in the
CALORIMETRY
213
spiral, and collects in the vessel V, where it can afterwards be weighed.
Let
b = boiling-point of liquid.
c = specific heat capacity of liquid,
m = mass of liquid condensed.
#! = initial temperature of water.
6 2 = final temperature of water, corrected for cooling.
m t = mass of water of specific heat c w .
C = thermal capacity of calorimeter + glassware below joint.
Then ml+mc(0 b -9 2 ) = (m 1 c w + C)(6 2 -e i ),
whence
/ = ^±5fc«-4-y
m
Electrical Method for Specific Latent Heat
A modern electrical method for the specific latent heat of evaporation
of water is illustrated in Fig. 9.11 below. The liquid is heated in a
vacuum-jacketed vessel U by the heating coil R. Its vapour passes down
K —
^
Fig. 9.11. Electrical method
for latent heat of
evaporation.
?N
214 ADVANCED LEVEL PHYSICS
the tube T, and is condensed by cold water flowing through the jacket K.
When the apparatus has reached its steady state, the liquid is at its
boiling-point, and the heat supplied by the coil is used in evaporating
the liquid, and in offsetting the losses. The liquid emerging from the
condenser is then collected for a measured time, and weighed.
If / and V are the current through the coil, and the potential differ-
ence across it, the electrical energy supplied in t seconds is IV t. And if
h is the heat lost from the vessel per second, and m the mass of liquid
collected in t seconds, then
IVt = ml+ht '. (1)
The heat losses h are determined by the temperature of the vessel,
which is fixed at the boiling-point of the liquid. Therefore they may
be eliminated by a second experiment with a different rate of evapora-
tion (cf. Callendar and Barnes, p. 206). If TV are the new current and
potential difference, and if m' grammes are evaporated in t seconds,
then
I'V't = m'l+ht.
Hence by subtraction from equation (1)
_ (iv-rv')t
(m — rri)
EXAMPLES
1. An electric kettle has a 750 W-240 V heater and is used on a 200 V mains.
If the heat capacity of the kettle is 400 JK" 1 and the initial water temperature
is 20°C, how long will it take to boil 500 g of water, assuming the resistance of
the heater is unaltered on changing to the new mains.
Firstly, find the new power absorbed on the 200 V mains. Since the resistance
R is constant and P = V 2 /R, it follows that P oc V 2 .
/2OOI 2
.'. new power = — n) x 75 ^ w = 520 W (approx.)
.'. heat supplied to water = 520 J per second . . . (1)
Secondly, assuming 100°C is the boiling point and 42 kJkg -1 K _1 (4-2 Jg _1
K~ 1 ) is the specific heat capacity of water,
heat gained by water and kettle = 500 x 4-2 x (100 - 20) + 400 x (100 - 20)
= (500 x 42+400)(100-20)
= 610x80 J.
From (1), .'. time, t = — — — = 385 seconds (approx.) = 6-4 mm.
2. Water flows at the rate of 150-0 g rnin - 1 through a tube and is heated by a
heater dissipating 25-2 W. The inflow and outflow water temperatures are
1 5-2°C and 17-4°C respectively. When the rate of flow is increased to 23 1 -8 g min~ 1
and the rate of heating to 37-8 W, the inflow and outflow temperatures are
unaltered. Find (i) the specific heat capacity of water, (ii) the rate of loss of heat
from the tube.
CALORIMETRY 215
Suppose c w is the specific heat of water in J g~ 1 K~ x and h is the heat lost in
J s~ 1 . Then, since 1 W = 1 J per second,
25-2 = ^c w (17-4-15-2)+fc . . . (1)
60
and 37-8 = ^c w (17-4-l 5-2) + h ... (2)
Subtracting (1) from (2),
.,37.8-252= &^c„(174- I5 .2)
.'. c. = ^|j|^ = 4-2 J g"' K- 1 = 4-2kIkg-' K-'.
Substituting for c w in (1),
.-. Ji = 25-2-i^x4-2x2-2 = 2-Us- 1 .
60
3. Define latent heat. Describe the measurement of the specific latent heat of
evaporation of water under school laboratory conditions.
A copper calorimeter of mass 70-5 g contains 1000 g of water at 19-5°C.
Naphthalene (MP. 79-9°C) is melted in a test tube, cooled to 80-0°C, and then
poured into the calorimeter. If the highest temperature reached by the water
after stirring is 28-7°C and the final mass of the calorimeter and its contents is
188-3 g calculate the latent heat of fusion of naphthalene. (Specific heat capacity
of copper 0*4, of naphthalene 1-3 kJ kg - 1 K~ 1 .) (L.)
First part. The specific latent heat of a substance is the heat required to change
unit mass of the solid at the melting-point to liquid at the same temperature
(fusion), or the heat required to change unit mass of the liquid at the boiling-point
to vapour at the same temperature (vaporization).
The measurement of the specific latent heat of evaporation of water requires the
following, among other points : (i) use of a steam trap, (ii) a rise in temperature of
the water in the calorimeter of about 10°C, (iii) a 'correction' to 100°C as the
steam temperature, if the barometric pressure is not 76 cm mercury, (iv) a cooling
correction.
Second part. The mass of naphthalene = 188-3— 170-5 = 17*8 g.
Heat lost by naphthalene = heat gained by water and calorimeter.
.-. 17-8/+ 17-8 x 1-3 x (79-9-28-7)
= 100 x 4-2 x (28-7 - 19-5)+ 705 x 0-4 x (28-7 - 19-5).
Solving, .-. /= 164 J g - - (approx.) = 164 kJ kg~ J .
4. In an X-ray tube, 10 18 electrons per second arrive with a speed of 2x 10 6
m s~ * at a metal target of mass 200 g and specific heat capacity 0-5 J g~ 1 K~ - . If the
mass of an electron is 9- 1 x 10" 3 1 kg, and assuming 98% of the incident energy is
converted into heat, find how long the target will take to rise in temperature by
50°C assuming no heat losses.
The kinetic energy of a moving object is \mv 2 joules, where m is the mass in kg
and v is the speed in m s~ 1 . Assuming the initial speed is zero,
.'. energy per second of incident electrons = \ x (10 18 x 91 x 10 ~ 31 ) x (2 x 10 6 ) 2 J
= 1-8 J (approx.).
Heat gained by target = 200 x 0-5 x 50 J = 50000 J
.'. time = = 2780 seconds (approx.) = 46-3 min.
1*8
216 ADVANCED LEVEL PHYSICS
EXERCISES 9
1. Describe, with the aid of a labelled diagram, how you would find the
specific heat capacity of a liquid by the method of continuous flow.
Discuss the advantages and disadvantages of the method compared with the
method of mixtures.
The temperature of 50 g of a liquid contained in a calorimeter is raised from
150°C (room temperature) to 45-0°C in 530 seconds by an electric heater dis-
sipating 100 watts. When 100 g of liquid is used and the same change in tem-
perature occurs in the same time, the power of the heater is 161 watts. Calculate
the specific heat capacity of the liquid. (N.)
2. Distinguish between specific heat capacity and latent heat capacity. With
what physical changes is each associated? Describe the processes involved in
terms of simple molecular theory.
A thin- walled tube containing 5 cm 3 of ether is surrounded by a jacket of water
calibrated so that changes in the volume of the water can be read off. The whole
apparatus is cooled down to 0°C and all the ether is then evaporated by blowing
a rapid stream of air pre-cooled to 0°C through it. The change of volume as ice
forms in the water is 0-35 cm 3 . Calculate the specific latent heat of evaporation
of the ether.
(Use the following values: specific latent heat of fusion of ice = 334 J g _1 .
Densities at 0°C: water, 1000 g cm -3 ; ice, 0-917 g cm"" 3 ; ether, 0-736 g cm" 3 .)
(O. & C.)
3. Give an account of an electrical method of finding the specific latent heat of
vaporisation of a liquid boiling at about 60°C. Point out any causes of inaccuracy
and explain how to reduce their effect.
Ice at 0°C is added to 200 g of water initially at 70°C in a vacuum flask. When
50 g of ice has been added and has all melted the temperature of the flask and
contents is 40°C. When a further 80 g of ice has been added and has all melted the
temperature of the whole becomes 10°C. Calculate the specific latent heat of
fusion of ice, neglecting any heat lost to the surroundings.
In the above experiment the flask is well shaken before taking each temperature
reading. Why is this necessary? (C)
4. The specific heat capacity of gallium metal is 0-33 kJ kg " * K ~ 1 . Explain care-
fully how this result may be determined experimentally. Indicate the sources of
error in your method and estimate the accuracy which could be achieved.
[Melting point of gallium = 30°C]
A ball of gallium is released from a stationary balloon, falls freely under gravity
and on striking the ground it just melts. Calculate the height of the balloon
assuming that the temperature of the gallium just before impact is 1°C and that
all the energy gained during its free fall is used to heat the gallium on impact.
Why are the conditions specified in this problem unrealistic?
[Specific latent heat of fusion of gallium = 79 kJ kg - 1 .] (O. & C.)
5. Explain what is meant by the specific latent heat of vaporization of a liquid,
and describe an experiment for an accurate determination of this quantity for
carbon tetrachloride, which boils at 77°C.
A thermally insulated vessel connected to a vacuum pump contains 10O g of
water at a temperature of 0°C. As air and water vapour are exhausted from the
vessel, it is observed that the water remaining in the vessel freezes. Explain why
this happens, and find the mass of water which is converted into ice.
[Specific latent heat of vaporization of water at 0°C = 2520 kJ kg" 1 ; specific
latent heat of fusion of ice at 0°C = 336 kJ kg" ^ -(C)
CALORIMETRY
217
6. Discuss the nature of the heat energy (a) of a solid, (b) of a gas, (c) of the sun
during the transmission of this energy to the earth.
265430 joules of heat are produced when a vehicle of total mass 1270 kg is
brought to rest on a level road. Calculate the speed of the vehicle in km per hr
just before the brakes are applied. (L.)
7. State Newton's law of cooling and describe how to obtain observations and
how to use them in order to test the validity of the law.
A solid of mass 250 g in a vessel of thermal capacity 67-2 J K _l is heated
to a few degrees above its melting point and allowed to cool in steady conditions
until solid again. Sketch the graph of its temperature plotted against time. If this
graph shows that immediately before solidification starts the rate of cooling is
3-2 deg C min -1 , .while immediately after solidification it is 4-7 deg C min~\
calculate the specific heat capacity of the solid taken by the solidifying process.
(The specific latent heat of fusion of the substance may be taken as 146-6 kJ kg - *
and its specific heat capacity in the liquid state as 1-22 kJ kg~ l K~ l .) (L.)
8. What do you understand by the specific heat capacity of a substance?
Describe how you would measure the specific heat capacity of a sample of rock,
describing the precautions that you would take to obtain an accurate result.
A room is heated during the day by a 1 kW electric fire. The fire is to be re-
placed by an electric storage heater consisting of a cube of concrete which is
heated overnight and is allowed to cool during the day, giving up its heat to the
room. Estimate the length of an edge of the cube if the heat it gives out in cooling
from 70°C to 30°C is the same as that given out by the electric fire in 8 hours.
[Density of concrete = 2700 kg m~ 3 ; specific heat capacity of concrete
= 0-85 kJ kg" * K~ ».] (O. & C)
9. An experiment was performed to determine the specific latent heat of
vaporization of a volatile liquid at the prevailing boiling point by the method of
electrical heating. The results are summarized in the following table :
Rate of supply of
Mass of liquid
energy to boiling
vaporized in
liquid
200 seconds
(watt)
(g)
10
1-6
20
6-4
30
11-2
40
160
Use the data to plot a graph (using the graph paper available) and hence
determine the latent heat of vaporization of the liquid and the rate of loss of heat
from the calorimeter containing the boiling liquid.
Draw a labelled diagram of a suitable apparatus for use in the experiment
and indicate how the above results would have been obtained.
Give two advantages of this method over the method of mixtures. (N.)
10. In 1948 the International Conference on Weights and Measures recom-
mended that the calorie should no longer be regarded as the basic unit of heat,
but that it should be replaced by the joule. Discuss the reasons for, and the
advantages and possible disadvantages of, this recommendation.
Give a labelled diagram of a continuous flow calorimeter suitable for the
determination of the specific heat capacity of a liquid. What measurements would
you make in such a determination, and how would the result be obtained from
them? State the precautions which you would adopt to ensure an accurate
result. (C.)
218 ADVANCED LEVEL PHYSICS
11. Give an account of a method of determining the specific latent heat of
evaporation of water, pointing out the ways in which the method you describe
achieves, or fails to achieve, high accuracy.
A 600 watt electric heater is used to raise the temperature of a certain mass of
water from room temperature to 80°C. Alternatively, by passing steam from a
boiler into the same initial mass of water at the same initial temperature the same
temperature rise is obtained in the same time. If 1 6 g of water were being evaporated
every minute in the boiler, find the specific latent heat of steam, assuming that
there were no heat losses. (0. & C.)
12. 300 g of a certain metal of density about 19 gem -3 is available in the
form of a coarse powder, together with a calorimeter of heat capacity about
336 J K _1 and volume about 160 cm 3 , and a 50°C thermometer reading to
ideg.
Using this and other necessary apparatus, how would you verify, by the method
of mixtures, that the specific heat of the metal is 013 kJ kg - 1 K~ *?
In the experiment you describe why is it (a) unnecessary to apply a correction
for heat exchange with the surroundings, (6) necessary to decide on a suitable
maximum temperature of the mixture? How would you ensure that such a
temperature is realized? (N.)
13. Describe an electrical method for the determination of the specific latent
heat of steam. State the probable sources of error in the experiment and suggest
how they may be minimised.
In one method for storing solar energy, Glauber's salt can be allowed to warm
up to 45 °C in the sun's rays during the day and the stored heat is used during the
night, the salt cooling down to 25°C. Glauber's salt melts at 32°C. Calculate the
mass of salt needed to store 1 million joules.
(Specific heat capacity of Glauber's salt, solid = 0-11 kJ kg -1 K _1 ; specific
heat capacity of Glauber's salt, molten = 016 kJ kg" 1 K~ l . Specific latent heat
of fusion = 14 kJ kg" *.) (O. & C.)
14. What is meant by the specific heat capacity of a substance? Give a brief
account of two methods, one in each case, which may be used to find the
specific heat capacity of each of the following: (a) a specimen of a metal in the
form of a block a few cm in linear dimensions, and (b) a liquid which is available
in large quantities. Indicate whether the methods you describe involve a cooling
correction.
An electrical fuse consists of a piece of lead wire 1-5 mm in diameter and 5 cm
long, which has a resistance of 6-5 x 10" 3 ohm. Owing to a fault a constant
current of 800 A passes through the fuse. If the wire is initially at 10°C and
melts at 330°C, find the time interval before it starts to melt, assuming that its
specific heat capacity and its electrical resistance are constant and that there are
no heat losses.
[Specific heat capacity of lead = 013 kJ kg" 1 K~ l . Density of lead = 11000
kgm- 3 .](0.£C.)
15. Describe the determination of the latent heat of fusion of ice by the method
of mixtures and, in particular, show how allowance is made for heat interchange
with the surroundings.
A calorimeter of heat capacity 84 J K ~ l contains 980 g of water supercooled to
- 4°C. Taking the latent heat of fusion of ice at 0°C as 336 kJ kg" *, find the amount
of ice formed when the water suddenly freezes. Calculate also the specific latent heat
of fusion at --4°C if the specific heat capacity of ice is 2-1 kJ kg -1 K _1 . (N.)
16. State Newton's law of cooling, and describe an experiment by which you
would verify it. A calorimeter containing first 40 and then 100 g of water is
heated and suspended in the same constant-temperature enclosure. It is found
CALORIMETRY 219
that the times taken to cool from 50° to 40°C in the two cases are 15 and 33
minutes respectively. Calculate the heat capacity of the calorimeter. (O. & C.)
17. Oil at 15-6°C enters a long glass tube containing an electrically heated
platinum wire and leaves it at 17-4°C, the rate of flow being 25 cm 3 per min and
the rate of supply of energy 1-34 watts. On changing the rate of flow to 15 cm 3 per
min and the power to 0-76 watt the temperature again rises from 15-6° to 17-4°C.
Calculate the mean specific heat capacity of the oil between these temperatures.
Assume that the density of the oil is 870 kg m~ 3 . (N.)
18. In the absence of bearing friction a winding engine would raise a cage
weighing 1000 kg at 10 m s~ \ but this is reduced by friction to 9 m s" 1 . How
much oil, initially at 20°C, is required per second to keep the temperature of the
bearings down to 70°C? (Specific heat capacity of oil = 21 kJ kg -1 K -1 ;
= 9-81ms- 2 .(O.<£C.)
19. A heating coil is embedded in a copper cylinder which also carries a
thermocouple. The whole is thermally equivalent to 25 g of copper. The cylinder
is suspended in liquid air until the thermocouple reading is constant. The cylinder
is taken out and rapidly transferred into a beaker of water at 0°C. A coating of
ice forms on the cylinder and when its temperature is again constant it is taken
out of the water and suspended in a space maintained at 0°C. The heating coil
is switched on at a steady energy dissipation of 24 watts. After 1 minute 5 seconds
the whole of the ice has just melted. What is the temperature of the liquid air?
What assumptions were made in carrying out the calculations? (Mean specific
heat capacity of copper is 0-336 kJ kg~ * K~ \) (L.)
20. Describe a continuous flow method of measuring the specific heat capacity
of a liquid. Explain the advantages of the method.
Use the following data to calculate the specific heat capacity of the liquid
flowing through a continuous flow calorimeter : Experiment I: Current 2 amp,
applied p.d. 3 volt, rate of flow of liquid 30 g min -1 , rise in temperature of
liquid 2-7°C. Experiment II. Current 2-5 amp, applied p.d. 375 volts, rate of flow
of liquid 48 g min~ *, rise in temperature of liquid 2-7°C. (L.)
chapter ten
Gases
In this chapter we shall be concerned with the relationship between
the temperature, pressure and volume of a gas. Unlike the case of a
solid or liquid this can be expressed in very simple laws, called the
Gas Laws, and reduced to a simple equation, called the Equation of
State. We shall also deal in this chapter with the specific heat capacities
of gases.
THE GAS LAWS AND THE EQUATION OF STATE
Pressure and Volume: Boyle's Law
In 1660 Robert Boyle — whose epitaph reads 'Father of Chemistry,
and Nephew of the Earl of Cork' — published the results of his experi-
ments on the natural spring of air. In the vigorous language of the
seventeenth century, he meant what we now tamely call the relationship
between the pressure of air and its volume. Similar results were pub-
lished in 1676 by Mariotte, who had not heard of Boyle's work. Boyle
trapped air in the closed limb of a U-tube, with mercury (Fig. 10.1 (a)).
He first adjusted the amount of mercury until its level was the same in
each limb, so that the trapped air was at atmospheric pressure. He next
Atmospheric pressure
76 cm
U
V
(a)
(b)
(c)
Fig. 10.1. Boyle's law apparatus.
220
GASES 221
poured in more mercury, until he had halved the volume of the trapped
air (Fig. 10.1 (b)). Then 'not without delight and satisfaction' he found
that the mercury in the open limb stood 740 mm above the mercury
in the closed limb. Since he knew that the height of the barometer was
about 740 mm of mercury, he realized that to halve the volume of his
air he had had to double the pressure on it.
We can repeat Boyle's experiment with the apparatus shown in
Fig. 10.1 (c); its form makes the pouring of mercury unnecessary. We
set the open limb of the tube at various heights above and below the
closed limb and measure the difference in level, h, of the mercury.
When the mercury in the open limb is below that in the closed, we
reckon h as negative. At each value of h we measure the corresponding
length / of the air column in the closed limb. To find the pressure of the
air we add the difference in level h to the height of the barometer, H ;
their sum gives the pressure p of the air in the closed limb :
p = gp(H + h)
where g is the acceleration of gravity and p is the density of mercury.
If S is the area of cross-section of the closed limb, the volume of the
trapped air is
V = IS.
To interpret our measurements we may either plot H + h, which is a
measure of p against 1// or tabulate the product (H + h)l. We find that
the plot is a straight line, and therefore
(tf + fc)a:j . . . . (1)
Alternatively, we find
(H + h)l = constant, ... (2)
which means the same as (1).
Since, g, p, and S are constants, the relationships (1) and (2) give
1
or pV = constant.
A little later in this chapter we shall see that the pressure of a gas
depends on its temperature as well as its volume. To express the results
of the above experiments, therefore, we say that the pressure of a given
mass of gas, at constant temperature, is inversely proportional to its
volume. This is Boyle's Law.
Mixture of Gases : Dalton's Law
Fig. 10.2 shows an apparatus with which we can study the pressure
of a mixture of gases. A is a bulb, of volume V t , containing air at
atmospheric pressure, p v C is another bulb, of volume V 2 , containing
222
ADVANCED LEVEL PHYSICS
V/
«-i
\l/lf
Evacuation tap
Inlet tap
M
Fig. 10.2. Apparatus for demonstrating Dalton's law of partial pressures.
carbon dioxide at a pressure p 2 . The pressure p 2 is measured on the
manometer M ; in millimetres of mercury it is
p 2 = h + H,
where H is the height of the barometer. (In the same units, the air
pressure, p x = H.)
When the bulbs are connected by opening the tap T, the gases mix,
and reach the same pressure, p ; this pressure is given by the new height
of the manometer. Its value is found to be given by
P = Pi
V,
+P
v l + v 2 "*v x + v 2
Now the quantity p x V x j{ V x + V 2 ) is the pressure which the air originally
in A would have, if it expanded to occupy A and C; for, if we denote
this pressure by p', then p'(V x + V 2 ) = p x V x . Similarly p 2 V 2 /(V x + V 2 ) is
the pressure which the carbon dioxide originally in C would have, if it
expanded to occupy A and C Thus the total pressure of the mixture is
the sum of the pressures which the individual gases exert, when they
have expanded to fill the vessel containing the mixture.
The pressure of an individual gas in a mixture is called its partial
pressure: it is the pressure which would be observed if that gas alone
occupied the volume of the mixture, and had the same temperature as
the mixture. The experiment described shows that the pressure of a
mixture of gases is the sum of the partial pressures of its constituents.
This statement was first made by Dalton, in 1801, and is called Dalton's
Law of Partial Pressures.
Volume and Temperature: Charles's Law
Measurements of the change in volume of a gas with temperature,
at constant pressure, were published by Charles in 1787 and by Gay-
Lussac in 1802. Fig. 10.3 shows an apparatus which we may use for
repeating their experiments. Air is trapped by mercury in the closed
limb C of the tube AC; a scale engraved upon C enables us to measure
the length of the air column, /. The tube is surrounded by a water-
bath W, which we can heat by passing in steam. After making the
temperature uniform by stirring, we level the mercury in the limbs
A and C, by pouring mercury in at A, or running it off at B. The air
GASES
223
Stirrer
Fig. 10.3. Charles' law experiment.
Fig. 10.4. Results of experiment.
in C is then always at atmospheric (constant) pressure. We measure
the length / and plot it against the temperature, (Fig. 10.4).
If S is the cross-section of the tube, the volume of the trapped air is
V=IS.
The cross-section S, and the distance between the divisions on which
we read /, both increase with the temperature 9. But their increases are
very small compared with the expansion of the gas, and therefore we
may say that the volume of the gas is proportional to the scale-reading
of /. The graph then shows that the volume of the gas, at constant
pressure, increases uniformly with its temperature. A similar result
is obtained with twice the mass of gas, as indicated in Fig. 10.4.
Expansivity of Gas (Volume Coefficient)
We express the rate at which the volume of a gas increases with
temperature by defining a quantity called its expansivity at constant
pressure, (Xp, or volume coefficient :
_ volume at fl°C- volume at 0°C 1
a " volume at 0°C X &
Thus, if V is the volume at 0°C, and V the volume at 0°C, then
- v-v
v e '
whence
or
V-V =V a p 6,
V=V (1 + ol p O).
224 ADVANCED LEVEL PHYSICS
The expansivity oc p has the dimensions
[volume] _ 1
[volume] x [temperature] [temperature]"
Its value is about 273 when the temperature is measured in °C, and
we therefore say that
«p = 273 P^ deg C, or 273 K~ *•
Charles, and Gay-Lussac, found that a p had the same value, 273, for
all gases. This observation is now called Charles's or Gay-Lussac's
Law : The volume of a given mass of any gas, at constant pressure, in-
creases by 273 of its value at 0°C, for every degree Centigrade rise in
temperature.
Absolute Temperature
Charles's Law shows that, if we plot the volume V of a given mass of
any gas at constant pressure against its temperature 9, we shall get a
straight line graph A as shown in Fig. 10.5. If we produce this line back-
-273 ° 6°c— -
Fig. 10.5. Absolute zero.
wards, it will meet the temperature axis at — 273°C. This temperature
is called the absolute zero. If a gas is cooled, it liquefies before it reaches
— 275° C, and Charles's Law no longer holds; but that fact does not
affect the form of the relationship between the volume and temperature
at higher temperatures. We may express this relationship by writing
Foe (273 + 9).
The quantity (273 + 9) is called the absolute temperature of the gas,
and is denoted by T. The idea of absolute temperature was developed
by Lord Kelvin, and absolute temperatures are hence expressed in
degrees Kelvin :
TK = (273 + 0)°C.
From Charles's Law, we see that the volume of a given mass of gas
at constant pressure is proportional to its absolute temperature, since
Vac (273 + 9) oc T.
Thus if a given mass of gas has a volume V t at 9 X °C, and is heated
at constant pressure to 9 2 °C, its new volume is given by
V x _ 273 + ^ _ T t
V 2 ~ 273 + 9 2 ~ T 2 '
GASES
225
Pressure and Temperature
The effect of temperature on the pressure of a gas, at constant
volume, was investigated by Amontons in 1702. His work was for-
gotten, however, and was re-discovered only after the work of Gay-
Lussac and Charles on the effect of temperature on volume.
An apparatus for measuring the pressure of a constant volume of
gas at various known temperatures is shown in Fig. 10.6 (a). The bulb B
-273
•<:-«►
(b)
Fig. 10.6. Pressure and temperature.
contains air, which can be brought to any temperature by heating
the water in the surrounding bath W. When the temperature is steady,
the mercury in the closed limb of the tube is brought to a fixed level D,
so that the volume of the air is fixed. The difference in level, h, of the
mercury in the open and closed limbs is then added to the height of the
barometer, H, to give the pressure p of the gas in cm of mercury.
If p, (h + H\ is plotted against the temperature, the plot is a straight
line (Fig. 10.6(b)).
The coefficient of pressure increase at constant volume, a v , known
as the pressure coefficient, is given by
a., =
P-Po
Pa**'
where p is the pressure at 0°C. The coefficient <x v , which expresses
the change of pressure with temperature, at constant volume, has
practically the same value for all gases : -£3 K" l . It is thus numerically
equal to the expansivity, a p . We may therefore say that, at constant
volume, the pressure of a given mass of gas is proportional to its
absolute temperature T, since
poc (273 + 0).
. Pl = 273 + l = T t
•p 2 273 + 2 T 2
Equality of Pressure and Volume Coefficients
If a gas obeys Boyle's Law, its coefficient of pressure change at
constant volume, ql v , and of volume change at constant pressure, a p ,
226
ADVANCED LEVEL PHYSICS
Fig. 10.7. Showing that a v = a p .
must be equal For let us suppose that a given mass of gas is warmed
at constant pressure, p , from 0°C to 0°C (Fig. 10.7 (a)). Its volume
expands from V to V, where
V=V (l+a p 6).
Now let us suppose that it is compressed, at constant temperature,
until its volume returns to V (Fig. 10.7 (b)). Then its pressure rises to
p, where
pV = P V
= p o F o (l+a p 0)
or p = p (l+oc p O) .... (3)
The condition of the gas is now the same as if it had been warmed at
constant volume from 0°C to 0°C (Fig. 10.7 (c)). Therefore
and, by equation (3), it follows that
cc v = a p .
We shall see later that gases do not obey Boyle's law exactly, although
at moderate pressures they do so very nearly. The difference between
ol p and OL v provides a sensitive test for departures from Boyle's Law.
The Equation of State
Fig 10.8 illustrates the argument by which we may find the general
relationship between pressure, volume and temperature of a given mass
of gas. This relationship is called the equation of state.
Pi
V
Pi
(a) (b)
Fig. 10.8. Changing temperature and pressure of a gas.
(c)
GASES
227
At (a) we have the gas occupying a volume V x at a pressure p v and
an absolute temperature T v We wish to calculate its volume V 2 at an
absolute temperature T 2 and pressure p 2 , as at (c). We proceed via (6),
raising the temperature of T 2 while keeping the pressure constant at
p x . If V is the volume of the gas at {b), then, by Charles's law :
V T-,
— = — • • • • (4)
We proceed now to (c), by increasing the pressure to p 2 , while keeping
the temperature constant at T 2 . By Boyle's law,
£ = ^ .... (5)
Eliminating V between equations (4) and (5), we find
V, T 1 'p 2
T 2 T x •
f = R .... (6)
or
In general therefore,
where R is a constant. This equation is often given in the form
pV= RT . . . (7)
Equation (7) is the equation of state for a perfect gas. The value of
the constant R depends on the nature of the gas — air, hydrogen, etc. —
and on the mass of the gas concerned. If we consider unit mass of a
gas, we can denote its volume by V and write
pV = RT . . (8)
R is called the gas-constant for unit mass of the gas. If p is the density
of the gas, at absolute temperature T and pressure p, then
1
P = V'
and equation (8) becomes — = RT . . (9)
The volume V of an arbitrary mass M of the gas, at absolute tem-
perature T and pressure p, is
V = MV;
therefore, by (8) pV=MRT; .... (10)
and, by (7), R = MR.
228 ADVANCED LEVEL PHYSICS
Magnitude of the Gas Constant
To calculate the constant R for a gas, we need to know the density
of the gas at a given temperature and pressure. Very often, in dealing
with gases, we specify the pressure not in newton per metre 2 (N m -2 )
but simply in millimetres of mercury. 1 mm mercury pressure is called
1 torr. We do so because we are concerned only with relative values.
A pressure of 760 mm of mercury, which is about the average pressure
of the atmosphere, is sometimes called 'standard' or 'normal' pressure.
A temperature of 0°C, or 273 K, is likewise called standard or normal
temperature. The conditions 273 K and 760 mm pressure are together
called standard temperature and pressure (s.t.p.). A pressure of 760 mm
mercury is given, in newton per metre 2 , by
P = gpH _
= 9-8 x 13600 x 0-76 = 1013 x 10 5 N m~ 2 ,
since p = mercury density = 13600 kg m -3 ; g = 9-8 m s -2 ; H =
0-76 m.
At s.t.p. the density of hydrogen is about 009 g/litre, or 009 kg m -3 .
The gas-constant for unit mass of hydrogen, from (9), is therefore
p _ 1013 x 10 5
~ pT ~ 0-09 x 273
= 416 x 10 3 , in the appropriate units.
We will now discuss the units in which R is expressed.
The Gas-constant Units: Work done in Expansion
The gas-constant R for an arbitrary mass of gas is defined by the
equation
pV = RT
or R = -=-•
Its unit is therefore that of
pressure x volume
temperature
In SI units, the pressure is in N m -2 , the volume in m 3 and the tem-
perature in K. If we are given values of p, V and T and work out the
value of R, we express it in the corresponding units. The constant per
unit mass, R, for 1 kg has the unit of
pressure x volume _ Nm~ 2 xm 3 _ Nmke -1 K _1
temperature x mass K x kg
= iJkg^K" 1 .
since 1 newton x 1 metre — 1 joule. The gas constant may thus be
expressed in the same units as specific heat capacity (p. 199).
The gas constant depends on the mass of gas. For 1 kg, the unit is
GASES 229
J kg -1 K _1 . For 1 mole, the unit is J mol -1 K _1 ; for 1 kmol, the
unit is Jkmol" 1 K _1 .
Work Done. The product of pressure and volume has the dimensions
of work. To see this, let us imagine some gas, at a pressure p, in a
cylinder fitted with a piston (Fig. 10.9).
Fig. 10.9. Work done in expansion.
If the piston has an area S, the force on it is
f = pS.
If we allow the piston to move outwards a distance 81, the gas will
expand, and its pressure will fall. But by making the distance very
short, we can make the fall in pressure so small that we may consider
the pressure constant. The force / is then constant, and the work
done is
8W=f.8l = pS.5l.
The product 5. 51 is the increase in volume, 8V, of the gas, so that
8W = p.SV .... (11)
The product of pressure and volume, in general, therefore represents work.
If the pressure p is in newton m -2 , and the area S is in m 2 , the force/
is in newtons. And if the movement SI is in m, the work f. 31 is in
newton x metre or joule (J). The increase of volume, 8V, is in m 3 . Thus
the product of pressure in N m~ 2 , and volume in m 3 , represents work
in joules.
Consequently, if we express the pressure of a gas in newton m -2
and its volume in m 3 , the gas-constant R, = pV/T, is in joule (J) per
degree. The constant for 1 kg, R, is in joule per kg per degree. The
value of R for hydrogen, which we calculated on p. 228, is
R = 416xl0 3 Jkg~ 1 K- 1 =416kJkg- 1 K" 1 .
A vogadro's Hypothesis: Molecular Weight
Amedeo Avogadro, with one simple-looking idea, illuminated
chemistry as Newton illuminated mechanics. In 1811 he suggested that
chemically active gases, such as oxygen, existed not as single atoms,
but as pairs: he proposed to distinguish between an atom, O, and a
molecule, 2 . Ampere proposed the same distinction, independently,
in 1814. Avogadro also put forward another idea, now called Avogadro' s
hypothesis: that equal volumes of all gases, at the same temperature
and pressure, contained equal numbers of molecules. The number of
molecules in lcm 3 of gas at s.t.p. is called Loschmidt's number; it is
2-69 xlO 19 .
230 ADVANCED LEVEL PHYSICS
Avogadro's hypothesis became accepted in the middle of the nine-
teenth century. Because molecules could not be observed, their masses
could not be measured directly; but they could be compared, by
chemical methods. The molecular mass of a substance, /i, was at
first defined as the ratio of the mass of its molecule, m, to the mass of
a hydrogen atom. Later, for the convenience of chemists, it was defined
as the ratio of the molecular mass to the mass of an imaginary atom,
this atom having -j^th the mass of a carbon atom 12 C :
_ mass of molecule _ m _ 12m
Y2 mass of C-atom i^m c — m c '
On this scale, the mass of a hydrogen atom is 1008 times the mass
of the imaginary atom. And since the hydrogen molecule contains
two atoms, its molar mass is
//„ 2 = 2016.
The unit of molecular mass, m c /12, is also the unit of atomic mass;
its value is 1-66 x 10 ~ 24 g.
The Mole: Molar Gas-constant
The amount of a substance which contains as many elementary units
as there are atoms in 00 12 kg (12 g) of carbon- 12 is called a mole,
symbol 'mol'. The number of molecules in a mole, N A , is given, if m
is the mass of a molecule in grammes, by
H = N A m,
, AT u, 12m/m r
whence N A = — = — -
A m m
= Y2
m c '
The number of molecules per mole is thus the same for all substances.
It is called Avogadro's constant, and is equal to 6-02x 10 23 mol -1 .
From Avogadro's hypothesis, it follows that the mole of all gases, at
the same temperature and pressure, occupy equal volumes. Experiment
confirms this; at s.t.p. 1 mole of any, gas occupies 22-4 litres. Conse-
pV
quently, if we denote by V the volume of 1 mole, then the ratio ^=r
is the same for all gases. We call it the molar gas constant, R, and
At s.t.p.
T
p = 760 mm mercury = 1013 x 10 5 N m~ 2 (p. 228)
T = 273 K.
1013 xl0 5 x 22-4 xlO -3
R =
273
= 8-31 J mol^K"
GASES 231
The value of R is the same for the moles of all substances. If n g is
the molar mass of a gas, the constant for 1 g is thus
R=^ . . • (12)
KINETIC THEORY
The kinetic theory of matter, which regards all bodies as assemblies
of particles in motion — either vibrating or flying about — is an old one.
Lucretius described it in the first century a.d. and Gassendi and Hooke
revived it in the seventeenth century. In 1738 D. Bernoulli applied it
in detail to a gas, and from it deduced Boyle's law, which was already
Joiown from experiment. Another century passed, however, before
the kinetic view of a gas was fully developed — mainly by Clausius
(1822-88), Boltzmann (1844-1906), and Maxwell (1831-79).
In the kinetic theory of gases, we seek to explain the behaviour of
gases by considering the motion of their molecules. In particular, we
suppose that the pressure of a gas is due to the molecules bombarding
the walls of its container. Whenever a molecule bounces off a wall, its
momentum at right-angles to the wall is reversed; the force which it
exerts on the wall is equal to the rate of change of its momentum.
The average force exerted by the gas on the whole of its container is
the average rate at which the momentum of its molecules is changed
by collision with the walls.
To find the pressure of the gas we must find this force, and then
divide it by the area of the walls. The following assumptions are made
to simplify the calculation :
(a) The attraction between the molecules is negligible.
(b) The volume of the molecules is negligible compared with the
volume occupied by the gas. t
(c) The molecules are like perfectly elastic spheres.
(d) The duration of a collision is negligible compared with the time
between collisions.
Calculation of Pressure
Consider for convenience a cube of side / containing N molecules of-
gas each of mass m. Fig. 10.10. A typical molecule will have a velocity
c at any instant and this will have components of u, v, w respectively
in the direction of the three perpendicular axes Ox, Oy, Oz as shown.
Thus c 2 = u 2 + v 2 + w 2 .
Consider the force exerted on the face X of the cube due to the
component u. Just before impact, the momentum of the molecule due
to u is mu. After impact, the momentum is — mu, since the momentum
reverses. Thus
momentum change on impact = mu — ( — mu) = 2mu.
232
ADVANCED LEVEL PHYSICS
The time taken for the molecule to move across the cube to the
Opposite face and back to X is 2l/u. Hence
momentum change per second = mo ^"tum change
time
2mu _ mu 2
2l/u ~ ~T
.'. force on X =
mu
_. force mu 2 mu 2
..pressure on X = ^- = ra , = - F (l )
1
z
1/
w Asi^
;___.
/
t /
/
/
/
/
Fig. 10.10. Calculation of gas pressure.
We now take account of the N molecules in the cube. Each has a
different velocity and hence a component of different magnitude in
the direction Ox. If these are represented by u lt u 2 , u 3 , . . ., u N , it follows
from (i) that the total pressure on X, p, is given by
mu 2 mu 2 2 mu 3 2 mu^
m /.. 2
= £(«l 2 + «2 2 + «3 2 + -+"N 2 )
(ii)
GASES 233
Let the symbol u 2 represent the average or mean value of all the
squares of the components in the Ox direction, that is,
-2 Ul 2 + U 2 2 +U 3 2 + ..
•W
u - N
Then
Nu* = u 1 2 +u 2 2 + u 3 2 + .
,.+M N 2
Hence, from (ii),
Nmu 2
n = ~ —
r - P ■ ■ m
Now with a large number of molecules of varying speed in random
motion, the mean square of the component speed in any one of the
three axes is the same.
/. u 2 = v 2 = w 2 .
But, for each molecule, c 2 = u 2 + v 2 + w 2 , so that the mean square c 2 is
given by c 2 = u 2 + v 2 + w 2 .
u 2 =kc 2 .
Hence, from (iii),
iJVmc
V = 3— jr
The number of molecules per unit volume, n, = N/l 3 . Thus we may write
p = \nmc 2 .... (13)
If n is in molecules per metre 3 , m is in kilogramme and c in metre per
second, then the pressure p is in newton per metre 2 (N m -2 ).
It should be carefully noted that the pressure p of the gas depends
on the 'mean square' of the speed. This is because (a) the momentum
change at a wall is proportional to u, as previously explained, and (b) the
time interval before this change is repeated is muerse/y-proportional
to u. Thus the rate of change of momentum is proportional to u -=- 1/m
or to u 2 . Further, the mean-square speed is not equal to the square of
the average speed. As an example, let us suppose that the speeds of
six molecules are, 1, 2, 3, 4, 5, 6 units. Their mean speed is
_ 1 + 2 + 3+4 + 5 + 6 21 ,,<-
c = 6^ - = -6~ = 35 '
and its square is (c) 2 = 3-5 2 = 12-25.
Their mean square speed, however is
-2 l 2 + 2 2 + 3 2 +4 2 + 5 2 + 6 2 91 ,.<,
c= = T =15 * 2 -
This differs by about 25 per cent from the square of the mean speed.
In our calculation^ we assumed that molecules of a gas do not collide
with other molecules as they move to-and-fro across the cube. If,
234 ADVANCED LEVEL PHYSICS
however, we assume that their collisions are perfectly elastic, both the
kinetic energy and the momentum are conserved in them. The average
momentum with which all the molecules strike the walls is then not
changed by their collisions with one another; what one loses, another
gains. The important effect of collisions between molecules is to
distribute their individual speeds ; on the average, the fast ones lose
speed to the slow. We suppose, then, that different molecules have
different speeds, and that the speeds of individual molecules vary with
time, as they make collisions with one another; but we also suppose
that the average speed of all the molecules is constant. These assump-
tions are justified by the fact that the kinetic theory leads to conclusions
which agree with experiment.
Root-mean-square (R.M.S.) Speed
In equation (13) the factor run is the product of the molecules per
unit volume and the mass of one molecule. It is therefore the total
mass of the gas per unit volume : its density p. Thus the equation gives
P = W ■ ■ ■ • (14)
or | = J?. ... . (15)
If we substitute known values of p and p in equation (15), we can
find c 2 ". For hydrogen at s.t.p.,
p = <M)9kgm
-3
The pressure in newton per m 2 , is p = gpH, where g = acceleration
of gravity = 9-81 m s -2 , p = density of mercury = 13600 kg m -3 ,
H = barometer height = 760 mm = 0-76 m.
• ~1 = 3p _ 3 x 9-81 x 13600 x 0-76
" C ~ p ~ 9xl0~ 2
= 3-37xl0 6 m 2 s- 2 .
The square root of c 2 is called the root-mean-square speed; it is of the
same magnitude as the average speed, but not quite equal to it Its
value is
7? = 7^37 x 10 3 = 1840 m s" l (approx.)
= 1-84 km s" 1
Molecular speeds were first calculated in this way by Joule in 1848;
they turn out to have a magnitude which is high, but reasonable. The
value is reasonable because it has the same order of magnitude as
the speed of sound (1-30 km s _1 in hydrogen at 0°C); the speed of
sound is the speed with which the molecules of a gas pass on a dis-
turbance from one to another, and this we may expect to be of the
same magnitude as the speeds of their natural motion.
GASES 235
Introduction of the Temperature
Let us consider a volume V of gas, containing N molecules. The
number of molecules per unit volume is
N
and therefore the pressure of the gas, by equation (13) is
p = \nmc 2 = fymc z
;. pV= ^Nmc 2 (16)
Equation (16) reminds us of the equation combining Boyle's and
Charles's laws :
pV = RT.
We can therefore make the kinetic theory consistent with the observed
behaviour of a gas, if we write
±Nmc~ 2 = RT . . . . (17)
Essentially,_we are here assuming that the mean square speed of the
molecules, c 2 , is proportional to the absolute temperature of the gas.
This is a reasonable assumption, because we have learnt that heat is a
form of energy; and the kinetic energy of a molecule, due to its random
motion within its container, is proportional to the square of its speed.
When we heat a gas, we expect to speed-up its molecules. See p. 237.
The kinetic energy of a molecule moving with a speed c is \mc 2 ; the
average kinetic energy of translation of the random motion of the
molecule of a gas is therefore \mc 2 . To relate this to the temperature,
we put equation (17) into the form
RT = \Nrnc 1 = fiV^mc 2 ),
~ 5 ^R*
•JV'
whence \mc 2 = f*T . . . . (18)
Thus, the average kinetic energy of translation of a molecule is propor-
tional to the absolute temperature of the gas.
The ratio R/N in equation (18) is a universal constant. To see that
it is, we have only to consider a mole. We have already seen that, for
a mole, the gas constant R, and number of molecules N, are universal
constants. If our arbitrary mass of gas is x moles, then R = xR, and
N = xN; therefore
R_R_ ,
N ~ N ~ k -
The constant k, the gas constant per molecule, is also a universal
constant ; it is often called Boltzmann's constant. In terms of k equation
(18) becomes
\m? = \kT . . . . (19)
236
ADVANCED LEVEL PHYSICS
Boltzmann's constant is usually given in joules per degree, since it
relates energy to temperature :
k =
mc
Its value is k = 1-38 x 10" 23 J K 1 .
Diffusion: Graham's Law
When a gas passes through a porous plug, a cotton-wool wad, for
example, it is said to 'diffuse'. Diffusion differs from the flow of a gas
through a wide tube, in that it is not a motion of the gas in bulk, but is
a result of the motion of its individual molecules.
Fig. 10.11 shows an apparatus devised by Graham (1805-69) to com-
pare the rates of diffusion of different gases. D is a
glass tube, closed with a plug P of plaster of Paris.
It is first filled with mercury, and inverted over
mercury in a bowl. Hydrogen is then passed into
it until the mercury levels are the same on each
side; the hydrogen is then at atmospheric pressure.
The volume of hydrogen, V H , is proportional to
the length of the tube above the mercury. The
apparatus is now left ; hydrogen diffuses out through
P, and air diffuses in. Ultimately no hydrogen
remains in the tube D. The tube is then adjusted
until the level of mercury is again the same on
each side, so that the air within it is at atmospheric
pressure. The volume of air, V A , is proportional to
the new length of the tube above the mercury.
The volumes V A and V H are, respectively, the
volumes of air and hydrogen which diffused through
the plug in the same time. Therefore the rates of
diffusion of the gases air and hydrogen are propor-
tional to the volumes V A and V H :
rate of diffusion of air V A
Fig. 10.11. Graham's
apparatus for
diffusion.
rate of diffusion of hydrogen V H '
Graham found in his experiments that the volumes were inversely
proportional to the square roots of the densities of the gases, p :
V A
thus
In general :
rate of diffusion of air
rate of diffusion of hydrogen
rate of diffusion oe —7= ;
Vp
Ph
Pa
and in words: the rate of diffusion of a gas is inversely proportional to
the square root of its density. This is Graham's Law.
GASES
237
Graham's law of diffusion is readily explained by the kinetic theory.
At the same Kelvin temperature T, the mean kinetic energies of the
molecules of different gases are equal, since
and k is a universal constant. Therefore, if the subscripts A and H
denote air and hydrogen respectively,
2 m A C A ~ 2 m ff C ff»
1. C A m H
whence ==- = — -.
At a given temperature and pressure, the density of a gas, p, is
proportional to the mass of its molecule, m, since equal volumes
contain equal numbers of molecules :
Therefore ^h = Ph
™a Pa
whence
<i _ Ph
l Pa
(20)
The average speed of the molecules of a gas is roughly equal to — and
strictly proportional to — the square root of its mean square speed.
Equation (20) therefore shows that the average molecular speeds are
inversely proportional to the square roots of the densities of the gases.
And so it explains why the rates of diffusion — which depend on the
molecular speeds — are also inversely proportional to the square roots
of the densities.
Thermal Agitations and Internal Energy
The random motion of the molecules of a gas, whose kinetic energy
depends upon the temperature, is often called the thermal agitation of
the molecules. And the kinetic energy of the thermal agitation is called
the internal energy of the gas. We must appreciate that this energy is
quite independent of any motion of the gas in bulk : when a cylinder of
oxygen is being carried by an express train, its kinetic energy as a
whole is greater than when it is standing on the platform; but the
random motion of the molecules within the cylinder is unchanged —
and so is the temperature of the gas. The same is true of a liquid ; in a
water-churning experiment to convert mechanical energy into heat,
baffles must be used to prevent the water from acquiring any mass-
motion — all the work done must be converted into random motion,
if it is to appear as heat. Likewise, the internal energy of a solid is
the kinetic energy of its atoms' vibrations about their mean positions :
throwing a lump of metal through the air does not raise its temperature,
but hitting it with a hammer does.
238
ADVANCED LEVEL PHYSICS
The internal energy of a gas depends on the number of atoms in its molecule.
A gas whose molecules consist of single atoms is said to be monatomic : for
example, chemically inert gases and metallic vapours, Hg, Na, He, Ne, A. A gas
(a)
Hg
•
Monatomic
Hg
(b) Diatomic
H,
(c)
H H
Polyatomic
H 2
Fig. 10.12. Types of gas molecule,
with two atoms to the molecule is said to be diatomic: 2 ,
H,
l2 , N 2 , Cl 2 , CO.
And a gas with more than two atoms to the molecule is said to be polyatomic:
H 2 0, 3 , H 2 S, C0 2 , CH 4 . The molecules of a monatomic gas we may regard as
points, but those of a diatomic gas we must regard as 'dumb-bells', and those of
a polyatomic gas as more complicated structures (Fig. 10.12). A molecule which
extends appreciably in space — a diatomic or polyatomic molecule — has an
appreciable moment of inertia. It may there-
fore have kinetic energy of rotation, as well
as of translation. A monatomic molecule,
however, must have a much smaller moment
of inertia than a diatomic or polyatomic; its
kinetic energy of rotation can therefore be
neglected.
Fig. 10.12a shows a monatomic molecule
whose velocity c has been resolved into its
components u, v, w along the x, y, z axes :
c 2 = u 2 + v 2 + w 2 .
The x, y, z axes are called the molecules'
degrees of freedom : they are three directions such that the motion of the molecule
along any one is independent of its motion along the others.
If we average the speed c, and the components u, v, w, over all the molecules in
a gas, we have
c 2 = u 2 + v 2 + w 2 .
And since the molecules do not pile up in any corner of the vessel containing the
gas, their average velocities in all directions must be the same. We may therefore
write
~H — ZJ — Z$
Fig. 10.12a. Components of velocity.
whence
u = v = w
3w 2
u 2 = v 2 = w 2 = \c 2 .
The average kinetic energy of a molecule of the gas is given by equation (19) :
Therefore the average kinetic energy of a monatomic molecule, in each degree
of freedom, is
\mu 2 = \mv 2 = jmw 2 — jkT.
Thus the molecule has kinetic energy ^kT per degree of freedom.
Rotational Energy
Let us now consider a diatomic or polyatomic gas. When two of its molecules
collide, they will, in general, tend to rotate, as well as to deflect each other. In
some collisions, energy will be transferred from the translations of the molecules
GASES
239
to their rotations; in others, from the rotations to the translations. We may
assume, then, that the internal energy of the gas is shared between the rotations
and translations of its molecules.
To discuss the kinetic energy of rotation, we must first extend the idea of
degrees of freedom to it. A diatomic molecule can have kinetic energy of rotation
about any axis at right-angles to its own. Its motion about any such axis can be
resolved into motions about two such axes at right-angles to each other (Fig.
10. 13 (a) ). Motions about these axes are independent of each other, and a diatomic
Fig. 10.13. Rotation of molecules.
molecule therefore has two degrees of rotational freedom. A polyatomic molecule,
unless it happens to consist of molecules all in a straight line, has no axis about
which its moment of inertia is negligible. It can therefore have kinetic energy of
rotation about three mutually perpendicular axes (Fig. 10.13 (b)). It has three
degrees of rotational freedom.
We have seen that the internal energy of a gas is shared between the translations
and rotations of its molecules. Maxwell assumed that the average kinetic energy
of a molecule, in each degree of freedom, rotational as well as translational,
was %kT. This assumption is called the principle of equipartition of energy; ex-
periment shows, as we shall find, that it is true at room temperature and above.
At very low temperatures, when the gas is near liquefaction, it fails. At ordinary
temperatures, then, we have :
3
average K.E. of monatomic molecule = -k T (trans.);
3 2 S
average K.E. of diatomic molecule = -kT (trans.) + -fc T (rot.) = ^k T ;
average K.E. of polyatomic molecule = -& T (trans.) +^7" (rot.) = ^kT.
Internal Energy of any Gas
From die average kinetic energy of its molecules, we can find the internal
energy of a mass Af of a gas. The number of molecules in this mass is, if m is the
mass of one molecule.
m
Its internal energy, U, is the total kinetic energy of its molecules' random motions ;
thus
U = Nx average K.E. of molecule.
For a monatomic gas, therefore,
3
U = ^NkT (monatomic).
240 ADVANCED LEVEL PHYSICS
The constant k is the gas-constant per molecule ; the product Nk is therefore the
gas-constant R for the mass M of the gas. Thus
3
U = -/?T(monatomic).
In particular, if R is the gas-constant per kg, the internal energy per kg is
Similarly, for a diatomic gas,
U = |«r(monatomic) (21)
V = ^NkT=^RT .... (22)
u = \rt
2
And for a polyatomic gas,
U = ^NkT=^RT .... (23)
2
Internal Energy and Volume
In our simple account of the kinetic theory of gases, we have implicitly
assumed that the molecules of a gas do not attract one another. If they
did, any molecule approaching the boundary of the gas would be
pulled towards the body of it, as is a molecule of water approaching
the surface (see Chapter 6, Surface Tension, p. 128). The attractions of
the molecules would thus reduce the pressure of the gas.
Since the molecules of a substance are presumably the same whether
it is liquid or gas, the molecules of a gas must attract one another
somewhat. But except for brief instants when they collide, the molecules
of a gas are much further apart than those of a liquid. In 1 cubic centi-
metre of gas at s.t.p. there are 2-69 x 10 19 molecules, and in 1 cubic
centimetre of water there are 3-33 x 10 22 ; there are a thousand times
as many molecules in the liquid, and so the molecules in the gas are ten
times further apart. We may therefore expect that the mutual attraction
of the molecules of a gas, for most purposes, can be neglected, as
experiment, in fact, shows.
The experiment consists in allowing a gas to expand without doing
external work; that is, to expand into a vacuum. Then, if the molecules
attract one another, work is done against their attractions, as they
move further apart But if the molecular attractions are negligible, the
work done is also negligible. If any work is done against the molecular
attractions, it will be done at the expense of the molecular kinetic
energies; as the molecules move apart, they will exert retarding forces
on one another. Thus the internal energy of the gas, and therefore its
temperature, will fall.
The expansion of a gas into a vacuum is called a 'free expansion'. If
a gas does not cool when it makes a free expansion, then the mutual
attractions of its molecules are negligible.
GASES
241
Joule's Experiments
Experiments on the free expansion of a gas were made in 1807 by
Gay-Lussac; they showed no fall in temperature. Joule repeated
these experiments with a better vacuum in 1845; he got the same nega-
tive result, and the greater accuracy of his experiments made them more
trustworthy. Joule used two forms of apparatus, as shown in Fig. 10.14
(a) and (b). Each consisted of a cylinder of air, R, at 22 atmospheres,
connected by a stop-cock S to an evacuated cylinder E. In the apparatus
(a) both cylinders stood in the same tin can C, which contained 16j lb
water. In (b) the cylinders stood in different cans, and the stop-cock in
a third, also containing water. When the stop-cock was opened, gas
expanded from R to E. With the apparatus (a) Joule found, after stirring
the water, that its temperature was unchanged. The expanding gas had
therefore neither liberated heat nor absorbed it. With the apparatus (b)
Joule found that heat was absorbed from the water round R, and given
to the water round S and E; the heat given out was equal to the heat
(a) (b)
Fig. 10.14. Joule's experiment on internal energy.
taken in. The heat taken in represented the work done by the gas from
R, expanding against the rising pressure of the gas in E and in the pipe
beyond S. The heat given out represented the work done on the gas in
E and S by the gas flowing in, against the rising pressure. The equality
of the two showed that the total mass of gas neither gained nor lost
energy in making its free expansion. Joule's experiments, therefore,
showed that the internal energy of a gas is independent of its volume.
From Joule's results we may argue back to show that the mutual
attractions of the molecules of the gas are negligible; in practice,
however, it is the property of the bulk gas which is important — the
fact that its internal energy does not depend on its volume.
Joule's experiments, though more reliable than Gay-Lussac's, were
crude; with so much water, a small amount of heat would not produce
a measurable temperature rise. Between 1852 and 1862, Joule worked
with William Thomson, later Lord Kelvin, on more delicate experi-
242 ADVANCED LEVEL PHYSICS
ments. They found that most gases, in expanding from high pressure
to low, do lose a little of their internal energy. The loss represents
work done against the molecular attractions, which are therefore not
quite negligible.
If the internal energy of a gas is independent of its volume, it is
determined only by the temperature of the gas. The simple expression
for the pressure, p = %pc 2 , then holds; and the gas obeys Boyle's and
Charles's laws. Its pressure coefficient, a v , is equal to its volume
coefficient, <x p . Such a gas is called an ideal, or perfect, gas. All gases,
when far from liquefaction, behave for most practical purposes as
though they were ideal.
Van der Waals' Equation
In deriving the ideal gas equation pV = RT from the kinetic theory
of gases, a number of assumptions were made. These are listed on
p. 231. Van der Waals modified the ideal gas equation to take account
that two of these assumptions may not be valid. Thus, as explained
on p. 127, to which the student should refer:
(1) The volume of the molecules may not be negligible in relation to
the volume V occupied by the gas.
(2) The attractive forces between the molecules may not be negligible.
Molecules have a particular diameter or volume because repulsive
forces occur when they approach very closely and hence they can not
be compressed indefinitely. The volume of the space inside a container
occupied by the molecules is thus not V but (V—b), where b is a factor
depending on the actual volume of the molecules.
If the attractive forces between molecules are not negligible, the
molecules approaching the container walls are attracted by the mole-
cules behind them. This reduces the momentum of the approaching
molecules and hence the pressure. The observed pressure p is thus less
than the ideal gas pressure, where there are no molecular forces, by a
pressure p'. Hence we write (p + p') in place of p in the ideal gas equation.
As explained on p. 127, the 'pressure defect' p' oc p 2 , where p is the
density of the gas, or p = a/V 2 , where a is a constant. Thus van der
Waals'' equation for real gases is :
[p+fyv-b) = RT.
At high pressures, when the molecules are relatively numerous and
close together, the volume factor b and pressure 'defect' a/V 2 both
become important. Conversely, at low pressures, where the molecules
are relatively few and far apart on the average, a gas behaves like an
ideal gas and obeys the equation pV = RT.
Critical Phenomena
A graph of pressure p v. volume V at constant temperature is called
an isotherm. Fig. 10.15 (i) shows a number of isotherms for an ideal gas,
which obeys the perfect gas law pV = RT. Fig. 10.15 (ii) shows a
GASES
243
number of isotherms for a gas which obeys van der Waals' equation,
(p + a/V 2 )(V-b) = RT.
At high temperatures the isotherms are similar. As the temperature
is lowered, however, the isotherms in Fig. 10.15 (ii) change in shape.
One curve has a point of inflexion at C, which corresponds to the
critical point of a real gas. The isotherms thus approximate to those
obtained by Andrews in his experiments on actual gases such as
carbon dioxide, described on p. 314.
Pi
High T
Critical
temperature
(0 (ii)
Fig. 10.15. Isothermals for ideal and van der Waals gases.
Below this temperature, however, isotherms such as EABF are
obtained by using van der Waals' equation. These are unlike the iso-
therms obtained with real gases, because in the region AB the pressure
increases with the volume, which is impossible. However, an actual
isotherm in this region corresponds to a straight line EF, as shown.
Here the liquid and vapour are in equilibrium (see p. 317) and the line
EF is drawn to make the shaded areas above and below it equal.
Thus van der Waals' equation roughly fits the isotherms of actual gases
above the critical temperature but below the critical temperature it
must be modified considerably. Many other gas equations have been
suggested for real gases but quantitative agreement is generally poor.
SPECIFIC HEAT CAPACITIES
Specific Heat Capacities at Constant Volume and Constant Pressure
When we warm a gas, we may let it expand or not, as we please. If
we do not let it expand — if we warm it in a closed vessel — then it
does no external work, and all the heat we give it goes to increase its
internal energy. The heat required to warm unit mass of a gas through
one degree, when its volume is kept constant, is called the specific heat
capacity of the gas at constant volume. It is denoted by c v , and is generally
expressed in J kg - 1 K~ 1 .
If we allow a gas to expand as we warm it, then it does external work.
The heat we give the gas appears partly as an increase to its internal
energy — and hence its temperature — and partly as the heat equivalent
of the work done. The work done depends on the increase in volume
of the gas, which in turn depends on the way in which we allow the
244
ADVANCED LEVEL PHYSICS
gas to expand We can get an important theoretical result by supposing
that the pressure is constant, and defining the corresponding specific
heat capacity. The specific heat capacity of a gas at constant pressure is
the heat required to warm unit mass of it by one degree, when its pressure
is kept constant. It is denoted by c p , and is expressed in the same units
as c v .
Specific Heat Capacities: .their Difference
Any number of heat capacities can be defined for a gas, according to
the mass and the conditions imposed upon its pressure and volume.
For unit mass, 1 kg or 1 g, of a gas, the heat capacities at constant
pressure c p , and at constant volume c v , are the specific heat capacities.
Fig. 10.16 shows how we can find a relationship between the specific
heat capacities of a gas. We first consider 1 kg of the gas warmed through
p constant
Piston fixed
">C
r i
T—-T + 1
y
II
T ~T + 1
X
>V 9
(a) (b)
Fig. 10.16. Specific heat capacity at constant volume and pressure.
1°C at constant volume, (a). The heat required is c v joules, and goes
wholly to increase the internal energy.
We next consider 1 kg warmed through 1°C at constant pressure, (b).
It expands from V t to V 2 , and does an amount of external work given by
W = p{ V 2 - VJ (equation (1 1), p. 229).
The work W is in joules if p is in newton m -2 , and the volumes in m 3 .
Thus the amount of heat in joules required for this work is
W = p(V 2 -V 1 ).
Further, since the temperature rise of the gas is 1°C, and the internal
energy of the gas is independent of volume, the rise in internal energy
is c v , the specific heat at constant volume. Hence, from 3Q = SU+
p.SV, the total amount of heat required to warm the gas at constant
pressure is therefore
c, = c v +piV 2 -V 1 ) . • • (24)
We can simplify the last term of this expression by using the equation
of state for unit mass :
pV = AT,
where T is the absolute temperature of the gas, and R is the gas-
GASES 245
constant for unit mass of it in J K _1 . If T x is the absolute temperature
before warming, then
pV x = .RT x .... (25)
The absolute temperature after warming is T ± + 1 ; therefore
pV 2 = R(T t + l), .... (26)
and on subtracting (25) from (26) we find
p(/ 2 -V 1 ) = R
Equation (24) now gives c p = c v + R
or c p -c v = R .... (27)
Equation (27) was first derived by Robert Mayer in 1842. He used it,
before Joule had done his water-churning experiments, to derive a
relation between heat and mechanical energy.
Ratio of Specific Heat Capacities
We have seen that the internal energy of a gas, at a given temperature, depends
on the number of atoms in its molecule. For a monatomic gas its value in joules
per kg is
U = |RT (i)
where U is the internal energy of the gas, R is the gas constant in J kg -1 K _1
and T is the absolute temperature of the gas.
The heat required to increase the internal energy of 1 kg of a monatomic gas,
when it is warmed through 1 degree, is therefore §R joule. But this is the specific
heat capacity at constant volume, and so
c v = 2 R
The specific heat capacity of a monatomic gas at constant pressure is therefore
3
c P = Cv+R = 2 R+R
-*R
Let us now divide c p by c v ', their quotient is called the ratio of the specific
heat capacities, and is denoted by y.
For a monatomic gas, its value is
c„
2'
Cy jR
= f=l-667.
Similarly, for a diatomic molecule,
U = |rT (ii)
246 ADVANCED LEVEL PHYSICS
This was shown on p. 240.
Hence c v = ~R
7
and Cp = c v + R = jR
c 7
Hence y = -^ = ^ = 1-40.
Cy J
And for a polyatomic molecule,
U = ^RT, . . . (iii)
2
and y = ^ = | = 1-33.
c K 6
In general, if the molecules of a gas have/degrees of freedom, the average kinetic
energy of a molecule is fx^kT (p. 238).
.-. U = jRT,
c p = c K + R = K+lR,
f - + l
and y = -J = ^ = l +7 . . . . (iv)
2
The ratio of the specific heat capacities of a gas thus gives us a measure of the
number of atoms in its molecule, at least when that number is less than three.
This ratio is fairly easy to measure, as we shall see later in this chapter. The poor
agreement between the observed and theoretical values of y for some of the
polyatomic gases shows that, in its application to such gases, the theory is over-
simple.
Measurement of c v .
Fig. 10.17 shows an apparatus for measuring the specific heat capacity
of a gas at constant volume, called a differential steam calorimeter.
The calorimeter consists of two copper globes, A and B, as nearly
identical as they can be made. They hang from the beam of a balance,
and are surrounded by a chest C into which steam can be admitted
at D. The sphere B is evacuated and A is filled with the gas whose
specific heat is required. By filling A to a high pressure, the mass of
gas can be made great enough to be accurately measurable on the
balance. Let its value be M. Steam is now admitted to the chest, and
condenses on both globes until they reach the temperature of the
GASES
247
steam. This will generally be about 100°C, but we shall denote it by B v
The balance measures the excess steam condensed on A, over that
condensed on B; let its mass by M y If the globes are identical, their
heat capacities are equal, and the masses of steam required to warm
the globes alone are equal. The excess steam condensed on A is then the
mass required to warm the gas within it. Therefore if 9 R is the room
temperature, and / the specific latent heat of evaporation of water,
the specific heat capacity c v for the gas is given by
whence
Mc v (d 1 -6 R ) = M s l;
MA
c v =
M{e,-e R )
(28)
The calorimeter is called 'differential' for the reason that it measures
the difference in mass of the steam on the two globes. This is an im-
portant feature of it, because the heat capacities of the globes may be
much greater than that of the gas. If a single globe were used, two
measurements would have to be made, one with and one without the
Fig. 10.17. Joly's differential steam calorimeter.
gas. The mass of steam condensed by the gas would then appear as the
difference of two nearly equal masses, and could not be determined
accurately. In practice the globes are not identical, and a control
experiment with both evacuated is made to find the difference in mass
of the steam condensed on them. This appears as a small correction to
M s in equation (28).
A small correction has also to be made to the result of the experiment,
because the volume of the gas is not quite constant: the globe A
expands when it is warmed.
The figure shows a few of the practical refinements of the apparatus.
S, S are shields to prevent drops of moisture, condensed oh the roof of
248
ADVANCED LEVEL PHYSICS
the chest, from falling on to the globes. P, P are pans to catch any
drops which, having condensed on A or B, might fall off. W, W are
platinum wires heated by an electric current, which prevent drops
forming in the holes through which the suspension wires pass out of
the chest.
Measurement of «
v
The method of mixtures was used to determine the specific heat
capacity of a gas at constant pressure by Regnault. Regnault was one
of the greatest experimenters of the nineteenth century — the reader
who sees pictures of his apparatus in other books should remember
that he worked before Bunsen had invented his famous burner — but
his method for c p is now outmoded. We shall describe here only a
continuous flow method, similar to Callendar and Barnes' for the
specific heat capacity of water. It is due to Swann (1909).
Fig. 10.18. Constant flow calorimeter for c P .
Gas from a cylinder, A in Fig. 10.18, flows out through a needle
valve N, which reduces its pressure to a little above atmospheric. If the
pressure in the cylinder is high, it will fall slowly during an experiment,
and the pressure of the emerging gas will be almost constant. Mano-
meters G t G 2 indicate the pressure of the gas in the cylinder, and of the
gas emerging. The gas passes through a coiled tube S, in a water bath,
GASES 249
which brings it to a uniform temperature. It then flows past a platinum
resistance thermometer, P l5 which measures the temperature, O^C.
From there it goes to a heating coil H, past a baffle B which enables it
to receive any heat that escapes from the neighbourhood of the coil.
Beyond the coil it passes through copper gauze D, which mixes the
stream of gas and so brings it to a uniform temperature. This tempera-
ture, 2 , is measured by the platinum resistance thermometer P 2 . A
vacuum jacket F makes the heat losses very small.
If M is the mass of gas flowing through the apparatus in t seconds,
then the heat received by it is Mc p (0 2 — 0J. If the heat losses are neg-
ligible, the heat supplied by the coil in t seconds is IVt joule, where /
is the current through it in ampere and V the potential difference
across it in volt. Then
ivt = Mc p {e 2 -e 1 ).
The mass of gas, M, is found from the fall of pressure in the cylinder.
If v is the volume of the cylinder, and p l the density of the gas at the
initial pressure p u then the mass initially in the cylinder is
M x = p^.
And if, after t seconds, the pressure has fallen to p 2 , and the density to
p 2 , the mass remaining in the cylinder is
M 2 = p 2 v.
The mass of gas which has escaped in then
M = M x -M 2 = {p 1 -p 2 )v.
The densities p x and p 2 can be readily calculated from the density p
at s.t.p. : if 3 is the temperature of the cylinder, and the pressures are
measured in mm mercury, then
Pi P 2 760
Pl (273 + 3 ) p 2 (273 + 3 ) 273p
The cylinder temperature 3 is kept constant by the water bath W.
CHANGES OF PRESSURE, VOLUME AND TEMPERATURE
In, for example, a steam engine or motor, gases expand and are
compressed, cool and are heated, in ways more complicated than those
which we have already described. We shall now consider some of these
ways.
Isothermal Changes
We have seen that the pressure p, and volume V of a given mass of
gas are related by the equation
pV=RT,
where T is the absolute temperature of the gas, and R is a constant.
250
ADVANCED LEVEL PHYSICS
If the temperature is constant the curve of pressure against volume is
a rectangular hyperbola,
pV = constant,
representing Boyle's law. Such a curve is called an isothermal for the
given mass of the given gas, at the temperature T. Fig. 10.19 shows a
family of isothermals, for 1 g of air at different temperatures. When a
gas expands, or is compressed, at constant temperature, its pressure
and volume vary along the appropriate isothermal, and the gas is said
to undergo an isothermal compression or expansion.
2000 r
18001-
1600
1400
2"1200r
E1000
E
E
800
600
400
200
200 400 600 800 1000 1200 1400 1600 1800 2000
V(cm 3 ) »►
Fig. 10.19. Isothermals for 1 g air.
When a gas expands, it does work — for example, in driving a piston
(Fig. 10.9, p. 229). The molecules of the gas bombard the piston, and
if the piston moves they give up some of their kinetic energy to it ;
when a molecule bounces off a moving piston, it does so with a velocity
less in magnitude than that with which it struck. The change in velocity
is small, because the piston moves much more slowly than the molecule ;
but there are many molecules striking the piston at any instant, and
their total loss of kinetic energy is equal to the work done in driving
the piston forward.
The work done by a gas in expanding, therefore, is done at the
expense of its internal energy. The temperature of the gas will conse-
quently fall during expansion, unless heat is supplied to it. For an
isothermal expansion, the gas. must be held in a thin- walled, highly
conducting vessel, surrounded by a constant temperature bath. And
the expansion must take place slowly, so that heat can pass into the
gas to maintain its temperature at every instant during the expansion.
GASES 251
External Work done in Expansion
The heat taken in when a gas expands isothermally is the heat
equivalent of the mechanical work done. If the volume of the gas
increases by a small amount SV, at the pressure p, then the work done is
dW = pSV (equation (11), p. 229).
In an expansion from V 1 to V 2 , therefore, the work done is
rv 2
W= \dW = pdV.
JVi
RT
By the gas equation, p = —=-,
whence W= I pdV = I RT^
fv 2 rv.
pdV =
Vi JVi
or W = RTlog 12
The heat required, Q, is therefore
yjr
where W is in joules if R is in J kg - * K~ 1 .
Now let us consider an isothermal compression. When a gas is
compressed, work is done on it by the compressing agent. To keep its
temperature constant, therefore, heat must be withdrawn from the
gas, to prevent the work done from increasing its internal energy. The
gas must again be held in a thin well-conducting vessel, surrounded
by a constant-temperature bath ; and it must be compressed slowly.
The conditions for an isothermal compression or expansion of a gas
are difficult to realize; heat cannot flow through the walls of the vessel
unless there is at least a small difference of temperature across them,
and therefore the temperature of the gas is bound to rise a little in
compression, or to fall a little in expansion.
Reversible isothermal change
Suppose a gas expands isothermally from p lt V t , T to p 2y V 2 , T.
If the change can be reversed so that the state of the gas is returned
from p 2 , V 2 , Tto p lf V lt T through exactly the same values of pressure
and volume at every stage, then the isothermal change is said to be
reversible. A reversible isothermal change is an ideal one. It requires
conditions such as a light frictionless piston, so that the pressure
inside and outside the gas can always be equalised and no work is
done against friction; very slow expansion, so that no eddies are
produced in the gas to dissipate the energy ; and a constant temperature
reservoir with very thin good-conducting walls, as we have seen. In a
reversible isothermal change, pV —constant = RT.
252
ADVANCED LEVEL PHYSICS
Equation for Reversible Adiabatic change
Let us now consider a change of volume in which the conditions are
at the opposite extreme from isothermal; no heat is allowed to enter
or leave the gas.
An expansion or contraction in which no heat enters or leaves the
gas is called an adiabatic expansion or contraction. In an adiabatic
expansion, the external work is done wholly at the expense of the
internal energy of the gas, and the gas therefore cools. In an adiabatic
compression, all the work done on the gas by the compressing agent
appears as an increase in its internal energy and therefore as a rise in
its temperature. We have already discussed a reversible isothermal
change. A reversible adiabatic change is an adiabatic change which can
be exactly reversed in the sense explained on p. 251. As noted there, a
reversible change is an ideal case.
The curve relating pressure and volume for a given mass of a given
gas for adiabatic changes is called an 'adiabatic'. In Fig. 10.20, the
heavy curve is an adiabatic for 1 g of air; it is steeper, at any point,
than the isothermal through that point. The curve AB is the isothermal
for the temperature T = 373 K, which cuts the adiabatic at the
point p V . If the gas is adiabatically compressed from V to V v its
2000
1800
1600
1400
>•
2 1200
E
I 1000
c{ 800
600
400
200
Adiabatic through p 0i v
Isothermals
P 2 .V 2
200 400 600 800 1000 1200 1400 1600 1800 2000 2200
V, cm 3 ■-
Fig. 10.20. Relationship between adiabatic and isothermals.
temperature rises to some value T v Its representative point p l5 V x
now lies on the isothermal for T 1? since p^ — RT^ Similarly, if the
gas is expanded adiabatically to V 2 , it cools .to T 2 and its representative
point p 2 , V 2 lies on the isothermal for T 2 . Thus the adiabatic through
GASES
253
any point— such as p , V — is steeper than the isothermal. We will
find its equation shortly.
The condition for an adiabatic change is that no heat must enter or
leave the gas. The gas must therefore be held in a thick-walled, badly
conducting vessel; and the change of volume must take place rapidly,
to give as little time as possible for heat to escape. However, in a rapid
compression, for example,, eddies may be formed, so that some of the
work done appears as kinetic energy of the gas in bulk, instead of as
random kinetic energy of its molecules. All the work done then does
not go to increase the internal energy of the gas, and the temperature
rise is less than in a truly adiabatic compression. If the compression
is made slowly, then more heat leaks out, since no vessel has perfectly
insulating, walls:
Perfectly adiabatic changes are therefore impossible; and so, we
have seen, are perfectly isothermal ones. Any practical expansion or
compression of a gas must lie between isothermal and adiabatic. It
may lie anywhere between them, but if it approximates to isothermal,
the curve representing it will always be a little steeper than, the ideal
(Fig. 10.21); if it approximates to adiabatic, the curve representing it
will never be quite as steep as the ideal.
• Ideal adiabatic
Practical approach
Ideal isothermal
Practical approach
V
Fig. 10.21. Ideal and real p-V curves for a gas.
Equation of Reversible Adiabatic
Before considering adiabatic changes in . particular,- let us > first;
consider a change of volume and temperature which takes place in an-
arbitrary manner. For simplicity, we consider .unit mass of; the gas,
and we suppose that its volume expands from V to V +SV, and that
an amount of heat 5Q is supplied to it In general, the internal energy
of the gas will increase by an amount SU. And the gas will do an amount
of external work equal to pdV, where p is its pressure. The heat supplied
is equal to the increase in internal energy, plus the external work done :
SQ = SU + pSV . . . . (29)
The increase in internal energy represents a temperature rise, ST. We
have seen already that the internal energy is independent of the volume,
254 ADVANCED LEVEL PHYSICS
and is related to the temperature by the specific heat capacity at
constant volume, c v (p. 244). Therefore
SU = c v 3T.
Equation (29) becomes
3Q = c v ST + pdV ... (30)
Equation (30) is the fundamental equation for any change in the state
of unit mass of a gas.
For a reversible isothermal change, ST = 0, and SQ = pSV.
For a reversible adiabatic change, 3Q = and therefore
c v ST + pdV = . . . . (31)
To eliminate 6T we use the general equation, relating pressure, volume and
temperature :
pV = RT,
where R is the gas constant for unit mass. Since both pressure and volume may
change, when we differentiate this to find <5Twe must write
pdV+VSp =* RST,
whence ST= pdV+ D Vdp .
K
Therefore, by equation (31),
c V R F +pdV =
or Cy(p5V+ Vdp)+ RpSV = 0.
Now we have seen, on p. 245, that
R = C p — Cy',
therefore Cy(p5V+ Vdp) + (c p - c v )pdV = 0.
Hence c v Vdp+c p pdV =
or Vdp+^pdV =
Cy
or Vdp + ypdV = J where y = —
Therefore +717 = 0.
Integrating, we find
p"V
(dp CdV n
or logeP+y loggV = A,
where A is a constant.
Therefore, p\T = C,
where C is also a constant. This is the equation of a reversible adiabatic ; the value
of C can be found from the initial pressure and volume of the gas.
GASES 255
If we have a mass M of the gas, its volume at any temperature and pressure is
V = MV,
where V is the volume of unit mass at the same temperature and pressure. There-
fore for any mass of gas, the equation of an adiabatic change is
pV y = constant . . . . . (32)
Equation for Temperature Change in an Adiabatic
If we wish to introduce the temperature, T, into equation (32), we
use the general gas equation
pV = RT.
Thus
P =
RT
and pV y = ~. V y = RTV y ~ x .
Thus equation (32) becomes
RTV yl = constant,
and since R is a constant for a given mass of gas, the equation for an
adiabatic temperature change becomes
TV 7 ' 1 = constant.
Measurement of y.
In books on Sound, it is shown that sound waves are propagated
through a gas by rapid compressions and rarefactions ; these changes
in pressure and volume are adiabatic. In consequence, the velocity of
sound in a gas depends upon the ratio of the specific heat capacities
of the gas, y; and the value of y can be found from measurements of
the velocity of sound in the gas. This is the most convenient way of
measuring y.
A direct measurement of y can be made by the method of Clement
and Desormes (1819). A large vessel — such as a carboy — contains the
gas, which in a teaching experiment is usually air (Fig. 10.22). The
CaCI
Fig. 10.22. Clement and Desormes experiment.
256
ADVANCED LEVEL PHYSICS
carboy is well lagged to minimize the exchange of heat with its sur-
roundings. It is attached to a manometer M, and, via a drying-tube,
to a bicycle pump. Its mouth has a large and well-fitting, flap-like,
lid, L. Air is blown in until its pressure is a little above atmospheric,
and time is allowed for the gas to settle down to room temperature.
When it has done so the manometer reading becomes steady, and the
pressure p x of the gas is recorded. The flap- valve is now sharply opened
and closed. The gas makes an adiabatic expansion, and its pressure p 2
is immediately read. With the flap still closed, the gas is then left; it
gradually returns to room
temperature, absolute tem-
perature T t , at constant vol-
ume, and its pressure rises
top 3 .
These changes are shown
in Fig. 10.23. Since some
gas escapes in the expan-
sion, we must consider unit
mass. Its state at the start
of the experiment is repre-
sented by the point A on
the isothermal for T x , its
volume being V v B repre-
AWPi V i
t
D,
i
i
p 3 v 2
• r, room
p 2 V 2
^ 2
V —
Fig. 10.23. Theory of Clement and Disomies.
T 2 , and expanded to V 2 per unit mass.
sents the end of the adiabatic,
when the gas has cooled to
DB is the isothermal for T 2 .
BC represents the return to room temperature. For the adiabatic AB,
we have
or
Pi v = pi vy,
3
Ei
Pi
(33)
2
Vi'
After the gas has returned to room temperature, its representative
point C lies on the same isothermal as A ; therefore
Pi Vi = Pi Vi,
Pi v-
or —
Pa
From equation (33), therefore,
El =IEl\
Pi (P 3 )
log e Pi -log c p 2 = y(log e Pi -log e p 3 )
log c P!-log e p 2
whence
and
y =
log^-log^'
If /i x is the difference in levels of M corresponding to the pressure
GASES
257
p lt and h 2 is the final difference in levels, corresponding to the pressure
p 3 , it can be shown that, to a good approximation, the formula for y
reduces to
y =
K
h 1 -h 2 '
A light oil is used in the manometer M.
VACUUM PUMPS AND GAUGES
The Filter Pump
The simplest pump for evacuating a
vessel is the filter pump, so-called because
it is used for speeding-up filtration (Fig.
10.24). It consists of a nozzle N surrounded
by a chamber C; water rushes from the
nozzle and out of the chamber at the
bottom. The layer of air around the jet
is dragged along with it, and carried out
of the chamber. The lowest pressure
which this pump can produce is the
saturation vapour-pressure of water, about
15 mm mercury at 18°C. (Chapter 14).
It cannot produce What we would nowa-
days call a 'good vacuum'.
The Piston Pump
A piston-type air pump is similar to the
common water pump, but more accurately
made. Its plunger has a greased leather
washer, W in Fig. 10.25, and its valves F t
and F 2 are flaps of oiled silk.
Fig. 10.24. Filter pump.
Fig. 10.25. Piston-type vacuum pump.
To develop a simple theory of the pump, we first assume that, when
the piston is pushed right in, there is no space between it and the
bottom of the barrel. We suppose that the pump is connected to a
vessel of volume V , that the piston displaces a volume V P , and that
the pressure in the vessel is p when the piston is right in, at the start
of the evacuation. When the piston is pulled right out, the volume of
the air originally in the vessel increases to V + V P . Since the action
258 ADVANCED LEVEL PHYSICS
is slow, we may assume that the expansion is isothermal; the pressure
p t after expansion is therefore given by
Pi(V +V P ) = p V ,
whence Pl = j^jp.
When the piston is pushed in again, the valve F t closes, and the air
in the vessel remains at the pressure p x . The second out-stroke then
reduces the pressure to
y o l„ / y o
H = W^W' = w^ p °-
Similarly, after n strokes, the pressure is reduced to
"" = \VoWr' "°-
According to this theory, the final pressure tends to zero as the
number of strokes tends to infinity. In practice, however, a pump has
a limiting pressure. This is due to the fact that the piston can never in
practice be brought right down to the valve F x , so that there is a
residual volume, or dead-space, v, between the piston and the bottom
of the barrel. Air can escape through the valve F 2 only when the pressure
in the volume v is greater than atmospheric. And air can pass from the
vessel through F x only when the pressure in the barrel is less than the
pressure in the vessel. Thus the limiting pressure, p^, is the pressure
which v cm 3 of gas, at atmospheric pressure, exert when expanded
to Vp cm 3
. That is to say
Poo^P = PatmosP,
whence
_ V
P co t/ Patmos.
The ratio v/V p may be about 1/1000, so that p^ is about 1 mm
mercury. Piston pumps of more elaborate designs can give better
vacua than this, about 001 mm mercury; they have all been made
obsolete, however, by the rotary pump.
Rotary Pumps
Fig. 10.26 illustrates one form of rotary vacuum pump. It consists of
a rotor, R, which turns eccentrically in a casing, C, being a close fit at
the ends and along the line A. The rotor carries two scraping blades
B ls B 2 , separated by a strong spring. The vessel to be avacuated is
connected to the inlet port I, and the outlet port O is fitted with a
valve N. As the rotor turns, the volume V x increases, so that air expands
from the vessel into the pump. When the blades are in the positions
shown, the air in the space V 2 is being compressed; when its pressure
rises to atmospheric, the valve N opens, and the air passes out through
O. As the blade B 2 crosses the seat of the valve N, the valve closes
GASES
259
Fig. 10.26. Rotary vacuum pump.
because the air in V 3 is below atmospheric pressure. Thus atmospheric
air cannot blow back into the pump. The lines of contact, A, D, E are
made airtight by a film of oil. All the working parts of the pump are
enclosed in a tank T containing oil. When the pump is at rest, the oil
seeps back through the outlet valve and fills the working space; but
the first revolutions of the pump sweep out the excess oil, and leave
just the necessary film over the metal surfaces.
A single rotary pump will give an ultimate pressure of about 001
mm mercury. Very often two pumps are housed in the same tank of
oil, and driven off the same shaft; they are connected in cascade, and
may give an ultimate pressure of less than 0001 mm mercury.
The MacLeod Gauge
The MacLeod Gauge is a gauge used for measuring pressures below
a few mm mercury; these are pressures which cannot be measured
accurately on a U-tube manometer. It consists of a bulb B, connected
to a mercury reservoir M and terminated in a capillary tube T (Fig.
10.27 (a)); just below the bulb a branch-tube, P, leads to the vacuum
system under test. It also carries a branch-tube, D, which is a capillary
of the same bore as T. A millimetre scale, S, lies underneath T and D.
To measure the pressure in a vacuum system, the reservoir M is
lowered until the mercury falls below the branch-point C. The air in
B is then at the unknown pressure, p. The reservoir is now slowly
raised. As soon as the mercury closes the branch at C, the air in B
starts to be compressed. M is raised until the mercury in B just reaches
the foot of the capillary T. The height h of the mercury in D, above
that in B, is then measured. The purpose of having equal bores for
T and D is to equalize surface tension effects in each.
If the pressure p is expressed in mm mercury then the pressure of
the air trapped in T is p + h. The volume of this air is the volume v
of the capillary T. At the moment when the mercury passed the point
C, this air had the pressure p, and the volume V+v, where V is the
260
ADVANCED LEVEL PHYSICS
Fig. 10.27. MacLeod gauge.
volume between C and the base of T. The compression is slow enough
to be isothermal, so that
Hence
and
p(V+v) = (p + h)v
pV = hv,
p = — h.
V
Another way of using the gauge is to raise the reservoir M until the
mercury in D is level with the top of T (Fig. 10.27 (b)). Then if s is the
cross-section of T, the volume of trapped air is hs. And if / is the whole
length of T, its volume is Is. Therefore, as before,
or
whence
p(V+ls) = {p+h)hs,
p(V+ls-hs) = h 2 s,
h 2 s
P =
V+(l-h)s
The term (l — h)s in the denominator is usually negligible compared
with V, so that
h 2 s
Because p is proportional to h 2 , the gauge can cover a wider range of
pressures when it is used in this way, than when it is used in the way
first described. But for the same reason it is less accurate. In practice
GASES 261
the second way is generally chosen, and the scale S is calibrated to
read the pressure p directly.
EXAMPLES
t The density of a gas is 10775 kg m~ 3 at 27°C and 10 5 N m -2 pressure
and its specific heat capacity at constant pressure is 0846 kJ kg" 1 K" 1 . Find the
ratio of its specific heat capacity at constant pressure to that at constant volume.
The gas constant per kg of gas is given by
pV 10 5 xl
R -T-l-775x300 Jkg K
since V = 1 m 3 /l-775, T = 273 + 27 = 300 K. Converting J to kJ,
• f? 1Q5xl 1 f Tkc- 1 K- 1
• K " 1-775 x 300 x 1000 kJKg '
Now c p —c v = R
.-. c v = 0-846-0188 = 0-658 kJkg -1 K _1
c £ = : 846 =
• • y c v 0-658 y -
This value for y suggests that the gas is polyatomic (see p. 246).
2. An ideal gas at 17°C has a pressure of 760 mm mercury, and is compressed
(i) isothermally, (ii) adiabatically until its volume is halved, in each case reversibly.
Calculate in each case the final pressure and temperature of the gas, assuming
c p = 21, <v= 1-5 kJkg-'KT 1 .
(i) Isothermally, pV = constant.
.-. px~ = 760xF
.'. p = 1520 mm mercury.
The temperature is constant at 17°C.
(ii) Adiabatically, pV y = constant, and y = 2-1/1-5 = 1-4.
M 1 ' 4
.-. pxl-l =760xK 1 ' 4
.. p = 760 x 2 1 ' 4 = 2010 mm mercury.
Since TV y ~ l = constant,
Wo -4
Txl^j =(273 + 17)xF°- 4
.-. T=290x2 04 = 383 K.
.'. temperature = 110°C.
3. State the laws of gases usually associated with the names of Boyle, Charles,
Dalton and Graham. Two gas containers with volumes of 100 cm 3 and 1000 cm 3
respectively are connected by a tube of negligible volume, and contain air at a
262 ADVANCED LEVEL PHYSICS
pressure of 1000 mm mercury. If the temperature of both vessels is originally
0°C, how much air will pass through the connecting tube when the temperature
of the smaller is raised to 100°C? Give your answer in cm 3 measured at 0°Cand
760 mm mercury. (L.)
First part. Boyle, Charles, Dalton, Graham, see text.
Second part. The pressure is 1000 mm mercury when the temperature is 0°C
(273 K). Let the density of air under these conditions be p v Let the volumes of
the large and small vessels be V and V 1 ; then the mass of air in the two vessels is
M = (V+V 1 )p 1 = (1000 + 100)^! = 1100 Pl . (i)
When the smaller vessel is heated, the pressure throughout the system rises to p,
say. Let p 2 be the density of the air in the smaller vessel; then, by equation (9),
p. 227:
A = Rx373;— = Rx273 . . . (ii)
Pi Pi y '
p 2 _ 273 p
■'■ p7 = 373 X .IOOO-
_273 p
■■ p2 ~m x woo Pv
In the larger vessel, the temperature of the air does not change ; therefore the
density of the air in the larger vessel, p$, is
_ P
P3_ Iooo pl -
The total mass of air, which is unchanged, is therefore
M = Vp 3 + V 1 p 2 = lOOO/^ + lOO^
= 1000 ^+ioo? 73 PPi
1000 373 1000
273
= 1+ \mo\ PPi
Hence, by equation (i),
noo Pl = [i + ^p Pl ,
a 3730x1100 ..__
and p = — — = 1025 mm mercury.
4003
The mass which flows out of the smaller vessel is
^(Pi-P2)= 100 Pl 11-^1
mp ^-W3*ma
--^-is-a • • • • *>
The volume of this mass, at 0°C and 760 mm mercury, is
V = —
GASES 263
where p 4 is the density of air at this temperature and pressure.
From the equation of state,
760 /
— ^=Rx273;
P4
therefore
or
Pl _ 1000
pX ~ 760 '
1 = 100
Pa 76pi"
t. /•■•* x, »» 100 x 100 f 273 1025
Hence, by (m), V = - = -^_(l-_x —
= 33 cm 3 .
4. Distinguish between isothermal and adiabatic changes. Show that for an
ideal gas the curves relating pressure and volume for an adiabatic change have a
greater slope than those for an isothermal change, at the same pressure.
A quantity of oxygen is compressed isothermally until its pressure is doubled.
It is then allowed to expand adiabatically until its original volume is restored.
Find the final pressure in terms of the initial pressure. (The ratio of the specific
heat capacities of oxygen is to be taken at 1-40.) (L.)
First part. An isothermal change is one made at constant temperature; an
adiabatic change is one made at constant heat, that is, no heat enters or leaves the
system concerned.
For a reversible isothermal change, pV = k, or p = k/V. By differentiation,
the slope, dp/dV, = -k/V 2 = -pV/V 2 = -p/V.
For a reversible adiabatic change, pV = c, or p = c/V y . By differentiation,
we find the slope, dp/dV, = —yp/V.
.'. ratio of adiabatic slope to isothermal slope = y.
Since y is always greater than 1, the adiabatic slope is greater than the isothermal
slope.
Second part. Let p , V = the original pressure and volume of the oxygen.
Since pV = constant for an isothermal change,
y
.'. new volume = -£- when new pressure is 2p .
Suppose the gas expands adiabatically to its volume V , when the pressure is p.
il-4
Then pxjy' -2p x
•'• P = 2p xl-J =0-8p .
5. Derive an expression for the difference between the specific heat capacities
of an ideal gas and discuss the significance of the ratio of these two specific heat
capacities for real gases.
Assuming that the ratio of the specific heat capacities of hydrogen is 1*41 and
that its density at s.t.p. is 0-0900 kg m~ 3 , find a value for its specific heat capacity
at constant volume in J kg - x K~ 1 .
What explanation can you suggest for the small difference between the specific
heat capacities ofa solid? (Standard atmospheric pressure = 1013x 10 5 Nm -2 .)
(N.)
First part. The expression required is c p —c v = R, discussed on p. 245. The
264 ADVANCED LEVEL PHYSICS
2
ratio, y, of these two specific heats = 1 + - on the kinetic theory of gases, where
n is the number of degrees of freedom of the molecules. For a monatomic gas
n = 3, so that y = 1-66; for a diatomic gas n = 5 usually, so that y = 1-4; for
triatomic gases y is less than 1-4, e.g. 1-29. Thus y gives information about the
number of atoms in the molecule of the gas.
Second part. c p -c v = R, where R may be in J kg -1 K _1 and c p , c v are in
the same units. Since 009 kg occupies 1 m 3 and p = 1013 x 10 5 newton m~ 2 ,
then, from pV = RT,
P V (1-013 x 10*) xl
K_ T ~ 273x009 Jkg K
= 412kJkg- 1 K~ 1
.-. c p -c K = 4-12 (i)
But ^ = 1-41 (ii)
.'. c p = 141 c v . Substituting for c p in (i),
.-. 1-41 Cy-Cy = 4-12 = 041 Cy.
'■ c v = ^= lOOkJkg^K- 1 .
Third part. The difference in the specific heat capacities of a solid is proportional
to the external work done in expansion. But the expansion of a solid is small.
Consequently the difference in specific heat capacities of the solid is small.
EXERCISES 10
Gas Laws— Specific Heat Capacities of Gases
1. State Boyle's law and Charles' 1 law, and show how they lead to the gas
equation PV = RT. Describe an experiment you would perform to measure the
thermal expansion coefficient of dry air.
What volume of liquid oxygen (density 1 140 kg m ~ 3 ) may be made by liquefying
completely the contents of a cylinder of gaseous oxygen containing 100 litres of
oxygen at 120 atmospheres pressure and 20°C? Assume that oxygen behaves as
an ideal gas in this latter region of pressure and temperature.
[1 atmosphere = 101 x 10 5 newton metre -2 ; gas constant = 8-31 joule
mol" l K~ * ; molecular weight of oxygen = 320.] (O. & C.)
2. Give brief accounts of experiments which illustrate the relationship
between the volume of a fixed mass of gas and (a) the pressure it exerts at a
fixed temperature, (b) the temperature on a centigrade mercury thermometer at a
fixed pressure. State the two 'laws' which summarise the results.
A gas cylinder contains 6400 g of oxygen at a pressure of 5 atmospheres. An
exactly similar cylinder contains 4200 g of nitrogen at the same temperature.
What is the pressure on the nitrogen? (Molecular weights : oxygen = 32, nitro-
gen = 28 ; assume that each behaves as a perfect gas.) (O. & C.)
3. State Boyle's Law.
Describe how you would verify the law for dry air over a range of pressures
from 0-5 to 1-5 atmospheres. Would the form of the apparatus you describe be
suitable if the working range of pressure was 0-5 to 10 atmospheres? Give reasons
for your answer.
Two glass vessels of equal volume are joined by a tube, the volume of which
may be neglected. The whole is sealed and contains air at S.T.P. If one vessel is
GASES 265
placed in boiling water at 100°C and the other is placed in melting ice, what will
be the resultant pressure of the air? (JV.)
4. The formula pv = mrT is often used to describe the relationship between
the pressure p, volume v, and temperature Tof a mass m of a gas, r being a constant.
Referring in particular to the experimental evidence how do you justify (a) the
use of this formula, (b) the usual method of calculating Tfrom the temperature t
of the gas on the centigrade (Celsius) scale?
Two vessels each of capacity 1-00 litre are connected by a tube of negligible
volume. Together they contain 0-342 g of helium at a pressure of 80 cm of mercury
and temperature 27°C. Calculate (i) a value for the constant r for helium, (ii)
the pressure developed in the apparatus if one vessel is cooled to 0°C and the
other heated to 100°C, assuming that the capacity of each vessel is unchanged. (N.)
5. State Boyle's law and Charles' law and show how they may be combined
to give the equation of state of an ideal gas.
Two glass bulbs of equal volume are joined by a narrow tube and are filled with
a gas at s.tp. When one bulb is kept in melting ice and the other is placed in a hot
bath, the new pressure is 87-76 cm mercury. Calculate the temperature of the
bath. (L.)
6. Describe experiments in which the relation between the pressure and
volume of a gas has been investigated at constant temperature over a wide
range of pressure. Sketch the form of the isothermal curves obtained.
Explain briefly how far van der Waals' equation accounts for the form of
these isothermals. (L.)
7. Describe, with a diagram, the essential features of an experiment to study
the departure of a real gas from ideal gas behaviour.
Give freehand, labelled sketches of the graphs you would expect to obtain on
plotting (a) pressure P against volume V, (b) PV against P for such a gas at its
critical temperature and at one temperature above and one below the critical
temperature.
Explain van der Waals' attempt to produce an equation of state which would
describe the behaviour of real gases.
Show that van der Waals' equation is consistent with the statement that all
gases approach ideal gas behaviour at low pressures. (O. & C.)
8. State Boyle's Law and describe how you would attempt to discover whether
air shows any deviations from the law. Draw an approximate set of curves to
show the way in which a gas deviates from Boyle's Law in the region close to
where the gas liquefies.
Two one-litre flasks are joined by a closed tap and the whole is held at a
constant temperature of 50°C. One flask is evacuated and the other contains air,
water vapour, and a small quantity of liquid water. The total pressure in the
latter flask is 200 mm Hg. The tap is then opened, and the system is allowed to
reach equilibrium, when some liquid water remains. Assuming that air obeys
Boyle's law, find the final pressure in the flasks.
[Vapour pressure of water at 50°C = 93 mmHg.] (O. & C.)
9. Explain why the specific heat capacity of a gas is greater if it is allowed to
expand while being heated than if the volume is kept constant. Discuss whether
it is possible for the specific heat capacity of a gas to be zero.
When 1 g of water at 100°C is converted into steam at the same temperature
2264 J must be supplied. How much of this energy is used in forcing back the
atmosphere? Explain what happens to the remainder of the energy. [ 1 g of
water 100°C occupies 1 cm 3 . 1 g of steam at 100°C and 76 cm of mercury occupies
1601 cm 3 . Density of mercury = 13600 kg m~ 3 .] (C.)
266 ADVANCED LEVEL PHYSICS
10. Define heat capacity and specific heat capacity.
Describe an experiment to determine the specific heat capacity of a gas either at
constant volume or at constant pressure. Point out likely sources of error and
indicate how they may be minimized.
Explain why it is necessary to specify the condition of constant pressure or
constant volume. (L.)
11. Explain why, when quoting the specific heat of a gas, it is necessary to
specify the conditions under which the change of temperature occurs. What
conditions are normally specified?
A vessel of capacity 10 litres contains 130 g of a gas at 20°C and 10 atmospheres
pressure. 8000 joule of heat energy are suddenly released in the gas and raise the
pressure to 14 atmospheres. Assuming no loss of heat to the vessel, and ideal gas
behaviour, calculate the specific heat of the gas under these conditions.
In a second experiment the same mass of gas, under the same initial conditions,
is heated through the same rise in temperature while it is allowed to expand
slowly so that the pressure remains constant. What fraction of the heat energy
supplied in this case is used in doing external work? Take 1 atmosphere =
10 5 newton metre" 2 . (O. & C.)
12. The two specific heat capacities in kJ kg -1 K _1 units for argon are
0-521 and 0-313 and for air are 1012 and 0-722.' Explain these statements and
discuss their significance in relation to (a) the atomicity of the molecules of the
two gases, (6) the relative values of the adiahatic elasticities of argon and air.
Describe an experiment to verify one of the above values of specific heat
capacities. (L.)
13. A litre of air, initially at 20°C and at 760 cm of mercury pressure, is heated
at constant pressure until its volume is doubled. Find (a) the final temperature,
(b) the external work done by the air in expanding, (c) the quantity of heat supplied.
[Assume that the density of air at s.t.p. is 1-293 kg m~ 3 and that the specific
heat capacity of air at constant volume is 0-714 kJ kg - * K~ 1 .] (L)
14. Describe an experiment to determine the specific heat capacity of water,
at about 15°C, deriving from first principles any equations used.
Deduce an expression for the difference between the specific heat capacities
of an ideal gas. If the specific heat capacity of air at constant pressure is 1013
kJ kg -1 K _1 and the density at s.t.p. is 1-29 kg m -3 estimate a value for the
specific heat capacity of air at constant volume. [Assume the density of mercury
at 0°C to be 13600 kg m" 3 .] (L.)
15. Distinguish between an isothermal change and an adiahatic change. In
each instance state, for a reversible change of an ideal gas, the relation between
pressure and volume.
A mass of air occupying initially a volume 2000 cm 3 at a pressure of 76 cm
of mercury and a temperature of 200°C is expanded adiabatically and reversibly
to twice its volume, and then compressed isothermally and reversibly to a volume
of 3000 cm 3 . Find the final temperature and pressure, assuming the ratio of the
specific heat capacities of air to be 1-40. (L.)
16. Explain why the specific heat of a gas at constant pressure is different
from that at constant volume.
The density of an ideal gas is 1-60 kg m -3 at 27°C and 1-00 x 10 5 newton
metre" 2 pressure and its specific heat capacity at constant volume is 0-312
kJkg -1 K 1 . Find the ratio of the specific heat capacity at constant pressure to
that at constant volume. Point out any significance to be attached to the result. (iV.)
17. Explain the meaning of the terms isothermal, adiahatic. What is the im-
portance of the ratio of the specific heat capacity of an ideal gas?
GASES 267
Air initially at 27°C and at 75 cm of mercury pressure is compressed isother-
mally until its volume is halved. It is then expanded adiabatically until its original
volume is recovered. Assuming the changes to be reversible, find the final pressure
and temperature.
[Take the ratio of the specific heat capacities of air as 1-40.] (L.)
Kinetic Theory of Gases
18. Explain what is meant by the root mean square velocity of the molecules
of a gas. Use the concepts of the elementary kinetic theory of gases to derive an
expression for the root mean square velocity of the molecules in terms of the
pressure and density of the gas.
Assuming the density of nitrogen at s.tp. to be 1-251 kg m -3 , find the root
mean square velocity of nitrogen molecules at 127°C. (L.)
19. State the postulates on which the simple kinetic theory of gases is based.
What modifications are made to the postulates in dealing with real gases? How
are these modifications represented in van der Waals' equation? (N .)
20. State the assumptions that are made in the kinetic theory of gases and
derive an expression for the pressure exerted by a gas which conforms to these
assumptions, in terms of its density (p) and the mean square velocity (c 2 ) of its
molecules.
Show (a) how temperature may be interpreted in terms of the theory, (b) how
the theory accounts for Dalton's law of partial pressures. (L.)
21. Calculate the pressure in mm of mercury exerted by hydrogen gas if the
number of molecules per cm 3 is 6-80 x 10 15 and the root mean square speed of
the molecules is 1-90 xlO 3 m s -1 . Comment on the effect of a pressure of this
magnitude (a) above the mercury in a barometer tube ; (b) in a cathode ray tube.
(Avogadro's Number — 6-02 x 10 23 . Molecular weight of hydrogen = 202.) (JV.)
22. Use a simple treatment of the kinetic theory of gases, stating any gump-
tions you make, to derive an expression for the pressure exerted by a gas on the
walls of its container. Thence deduce a value for the root nean square speed of
thermal agitation of the molecules of helium in a vessel at 0°C. (Density of
helium at s.tp. = 01785 kg m -3 ; 1 atmosphere = 1013 x 10 5 N m~ 2 .)
If the total translational kinetic energy of all the molecules of helium in the
vessel is 5 x 10 ~ 6 joule, what is the temperature in another vessel which contains
twice the mass of helium and in which the total kinetic energy is 10 -5 joule?
(Assume that helium behaves as a perfect gas.) (O. & C.)
23. Explain the meaning of the terms ideal gas and molecule.
What properties of a gas such as carbon dioxide distinguish it from an ideal
gas and how may these differences from 'ideal' be demonstrated experimentally?
According to_simple kinetic theory the pressure exerted by a gas of density p
is j pc 2 where c 2 is the mean square molecular velocity. Show how this relation
may be correlated with the equation of state for an ideal gas, PV = RT, explaining
clearly what further assumptions you have to make. (O. & C.)
24. Derive an expression for the pressure (p) exerted by a mass of gas in terms
of itsmolecular velocities, stating the assumptions made. What further assumption
regarding the absolute temperature (T) of the gas is necessary to show that the
expression is consistent with the equation pv = kT where v is the volume of the
mass of gas and k a constant? Which of the assumptions referred to did van der
Waals modify to bring the equation pv = kT more closely in agreement with the
behaviour of real gases and what equation did he deduce? (L.)
25. Explain the following in terms of the simple kinetic theory without mathe-
matical treatment :
268 ADVANCED LEVEL PHYSICS
(a) A gas fills any container in which it is placed and exerts a pressure on the
walls of the container.
(b) The pressure of a gas rises if its temperature is increased without the mass
and volume being changed.
(c) The temperature of a gas rises if it is compressed in a vessel from which
heat cannot escape.
(d) The pressure in an oxygen cylinder falls continuously as the gas is taken
from it, while the pressure in a cylinder containing chlorine remains constant
until very nearly all the chlorine has been used. The contents of the cylinder are
kept at room temperature in both cases.
[Critical temperature of chlorine = 146°C] (C.)
26. (a) State the assumptions of the kinetic theory of gases. How does the
theory represent the temperature of a gas, and how does it account for the fact
that a gas exerts a pressure on the walls of its container? (The derivation of an
expression for the pressure is not required.)
(b) Near K the specific heat capacity of silver, c, is not constant, but obeys
the relation
where T is the absolute temperature and a and ft are constants typical of silver
given by a = 1512 x lO^kJkg' 1 deg~ 4 and£ = 5-88 x lO^kJkg -1 deg -1 . By-
means of a graph, or otherwise, find the heat required to raise the temperature
of 50 g of silver from 1 K to 20 K. (C.)
27. Explain in terms of the kinetic theory what happens to the energy supplied
to a gas when it is heated (a) at constant volume, (b) at constant pressure.
Deduce the total kinetic energy of the molecules in 1 g of an ideal gas at 0°C
if its specific heat capacity at constant volume is 0-60 k J kg -1 K _1 .
An iron rod 1 metre long is heated without being allowed to expand length-
wise. Wjaen the temperature has been raised by 50Q°C the rod exerts a force of
1-2 x 1(F newtons on the walls preventing its expansion. How much work could
be obtained if it wen* possible to maintain it at this temperature and allow it to
expand gradually until free from stress? [Linear expansivity of iron = 10 x
10- 5 K _1 .](C.)..
chapter eleven
Thermal expansion
In this chapter we shall discuss the thermal expansion of solids and
liquids.
SOLIDS
Linear Expansion
Most solids increase in length when
they are warmed. Fig. 11.1 shows a
simple apparatus with which we can
measure the linear expansion of a metal
tube A. We first measure the length of
the tube, l x , at room temperature, t ;
then we screw the spherometer S against
the end of the tube and take its reading,
S v We next heat the tube by passing
steam through it. At intervals we re-
adjust the spherometer; when its reading
becomes constant, the temperature of
the rod is steady. We measure the
temperature 2 on the thermometer B,
and take the new reading of the sphero-
meter, S 2 . The expansion of the tube is
e = S 2 ~S V
The increase in length, A, of unit length
of the material for one degree tempera-
ture rise is then given by
Steam
Fig. 11.1. Linear -expansivity..
x =
expansion
original length x temperature rise /i(^ 2 ~^i)
The quantity k is called the mean linear.expartsivity of the metal, over
the range X to 2 . If this range is not too great — say less than 100°C —
the quantity X may, to a first approximation, be taken as constant. .
The linear expansivity of a solid, like the pressure and volume
coefficients of a gas, has the unit deg C _1 or 'K -1 ' in SI units; its
dimensions- are
w =
[length]
[length] x [temp]
269
= [temp.]
270
ADVANCED LEVEL PHYSICS
From the definition of X, we can estimate the new length of a rod,
/ 2 , at a temperature 2 from the equation
h-hV+Wi-Ox)}, . (1)
where l t is the length of the rod at the temperature lt and X is the
mean value over a range which includes 2 and V
For accurate work, however, the length of a solid at a temperature
must be represented by an equation of the form
/ = l (l + aO + b0 2 + c0 3 + ...), . . . (2)
where / is the length at 0°C, and a,b,c are constants. The constant a
is of the same order of magnitude as the mean coefficient X ; the other
constants are smaller.
MEAN LINEAR EXPANSIVITY
(Near room temperature)
Substance
X,K~ L
Substance
^K 1
xl(T 6
xlO" 6
Copper
17
Bakelite
. 22
Iron ....
12
Brick .
9-5
Brass ....
19
Glass (soda)
8-5
Nickel.
13
Quartz (fused) (0-30°C)
042
Platinum
9
Pine — across grain
c. 0-34
Invar (36% nickel-steel) .
c.0-1
Pyrex .
3
When a solid is subjected to small changes of temperature, about a mean
value 0, its linear expansivity X e in the neighbourhood of 6 may be defined
by the equation
i -±£L
Ae ~ IdO'
where I is the length of the bar at the temperature 9. The following table shows
how the coefficient varies with temperature.
VALUES OF k e COPPER
e
-87
100
400
600
°C
A g
14-1
161
16-9
19-3
20-9
xl0 -o K -i
Accurate Measurement of Expansion
An instrument for accurately measuring the length of a bar, at a
controlled temperature, is called a comparator (Fig. 11.2). It consists
of two microscopes, M v M 2 , rigidly attached to steel or concrete
pillars P l5 P 2 . Between the pillars are rails R l5 R 2 , carrying water-
baths such as B. One of these baths contains the bar under test, X,
which has scratches near its ends; the scratches are nominally a metre
apart. Another water bath contains a substandard metre. The eye-
pieces of the microscopes are fitted with cross-webs carried on micro-
meter screws, m l9 m 2 .
THERMAL EXPANSION
271
First the substandard metre is run under the microscopes, and the
temperature of its bath is adjusted to that at which the bar was cali-
brated (usually about 18°C).
When the temperature of the bar is steady, the eyepiece webs are
adjusted to intersect the scratches on its ends (Fig. 11.2 (b)), and their
micrometers are read The distance between the cross-webs is then
1 metre.
Webs
Bar
Scratch
(b) Field of view
]
[
w
Mi
rri2
Glass
I i — Air (
-:P xT \
1
M
I, .1
n~n
u.
±±- &E±r~£==:=3rfer- -^C^-^^W
\
n
(a) General view
Fig. 1 1 .2. The comparator.
The substandard is now removed, and the unknown bar put in its
place; the temperature of the bar is brought to 0°C by filling its bath
with ice- water. When the temperature of the bar is steady, the eyepiece
webs are re-adjusted to intersect the scratches on its ends, and their,
micrometers are read If the right-hand web has been shifted x mm to
the right, and the left-hand y mm, also to the right, then the length of
the bar at 0°C is
/ = 1 metre + (x — y) mm.
The bath is now warmed to say 10°C, and the length of the bar
again measured. In this way the length can be measured at small
intervals of temperature, and the mean linear expansivity, or the
coefficients a, b, c in equation (2), can be determined.
Expansion at High Temperatures
Fig. 11.3 illustrates the principle of a method for measuring the
expansion of a solid at high temperatures. A is a tube of fused silica,
272
ADVANCED LEVEL PHYSICS
\
Uek
Fig. 11.3. Appar-
atus for measur-
ing expansion at
high tempera-
tures.
having a scale S engraved on the edge of an opening,
H, in its side. B is the specimen, and C is a rod of fused
silica with a second scale S' engraved on it.
The thermal expansion of fused silica is much less
than that of most solids, over a given temperature
range, and has been accurately measured by a method
depending on optical interference. When the apparatus
shown is placed in a high-temperature bath, the
rod C rises by an amount equal to the difference in
expansion of the specimen, and an equal length of
fused silica Its rise is measured by observing the
displacement of the scale S' relative to S through a
microscope.
Force set up when Expansion is Resisted
Consider a metal rod between two supports P, which
we suppose are immovable (Fig. 11.4). Let /j be the
distance between the supports, and X the temperature
at which the rod just fits between them. If the rod
is heated to 2 , it will try to expand to a greater
length l 2 , but will not be able to do so.
The value of l 2 would be
<i
Z
V
Fig. 11.4. Force in bar.
i 2 = i l {i+W 2 -o 1 )},
and the expansion would be
e = l 2 -h = hWi-Oil
where X is the mean linear expansivity of
the rod.
The force which opposes the expansion is the force which would
compress .a bar of natural length l 2 by the amount e. Its magnitude F
depends on the cross-section of the bar, A, and the Young's modulus
of its material, E :
A L l 2
To a very good approximation we may replace l 2 by l x , because
their difference is small compared with either of them. Thus
M 1 {0 2 — 0i)
A \~
h
= EX(0 2 — x ),
:. F = EAX(0 2 -0 1 ).
For steel, A = 12 x 10" 6 K _1 and E = 2x 10 11 newton per m 2 .
If the temperature difference, 2 — x , is 100°C, then, for a cross-
sectional area 5 of 4 cm 2 or 4 x 10 ~ 4 m 2 ,
F = 2 x 10 11 x 12 x 10 -6 x 4 x 10 -4 x 100 newton
= 9-6 x 10 4 newton.
On converting to kgf we find F is nearly 10000 kgf.
THERMAL EXPANSION
273
Expansion of a Measuring Scale
A scale, such as a metre rule, expands
with rise in temperature ; its readings,
therefore, are correct at one tempera-
ture, t say. When the temperature of
the scale is greater than t , the dis-
tance between any two of its divisions
increases, and its reading is therefore
too low (Fig. 11.5); when the scale is
below X , its reading is too high. Let
us suppose that, at lt the distance
between any two points P, Q, on the
scale is l x cm. At 2 it is
l 2 = l l {\+X{0 2 -0 l )},
where X is the mean linear expansivity
of the material of this scale. According
to the divisions on the scale, however,
the distance between P and Q will still
be l x cm. Thus
Cool
4
Warm
4
Correct
4
1
Q
A
OJ [0
Fig. 11.5. Expansion of a
measuring scale.
true distance at 2 = scale value x {1 + X(0 2 — 0\)}
(3)
zZA
If a sheet of material with a hole in it is warmed, it expands, and
the hole expands with it. In Fig. 11.6, A
represents a hole in a plate, and A' repre-
sents a plug, of the same material, that fits
the hole. If A and A' are at the same tem-
perature, then A' will fit A, whatever the
value of that temperature ; for we can always
imagine A' to have just been cut out, with-
out loss of material. It follows that the
expansion of the hole A, in every direction, is
the same as the expansion of the solid plug A'.
Fig.
11.6. Expansion of a
hole.
Differential Expansion
The difference in the expansions of dif-
ferent materials is used in practical arrange-
ments discussed shortly. Fig. 11.7 (a) shows
two rods AB, AB', of different metals,
rigidly connected at A. If / t , l\, are their
lengths at a temperature lf their difference
l 2i
is
d l = BB' = l x -V x
B
f-i
If X and X' are the mean linear expansivities
of the materials of the rods, then the lengths
of the rods at (9, are
*i
"}
B'
Mi
Fig. 11.7a. Differential
expansion.
274 ADVANCED LEVEL PHYSICS
i 2 = ii{i+W 2 -Oi)h
r 2 = r 1 {i+x'{$ 2 -e 1 )}.
The distance between their ends is now
d 2 = i 2 -i> 2 = i 1 -r 1 +(i^-r 1 x , )(e 2 -e 1 ) t
or d 2 = d l +(l l A-l' l X)(0 2 -0i) .... (4)
By a suitable choice of lengths and materials, the distance BB' can
be made to vary with temperature in any one of the following ways :
(1) The bar AB' is made of invar, a nickel-steel whose linear expan-
sivity is very small (p. 270). The point B' then does not move with
changes in temperature. In equation (4) we neglect X', and find :
d 2 = d l + l 1 X(9 2 -0 1 ).
Thus the short distance BB' expands by the same amount as the long
distance AB. Consequently,
Invar the relative expansion of BB'
-f 1 with temperature is much
\
greater than that of a bar
J \b- ^ B of length d v
Gas' Copper tube (2) AB is made of invar, so
Fig. 11.7b. Thermostat principle. that X is negligible. The point
B then does not move, and
the distance BB' shrinks rapidly as the temperature rises. This principle
is used in the thermostats used to maintain gas ovens at constant
temperatures (Fig. 11.7 (b)).
(3) The lengths and materials are chosen so that
/; ~ A'
or l x X = l\X.
Then, by equation (4), d x or BB' does not change with temperature.
This principle is used in compensating clock pendula for temperature
changes (p. 276).
Bimetal Strip
Fig. 11.8 (a) shows two strips of different metals, welded together
along B, called a bimetal strip. The metal M t has a greater linear
expansivity than the metal M 2 . Therefore, when the strip is heated,
Mi will expand more than M 2 , and the strip will curl with M x on
the outside. The reverse is true when the strip is cooled, as M x then
shrinks more than M 2 (see Fig. 11.8(a)).
Bi-metal strips are used in electrical thermostats for ovens, irons,
laboratories, etc. The strip carries a contact, K in Fig. 11.8 (b), which
presses against another contact K' on the end of an adjusting screw S.
When the strip warms, it tends to curl away from K'; the temperature
at which the contact is broken can be set by turning the screw. When
THERMAL EXPANSION
Cold
275
Mi
B
rr>
l fi
(a) Principle
Leads to circuit
(a)
a
Y
,_L
(b) Application to thermostat
Fig. 11.8. Principle of gas thermostat:
the contacts open, they switch off the heating current. If the heating
current is too great to be controlled by the contacts KK', it is switched
off by a relay, which is controlled by the contacts on the bimetal.
Let us now estimate the *"* ' *"•
deflection of a heated bimetal
strip. We suppose that the
component strips have the
same thickness d, and the
same length / at a tempera-
ture t (Fig. 11.9 (a)). When
they are heated, they are dis-
torted, but to a first approxi-
mation we may assume that
the mid-line of each, shown
dotted, has the length which
it would naturally have (Fig.
11.9 (b)). The difference in
length of the mid-lines, p, is
then the difference in their
expansions. At a temperature
p = tt(e 2 -e 1 )-w(d 2 -e 1 )
The difference is taken up by
the curvature of the strip. If
a is the angle through which
it bends, then, from the figure,
<x = ^ = t(X-X')(0 2 -0i)-
Fig. 11.9. Expansion of bimetal strip
(exaggerated).
(The expansion of d is negligible, to a very good approximation.)
To find the radius r of the arc formed by the strip, we assume that the length of
the arc is /; this we may do because the expansions in length are all small. Then
I
a = -;
r
1 = L
r d
{k-x%e 2 -e 1 x
276 ADVANCED LEVEL PHYSICS
and, by the above equation,
whence r =
{k-k%e 2 -e 1 y
The deflection y of the end of the strip is given by the approximate equation
2ry = I 2 ,
from the geometry of the circle.
Thus
y =2r =
i 2 i 2 (k-A')(e 2 -di)
2d
Temperature Compensation
The rate of a clock varies considerably with temperature, unless
arrangements are made to prevent its doing so. If the clock is governed
by a pendulum, the length of the pendulum increases with temperature,
its period therefore also increases, and the clock loses. A clock governed
by a balance-wheel and hair-spring also loses as the temperature rises.
For, as the temperature rises, the spring becomes less stiff, and the
period of the balance-wheel increases. Also the spokes of the wheel
expand a little, increasing the moment of inertia of the wheel, and
thus further increasing its period.
A balance-wheel clock may be compensated against temperature
changes by making the circumference of the
wheel in the form of two or three bimetal
strips, as shown in Fig. 11.10. The strips
carry small weights W, to give the wheel
the necessary moment of inertia. As the
temperature rises, the strips curl inwards,
and bring the weights nearer to the axle;
thus the moment of inertia of the wheel
decreases. In a correctly designed timing-
system, the decrease in moment of inertia
just offsets the decrease in stiffness of the
spring, and then the period of the balance-
wheel does not change with temperature.
Many modern watches are not compensated. Their balance-wheels
are made of invar, which, as we have seen, has a very small coefficient
of expansion. Another nickel-steel, of slightly different composition,
is used for their hair-springs. This alloy changes its elasticity very
little with temperature, and is called elinvar. The combination of
invar balance-wheel and elinvar hair-spring gives a rate which is
nearly enough independent of temperature for everyday purposes.
Pendula
To compensate a pendulum clock against changes of temperature,
the pendulum must be so made that its effective length remains con-
Fig. 11.10. Bimetal balance
wheel.
THERMAL EXPANSION
277
stant. Fig. 11.11 (a) shows one way of doing this, in the so-called grid-
iron pendulum (Harrison, 1761). Brass and steel rods are arranged so
that the expansion of the brass rods raises the bob B of the pendulum,
while the expansion of the steel rods lowers it. As explained on p. 274,
the expansions can be made to cancel if
here l B and / s are the total lengths of brass and steel respectively, and
A g , A s are their linear expansivities.
Fig. 11,11 (b) shows the same principle applied to a pendulum with a
wooden rod and a cylindrical metal bob. To a first approximation the
effective length of the pendulum, /, is the distance from its support
to the centre of gravity, G, of the bob. The condition for constant
period is then
length of rod A wood
\ length of bob X metaf
The bob may be made of lead or zinc, either of which is much more
expansible than is wood, along its grain.
l
-Brass
^ Steel
.Wood
B
(a) Grid-iron (b) Wood and metal
Fig. 11.11. Compensated pendula.
Metal-Glass Seals
In radio valves and many other pieces of physical apparatus, it is
necessary to seal metal cones into glass tubes, with a vacuum-tight
joint The seal must be made at about 400°C, when the glass is soft;
as it cools to room temperature, the glass will crack unless the glass
and metal contract at the same rate. This condition requires that
the metal and the glass have the same linear expansivity at every
temperature between room temperature and the melting-point of
278 ADVANCED LEVEL PHYSICS
glass. It is satisfied nearly enough by platinum and soda glass (mean
X = 9 and 8-5 x 10 ~ 6 per deg Q respectively), and by tungsten and
some types of hard glass similar to pyrex (mean linear expansivity
X = 3-4xl(T 6 perdegC).
Modern seals through soft glass are not made with platinum, but
with a wire of nickel-iron alloy, which has about the same linear
coefficient as the glass. The wire has a thin coating of copper, which
adheres to glass more firmly than the alloy. Also, being soft, the copper
takes up small differences in expansion between the alloy and the glass.
In transmitting valves, and large vacuum plants, glass and metal
tubes several centimetres in diameter must be joined end-to-end. The
metal tubes are made of copper, chamfered to a fine taper at the end
where the joint is to be made. The glass is sealed on to the edge of the
chamfer ; the copper there is thin enough to distort, with the difference
in contraction, without cracking the glass.
Superficial Expansion
The increase in area of a body with temperature change is called the
superficial expansion of the body. A rectangular plate, of sides a, b, at
a given temperature, has an area
Sj = ab.
If its temperature is increased by its sides become a(l + X0), b{\ + X0)
where X is its mean linear expansivity. Thus its area becomes
S 2 = a(l+X0)b(l + X0)
= S 1 (1+X0) 2
= S l (l+2X0 + X 2 2 ).
In this expression, the term X 2 2 is small compared with 2X0; if X is
of the order of 10" 5 , and of the order of 100, then X0 — 10 -3 , and
X 2 2 ^= 10~ 6 . Therefore we may neglect X 2 2 and write
S 2 = S 1 (l + 2;i0).
The superficial expansivity of the material of the plate is defined as
increase of area _ S 2 — S t
original area x temp, rise S t
Its value is hence equal to 2X, twice the linear expansivity. A hole in a
plate changes its area by the same amount as would a plug that fitted
the hole.
Cubic Expansivity
Cubic expansivity is expansion in volume. Consider a rectangular
block of sides a, b, c, and therefore of volume
V t = abc.
THERMAL EXPANSION
279
If the block is raised in temperature by 6 its sides expand, and its
volume becomes
V 2 = a(l +W) b(l + Xe) c(l+Xd) = abcil+XO?
= abc(\ + 3>X6 + 3X 2 2 + A 3 3 ).
Since X 2 6 2 and A 3 3 are small compared with kd, we may in practice
neglect them. We then have
V 2 = abc(l + 3W) = V x {\ + ZX0).
The cubic expansivity of the solid is defined as
_ increase in volume
original volume x temp, rise
= 3A, to a very good approximation.
Thus the cubic expansivity is three times the linear expansivity.
By imagining a block cut out of a larger block, we can see that the
cubical expansion of a hollow vessel is the same as that of a solid plug
which would fit into it.
LIQUIDS
Cubic Expansivity
The temperature of a liquid determines its volume, but its vessel
determines its shape. The only expansivity which we can define for
a liquid is therefore its cubic expansivity, y. Most liquids, like most
solids, do not expand uniformly, and y is not constant over a wide
range of temperature. Over a given range 9 X to 9 2 , the mean coefficient
y is defined as
V 2 -V x
7 =
v x {6 2 -ej
where V x and V 2 are the volumes of a given mass of liquid at the tem-
peratures t and 2 .
MEAN EXPANSIVITIES OF LIQUIDS
(Near room temperature)
Liquid
7(K _1 )
Liquid
7(K _1 )
xl(T 4
xl(T 4
Alcohol (methyl)
12-2
Water: 5-10°C
053
Alcohol (ethyl) .
110
10-20°C
1-50
Aniline
8-5
20-40°C
302
Ether (ethyl)
16-3
4O-60°C
4-58
Glycerine .
5-3
60-80°C
5-87
Olive oil .
7-0
Mercury: 0-30°C
1-81
Paraffin oil
90
0-100°C
1-82
Toluene
10-9
0-300°C
1-87
280
ADVANCED LEVEL PHYSICS
True and Apparent Expansion: Change of Density
If we try to find the cubic expansivity of a liquid by warming it
in a vessel, the vessel also expands. The expansion which we observe
is the difference between the increases in volume of the liquid and the
vessel. This is true whether we start with the vessel full, and catch the
overflow, or observe the creep of the liquid up the vessel (Fig. 11.12).
The expansion we observe we call the apparent expansion ; it is always
less than the true expansion of the liquid.
Apparent
expansion
6
5
4
•*_=-
-"■!-
3
~-^— :
~T™-.
2
~tL
-- -
1
': —
-_r
T'i
I
\
Apparent
expansion
(a) Overflow
(b) Rise in vessel
Fig. 11.12. Apparent expansion of a liquid.
Most methods of measuring the expansion of a liquid, whether true
or apparent, depend on the change in density of the liquid when it
expands. We therefore consider this change, before describing the
measurements in detail. The mean true or absolute expansivity of a
liquid y, is defined in the same way as the mean cubic expansivity of a
solid :
y =
increase in volume
initial volume x temperature rise'
Thus, if V x and V 2 are the volumes of unit mass of the liquid at 6 l
and 2 , then
v 2 = Pi{i + y(0 2 -0i)}.
The densities of the liquid at the two temperatures are
J_
1
Pi =
Pi
V 2 '
so that
J_ = J_ { l + y( 2 _0 i)}
Pi Pi
or
Pi =
Pi
1 + 7(02-0!)
(5)
Measurement of True (Absolute) Expansivity
The first measurement of the true expansion of a liquid was made
by Dulong and Petit in 1817. A simple form of their apparatus is
THERMAL EXPANSION
281
shown in Fig. 11.13. It consists
of a glass tube ABCD, a foot
or two high, containing mer-
cury, and surrounded by glass
jackets XY. The jacket X con-
tains ice-water, and steam is
passed through the jacket Y.
For the mercury to be in
equilibrium, its hydrostatic
pressure at B must equal its
hydrostatic pressure at G Let
h be the height of the mercury
in the limb at 0°C and p
its density; and let h and p
be the corresponding quanti-
ties at the temperature of
the steam.
Stirrer
Fig. 11.13. Apparatus for true (absolute)
expansivity of mercury.
Then gp h = gph,
where g is the acceleration of gravity.
P K
Hence
But, by equation (5),
Po
P_
Po
1
l + y0'
. K
'• h
1
l + y0
he
+ h o y0
V
= h.
h-h
or
• ' ' h e ■
The height h—h is measured with a cathetometer (a travelling tele-
scope on a vertical column).
This simple apparatus is inaccurate because :
(i) the expansion of CD throws BC out of the horizontal ;
(ii) the wide separation of A and D makes the measurement of
(h — h ) inaccurate ;
(iii) surface tension causes a difference of pressure across each free
surface of mercury; and these do not cancel one another,
because the surface tensions are different at the temperatures
of the hot and cold columns.
Regnault got round these difficulties with the apparatus shown,
somewhat simplified, in Fig. 11.14. The points B and G are fixed at the
same horizontal level, the join DE is made of flexible iron tubing, and
282
ADVANCED LEVEL PHYSICS
the difference in height
between its ends, h 2 , is
measured. The parts
AB, GH are at room
temperature 6 X ; and to
a fair approximation,
the average tempera-
ture of DE is also V
Suppose is the steam
temperature.
If the density of mer-
cury at t , 6 is p lt p, re-
spectively then equating
the pressures on both
sides at the horizontal
level of E, we have
gpihi+gpoho + gp^
= gph + gp t h 3 .
Therefore*
Po h i
Room Q y °C
Fig. 11.14. Modified apparatus for true expansivity
+ PoK +
Po h :
Po h , Po h 2
■+
(6)
1 + 70! " ™" u " l+y0! l+7# l + y0i "
The uncertainty in the temperature of DE is not important, since the
height h 2 is very small. Equation (6) gives
h h 3 — h x
+
(7)
1+ydj. 1+70 ' 1+70J '
Equation (7) can be solved for y: the quantities which need to be
known accurately are h , h, and the difference h 3 — h v This difference
can be measured accurately, because AB and GH are close together;
and because they are at the same temperature there is no error due to
surface tension. The heights h and h are 1 or 2 metres, and so are
easy to measure accurately.
Callender and Moss used six pairs of hot and cold columns, each
2 metres long, to increase the difference in level of the liquid. In this
way they avoided the complication due to density change of the liquid
under high pressure. All the hot columns were beside one another in
a hot oil bath kept at a constant high temperature, while the cold
columns were similarly placed in a bath of melting ice. Platinum
resistance thermometers were used to measure the temperatures, and
a cathetometer to measure the heights and difference in level.
Strictly, equation (6) should be written
Po h i a.nU j. Po h z
+ p h +-
- Po h I P0 h 3
1+yA ' '" " l + yA 1+yO 1+yA
where y x is the mean coefficient between 0°C and d u and y is that between 0°C and 6.
This equation can be solved for y by successive approximations. (See Roberts-Miller,
Heat and Thermodynamics, Chapter X. Blackie.)
THERMAL EXPANSION
283
Apparent Expansion: Weight Thermometer Method
The method of balancing columns for the absolute expansivity of a
liquid is slow and awkward; it has only been applied to mercury.
Routine measurements are more conveniently made by measuring the
apparent expansion; from this, as we shall see, the absolute expansion
can be calculated.
A weight thermometer is a bulb of fused quartz fitted with a fine
stem (Fig. 11.15). It is filled with liquid at a low temperature, and then
warmed; the liquid
G
HK
(a)
Common form
(b)
?
YJ
Pyknometer form
filled by suction
Fig. 11.15. Weight thermometers.
ex-
pands, and from the mass
which flows out the ap-
parent expansion of the
liquid can be found. The
weight thermometer is fil-
led by warming it to expel
air, and then dipping the
stem into the liquid. The
process has to be repeated
many times; and for ac-
curate work the liquid in
the thermometer should be boiled at intervals during the filling, to
expel dissolved air. In a laboratory experiment, a glass density bottle
may be used, filled in the usual way.
The weight thermometer must be filled at a temperature slightly
below the lower limit, 9 X , of the range over which the expansion is to
be measured. It is then kept in a bath at 9 X , until no more liquid flows
out. Next it is weighed, and from its known mass when empty the
mass of liquid in it is found. Let this be m x . The weight thermometer is
then placed in a bath at the higher temperature of the range, 2 , and the
mass remaining in it, m 2 , is found by weighing.
If V x and V 2 are the volumes of the weight thermometer at 6 t and 2 ,
then
V 2 = V X {1 + U(9 2 -9 X )},
where X is the linear expansivity of quartz, the material of the weight
thermometer. Now, if p denotes the density of the liquid, whose
cubic expansivity is y, we have
m x = V lPx ,
Ml = V2P2,
and
p 2 1
Pi l+y(9 2 -9 x Y
Therefore
m 2 _ V 2P2
m x V x p x
1+3/1(02-0!)
i + y(0 2 -0i)
284 ADVANCED LEVEL PHYSICS
Hence m 2 + m 2 y(0 2 -OJ = m t + 2>km 1 {0 2 - OJ
or m 2 y(6 2 — 6 1 ) = m 1 —m 2 + 3A,m 1 (O 2 — l )
and y=— j- a — 7t\ + 3X • • ( u )
If we ignore the expansion of the solid, we obtain the apparent
expansivity of the liquid, y a , relative to the solid.
m 1 -m 2
llu " y " = ^k^7) • • ■ • < l2 >
The expression for the apparent expansivity may be expressed in words :
mass expelled
y = — .
mass left in x temp, rise
Attention is often drawn to the fact that the mass in the denominator
is the mass at the higher temperature, and not the lower.
From equation (11),
m i MO X
y = ya+-^y g .... (i3a)
where y g = 3 A = the cubical coefficient of the solid. Since m t and m 2
are nearly equal, then, to a good approximation,
y = y a +y 9 .... (i3b)
Thus after the apparent coefficient has been calculated, the absolute
coefficient is given to a good approximation by
absolute expansivity = apparent expansivity + cubic expansivity of vessel.
Coefficient of Weight-thermometer Material
The cubic expansivity of a glass weight-thermometer may not be
accurately known, because glasses differ considerably in their physical
properties. Also the expansivity may be changed when the glass is
heated in the blowing of the bulb. The expansivity can conveniently be
measured, however, by using the weight thermometer to find the
apparent value for mercury; and then subtracting the value found
from the known value of the absolute expansivity of mercury.
Expansion of a Powder
The weight thermometer can be used to find the cubic expansivity
of a granular or powdery solid, such as sand. The procedure is the
same as in finding the relative density of the solid, but is gone through
at two known temperatures. From the change in relative density, and
a knowledge of the expansions of the liquid and the weight thermo-
meter, the change in absolute density of the powder can be found,
and hence its cubic expansivity.
THERMAL EXPANSION 285
The Dilatometer
A dilatometer is an instrument for rapidly — but roughly — measuring
the expansion of a liquid. It consists of a glass bulb B, with a graduated
stem S (Fig. 11.16). The volume V b of the bulb, up to the zero of S, is
known, and S is graduated in cubic millimetres or other small units.
The volume of the bulb, and the value of one scale division, vary with
temperature ; the dilatometer therefore measures apparent expansion.
Fig. 11.16. Dilatometer.
Fig. 11.17. Compensated dilatometer.
The dilatometer is filled with the liquid under test to a point just
above the zero of the stem, at a temperature V The volume V t of the
liquid is found by adding V b to the stem-reading v v Next the dilato-
meter is warmed to 2 , and the liquid rises to v 2 . Then (v 2 — vj is the
apparent expansion of the liquid, and hence
If y g is the cubical coefficient of the glass, then
y = y a +y 9 -
For demonstration work, a dilatometer can be compensated so that
it shows roughly the true expansion of a liquid. Mercury is introduced
into the bulb, until it occupies l/7th of the bulb's volume (Fig. 11.17).
The expansion of the mercury is then about equal to the expansion of
the glass, so that the free volume in the bulb is roughly constant. The
cubic expansivities of mercury and glass are given respectively by
y Hg = 18-lxlO- 5 K- x
and y g = 3A g = 3x8-5xl0 -6 = 2-55 xl(T 5 K _1 .
Thus ^ = ^ = 7-1.
y 9 2-55
Thus the expansion of the mercury offsets that of the glass, within
about \\ per cent.
The space above the mercury, whose volume is constant, is filled
286
ADVANCED LEVEL PHYSICS
with the liquid to be examined When the bulb is warmed, the move-
ment of the liquid up the stem shows the liquid's true expansion. This
device may be used to show the anomalous expansion of water (p. 288).
Correction of the Barometer
The hydrostatic pressure of a column of mercury, such as that in a
barometer, depends on its density as well as its height. When we speak
of a pressure of 760 mm mercury, therefore, we must specify the tem-
perature of the mercury; we choose 0°C. In practice barometers are
generally warmer than that, and their readings must therefore be
(a) Warm mercury,
warm scale
(b) Warm mercury,
cold scale
(c) Cold mercury,
cold scale
Fig. 11.18. Reduction of barometer height to 0°C.
reduced to what they would be at 0°C. Also we must allow for the
expansion of the scale with which the height is measured.
The scale of a barometer may be calibrated at 0°C. At any higher
temperature, 8, the height which it indicates, h scale , is less than the true
height, h true , of the mercury meniscus above the free surface in the
reservoir (Fig. 11.18 (a) (b)). The true height is given by equation (3)
of p. 273.
h t rue = Kcale(l+M) • ■ • (14)
where X is the linear expansivity of the scale. If p e is the density of
mercury at 6, then the pressure of the atmosphere is
P = gPeKue-
The height of a mercury column at 0°C which would exert the same
pressure is called the corrected height, h cor . (Fig. 11.18 (b) (c)). It is
given by
P = gPo h cor.>
where p is the density of mercury at 0°C.
THERMAL EXPANSION 287
Therefore
PoKrue = PoKor.-
If y is the coefficient of expansion of mercury, then
Po
Pe =
l+y&'
Hence T+yfl"' 1 '™ = p ° h ™
and h„ = htrue
1+yO'
Therefore, by equation (13),
1 + yO '
Let us write this as
Kor. = h scale (l+M)(l + yOr 1 .
l scale
.2/32
Then, if we ignore yd and higher terms, we may write
Kor. = KaleV+mi-yO)-
Hence h cor . = h^ (l+AO-yd+yW 2 ),
and ignoring yW 2 we find
Kor. = Kale{l+tt-y)0}-
The coefficient y is greater than X, and the corrected height is less than
the scale height. It is convenient therefore, to write
Kor. = Kcale{l-(y-W}-
The reader should notice that the correction depends on the difference
between the cubic expansivity of the mercury, and the linear expan-
sivity of the scale ; as in Dulong and Petit's experiment, there is no
question of apparent expansion.
*
The Anomalous Expansion of Water
If we nearly fill a tall jar with water, and float lumps of ice on it, the
water at the base of the jar does not cool below 4°C, although the
water at the top soon reaches 0°C. At 4°C water has its greatest density ;
as it cools to this temperature it sinks to the bottom. When the water
at the top cools below 4°C, it becomes less dense than the water below,
and stays on the top. Convection ceases, and the water near the bottom
of the jar can lose heat only by conduction. Since water is a bad con-
ductor, the loss by conduction is extremely small, and in practice the
water at the bottom does not cool below 4 C. The same happens in a
pond, cooled at the top by cold air. Ice forms at the surface, but a little
below it the water remains at 4°C, and life in the pond survives. It
288
ADVANCED LEVEL PHYSICS
could not if the water contracted in volume continuously to 0°C; for
then convection would always carry the coldest water to the bottom,
and the pond would freeze solid. In fact, lakes and rivers, unless they
are extremely shallow, never do freeze solid; even in arctic climates,
they take only a crust of ice.
Fig. 11.19 shows how the volume of 1 g of water varies with tem-
perature. The decrease from 0°C to 4°C is called anomalous expansion;
it can be shown with a compensated dilatometer (p. 285). Water also
has the unusual property of expanding when it freezes, hence burst
1 TO
Fig. 11.19. Volume variation of 1 g of water.
pipes. As the figure shows, the expansion is about 9 per cent. If ice
were not less dense than water, it would sink, and so, despite the
anomalous expansion of water, lakes and rivers would freeze solid in
winter.
The following table gives the densities of ice and water at various
temperatures. Accurate experiments show that the temperature of
maximum density is 3-98°C.
DENSITY OF ICE AND WATER
Temperature °C
2 4 6 8
p water kgm~ 3
999-84 999-94 999-97 999-94 999-85
Pue kg m d
9160 (volume of 1 g ice at 0°C = 1092 cm 3 )
Temperature °C
10 20 40 60 80 100
p wa(e ,kgrrT 3
999-70 998-20 992-2 983-2 971-8 958-4
THERMAL EXPANSION
289
Ice and
salt
The temperature of maximum density was first measured in 1804,
by Hope, whose apparatus is shown in Fig. 11.20. Both vessels are made
of metal; the inner one contains
water, and the outer one contains a
mixture of ice and salt, which has
a temperature below 0°C. After
several hours, the water at the top
cools to 0°C— it may even freeze —
but the water at the bottom does
not fall below 4°C.
The contraction of water from
to 4°C is explained by supposing
that the molecules form clusters,
such as H 4 2 , H 6 3 . At first, the
contraction due to the formation of
these more than offsets the expansion
due to the rise of temperature ; but above 4°C the expansion prevails.
The metal antimony behaves like water : it expands on freezing, and
contracts at first when warmed above its melting-point. Because it
expands on freezing, it makes sharp castings, and is used as a con-
stituent of type-metal.
Fig. 1 1 .20. Hope's experiment.
EXAMPLES
1. Describe how you would use a specific gravity bottle to find the coefficient
of expansion of paraffin oil relative to glass between 0° and 50°C.
A specific gravity bottle contains 44-25 g of a liquid at 0°C and 42-02 g at 50°C
Assuming that the linear expansivity of the glass is 000001 K _1 , (a) compare
the densities of the liquid at 0° and 50°C, (b) deduce the real expansivity of the
liquid. Prove any formula employed. (N.)
First part. See text.
Second part. The apparent expansivity of the liquid, y a , is given by
y a
y a =
mass expelled
mass left x temp, rise
2-23
44-25-42-02
4202x50 '
42-02 x 50
= 0-00106 K
Now cubic expansivity, y g , of glass = 3 x 000001 = 000003 K 1 .
:. real coefficient, y, = y a +y 9 = 000106+ 000003 = 000109 K
Also, if p , p t are the densities at 0°C, 50°C respectively,
Po
Pt
= l + yt
= 1+0-00109x50
= 1055.
2. Define the linear expansivity of a solid, and show how it is related to the
cubic expansivity. Describe an accurate method of determining the coefficient
of expansion of a solid.
In order to make connexion to the carbon anode of a transmitting valve it is
290 ADVANCED LEVEL PHYSICS
required to thread a copper wire (2 mm in diameter at 20°C) through a smaller
hole in the carbon block. If the process can just be carried out by immersing both
specimens in dry ice (solid C0 2 ) at - 80°C and the coefficients of linear expansion
of copper and carbon are 17 x 10~ 6 and 5 x 10~ 6 K~ 1 , calculate the size of the
hole at 20°C. (0. & C.)
First part. The linear expansivity is the increase in length per unit length
per degree rise in temperature. It is one-third of the cubic expansivity, proved on
p. 279. An accurate method of measuring the linear expansivity is the com-
parator method, described on p. 270.
Second part. Suppose r cm is the diameter of the hole at 20°C. Then, from the
formula
l 2 = h\l+tt6 2 -0 l )]. where (^-ej = -80 -20 = -100°C,
diameter at -80°C = r(l -5 x 10" 6 x 100) = 0-9995 r cm.
Similarly, the diameter of the copper wire at — 80°C
= 0-2(1-17 x 10~ 6 x 100) = 019966 cm.
.-. 0-9995 r = 019966
' • r = Vims = 019976 cm = i" 8 mm -
3. Describe in detail how the cubic expansivity of a liquid may be deter-
mined by the weight thermometer method.
The height of the mercury column in a barometer provided with a brass scale
correct at 0°C is observed to be 749-0 mm on an occasion when the temperature
is 15°C. Find (a) the true height of the column at 15 C C, (b) the height of a column
of mercury at 0°C which would exert an equal pressure. Assume that the cubic
expansivities of brass and of mercury are respectively 0000054 and 00001 81
k-Mjv.)
First part. The weight thermometer method gives the apparent expansivity of
the liquid, which is calculated from :
mass expelled
mass left x temperature rise
The true expansivity is obtained by adding the cubic expansivity of the container's
material to the apparent expansivity.
Second part, (a) The linear expansivity of brass, A,
= 3 x volume expansivity
.-. true height, h e , = 7490 (1 + A0) - 7490 (1 + 0-000018 x 15)
= 749-2 mm.
(b) Suppose h is the required height at 0°C. Then, since pressure = hpg,
h Po9 = KPed
Po
THERMAL EXPANSION 291
Pe 1 1 *
But
p 1+yO 1+0-000181x15 10027
. nAan 100027
ho = 7490x Too27
= 749 [1 - (00027 - 0-00027)]
= 747-2 mm.
EXERCISES 11
1. Describe and explain how the absolute expansivity of a liquid may be
determined without a previous knowledge of any other expansivity.
Aniline is a liquid which does not mix with water, and when a small quantity
of it is poured into a beaker of water at 20°C it sinks to the bottom, the densities
of the two liquids at 20°C being 1021 and 998 kg m~ 3 respectively. To what
temperature must the beaker and its contents be uniformly heated so that the
aniline will form a globule which just floats in the water? (The mean absolute
expansivity of aniline and water over the temperature range concerned are
000085 and 000045 K~ *, respectively.) (L.)
2. Define the linear expansivity of a solid, and describe a method by which
it may be measured.
Show how the superficial expansivity can be derived from this value.
A 'thermal tap' used in certain apparatus consists of a silica rod which fits
tightly inside an aluminium tube whose internal diameter is 8 mm at 0°C. When
the temperature is raised, the fit is no longer exact. Calculate what change in
temperature is necessary to produce a channel whose cross-section is equal to
that of a tube of 1 mm internal diameter. [Linear expansivity of silica = 8 x 10~ 6
K -1 . Linear expansivity of aluminium = 26 x 10" 6 K -1 .] (O. & C.)
3. Define the cubic expansivity of a liquid. Find an expression for the variation
of the density of a liquid with temperature in terms of its expansivity.
Describe, without experimental details, how the cubic expansivity of a liquid
may be determined by the use of balanced columns.
A certain Fortin barometer has its pointers, body and scales made from brass.
When it is at 0°C it records a barometric pressure of 760 mm Hg. What will it
read when its temperature is increased to 20°C if the pressure of the atmosphere
remains unchanged?
[Cubic expansivity of mercury = 1-8 x 10 -4 K~ l ; linear expansivity of brass
= 2xl0 _5 K- 1 .](O. &C.)
4. Describe an experiment to determine the absolute (real) expansivity of
paraffin between 0°C and 50°C, using a weight thermometer made of glass of
known cubic expansivity. Derive the formula used to calculate the result.
A glass capillary tube with a uniform bore contains a thread of mercury
1000 cm long, the temperature being 0°C. When the temperature of the tube
and mercury is raised to 100°C the thread is increased in length by 1-65 cm. If the
mean coefficient of absolute expansion of mercury between 0°C and 100°C is
0-000182 K _1 , calculate a value for the linear expansivity of glass (N.)
5. Define the linear and cubic expansivity, and derive the relation between
them for a particular substance. Describe, and give the theory of, a method for
finding directly the absolute coefficient of expansion of a liquid. The bulb of a
mercury-in-glass thermometer has a volume of 0-5 cm 3 and the distance between
progressive degree marks is 2 mm. If the linear expansivity of glass is 10 ~ 5 K~ 1 ,
292 ADVANCED LEVEL PHYSICS
and the cubic expansivity of mercury is 18 x 10"" 4 K~Vfind the cross-sectional
area of the bore of the stem. (C.)
6. Define the terms 'apparent' and 'absolute' expansivity of a liquid, and
show how the former is found by means of a weight thermometer. A litre flask,
which is correctly calibrated at 4°C, is filled to the mark with water at 80°C.
What is the weight of water in the flask? [Linear expansivity of the glass of the
flask = 8-5 x 10 -6 K _1 ; mean cubic expansivity of water = 50 x 1(T 4 K -1 .]
{O. & C.)
7. Describe how to measure the apparent expansivity of a liquid using a
weight thermometer. Show how the result can be calculated from the observa-
tions.
A specific gravity bottle of volume 500 cm 3 at 0°C is filled with glycerine at
20°C. What mass of glycerine is contained in the bottle if the density of glycerine
at 0°C is 1-26 g per cm 3 , and its real expansivity is 5-2 x 10 ~ 4 K~ *? Assume that
the linear expansivity of the glass is 8 x 10 -6 K' 1 . (N.)
8. Describe in detail how a reliable value for the expansivity of mercury may
be found by a method independent of the expansion of the containing vessel.
Give the necessary theory.
A silica bulb of negligible expansivity holds 3400 g of mercury at 0°C when
full. Some steel balls are introduced and the remaining space is occupied at
0°C by 2550 g of mercury. On heating the bulb and its contents to 100°C, 4-800
g of mercury overflow. Find the linear expansivity of the steel.
[Assume that the expansivity of mercury is 180 x 10~ 6 K -1 .] (L.)
9. Describe an accurate method for determining the linear expansivity of a
solid in the form of a rod. The pendulum of a clock is made of brass whose
linear expansivity is 1-9 x 10 ~ 5 per deg C. If the clock keeps correct time at
15°C, how many seconds per day will it lose at 20°C? (0. & C.)
10. A steel wire 8 metres long and 4 mm in diameter is fixed to two rigid sup-
ports. Calculate the increase in tension when the temperature falls 10°C. [Linear
expansivity of steel = 12xl0 -6 K _1 , Young's modulus for steel = 2xlO u
Nm- 2 .](OJC.)
11. How can you show that the density of water does not fall steadily as the
temperature is raised from 0°C to 100°C? What does your experiment indicate
about the expansion of water? What importance has this result in nature? (C.)
12. Why is mercury used as a thermometric fluid? Compare the advantages
and disadvantages of the use of a mercury-in-glass thermometer and a platinum
resistance thermometer to determine the temperature of a liquid at about 300°C.
A dilatometer having a glass bulb and a tube of uniform bore contains 150 g of
mercury which extends into the tube at 0°C. How far will the meniscus rise up
the tube when the temperature is raised to 100°C if the area of cross-section of
the bore is 0-8 mm 2 at 0°C? Assume that the density of mercury at 0°C is 13-6 g
cm^ 3 , that the expansivity of mercury is 1-82 x 10 -4 K _1 , and that the linear
expansitivity of glass isMxlO _5 K _1 . (N.)
13. Describe in detail how you would determine the linear expansivity of a
metal rod or tube. Indicate the chief sources of error and discuss the accuracy
you would expect to obtain.
A steel cylinder has an aluminium alloy piston and, at a temperature of 20°C
when the internal diameter of the cylinder is exactly 10 cm, there is an all-round
clearance of 0-05 mm between the piston and the cylinder wall. At what tempera-
ture will the fit be perfect? (The linear expansivity of steel and the aluminium
alloy are 1-2 x 10" 5 and 1-6 x 10" 5 K -1 respectively.) (0. & C.)
THERMAL EXPANSION 293
14. Explain the statement: the absolute expansivity of mercury is 1 -81 x 10~ 4
K~ 1 . Describe an experiment to test the accuracy of this value. Why is a know-
ledge of it important?
Calculate the volume at 0°C required in a thermometer to give a degree of
length 015 cm on the stem, the diameter of the bore being 0-24 mm. What would
be the volume of this mercury at 100°C? [The linear expansivity of the glass
may be taken as 8-5 x 10~ 6 K -1 .] (L.)
chapter twelve
Changes of State. Vapours
SOLID TO liquid: fusion
The Solid State
Substances exist in the solid, liquid or gaseous state. In the solid
state, a body has a regular, geometrical structure. Sometimes this
structure gives the body a regular outward form, as in a crystal of
alum; sometimes, as in a strand of wool, it does not. But X-rays can
reveal to us the arrangement of the individual atoms or molecules in
a solid; and whether the solid is wool or alum, we find that its atoms
or molecules are arranged in a regular pattern. This pattern we call a
space-lattice; its form may be simple, as in metals, or complicated, as
in wool, proteins, and other chemically complex substances.
We consider that the atoms or molecules of a solid are vibrating
about their mean positions in its space-lattice. And we consider that
the kinetic energy of their vibrations increases with the temperature
of the solid : its increase is the heat energy supplied to cause the rise
in temperature. When the temperature reaches the melting-point, the
solid liquefies. Lindemann has suggested that, at the melting-point,
the atoms or molecules vibrate so violently that they collide with one
another. The attractive forces between them can then no longer hold
them in their pattern, the space-lattice collapses, and the solid melts.
The work necessary to overcome the forces between the atoms or
molecules of the solid, that is, to break-up the space-lattice, is the
latent heat of melting or fusion.
The Liquid State
In the liquid state, a body has no form, but a fixed volume. It adapts
itself to the shape of its vessel, but does not expand to fill it. We consider
that its molecules still dart about at random, as in the gaseous state,
and we consider that their average kinetic energy rises with the liquid's
temperature. But we think that they are now close enough together
to attract one another — by forces of a more-or-less gravitational nature.
Any molecule approaching the surface of the liquid experiences a
resultant force opposing its escape (see p. 128, Surface Tension) Never-
theless, some molecules do escape, as is shown by the fact that the
liquid evaporates : even in cold weather, a pool of water does not last
for ever. The molecules which escape are the fastest, for they have the
greatest kinetic energy, and therefore the greatest chance of over-
coming the attraction of the others. Since the fastest escape, the slower,
which remain, begin to predominate : the average kinetic energy of the
molecules falls, and the liquid cools. The faster a liquid evaporates, the
colder it feels on the hand — petrol feels colder than water, water feels
294
CHANGE OF STATE
295
colder than paraffin. To keep a liquid at constant temperature as it
evaporates, heat must be supplied to it; the heat required is the latent
heat of evaporation.
Melting and Freezing
When a solid changes to a liquid, we say it undergoes a change of
state or phase. Pure crystalline solids melt and freeze sharply. If, for
example, paradichlorbenzene is warmed in a test tube until it melts,
and then allowed to cool, its temperature falls as shown in Fig. 12.1 (a).
100 T
100 t
o
Time —
(a) Paradichlorbenzene (b) Paraffin wax
Fig. 12.1. Cooling curves, showing freezing.
Time — »-
Paradi ch I or benzene
100
A well-defined plateau in the cooling curve indicates the freezing (or
melting) point. While the substance is freezing, it is evolving its latent
heat of fusion, which compensates for the heat lost by cooling, and its
temperature does not fall. An impure substance such as paraffin wax,
on the other hand, has no definite plateau on its cooling curve ; it is a
mixture of several waxes, which freeze out from the liquid at slightly
different temperatures (Fig. 12.1 (b)).
Supercooling
If we try to find the melting-point of hypo from its cooling curve, we
generally fail; the liquid goes on cooling down to room temperature.
But if we drop a crystal of
solid hypo into the liquid
the temperature rises to the
melting-point of hypo, and
the hypo starts to freeze.
While the hypo is freezing,
its temperature stays con-
stant at the melting-point;
when all the hypo has frozen,
ite temperature starts to fall
again (Fig. 12.2).
The cooling of a liquid
below its freezing-point is Time — *-
called supercooling; the Fig. 12.2. Cooling curve of hypo.
296
ADVANCED LEVEL PHYSICS
molecules of the liquid lose their kinetic energy as it cools, but do
not take up the rigid geometric pattern of the solid. Shaking or stirring
the liquid, or dropping grit or dust into it, may cause it to solidify;
but dropping in a crystal of its own solid is more likely to make it
solidify. As soon as the substance begins to solidify, it returns to its
melting (or freezing) point. The melting-point is the only temperature at
which solid and liquid can be in equilibrium.
No one has succeeded in warming a solid above its melting-point —
or, if he has, he has failed to report his success. We may therefore
suppose that to superheat a solid is not possible ; and we need not be
surprised For the melting-point of a solid is the temperature at which
its atoms or molecules have enough kinetic energy to break up its
crystal lattice: as soon as the molecules are moving fast enough, they
burst from their pattern On the other hand, when a liquid cools to its
melting-point, there is no particular reason why its molecules should
spontaneously arrange themselves. They may readily do so, however,
around a crystal in which their characteristic pattern is already set up.
Pressure and Melting
The melting-point of a solid
is affected by increase of pres-
sure. If we run a copper wire
over a block of ice, and hang
a heavy weight from it, as in
Fig. 12.3, we find that the wire
slowly works through the
block. It does not cut its way
through, for the ice freezes up
behind it; the pressure of the
wire makes the ice under it
melt, and above the wire, where
the pressure is released, the ice
freezes again. The freezing
again after melting by pressure
is called r eg elation.
This experiment shows that increasing the pressure on ice makes it
melt more readily; that is to say, it lowers the melting-point of the ice.
We can understand this effect when we remember that ice shrinks
when it melts (see p. 209) ; pressure encourages shrinking, and therefore
melting.
The fall in the melting-point of ice with increase in pressure is small :
00072°C per atmosphere. It is interesting, because it explains the
slipperiness of ice; skates for example, are hollow ground, so that the
pressure on the line of contact is very high, and gives rise to a lubricating
film of water. Ice which is much colder than 0°C is not slippery, because
to bring its melting-point down to its actual temperature would
require a greater pressure than can be realized. Most substances swell
on melting; an increase of pressure opposes the melting of such sub-
stances, and raises their melting-point.
Fig. 12.3. Melting of ice under pressure.
CHANGE OF STATE 297
Freezing of Solutions
Water containing a dissolved substance freezes below 0°C. The
depression of the freezing-point increases at first with the concentra-
tion, but eventually reaches a maximum. The lowest freezing-point of
common salt solution is — 22°C, when the solution contains about
one-quarter of its weight of salt. When a solution does freeze, pure ice
separates out; an easy way of preparing pure water is therefore to
freeze it, remove the ice, and then melt the ice. The water which is
mixed with the ice in determining the ice-point of a thermometer must
be pure, or its temperature will not be 0°C.
When ice and salt are mixed, the mixture cools below 0°C, but
remains liquid. As the proportion of salt is increased, the temperature
of the mixture falls, until it reaches a minimum at — 22°C. A mixture of
ice and salt provides a simple means of reaching temperatures below
0°C, and is called a 'freezing mixture'.
The phenomena of the freezing of solutions are important in
chemistry, and particularly in metallurgy. We shall give a brief explana-
tion of them later in this chapter.
LIQUID TO GAS! EVAPORATION
Evaporation differs from melting in that it takes place at all tem-
peratures; as long as the weather is dry, a puddle will always clear up.
In cold weather the puddle lasts longer than in warm, as the rate of
evaporation falls rapidly with the temperature.
Solids as well as liquids evaporate. Tungsten evaporates from the
filament of an electric lamp, and blackens its bulb; the blackening
can be particularly well seen on the headlamp bulb of a bicycle dynamo
set, if it has been frequently over-run through riding down-hill. The
rate of evaporation of a solid is negligible at temperatures well below
its melting-point, as we may see from the fact that bars of metal do not
gradually disappear.
Saturated and Unsaturated Vapours
Fig. 12.4 (a) shows an apparatus with which we can study vapours
and their pressures. A is a glass tube, about a metre long, dipping in a
mercury trough and backed by a scale S. Its upper end carries a bulb
B, which is fitted with three taps T, of which T x and T 2 should be as
close together as possible. Above Tj is a funnel F. With T 1 closed but
T 2 open, we evacuate the bulb and tube through T 3 , with a rotary
pump. If the apparatus is clean, the mercury in A rises to the barometer
height H. Meanwhile we put some ether in the funnel F. When the
apparatus is evacuated, we close T 3 and T 2 . We now open and close T t ,
so that a little ether flows into the space C. Lastly, we open T 2 , so that
the ether evaporates into the bulb B. As it does so, the mercury in A
falls, showing that the ether- vapour is exerting a pressure (Fig. 12.4 (b)).
If h is the new height of the mercury in A, then the pressure of the vapour
in mm of mercury is equal to H — h.
By closing T l5 opening and closing T 2 , and then opening T t again,
298
ADVANCED LEVEL PHYSICS
Pump
Saturated
(a)
Vacuum
(b)
Unsaturated
(c)
Saturated
Fig. 12.4. Apparatus for studying vapours.
we can introduce more ether into the space B. At first, we find that,
with each introduction, the pressure of the vapour, H — h, increases.
But we reach a point at which the introduction of more ether does not
increase the pressure, the height of the mercury column remains
constant at h'. At this point we notice that liquid ether appears above
the mercury in A (Fig. 12.4 (c)). We say that the vapour in B is now
saturated; a saturated vapour is one that is in contact with its own liquid.
Before the liquid appeared in the above experiment, the pressure of
the vapour could be increased by introducing more ether, and we say
that the vapour in B was then unsaturated.
Behaviour of Saturated Vapour
To find out more about the saturated vapour, we may try to expand
or compress it. We can try to compress it by raising the mercury reser-
voir M. But when we do, we find that the height h' does not change : the
pressure of the vapour, H — h', is therefore constant (Fig. 12.4 (d)).
The only change we notice is an increase in the volume of liquid above
CHANGE OF STATE
299
the mercury. We conclude, therefore, that reducing the volume of a
saturated vapour does not increase its pressure, but merely makes
some of it condense to liquid.
Similarly, if we lower the reservoir M, to increase the volume of the
vapour, we do not decrease its pressure. Its pressure stays constant,
but the volume of liquid above the mercury now decreases; liquid
evaporates, and keeps the vapour saturated. If we increase the volume
of the vapour until all the liquid has evaporated, then the pressure of
the vapour begins to fall, because it becomes unsaturated (see Fig.
12.5 (a)).
Effect of Temperature: Validity of Gas-laws for Vapours
We cannot heat the apparatus of Fig. 12.4 through any great rise of
temperature. But we can warm it with our hands, or by pointing an
electric fire at it If we do warm it, we find that the ether above the
mercury evaporates further, and the pressure of the vapour increases.
Experiments which we shall describe later show that the pressure of a
saturated vapour rises, with the temperature, at a rate much greater
than that given by Charles's law. Its rise is roughly exponential.
To Boyle's law, saturated vapours are indifferent: their pressure is
independent of their volume. Unsaturated vapours obey Boyle's law
roughly, as they also obey roughly Charles's law. Fig. 12.5 (a). Vapours,
saturated and unsaturated, are gases in that they spread throughout
their vessels; but we find it convenient to distinguish them by name
from gases such as air, which obey Charles's and Boyle's laws closely.
We shall elaborate this distinction later.
PROPERTIES OF SATURATED VAPOURS
Water
Mercury
Ethyl Ether
Temperature
°C
Pressure,
Pressure,
Pressure,
mm mercury
mm mercury
mm mercury
-20
0-784 (ice)
-10
1-96 (ice)
112
4-56
0-00016
185
10
9-20
0-00043
291
20
17-5
0-0011
440
30
31-7
00026
34-6
760
40
551
0-0057
921
50
92-3
00122
60
149
00246
1734
70
234
80
355
00885
2974
90
526
100
760
0-276
4855
150
3569
2-88
200
11647
17-8
250
29770
75-8
300
67620
249
356-7
760
300
ADVANCED LEVEL PHYSICS
Fig. 12.5 (b) shows the effect of heating a saturated vapour. More
and more of the liquid evaporates, and the pressure rises very rapidly.
As soon as all the liquid has evaporated, however, the vapour becomes
unsaturated, and its pressure rises more sedately.
\ Approx.
VBoyles law
/S.V.P. curve
(a)
V. litres
Fig. 12.5. Effect of volume and temperature on pressure of water vapour.
Fig. 12.6 (a) shows isothermals for a given mass of liquid and vapour
at two temperatures, Q 1 = 10°C, and 6 2 = 21°C. The temperatures
are chosen so that the saturated vapour pressure at 9 2 is double that
at V The absolute temperatures are 7\ = 273 + 9 t = 283 K, and
T 2 = 273 + 2 = 294 K. Because the saturated vapour pressure rises
4 8 12
V, litre — »
Fig. 12.6. Relationship between pressure, temperature, total mass,
and volume, for water-vapour and liquid.
so rapidly with temperature, the absolute temperature T 2 is not nearly
double the absolute temperature T v Consequently the isothermals for
the unsaturated vapour are fairly close together, as shown; and the
transition from saturated to unsaturated vapour occurs at a smaller
volume at the higher temperature.
Fig. 12.6 (b) shows pressure-temperature curves for a vapour, initially
in contact with different amounts of liquid, in equal total volumes.
The more liquid present, the greater is the density of the vapour when
it becomes unsaturated, and therefore the higher the pressure and
temperature at which it does so.
CHANGE OF STATE 301
Kinetic Theory of Saturation
Let us consider a vapour in contact with its
liquid, in an otherwise empty vessel which is closed
by a piston (Fig. 12.7). The molecules of the vapour,
we suppose, are rushing randomly about, like the
molecules of a gas, with kinetic energies whose
average value is determined by the temperature of
the vapour. They bombard the walls of the vessel,
giving rise to the pressure of the vapour, and they
also bombard the surface of the liquid.
The molecules of the liquid, we further suppose, Fig. 12.7. Dynamic
are also rushing about with kinetic energies deter- equilibrium,
mined by the temperature of the liquid. The fastest of them escape from
the surface of the liquid. At the surface, therefore, there are molecules
leaving the liquid, and molecules arriving from the vapour. To complete
our picture of the conditions at the surface, we suppose that the vapour
molecules bombarding it are not reflected — as they are at the walls of
the vessel — but are absorbed into the liquid. We may expect them to
be, because we consider that molecules near the surface of a liquid are
attracted towards the body of the liquid.
We shall assume that the liquid and vapour have the same tempera-
ture. Then the proportions of liquid and vapour will not change, if
the temperature and the total volume are kept constant. Therefore, at
the surface of the liquid, molecules must be arriving and departing
at the same rate, and hence evaporation from the liquid is balanced
by condensation from the vapour. This state of affairs is called a
dynamic equilibrium. In terms of it, we can explain the behaviour of a
saturated vapour.
The rate at which molecules leave unit area of the liquid depends
simply on their average kinetic energy, and therefore on the tempera-
ture. The rate at which molecules strike unit area of the liquid, from
the vapour, likewise depends on the temperature; but it also depends
on the concentration of the molecules in the vapour, that is to say,
on the density of the vapour. The density and temperature of the
vapour also determine its pressure ; the rate of bombardment therefore
depends on the pressure of the vapour.
Now let us suppose that we decrease the volume of the vessel in
Fig. 12.7 by pushing in the piston. Then we momentarily increase the
density of the vapour, and hence the number of its molecules striking
the liquid surface per second. The rate of condensation thus becomes
greater than the rate of evaporation, and the liquid grows at the
expense of the vapour. As the vapour condenses its density falls, and
so does the rate of condensation. The dynamic equilibrium is restored
when the rate of condensation, and the density of the vapour, have
returned to their original values. The pressure of the vapour will then
also have returned to its original value. Thus the pressure of a saturated
vapour is independent of its volume. The proportion of liquid to vapour,
however, increases as the volume decreases.
302
ADVANCED LEVEL PHYSICS
Let us now suppose that we warm the vessel in Fig. 12.7, but keep the
piston fixed. Then we increase the rate of evaporation from the liquid,
and increase the proportion of vapour in the mixture. Since the volume
is constant, the pressure of the vapour rises, and increases the rate
at which molecules bombard the liquid. Thus the dynamic equilibrium
is restored, at a higher pressure of vapour. The increase of pressure
with temperature is rapid, because the rate of evaporation of the liquid
increases rapidly — almost exponentially — with the temperature. A
small rise in temperature causes a large increase in the proportion of
molecules in the liquid moving fast enough to escape from it.
Boiling
A liquid boils when its saturated vapour pressure is equal to the atmo-
spheric pressure. To see that this is true, we take a closed J-shaped
tube, with water trapped in its closed limb (Fig. 12.8 (a)). We heat the
tube in a beaker of
water, and watch the
water in the J-tube. It
remains trapped as at
(a) until the water in
the beaker is boiling.
Then the water in the
J-tube comes to the
same level in each limb,
showing that the pres-
sure of the vapour in
the closed limb is equal
to the pressure of the
air outside (Fig. 12.8
(b)).
The J-tube gives a
simple means of mea-
suring the boiling-point
of a liquid which is in-
flammable, or which has
a poisonous vapour, or
(a)
(b)
(c)
Fig. 12.8. Use of J-tube for boiling-point.
of which only a small quantity can be had. A few drops of the liquid
are imprisoned by mercury in the closed limb of the tube, all entrapped
air having been shaken out (Fig. 12.8 (c)). The tube is then heated in a
bath, and the temperature observed at which the mercury comes to the
same level in both limbs. The bath is warmed a little further, and then
a second observation made as the bath cools; the mean of the two
observations is taken as the boiling-point of the liquid.
Boiling differs from evaporation in that a liquid evaporates from its
surface alone, but it boils throughout its volume. If we ignore the
small hydrostatic pressure of the liquid itself, we may say that the
pressure throughout a vessel of liquid is the atmospheric pressure.
Therefore, when the saturated vapour pressure is equal to the atmo-
spheric pressure, a bubble of vapour can form anywhere in the liquid.
CHANGE OF STATE
303
Generally the bottom of the liquid is the hottest part of it, and bubbles
form there and rise through the liquid to the surface. Just before the
liquid boils, its bottom part may be at the boiling-point, and its upper
part below. Bubbles of vapour then form at the bottom, rise in to the
colder liquid, and then collapse. The collapsing gives rise to the singing
of a kettle about to boil.
Further Consideration of Boiling
The account of boiling which we have just given is crude because
in it we ignored the effect of surface-tension. Because
of surface tension, a bubble can exist in a liquid only
if there is an excess pressure inside it. If y is the sur-
face-tension of the liquid, and r the radius of the
bubble, the excess pressure is 2y/r. If the bubble is
formed at a depth h below the surface of the liquid,
as in Fig. 12.9, the external pressure acting on it is
P = P a + hpg,
where P a = atmospheric pressure, p = density of
liquid, g = acceleration of gravity.
Therefore a bubble, of radius r, can form at a depth
h only if its vapour pressure, p, satisfies the equation
' a
1
X
Fig. 12.9. Boiling.
P = P +
2y
= P a +hpg +
2y
If the radius r is small, the term 2y/r is great, and the bubble cannot
form unless the vapour pressure is considerably above atmospheric.
In fact the equation shows that a bubble can never start from zero
radius, because it would require an infinite vapour pressure to do so.
Bubbles actually form on roughnesses in the vessel, or specks of solid
suspended in the liquid. Very clean water in a very smooth beaker may
not boil until it is well above 100°C; its bubbles then grow violently,
and the liquid 'bumps' in the beaker. A piece of broken pipe-clay pre-
vents bumping, by presenting fine points for the bubbles to form on.
Thus the temperature of a boiling liquid is not definite — it depends
on the conditions of boiling. But the temperature of the vapour is
definite. The vapour escaping is in equilibrium with the liquid at the
surface, and is at atmospheric pressure. Its temperature, therefore, is
the temperature at which the saturated vapour pressure is equal to the
atmospheric pressure. This idea is important in defining the upper
fixed point of the temperature scale (p. 190). We say that the upper
fixed point is the temperature of the steam from water boiling under
a pressure of 760 mm mercury. We must not refer to the temperature
of the water, and we must specify the atmospheric pressure because
as we have just seen, it determines the temperature of the steam.
304 ADVANCED LEVEL PHYSICS
TEMPERATURE OF SATURATED STEAM AT PRESSURES
NEAR NORMAL ATMOSPHERIC
Barometer height, H, mm
Temperature, 8, "C
680
96-910
690
97-312
700
•709
710
98102
720
■490
730
•874
740
99-254
750
•629
751
•666
Barometer height, H, mm
Temperature, 8, °C
752
•704
753
•741
754
■778
755
•815
756
•852
757
■889
758
•926
759
•963
760
100000
761
•037
Barometer height, H, mm
Temperature, 0, °C
762
•074
763
■110
764
•147
765
•184
766
•220
767
•257
768
•294
769
•330
770
•367
780
•729
In general, around H = 760 mm, the temperature is given by
d = 100 + 0O36 (H - 760) - 2-3 x 10" 5 (H - 760) 2 .
Variation of Saturated Vapour Pressure with Temperature
We can now see how the relationship between the pressure of a
saturated vapour and its temperature can be measured. We must
apply various known air pressures to the liquid, heat the liquid, and
measure the temperature of its vapour. Fig. 12.10 shows a suitable
apparatus, due to Regnault. The flask F contains the liquid, water in
Fig. 12.10. Apparatus for variation of S.V.P. with temperature.
a laboratory experiment, and the flask R is an air reservoir. The
pressure of the air in R is shown by the mercury manometer M ; if its
height is h, the pressure in mm mercury is
p = H-h,
where H is the barometer height.
We first withdraw some air from R through the tap T, with a filter
pump, until p is about 700 mm. We then close T and heat the water
gently. The water vapour condenses in the condenser, and runs back
to the flask. After a few minutes the water boils steadily. The temperature
of the vapour, 6, and the pressure, p, become constant and we record
their values. We next remove the flame from the flask F, and let the
apparatus cool for a minute or two. Then we withdraw some more air
from R, close T again, and repeat the observations.
If we wish to find the saturated vapour pressure when it is above
CHANGE OF STATE
305
atmospheric, that is to say, when the temperature is above the normal
boiling-point of the liquid, air is pumped into the reservoir R — with a
bicycle pump — instead of drawing it out. The manometer M then
shows the excess pressure, and
p = H + h.
With simple glass apparatus we cannot go far in this direction.
Boiling Point of a Solution
At a given pressure, the boiling-point of water containing a dissolved
substance is higher than that of pure water. The temperature of the
steam evolved from the solution, however, is the temperature of
saturated steam at the prevailing pressure. Traces of dissolved sub-
stances in the water therefore do not affect the steam point in thermo-
metry.
Since a liquid boils when its saturated vapour pressure is equal to
the atmospheric pressure, we must conclude that dissolving a substance
in water lowers its saturated vapour pressure, at a given temperature
(Fig. 12.11). We may explain this by supposing that the molecules of
the dissolved substance, which do not evaporate, hinder the escape of
the molecules of the water.
760
mm.Hg
Elevation
of B.P.
100°C
Fig. 12. 1 1 . Effect of solute on saturated vapour pressure.
The lowering of the vapour pressure of water by a dissolved solid
gives striking support to the kinetic theory of evaporation. For mea-
surements of the vapour pressure show that reduction does not depend
on the nature of the solute ; it depends only on the number of dissolved
particles in the solution expressed as a fraction of the total number of
particles (solute plus water molecules). In fact, if there are n solute
particles to every 100— n water molecules, then
Ap _ n
y ~ 100'
where p is the saturated vapour pressure of water, and Ap is the lowering
by solution. Thus the lowering simply depends on the number of
particles hindering evaporation.
306 ADVANCED LEVEL PHYSICS
Effect of Altitude on Boiling Point
The pressure of the atmosphere decreases with increasing height
above the earth's surface, because the thickness, and therefore the
weight, of the belt of air above the observer decreases. The rate of
fall in pressure is almost uniform over fairly small heights — about
85 mm mercury per km. But at great altitudes the rate of fall diminishes.
At the height of Everest, 9000 m, the atmospheric pressure is about
280 mm of mercury. On account of the fall in atmospheric pressure, the
boiling point of water falls with increasing height. Cooking-pots for
use in high mountainous districts, such as the Andes, are therefore
fitted with clamped lids. As the water boils, the steam accumulates in
the pot, and its pressure rises above atmospheric. At about 760 mm
mercury a safety valve opens, so that the pressure does not rise above
that value, and the cooking is done at 100°C.
The fall in the boiling-point with atmospheric pressure gives a simple
way of determining one's height above sea-level. One observes the
steam point with a thermometer and hypsometer (p. 190). Knowing
how the steam point falls with pressure, and how atmospheric pressure
falls with increasing height, one can then find one's altitude. The hypso-
meter was, in fact, devised for this purpose, and takes its name from it ;
hypsos is Greek for height. Hypsometers have been carried up Hima-
layan peaks; and one was found by Scott and his companions in
Amundsen's abandoned tent at the South Pole.
Variation of Latent Heat with Temperature
When we speak of the latent heat of evaporation of a liquid, we
usually mean the heat required to vaporize unit mass of it at its normal
boiling-point, that is to say, under normal atmospheric pressure.
But since evaporation takes place at all temperatures, the latent heat
has a value for every temperature. Regnault measured the latent heat
of steam over a range of temperatures, by boiling water at controlled
pressures, as in measuring its saturated vapour pressure. His apparatus
was in principle similar to Berthelot's (Fig. 9.10); but he connected the
outlet tube to an air reservoir, manometer, and pump, as in Fig. 12.10.
Modern measurements give, approximately,
/ = 2520-2-50
where / is the specific latent heat in kJ kg - 1 at 0°C.
Internal and External Latent Heats
The volume of 1 g of steam at 100°C is 1672 cm 3 . Therefore when
1 g of water turns into steam, it expands by 1671 cm 3 ; in doing so,
it does work against the atmospheric pressure. The heat equivalent of
this work is that part of the latent heat which must be supplied to the
water to make it overcome atmospheric pressure as it evaporates; it is
called the 'external latent heat'. The rest of the specific latent heat —
the internal part — is the equivalent of the work done in separating the
molecules, against their mutual attractions.
CHANGE OF STATE
307
The work done, W, in the expansion of 1 g from water to steam is
the product of the atmospheric pressure p and the increase in volume
AV:
W = p.AV.
Normal atmospheric pressure corresponds to a barometer height H
760 mm. Hence, as on p. 228>
p = gp H = 9-81 x 13600 x 0-76
= 1-013 xl0 5 Nm _2
and W = p. AV = 1013 x 10 5 x 1671 x 10 -6 joule.
The external specific latent heat in joules is therefore
l ex = 1013 x 10 5 x 1671 x 10~ 6
= 170Jg _1 = 170kJkg _1 .
This result shows that the external part of the specific latent heat is
much less than the internal part. Since the total specific latent heat / is
2270 joule g _1 , the internal part is
/„ = /-/„ = 2270-170
= 2100Jg- 1 =2100kJkg" 1 .
Density of a Saturated Vapour
In any experiment to measure the density
of a saturated vapour, the vessel containing
the vapour must also contain some liquid, to
ensure that the vapour is saturated. The
problem is therefore to find how much of the
total mass is vapour, and how much is liquid.
Fig. 12.12 shows one method of solving this
problem, due to Cailletet and Mathias.
A, B are two glass tubes, which have been
calibrated with volume scales, and then
evacuated. Known masses m u m 2 of liquid
are introduced into the tubes, which are then
sealed off. The tubes are warmed to the same
temperature in a bath, and the volumes of
liquid V h , V h and of vapour V Vl V V2 are observed. Then if p, and p v are the densities
of liquid and vapour respectively :
™i = PiVh + PvKt
™ 2 = P t V h + p v V V2 .
From these equations p, can be eliminated, and p v found. The equations can
also, of course, be made to give p, ; this method is useful for finding the density of
a liquid 'gas' — e.g. liquid oxygen (p. 324).
Density of an Unsaturated Vapour
We have seen that the molecular weight of a gas, p, is given very nearly by
p = 2A, where A is the density of the gas relative to that of hydrogen (p. 230).
Fig. 12.12. Determination of
density of a saturated vapour.
308
ADVANCED LEVEL PHYSICS
The proof that ft = 2A depends on Avogadro's principle, which says that equal
volumes of all gases at the same temperature and pressure contain equal numbers
of molecules. This principle is true only of those gases which we normally call
'perfect' — which obey Boyle's and Charles's laws accurately (p. 225). It is not
true of saturated vapours, but it is roughly true of vapours which are far from
saturation. To find die molecular weight of a substance which is liquid at room
temperature, therefore, we must vaporize it, and measure the density of its vapour
when it is as far from saturation as we can conveniently get it.
Fig. 12.13. Determination of density of an unsaturated vapour.
Several methods have been devised for doing this, one of which, Dumas 's of
1827, is illustrated in Fig. 12.13. A glass bulb B, with a long thin stem, is weighed
and then partly filled with liquid by warming and dipping. The amount of liquid
introduced must be great enough to ensure that all the air in the bulb will be
-driven out by vapour when the liquid evaporates. The liquid is made to evaporate
by plunging the bulb into a bath at a temperature about 40°C above its boiling-
point. It then evaporates rapidly, and its vapour sweeps the air out of the bulb.
When vapour has stopped coming out, the stem is sealed with a flame: the bulb
now contains nothing but the vapour, at the temperature 9 and under atmospheric
pressure.
The bulb is removed, allowed to cool, dried, and weighed. The stem is then
broken at the tip under water; since nearly all the vapour has condensed, at the
room temperature, water rushes in and fills the bulb.
Let m t = mass of bulb full of air, at room temperature.
m 2 = mass of bulb full of vapour.
m 3 =; mass of bulb full of water.
Since the mass of air in the bulb is negligible compared with that of water, we
have, numerically,
Volume of bulb in cm 3 , V lt = m 3 — m 1 'mg.
The mass of air in the bulb at room temperature is
™ a = yp a
CHANGE OF STATE 309
where p a is the density of air at room temperature and atmospheric pressure.
The mass of the bulb itself is
m b = m 1 —m 9
and the mass of vapour which filled it when hot is
m v = m t -m h .
This mass of vapour occupied the volume V y at the temperature 6; its density
was therefore
m v
Pv = y.
Since the temperature was well above the boiling-point, the vapour was far from
saturated ; Boyle's and Charles's laws can therefore be used to reduce its density
to s.t.p., for comparison with that of hydrogen.
WATER-VAPOUR IN THE ATMOSPHERE!
HYGROMETRY
The water vapour in the atmosphere is important because it affects
our comfort Except in cold weather, we sweat continuously: the
water in the sweat evaporates, draws its latent heat of evaporation
from the skin, and so keeps us cool. Beads of sweat appear only when
the water cannot evaporate as fast as it reaches the surface of the skin ;
we then feel uncomfortably hot.
On the other hand, if water evaporates from the skin too rapidly,
the skin feels parched and hard; around the mucous membranes — at
the mouth and nose — it tends to crack.
The rate at which water evaporates, from the skin or anywhere else,
depends on the pressure of the water vapour surrounding it. If the
water vapour above the skin is far from saturated, evaporation is swift.
If the vapour is already saturated, water reaching the skin comes
immediately into dynamic equilibrium with it; individual molecules
are exchanged between liquid and vapour, but no mass of liquid is lost,
and water accumulates.
The Partial Pressure of Atmospheric Water
The atmosphere contains other gases besides water-vapour, such as
oxygen and nitrogen In speaking of the water-vapour, therefore, we
must refer to its 'partial pressure', as explained on p. 222.
Water-vapour in the atmosphere is also important because it affects
the weather. Let us suppose that the atmosphere has a temperature
of 20°C — a warm day — and that the water vapour in it has a partial
pressure of 12 mm mercury. It will have a density of about 12 mg per
litre. The density of saturated water vapour at 20°C is 17-3 mg per litre,
and its pressure 17-5 mm mercury. The water vapour in the atmosphere
is therefore not saturated.
Now let us suppose that the atmosphere cools to 14°C, without
changing its composition The 6°C fall in temperature will hardly affect
the density of the water vapour, but it will bring the atmosphere to
310 ADVANCED LEVEL PHYSICS
saturation. For the pressure of saturated water vapour at 14°C is 12 mm
mercury, and its density about 12 mg per litre. If the atmosphere cools
any further, water vapour will condense out of it, forming drops of
liquid water — that is, of fog or cloud.
Relative Humidity
The dampness of the atmosphere, besides affecting the weather and
our comfort, is important also in storage and manufacture of many
substances — tobacco and cotton, for example. From what we have
said already, we can see that the important factor is not the actual
proportion of water vapour in the atmosphere, but its nearness to
saturation In the above example, the density of the vapour remained
almost constant, but we would have felt the atmosphere becoming
much damper as it cooled from 20°C to 14°C.
The dampness of the atmosphere is expressed by its relative humidity,
R.H., which is defined as follows :
mass of water- vapour in a given volume of atmosphere
R.H. = — — f (1)
mass of an equal volume of saturated water-vapour at the same temperature
In other words,
density of water-vapour in atmosphere
R.H. =
density of saturated water-vapour at the same temperature
Because an unsaturated vapour roughly obeys Boyle's law, its density
is roughly proportional to its pressure ; the relative humidity as defined
above is therefore roughly given by
partial pressure of water-vapour present
S.V.P. at temperature of atmosphere
where S.V.P. stands for 'saturated vapour pressure'.
Before describing the methods of measurement, we must warn the
reader against thinking that the atmosphere 'takes up' water vapour.
The atmosphere is not a sponge. Water-vapour exists in it in its own
right; and our knowledge of vapours makes us feel sure that, if we
could live in an atmosphere of water-vapour alone, we would have
just the same experiences of humidity as we now have in our happily
richer surroundings.
Dew-point
In the evening, the earth cools more rapidly, by radiation, than the
air above it Then, on smooth surfaces such as metals, we often find
a thin film of moisture. The surface has cooled to such a temperature
that the water vapour in contact with it has become saturated, and has
begun to condense. No fog has formed because the atmosphere in
general is warmer than the cold solid, and the vapour in it is not satu-
rated. The temperature of a cold surface on which dew just appears is
called the dew-point : it is the temperature at which the saturated vapour-
pressure of water is equal to the partial pressure of the water-vapour
present in the atmosphere.
CHANGE OF STATE
311
If we know the dew-point, we can find the corresponding pressure
of saturated water-vapour, p v from tables. From the same tables we
can find the pressure of saturated water-vapour at the temperature of
the atmosphere. If this is p 2 , then, from p. 310, the relative humidity,
R.H ., is given by
R.H. =
Ei
Pi
(2)
Dew-point Hygrometers
Hygrometery is the measurement of relative humidity, and a hygro-
meter is an instrument for measuring it. Many forms of dew-point
hygrometer have been devised;
we shall describe only the one
due, like so much apparatus in
Heat, to Regnault. It consists of
two glass tubes, A, B, with silver-
plated thimbles C, D cemented
on to their lower ends. C contains
ether, into which dips a thermo-
meter T and a glass tube R (Fig.
12.14). To use the hygrometer, we
first place a sheet of glass between
it and ourselves, to prevent our
breath from adding to the humi-
dity of the air around it. We then
gently blow air through the ether,
by means of a scent-spray bulb
connected through a rubber tube
to R. (Alternatively, we may pass
in a gentle stream of coal-gas, and burn it at the end of a long tube
connected to the outlet S.) The gas or air passing through the ether
carries away its vapour, and makes it evaporate continuously. In
doing so it cools, since it must provide its latent heat of evaporation
from its own heat content When the ether has cooled to the dew-point,
a film of mist appears on the thimble C; the thimble D enables us to
notice this more sharply, by contrast At the moment when the dew
appears, the thermometer T is read. The flow of air or gas through the
ether is stopped, and the ether allowed to warm up. The temperature
at which the dew vanishes is noted, and the mean of this and the tem-
perature at which it appeared is taken as the dew-point #i say. By so
doing we correct reasonably well for any difference of temperature
between the ether and the outer surface of the tube. Lastly, we take the
temperature of the room, 2 , often from a thermometer (not shown) in
B, and look up the saturation pressures at U 6 2 respectively. These
are given in the following table. Then
Fig. 12.14. Regnault's hygrometer.
relative humidity =
S.V.P. at fl t
S.V.P. at 0,
x 100 per cent.
312
ADVANCED LEVEL PHYSICS
PRESSURE AND DENSITY OF SATURATED WATER VAPOUR
e°c
2 4 6 8 10 12 14 16 18 20
p, mm mercury
4-58 5-29 610 701 8-04 9-21 10-5 120 13-6 15-5 17-5
p, mg/litre (or gm" 3 )
4-84 5-54 6-33 7-22 8-21 9-33 10-6 12-0 13-5 15-2 171
6»°C
22 24 26 28 30 32 34 36 38 40
p, mm mercury
19-8 22-3 251 28-3 31-7 35-5 39-8 44-4 49-5 551
p, mg/litre (or gm" J )
19-2 21-5 241 26-9 30-0 33-5 37-2 41-3 45-8 511
The Wet-and-dry-bulb Hygrometer
A piece of wet cloth feels cold, because the moisture evaporating
from it takes latent heat, and cools the remaining liquid. This effect is
used in the wet-and-dry-bulb hygrometer. A piece of muslin is tied
round the bulb of a thermometer, and allowed to dip into a small
jar of water. It is mounted, with a second,
dry-bulb, thermometer, in a louvred draught-
shield (Fig. 12.15). The rate at which water
evaporates from the muslin increases as the
relative humidity of the atmosphere falls;
the cooling of the wet bulb therefore also
increases. The greater the difference in
reading of the two thermometers, the less
is the relative humidity. By calibration
against a chemical hygrometer, tables and
charts have been prepared which give the
Fig. 12.15. Wet-and-dry relative humidity in terms of the thermo-
bulb hygrometer. meter readings.
r—r *
/
\
/
\
/
\
/
\
' i 1 IDepressibn
^
/ Dryl
Wet\
, • i
1 N
/ J
L \
^ ft^A
WET-AND-DRY-BULB HYGROMETER TABLE
(Percentage relative humidity)
Dry-bulb
Difference (depression of
wet-bulb), °C
reading
°C
1
2
3 4 5 6
8 10
12
14
82
65
48 31%
5
85
72
58 45 32%
10
88
76
65 54 44 34%
15
90
80
71 61 52 44
27 12%
20
91
83
74 66 ^9 51
37 24
12%
25
92
84
77 70 63 57
44 33
22
12%
30
86
79 73 67 61
50 39
30
21%
The wet-and-dry-bulb hygrometer is not very reliable when used in
a simple screen. It is more accurate if a steady stream of air is driven
past it by a fan, or by whirling the thermometers around in a frame
like a football-fan's rattle. The hygrometer is then said to be ventilated.
The Hair Hygrometer
Human hair expands in length in damp air. A hair hygrometer is
one consisting of a bundle of hairs fixed to a spring at one end, and
CHANGE OF STATE
313
wrapped round a spindle at the other. The expansion of the hair turns
the spindle and moves a pointer over a scale, which is directly calibrated
in relative humidities. Such instruments need to be recalibrated
frequently, because the hair shows elastic fatigue.
THE BEHAVIOUR OF REAL GASES;
CRITICAL PHENOMENA
A perfect, or ideal, gas is one which obeys Boyle's and Charles's law
exactly, and whose internal energy is independent of its volume. No
such gas exists, but at room temperature, and under moderate pres-
sures, many gases approach the ideal closely enough for most purposes.
We shall consider now the departures of gases from perfection; in
doing so we shall come to appreciate better the relationship between
liquid, vapour, and gas, and we shall see how gases such as air can be
liquefied.
Departures from Boyle's Law
In 1 847 Regnault measured the volumes of various gases at pressures
of several atmospheres. Using the apparatus of Fig. 12.16, he found
that, to halve the volume of the gas, he did not have quite to double
the pressure on it. The product pV, therefore, instead of being constant,
decreased slightly with the pressure. He found one exception to this
Mercury
Temperature
bath
Gas
Loaded piston
Fig. 12.16. Regnault's apparatus for isothermals at high pressure.
rule : hydrogen. By compressing the gases further, Regnault found the
variation of pV with p at constant temperature, and obtained results
which are represented by the early parts of the curves in Fig. 12.17.
314
ADVANCED LEVEL PHYSICS
12
| 10
2 08
2
0-4
co 2
pV= const.
100
200
300
400
500
p, atmospheres — »-
Fig. 12.17. Isothermals for various gases, at room temperature and high
pressure.
The complete curves in the figure show some of the results obtained
by Amagat in 1 892. Amagat's apparatus
for nitrogen is shown in Fig. 12.18.
To get high pressures, he put the ap-
paratus at the bottom of a coal-mine,
and made the manometer tube out of
I rifle barrels, screwed together and stand-
ing up the shaft. He reached a pressure
of 3000 atmospheres.
Having found the volume-pressure
relationship for nitrogen, Amagat used
it to measure high pressures in the
laboratory, without having to resort to
the mine. His method was similar to
that of Andrews, which we are about to
describe ; by means of it he found the
pressure-volume relationships for other
gases.
Fig. 12.18. Amagat's apparatus
for isothermals at high
pressure.
Andrews' Work on Carbon Dioxide
In 1863 Andrews made experiments on carbon dioxide which have
become classics. Fig. 12.19 shows his apparatus. In the glass tube A
he trapped carbon dioxide above the pellet of mercury X. To do this,
he started with the tube open at both ends and passed the gas through
it for a long time. Then he sealed the end of the capillary. He introduced
the mercury pellet by warming the tube, and allowing it to cool with
the open end dipping into mercury. Similarly, he trapped nitrogen in
the tube B.
Andrews then fitted the tubes into the copper casing C, which con-
tained water. By turning the screws S, he forced water into the lower
parts of the tubes A and B, and drove the mercury upwards. The wide
parts of the tubes were under the same pressure inside and out, and so
were under no stress. The capillary extensions were strong enough to
CHANGE OF STATE
315
I
V
Fig. 12.19. Andrews's apparatus for isothermals of C0 2 at high pressures.
withstand hundreds of atmospheres. Andrews actually reached 108
atmospheres.
When the screws S were turned far into the casing, the gases were
forced into the capillaries, as shown on the right of the figure, and
greatly compressed. From the known volumes of the wide parts of the
tubes, and the calibrations of the capillaries, Andrews determined the
258°C
5-
100 150 200
p, atmospheres — »-
Fig. 12.20. Isothermals forC0 2 , as pV/p curves, at various temperatures.
The small dotted loop passes through the ends of the vertical parts ; the
large dotted loop is the locus of the minima of pV.
316
ADVANCED LEVEL PHYSICS
"_-. -t£t=^-1 - Water
volumes of the gases. He estimated the pressure from the compression
of the nitrogen, assuming that it obeyed Boyle's law.
For work above and below room temperature, Andrews surrounded
the capillary part of A with a water bath, which he maintained at a
constant temperature between about 10°C and 50°C.
Fig. 12.20 shows some of Andrews's results, corrected for the de-
parture of nitrogen from Boyle's law ; it also shows the results of similar
experiments over a wider range of temperature, by Amagat in 1893.
Critical Temperature
Before we can interpret Andrews's results for carbon dioxide, we
must describe a simple experiment, made by Cagniard de la Tour in
1822. De la Tour made a tube of
strong glass, as shown in Fig. 12.21.
In the bulb he had water, round the
bend mercury, and at the top —
where the tube was sealed off — air.
He heated the tube in a bath to over
300°C. The expansion of the liquids
was taken up by the compression of
the air, from which de la Tour
estimated the pressure; it went be-
yond 100 atmospheres. Above about
100°C he observed what we would
expect; that a meniscus formed in
bulb, showing that steam was pre-
sent as well as water. But above
about 300°C he noticed that the
meniscus vanished : that there was no observable distinction between
liquid and vapour. The temperature at which the meniscus vanished
he called the critical temperature.
If we consider the nature of a saturated vapour, the phenomenon
of the critical temperature need not surprise us. For as its temperature
rises a saturated vapour becomes denser, whereas a liquid becomes less
dense. The critical temperature is, we may suppose, the temperature
at which liquid and saturated vapour have the same density. Fig. 12.22
supports this view : it shows the results of measurements made on
liquid oxygen by the method of Cailletet and Mathias (p. 307).
Behaviour of Carbon Dioxide near the Critical Point
Now let us turn to Andrews' isothermals for carbon dioxide. These
are shown again, this time as a simple pressure-volume diagram, in
Fig. 12.23. Let us consider the one for 21-5°C, ABCD. Andrews noticed
that, when the pressure reached the value corresponding to B, a
meniscus appeared above the mercury in the capillary containing the
carbon dioxide. He concluded that the liquid had begun to form.
From B to C, he found no change in pressure as the screws were turned,
but simply a decrease in the volume of the carbon dioxide. At the same
time the meniscus moved upwards, suggesting that the proportion of
Fig. 12.21. Cagniard de la Tour's
experiment.
CHANGE OF STATE
317
liquid was increasing. At C the meniscus disappeared at the top of
the tube, suggesting that the carbon dioxide had become wholly liquid.
Beyond C the pressure rose very rapidly; this confirmed the idea that
the carbon dioxide was wholly liquid, since liquids are almost in-
compressible.
Thus the part CBA of the isothermal for 21-5°C is a curve of volume
-120
Crit. temp.
-119*C
Fig. 12.22. Densities of liquid oxygen, and its saturated vapour.
against pressure for a liquid and vapour, showing saturation at B;
it is like the isothermal for water given in Fig. 12.5 (a), p. 300. And the
curve GFE is another such isothermal, for the lower temperature
131°C ; the two curves are like the two in Fig. 12.6 (a), p. 300.
The isothermal for 311°C has no extended plateau ;,it merely shows
P 100x
V, arbitrary units
Fig. 12.23. Andrews's isothermals for C0 2 .
a point of inflection at X. At that temperature, Andrews observed no
meniscus ; he concluded that it was the critical temperature. The iso-
thermals for temperature above 311°C never become horizontal, and
show no breaks such as B or F. At temperatures above the critical, no
transition from gas to liquid can be observed.
The isothermal for 48-l°C conforms fairly well to Boyle's law; even
318
ADVANCED LEVEL PHYSICS
when the gas is highly compressed its behaviour is not far from ideal.
The point X in Fig. 12.23 is called the critical point. The pressure and
volume (of unit mass) corresponding to it are called the critical pressure
and volume; the reciprocal of the critical volume is the critical density.
CRITICAL CONSTANTS OF
GASES AND BOILING POINTS
(At atmospheric pressure)
Substance
Critical
Boiling-
Temperature
Pressure,
Density
°C
atmospheres
kgm" 3
point
°C
Argon .
-122
48
0-53 x 10 3
-186
Neon .
-229
27
0-48
-246
Helium .
-268
2-26
0069
-269
Chlorine
146
76
0-57
- 34
Hydrogen
-240
12-8
0-031
-253
Nitrogen
-146
33
0-31
-196
Oxygen.
-118
50
0-43
-183
Air
-140
39
0-35
Ammonia
130
115
0-24
- 33-5
Carbon dioxide
311
73
0-46
- 78-2
Ethylene
10
52
0-22
-102-7
Freon, CC1 2 F 2
112
40
0-56
Sulphur dioxide
155
79
0-52
- 10-8
Water .
374
219
0-4
100
The above account of the phenomena near the critical point is over-simple,
and may create the impression that these phenomena are fully understood. They
are not; but this is not the place to say much about the matter. We may just
point out that, even at temperature well above the critical, and when no meniscus
can be seen, considerable differences of density can be found in a so-called gas.
They have been shown by including, in a sealed tube of liquid and vapour, a
number of small glass balls of different densities. When the tube was heated
above the critical temperature, each ball floated at a point where the substance
had a density equal to that of the ball.
Gases and Vapours
A gas above its critical temperature cannot be liquefied. Early
attempts to liquefy gases such as air, by compression without cooling,
failed ; and the gases were wrongly called 'permanent' gases. We still,
for convenience, refer to a gas as a vapour when it is below its critical
temperature, and as a gas when it is above it. But the distinction is not
the same as that between an ideal gas and one which is far from ideal.
For a gas which is near its critical point, though it may be a little above
its critical temperature, does not obey Boyle's law, as Fig. 12.23 shows.
On the other hand, a vapour which is far from saturation obeys Boyle's
law fairly well.
Refrigeration
The action of a refrigerator depends on the absorption of its latent
heat by a liquid — the working substance— in evaporating. The working
substance must be one whose vapour has a critical temperature above
CHANGE OF STATE
319
normal atmospheric temperatures, so that it can be liquefied by com-
pression alone. Common working substances are ammonia, carbon
dioxide, sulphur dioxide, and spe-
cially developed compounds such as
the two varieties of Freon : CC1 2 F 2 ,
and C 2 C1 2 F 4 . The working substance
is compressed by a pump, P, in Fig.
12.24, and passed through a metal
pipe C ; there the heat of compression
is carried away by circulating water,
and the substance liquefies. The
liquid passes to a reservoir R. From
the reservoir, liquid escapes through
a throttle valve V into the coil D,
which is connected to the low pres-
sure side of the pump. The coil D
lies round the walls of the space to
be cooled (not shown).
When the liquid escapes from the reservoir, it starts to evaporate,
because of the low pressure. It draws its latent heat from its own heat
content, and cools. Not all of the liquid evaporates as it emerges, and
the mixture of cool liquid and vapour passes round the metal coil D.
If the atmosphere in the chamber containing D is warmer than the
liquid, the liquid evaporates further. The latent heat which it requires
is furnished by the surroundings of D, which are therefore cooled.
Fig. 12.24. Refrigeration.
THE EQUILIBRIUM OF SOLID, LIQUID, VAPOUR
We have pointed out that solids as well as liquids evaporate (p. 297). A solid
thus has saturated vapour over it, just as a liquid has, and the pressure of the
saturated vapour depends on the temperature. The table on p. 299 shows the
pressure of saturated water vapour over ice, at — 10°C and — 20°C.
The Triple Point
In Fig. 12.25, the curve AP relates the saturated vapour pressure of ice to its
temperature ; at any point on the curve, ice and water-vapour are in equilibrium.
BP is the saturated- vapour-pressure curve of water ; at any point on it, water
and water-vapour are in equilibrium. CP is the curve relating the melting-point
of ice with the pressure: at any point on it, ice and water are in equilibrium. The
three curves meet at the point P, whose co-ordinates are p = 4-6 mm mercury,
9 = 0-01 °C. These are the only conditions in which ice, water, and water-vapour
can exist together : if either the temperature or pressure is altered, at least one
phase vanishes. If, when the pressure and temperature are altered, their new
values happen to lie on one of the curves, then the two corresponding phases
survive, liquid and solid along PC, for example. But if the new conditions lie in
one of the three sectors of the diagram — say in PAC — then the only phase
which survives is the one corresponding to that sector : solid, in PAC. The point
P is called the triple point.
The curve AP, which gives the saturated vapour pressure of ice, is steeper at
P than the curve BP for water. It is steeper because a solid evaporates less readily
than a liquid — molecules escape from it less easily. Therefore the saturated-
vapour-pressure of the solid falls more rapidly with the temperature.
320
ADVANCED LEVEL PHYSICS
Fig. 12.26 shows the triple point for carbon dioxide. Its co-ordinates are
p = 3800 mm mercury, 6 = — 56-6°C. At atmospheric pressure, 760 mm mercury,
therefore, solid carbon dioxide (C0 2 ) can be in equilibrium with its vapour, but
not with liquid C0 2 . It is therefore dry, and in America it is called 'dry ice'; in
England it is called 'carbon-dioxide snow'. At atmospheric pressure its tem-
perature is — 78-5°C, and it is much used as a coolant — in ice-cream trucks,
for example.
Solid carbon dioxide is prepared by simply opening the valve of a cylinder
containing carbon dioxide at high pressure. The gas rushes out, and does work
4-6 mm
mercury
vapour
001 *C
e
Fig. 12.25. Triple point for water {not to scale).
in acquiring kinetic energy of mass motion. Since the expansion is rapid, it is
adiabatic, and the gas cools. As it does so, it goes over directly to the solid phase.
When solid carbon dioxide is warmed, it goes over directly into vapour. So,
incidentally, do solid iodine and a few other substances, at atmospheric pressure.
The change from solid to vapour is called sublimation. As the diagram shows,
liquid C0 2 cannotexist at any temperature at all, if the pressure is below 3800 mm
mercury (51 atmospheres).
CD
X
E
E
3880
-785
-566
Fig. 12.26. Triple point for carbon dioxide {not to scale).
CHANGE OF STATE 321
Freezing of Solutions
We have seen that dissolving a solid in water lowers its vapour pressure, and
also its freezing-point (p. 297). To explain the lowering of the freezing-point, let
us draw, as in Fig. 12.27, the curves of the saturated vapour pressures of ice,
e — *-c
Fig. 12.27. Equilibrium between ice and a solution in water.
water and solution. We see that the curve for the solution cuts the ice curve at a
point Q which corresponds to a temperature l5 below 0°C. This is the only
temperature at which ice and solution can be in equilibrium. At a higher tem-
perature, 8 2 , ice has the higher vapour pressure: it therefore sublimes faster than
water evaporates from the solution, and, on the whole, vapour from the ice
condenses into the solution. At a lower temperature 3 , the solution has the
higher vapour pressure; water therefore evaporates from it faster than the ice
sublimes, and, on the whole, water from the solution condenses on the ice. Thus
the temperature t is the freezing-point of the solution. It is the temperature at
which solution and ice exchange water molecules one for one, and neither grows
at the expense of the other.
We can now see why ice and salt, for example, form a freezing mixture. When
salt is mixed with ice, it dissolves in the water clinging to the ice, and forms a
solution. Since this is above its melting-point, being at 0°C, it has a lower satura-
tion vapour pressure than the ice (Fig. 12.27). Therefore the ice sublimes and
condenses in the solution. In effect, the ice becomes water. And in doing so it
abstracts its latent heat of fusion from its surroundings. Thus the mixture changes
from solid ice and salt to a liquid solution of salt, and its temperature falls below
0°C.
LIQUEFACTION OF GASES
If one of the so-called permanent gases, hydrogen or nitrogen, is to
be liquefied, it must first be cooled below its critical temperature. There
are three principal ways of doing this : (i) the gas may be passed through
a cold bath containing a more easily liquefied gas, which is boiling at a
reduced pressure and therefore has a very low temperature ; (ii) the gas
may be allowed to expand adiabatically and do work, losing its heat-
energy in the process ; (in) the gas may be cooled by a method depending
on the fact that, for a real gas, the internal energy is not independent
of the volume.
The third of these is the commonest nowadays, and the only one
322
ADVANCED LEVEL PHYSICS
which we shall describe. First we must explain the phenomenon on
which it depends.
The Joule-Kelvin Effect
We have already described, on p. 241, Joule's crude experiments on
the expansion of a gas into a vacuum — a 'free expansion', as it is
called. These experiments suggested that in such an expansion the
Plug
Fig. 12.28. Joule-Kelvin apparatus.
gas lost no internal energy, and therefore did no work. We concluded
that the molecules of a gas had negligible attraction for one another,
since otherwise work would have had to be done against their attrac-
tions whenever the gas expanded.
In 1852, Joule and Kelvin made more delicate experiments of
essentially the same kind. They allowed a gas at high pressure to
expand into a vacuum through a plug of cotton wool (Fig. 12.28). The
plug prevented eddies from forming in the gas, so that the gas did
not acquire any kinetic energy of motion in bulk. Neither did the gas
do any external work, since it pushed back no piston. Nevertheless,
Joule and Kelvin found that the gas was cooled slightly in its passage
through the porous plug. Therefore work must have been done in
separating its molecules ; and this work must have been done at the
expense of their kinetic energy, the heat-energy of the gas.
The magnitude of the cooling in the Joule-Kelvin effect depends on
the temperature at which the gas enters the plug ; for air at room tem-
perature it has the order of 0-1 °C per atmosphere pressure difference.
It is not essential for the gas to expand into a vacuum. Whenever a
gas expands from high pressure to low, its volume increases, and some
work is done against its inter-molecular attractions. If heat cannot
CHANGE OF STATE
323
enter the gas, the work is done at the expense of the gas's internal
energy, and the gas cools. The cooling is analogous to that which takes
place in an adiabatic expansion, but in a normal adiabatic expansion
most of the work is done externally against a piston (compare p. 251).
An ideal gas would cool in an adiabatic expansion with external work,
but not in a free expansion — it would show no Joule-Kelvin effect.
The Linde Process
The cooling of a gas in a free expansion is small, but Linde devised
an ingenious arrangement for making it cumulative, and so producing
a great temperature fall. His apparatus is shown diagrammatically in
Fig. 12.29. When air is to be liquefied, it must first be freed of carbon
dioxide and water, which would solidify and choke the pipes ; both are
removed by solid caustic soda in a vessel not shown in the figure. The
pure air is compressed to about 150 atmospheres by the pump P,
and the heat of compression is removed in the water-cooled copper
coil C. The air then passes down the copper coil D, which runs within
another copper coil E. It emerges through the nozzle N whose opening
can be adjusted from outside. The nozzle lies inside a Dewar vessel
or thermos flask F. The air expands on emerging, and is cooled by the
Joule-Kelvin effect. It then passes upwards through the outer coil E,
and as it does so, cools the incoming gas. The incoming gas is thus
cooled before making its expansion, and after its expansion becomes
cooler still. On escaping through E it cools the following gas yet
further. Thus the cooling of the escaping gas continuously helps the
cooling of the arriving gas, and the cooling is said to be regenerative.
Eventually the gas emerging from the nozzle cools below the critical
temperature; and since the actual pressure, 150 atmospheres, is well
^U
777ZZZZZZZZZZZZZ&,
Fig. 12.29. Liquefaction of air.
324 ADVANCED LEVEL PHYSICS
above the critical pressure, 39 atmospheres, the gas liquefies and
collects in the flask. The liquefier is heavily lagged with insulating
material, G, to prevent heat coming in from the outside.
The reader should appreciate that the regenerative cooling takes
place only in the double coil. At all stages in the process the air enters
the inner coil at the temperature of the cooling water around C. But
as time goes on it passes through ever-cooler gas coming up from the
nozzle, until liquid begins to form and the system reaches equilibrium.
Liquid Nitrogen and Oxygen
As the table on p. 318 shows, the boiling-point of nitrogen, at
atmospheric pressure, is — 196°C, whereas that of oxygen is — 183°C.
When liquid air is exposed to the atmosphere, therefore, the nitrogen
boils off faster and the proportion of liquid oxygen increases. The
so-called liquid air sold commercially is mostly liquid oxygen. It is
more dangerous than true liquid air, particularly if there is hydrogen
about.
Hydrogen and Helium
At orinary temperatures, the Joule-Kelvin effect is reversed for
helium and hydrogen, that is, a free expansion causes warming. We
cannot go into the explanation of that here, but we may say that it is
connected with the fact that the pV/p curves of hydrogen and helium
rise with increasing pressure, instead of falling at first (Fig. 12.17,
p. 314).
These gases show a Joule-Kelvin cooling, however, if they are
sufficiently cooled before the expansion. Hydrogen must be cooled
below — 83°C and helium below — 240°C; these temperatures are
called the inversion temperatures of the gases.
Hydrogen can be cooled below its inversion temperature by passing
it through a coil in liquid air before it enters the double coil of the
liquefier. Helium must be passed through a coil in liquid hydrogen,
boiling under reduced pressure.
EXAMPLES
1. Describe an experiment which demonstrates that the pressure of a vapour
in equilibrium with its liquid depends on the temperature.
A narrow tube of uniform bore, closed at one end, has some air entrapped by a
small quantity of water. If the pressure of the atmosphere is 760 mm of mercury,
the equilibrium vapour pressure of water at 12°C and at 35°C is 10-5 mm of
mercury and 42-0 mm of mercury respectively, and the length of the air column
at 12°C is 10 cm, calculate its length at 35 °C. (L.)
First part. See text.
Second part. For the given mass of air,
PiVi = P2V2
CHANGE OF STATE 325
Pl = 76- 105 = 74-95 cm, V x = 10, T = 273+ 12 = 285 K;
p 2 = 76-4-2 = 71-8 cm, T 2 = 273 + 35 = 308 K;
. 74-95 x 10 _ 71-8 V 2
" * 285 ~ 308
_ 74-95 x 10 x 308 _ « « ,
• * Vl ~ 285x71-8 " Lli -
2. State Dalton's law of partial pressures ; how is it explained on the kinetic
theory? A closed vessel contains air, saturated water-vapour, and an excess of
water. The total pressure in the vessel is 760 mm of mercury when the temperature
is 25°C ; what will it be when the temperature has been raised to 100°C? (Satura-
tion vapour pressure of water at 25°C is 24 mm of mercury.) (C.)
First part. See text.
Second part. From Dalton's law, the pressure of the air at 25°C = 760-24 =
736 mm of mercury. Suppose the pressure is p mm at 100°C. Then, since pressure
is proportional to absolute temperature for a fixed mass of air, we have
p 373
736 ~ 298'
from which p = 921 mm.
Now the saturation vapour pressure of water at 100°C = 760 mm.
... total pressure in vessel = 921 + 760 = 1681 mm mercury.
3. Define relative humidity and dew-point. Describe an instrument with which
the dew-point can be determined. The relative humidity in a closed room at
15°C is 60 per cent. If the temperature rises to 20°C, what will the relative humidity
become? On what assumptions is your calculation based? (Saturation vapour
pressure of water-vapour at 15°C = 12-67 mm of mercury, at 20°C = 17-36 mm.)
(L.)
First part. See text.
Second part. Suppose p is the actual water vapour pressure in mm mercury in
the air at 15°C.
Then
P
12-67
= relative humidity = 60 per cent.
.'. p = y^-x 12-67 = 7-60 mm.
Assuming the pressure of the water- vapour is proportional to its absolute tem-
perature, the pressure p r at 20°C is given by
Pl _ 273 + 20
7-60 "273 + 15"
. n 7-60x293 77 ~ m
. . P = z^t: = 7*73 mm.
* 288
7-734
.*. relative humidity at 20°C = -^=-— x 100 per cent = 45 per cent.
4. What is meant by saturation pressure of water vapour, dew-pointl Describe
briefly the principles underlying two different methods for the determination of
the relative humidity in the laboratory.
326 ADVANCED LEVEL PHYSICS
A barometer tube dips into a mercury reservoir and encloses a mixture of air
and saturated water vapour above the mercury column in the tube, the height
of the column being 70 cm above the level in the reservoir. If the atmospheric
pressure and the saturation pressure of water vapour are respectively 76 cm
and 1 cm of mercury, determine the height of the column when the tube is de-
pressed in the reservoir to reduce the air volume to half its initial value. (L.)
First part. The saturation pressure of water vapour is the pressure of water
vapour in contact with water in a closed space. The dew-point is the temperature
at which the air is just saturated with the water- vapour present. The different
methods for measuring relative humidity concern the dew-point (Regnault)
hygrometer and the wet-and-dry bulb hygrometer, discussed on pp. 311-2.
Second part. We apply the gas laws to the air only, as the mass of the air remains
constant. From Dalton's law.
pressure of air = total pressure— pressure of water-vapour
= (76— 70)— 1 = 5 cm mercury.
The volume of the air changes from V, say, to V/2. Hence the new pressure, p,
of the air is given, from Boyle's law, by
5xV= pxj
.'. p = 10 cm.
.*. new total pressure of mixture of gases = 10+1 = 11 cm.
.". new height of mercury column = 76—11 = 65 cm.
5. What is meant by the relative humidity of the air? Describe in detail a good
method for finding it.
Air at 19-5°C has a relative humidity of 75 per cent. Calculate its dew-point
and the mass of water vapour contained in 1 litre, being given that the boiling-
points of water under pressure of 12 mm, 14 mm, 16 mm and 18 mm of mercury are
14°C, 16-45°C, 18-55°C and 20-45°C respectively. Assume that water vapour
behaves as an ideal gas, that its density at s.t.p is 000080 g per cm 3 . (N.)
First part. The relative humidity of the air is the ratio of the mass of water-
vapour in a given volume of the air to the mass of water-vapour required to
saturate that volume. A good method for finding it is by the dew-point (Regnault)
hygrometer, described on p. 311.
Second part.
_ , . , ... s.v.p. of water at dew-point A n ^
Relative humidity = — - ^ Z n/r , x 100 per cent.
J s.v.p. of water at 19-5°C
, 75 = s.y p. at dew-point xl(X)
17 mm mercury
.". s.v.p. at dew-point — f x 17 = 12-75 mm mercury.
.'. dew-point = 14°C+~x245°C
= 14-9°C,
as s.v.p. at 14°C = 12 mm, at 16-45°C = 14 mm.
To find the mass of water-vapour in 1 litre. The pressure of water- vapour =
CHANGE OF STATE 327
s.v.p. at dew-point = 12-75 mm = 1 -275 can mercury ; the absolute temperature =
273 + 19-5 = 292-5 K.
.'. vol. in cm 3 at s.t.p. = 1,000 x
1-275
X
273
76
292-5
1-275
X
273
.*. mass of water- vapour = 1,000 x x x 00008
76 292-5
= 0013 g.
EXERCISES 12
1. Distinguish between a saturated and an unsaturated vapour. What is
meant by saturation vapour pressure!
Describe an experiment to measure the saturation vapour pressure of water
vapour for temperatures between 20°C and 100°C. Indicate graphically the
results which would be obtained from such an experiment. (L.)
2. Explain concisely four of the following in terms of the simple kinetic theory
of matter :
(a) energy must be supplied to a liquid to convert it to a vapour without
change of temperature ;
(b) when some water is introduced into an evacuated flask, some of the water
at first evaporates, but subsequently, provided the temperature of the flask is
kept constant, the volume of the water present remains unchanged ;
(c) gases are generally poor conductors of heat compared with solids ;
(d) when a gas, which is enclosed in a thermally insulated cylinder provided
with a piston, is compressed by moving the piston, the temperature of the gas is
raised ;
(e) water can be heated by stirring. (O. & C.)
3. Distinguish between a saturated and an unsaturated vapour. Describe an
experiment to investigate the effect of pressure on the boiling point of water and
draw a sketch graph to show the general nature of the results to be expected.
A column of air was sealed into a horizontal uniform-bore capillary tube by a
water index. When the atmospheric pressure was 762-5 torr (mm of Hg) and the
temperature was 20°C, the air-column was 15-6 cm long; with the tube immersed
in a water bath at 50°C, it was 191 cm long, the atmospheric pressure remaining
the same. If the s.v.p. of water at 20°C is 17-5 torr, deduce its value at 50°C.
(O. & C.)
4. What is meant by saturation vapour pressure! Describe an experiment to
investigate the variation of the saturation pressure of water vapour with tem-
perature.
Sketch the isothermal curve relating pressure and volume (a) for a mass of dry
air at room temperature, (b) for water vapour at 100°C. (L.)
5. Explain the physical principles of a domestic refrigerator employing an
evaporating liquid and give a labelled diagram showing its essential components.
How may the temperature of the main storage compartment be regulated and
what factors determine the lowest attainable temperature?
A certain refrigerator converts water at 0°C into ice at a maximum rate of 5 g
per minute when the exterior temperature is 1 5°C. Assuming that the rate at which
heat leaks into the refrigerator from its surroundings is proportional to the
temperature difference between the exterior and interior and is 2-5 watt deg C~ 1 ,
what is the maximum exterior temperature at which this refrigerator could just
maintain a temperature of 0°C in the interior? [Specific latent heat of fusion of
iee = 330kJkg _1 .](O. &C.)
328 ADVANCED LEVEL PHYSICS
6. Use the simple kinetic theory of matter to answer the following questions :
(a) How do gases conduct heat?
(b) Why does a liquid tend to cool when it evaporates?
(c) Why does the boiling point of a liquid depend upon the external pressure?
Show that the pressure p, the density p, and the mean square molecular velocity
c 2 of an ideal gas are related by p = \p<?, stating any assumptions at the points
where they become necessary in the proof. (O. & C.)
7. State Boyle's law and Daltori's law of partial pressures.
The space above the mercury in a Boyle's law apparatus contains air together
with alcohol vapour and a little liquid alcohol. Describe how the saturation
vapour pressure of alcohol at room temperature may be determined with this
apparatus.
A mixture of air and saturated alcohol vapour in the presence of liquid alcohol
exerts a pressure of 12-8 cm of mercury at 20°C. When the mixture is heated at
constant volume to the boiling point of alcohol at standard pressure (i.e. 78 °C),
the vapour remaining saturated, the pressure becomes 860 cm of mercury. Find
the saturation vapour pressure of alcohol at 20°C. (L.)
8. The saturation pressure of water vapour is 1-2 cm of mercury at 14°C and
24 cm of mercury at 25°C. Describe and explain the experiment you would
perform to verify these data.
Explain qualitatively in terms of the kinetic theory of matter (a) what is meant
by saturation vapour pressure, (b) the relative magnitudes of the saturation
vapour pressures quoted.
Sketch a graph showing how the saturation pressure of water vapour varies
between 0°C and 110°C. (N .)
9. Draw p against v curves for temperatures above, at, and below the critical
temperature, for a real gas. Explain the significance of critical temperature.
Sketch graphs of pv against p at the same temperatures for a real gas. What
would be the form of the pv against p curves for an ideal gas?
Describe briefly experiments which provide the data on which these curves
for a real gas are based. (O. & C.)
10. Compare the properties of saturated and unsaturated vapours. By means
of diagrams show how the pressure of (a) a gas, and (b) a vapour, vary with change
(i) of volume at constant temperature, and (ii) of temperature at constant volume.
The saturation vapour pressure of ether vapour at 0°C is 185 mm of mercury
and at 20°C it is 440 mm. The bulb of a constant volume gas thermometer contains
dry air and sufficient ether for saturation. If the observed pressure in the bulb is
1000 mm at 20°C, what will it be at 0°C? (L.)
11. Explain the terms relative humidity, dew-point. Describe in detail an
experiment to determine the dew-point.
Find the mass of air and water- vapour in a room of 20 x 10 x 5 cubic metres
capacity, the temperature being 20°C and the pressure 750 mm of mercury.
Assume that the saturation pressure of water-vapour at the dew-point is 90 mm
of mercury ; that the density of dry air at s.t.p. is 1-30 kg m" 3 ; that the density of
water vapour is f that of air under the same conditions of pressure and tem-
perature. (L.)
12. Define pressure of a saturated vapour, critical temperature. Give an account
of the isothermal curves for carbon dioxide at temperatures above and below its
critical temperature.
Some liquid ether is sealed in a thick-walled glass tube, leaving a space con-
taining only the vapour. Describe what is observed as the temperature of the
tube and its contents is raised above 197°C, which is the critical temperature for
CHANGE OF STATE 329
ether. (Assume that the vessel is strong enough to withstand the internal pressure.)
(L.)
13. Define dew-point and explain what is meant by relative humidity. Describe
how you would determine the dew-point of the atmosphere.
What is the relative humidity of an atmosphere whose temperature is 16-3°C
if its dew-point is 12-5°C? The following table gives the saturation pressure, p,
of water-vapour in mm of mercury at various temperature, t.
t°C
10
12
14
16
18
20
pmm
9-20
10-51
11-98
13-62
15-46
17-51
(AT.)
14. Give briefly the principles of one method each for liquefying (a) chlorine,
(b) hydrogen.
Describe a suitable container for liquid air. Point out carefully the physical
principles involved. (C.)
15. Give an account of three methods used to obtain temperatures below 0°C.
Describe a method for liquefying air. How must the procedure be modified in
order to liquefy hydrogen? (N.)
16. Describe, in detail the method you would use to find (a) the melting-point
of lead, and (b) the boiling-point of brine.
An alloy of copper and silver is made with different percentage composition.
For each mix the melting-point is measured, with the following results : —
M.P.degC . 960 810 760 740 760 790 900 1080
Per cent of copper
in alloy . . 20 30 40 50 60 80 100
Plot a curve showing the relation between the melting-point in degrees Centi-
grade and the percentage of copper present. Comment on the curve. (L.)
chapter thirteen
Transfer of Heat
Conduction
If we put a poker into the fire, and hold on to it, then heat reaches
us along the metal. We say the heat is conducted ; and we soon find that
some substances — metals — are good conductors, and others — such as
wood or glass — are not. Good conductors feel cold to the touch on a
cold day, because they rapidly conduct away the body's heat.
Temperature Distribution along a Conductor
In order to study conduction in more detail consider Fig. 13.1 (a),
which shows a metal bar AB whose ends have been soldered into the
jj (b) Lagged
Fig. 13.1. Temperature fall along lagged and unlagged bars.
walls of two metal tanks, H, C ; H contains boiling water, and C contains
ice-water. Heat flows along the bar from A to B, and when conditions
are steady the temperature 6 of the bar is measured at points along its
length; the measurements may be made with thermojunctions, not
shown in the figure, which have been soldered to the rod. The curve in
330
TRANSFER OF HEAT
331
the upper part of the figure shows how the temperature falls along the
bar, less and less steeply from the hot end to the cold.
The figure 13.1 (b) shows how the temperature varies along the bar,
if the bar is well lagged with a bad conductor, such as asbestos wool.
It how falls uniformly from hot to cold.
The difference between the temperature distributions is due to the
fact that, when the bar is unlagged, heat escapes from its sides, by
convection in the surrounding air. Thus the heat flowing past D per
second, is less than that entering the bar at A by the amount which
escapes from the surface AD. The arrows in the figure represent the
heat escaping per second from the surface of the bar, and the heat
flowing per second along its length. The heat flowing per second along
the length decreases from the hot end to the cold. But when the bar is
lagged, the heat escaping from its sides is negligible, and the flow per
second is constant along the length of the bar.
We thus see that the temperature gradient along a bar is greatest
where the heat flow through it is greatest. We also see that the tempera-
ture gradient is uniform only when there is a negligible loss of heat
from the sides of the bar.
Thermal Conductivity
Let us consider a very large thick bar, of which AB in Fig. 13.2 (i) is
a part, and along which heat is flowing steadily. We suppose that the
&2 01
e 2 01
Fig. 13.2. Definition of thermal conductivity.
loss of heat from the sides of the bar is made negligible by lagging.
XY is a slice of the bar, of thickness /, whose faces are at temperatures
2 and X . Then the temperature gradient over the slice is
02 — fli
/ *
We now consider an element abed of the slice of unit cross-sectional
area, and we denote by Q the heat flowing through it per second.
The value of Q depends on the temperature gradient, and, since some
332 ADVANCED LEVEL PHYSICS
substances are better conductors than others, it also depends on the
material of the bar.
We therefore write
where k is a factor depending on the material.
To a fair approximation the factor k is a constant for a given
material ; that is to say, it is independent of 6 2 , lt and /. It is called the
thermal conductivity of the material concerned. To put its definition
into words, we let 6 2 — B\ be 1°C, and / be 1 m, so that
Q = k.
We then say :
Consider a cube of material, whose faces are 1 m apart, and have a
temperature difference of I deg C. If heat flows in the steady state through
the cube at right angles to its faces, and none is lost from its sides, then
the heat flow per unit area is numerically equal to the conductivity of the
material.
This definition leads to a general equation for the flow of heat through
any parallel-sided slab of the material, when no heat is lost from the
sides of the slab. As in Fig. 13.2 (ii), we denote the cross-sectional area
of the slab by A, its thickness by /, and the temperature of its faces by
d x and 2 . Then the heat Q flowing through it per second is
Q = k -W±=M .... (1)
A useful form of this equation is
|=^ .... (2)
or
heat flow per m 2 per second = conductivity x temperature gradient.
(2a)
In terms of the calculus, (2) may be re- written
Adt K dl {i)
the temperature gradient being negative since diminishes as / increases.
Units and Magnitude of Conductivity
Equation (2) enables us to find the unit of thermal conductivity.
We have
Q/AjJm- 2 *- 1 )
{Ot-ej/lQLm- 1 )'
Thus the unit of thermal conductivity = J s _1 m _1 K _1 , or since
joule second" i = watt (W), the unit of k isWm _1 K _1 .
TRANSFER OF HEAT
333
THERMAL CONDUCTIVITIES
SOLIDS
(Mean values, c. O-100°C)
Substance
k
Substance
k
Substance
k
Ag
420
Asbestos
013
Ice
21
Al
210
Brick .
013
Marble
30
Cu
380
Cardboard
0-21
(white)
Fe pure .
76
Cork .
0-42
Mica
0-76
wrought
59
Cotton .
0-22
Paraffin wax .
0-25
Hg .
8
Cotton wool .
0025
Silica (fused) .
1-4
Ni
87
Ebonite
017
Rubber (para)
019
Pb
35
Felt
0038
Sand .
0-054
Pt .
71
Flannel
0097
Silk
0O92
Brass
109
Glass .
11
Slate .
2-0
Duralumin
130
(window)
Wood .
c.0-21
Steel
46
Graphite
130
LIQUIDS
AND GASES
Liquid
k
Gas
k
Alcohol (25°C) .
018
Air (0°C).
0024
Glycerine (20°C) .
0-29
(100°C) .
0-032
Olive oil (0°C) .
017
C0 2 (0°C) . . .
0015
Paraffin oil (0°C) .
013
H 2 (0°C).
017
Water (10°C)
0-62
N 2 (0°C).
0024
(80°C)
0-67
Q 2 (0°C).
0-024
To a rough approximation we may say that the conductivities of
metals are about 1000 times as great as those of other solids, and of
liquids ; and they they are about 10000 times as great as those of gases.
Effect of Thin Layer of Bad Conductor
Fig. 13.3 shows a lagged copper bar AB, whose ends are pressed
against metal tanks at 0° and 100°C, but are separated from them by
layers of dirt. The length of the bar is 10 cm or 01 m, and the dirt
layers are 01 mm or 01 x 10~ 3 rri thick. Assuming that the conductivity
of dirt is 1/1000 that of copper, let us find the temperature of each
end of the bar.
Suppose k = conductivity of copper,
A = cross-section of copper,
d 1 ,0 1 = temperature of hot and cold ends.
Since the bar is lagged, the heat flow per second Q is constant from
end to end. Therefore,
Q =
k A 100-
-0o
1000 01 x 10
= kA
01
t -O
ToocToi x 10"
zA
334
ADVANCED LEVEL PHYSICS
100X-
I 66-7-
33-3-
ore ^\)
=yjobJc J=
---£_. OX -£^-
\
\ A B
i
i +~ •
i— __!
Vii
y
I mm 0-1 mm
Fig. 13.3. Temperature gradients in good and bad conductors.
Dividing through k, these equations give
100-02 02-0i ^l
01 01 01'
or
lOO-0 2 = 2 -0 1 = 1?
whence
2 = 66-7°C,
t = 33-3°C.
Thus the total temperature drop, 100°C, is divided equally over the
two thin layers of dirt and the long copper bar. The heavy lines in the
figure show the temperature distribution ; the broken line shows what
it would be if there were no dirt.
Good and Bad Conductors
The foregoing example shows what a great effect a thin layer of a
bad conductor may have on thermal conditions ; 01 mm of dirt causes
as great a temperature fall as 10 cm of copper. We can generalize this
result with the help of equation (2a) :
heat flow/m 2 s = conductivity x temperature gradient.
The equation shows that, if the heat flow is uniform, the temperature
gradient is inversely proportional to the conductivity. If the conduc-
tivity of dirt is 1/1000 that of copper, the temperature gradient in it is
1000 times that in copper ; thus 1 mm of dirt sets up the same tempera-
ture fall as 1 m of copper. In general terms we express this result by
saying that the dirt prevents a good thermal contact, or that it provides
a bad one. The reader who has already studied electricity will se an
obvious analogy here. The flow of heat can, in fact, be treated mathe-
matically in the same way as the flow of electricity ; we may say that a
dirt layer has a high thermal resistance, and hence causes a great
temperature drop.
Boiler plates are made of steel, not copper j although copper is about
TRANSFER OF HEAT
335
eight times as good a conductor of heat. The material of the plates
makes no noticeable difference to the heat flow from the furnace out-
side the boiler to the water inside it, because there is always a layer of
gas between the flame and the boiler-plate. This layer may be very
thin, but its conductivity is about 1/10000 that of steel ; if the plate is a
centimetre thick, and the gas-film 1/1000 centimetre, then the tem-
perature drop across the film is ten times that across the plate. Thus the
rate at which heat flows into the boiler is determined mainly by the gas.
If the water in the boiler deposits scale on the plates, the rate of heat
flow is further reduced. For scale is a bad conductor, and, though it
may not be as bad a conductor as gas, it can build up a much thicker
layer. Scale must therefore be prevented from forming, if possible ;
and if not, it must from time to time be removed.
Badly conducting materials are often called insulators. The import-
ance of building dwelling-places from insulating materials hardly
needs to be pointed out. Window-glass is a ten-times better conductor
than brick, and it is also much thinner; a room with large windows
therefore requires more heating in winter than one whose walls are
more modestly pierced. Wood is as bad a conductor (or as good an
insulator) as brick, but it also is thinner. Wooden houses therefore
have double walls, with an air-space between them ; air is an excellent
insulator, and the walls prevent convection. In polar climates, wooden
huts must not be built with steel bolts going right through them;
otherwise the inside ends of the bolts grow icicles from the moisture
in the explorer's breath.
Measurement of High Conductivity: Metals
When the thermal conductivity of a metal is to be measured, two
conditions must usually be satisfied : heat must flow through the
specimen at a measurable rate, and the temperature gradient along the
specimen must be measurably steep. These conditions determine the
form of the apparatus used.
When the conductor is a metal, it is easy to get a fast enough heat
flow ; the problem is to build up a temperature gradient. It is solved by
Steam
.Water
Fig. 13.4. Apparatus for thermal conductivity of a metal.
336 ADVANCED LEVEL PHYSICS
having as the specimen a bar long compared with its diameter. Fig. 13.4
shows the apparatus, which is due to Searle. AB is the specimen,
about 4 cm diameter and 20 cm long. In one form of apparatus it is
heated by steam at A, and cooled by circulating water at B. The whole
apparatus is heavily lagged with felt. To measure the temperature
gradient, thermometers are placed in the two mercury-filled cups C, D ;
the cups are made of copper, and are soldered to the specimen at a
known distance apart. Alternatively, thermometers are placed in holes
bored in the bar, which are filled with mercury. In this way errors due
to bad thermal contact are avoided.
The cooling water flows in at E, round the copper coil F which is
soldered to the specimen, and out at G. The water leaving at G is
warmer than that coming in at E, so that the temperature falls con-
tinuously along the bar: if the water came in at G and out at E, it
would tend to reverse the temperature gradient at the end of the bar,
and might upset it as far back as D or C.
The whole apparatus is left running, with a steady flow of water, until
all the temperatures have become constant : the temperature 2 and U
at C and D in the bar, and 4 and 3 of the water leaving and entering.
The steady rate of flow of the cooling water is measured with a measuring
cylinder and a stop-clock.
If A is the cross-sectional area of the bar and k its conductivity,
then the heat flow per second through a section such as S is
- This heat is carried away by the cooling water ; if a mass m of specific
• heat capacity e w , flows through F in 1 second, the heat carried away is
mc w (0 4 -0 3 ).
Therefore kA 02 ~ 01 = mc w (0 4 -0 3 ).
With this apparatus we can show that the conductivity k is a
constant over small ranges of temperature. To do so we increase the
flow of cooling water, and thus lower the outflow temperature 4 .
The gradient in the bar then steepens, and (0 2 — 0i) increases. When the
new steady state has been reached, the conductivity k is measured as
before. Within the limits of experimental error, it is found to be
unchanged.
Measurement of Low Conductivity: Non-metallic Solids
In measuring the conductivity of a bad conductor, the difficulty is to
get an adequate heat flow. The specimen is therefore made in the form
of a thin disc, D, about 10 cm in diameter and a few millimetres thick
(Fig. 13.5 (a)). It is heated by a steam-chest C, whose bottom is thick
enough to contain a hole for a thermometer.
The specimen rests on a thick brass slab B, also containing a thermo-
meter. The whole apparatus is hung in mid air by three strings attached
toB.
TRANSFER OF HEAT
337
(b) (C) Time —
Fig. 13.5. Apparatus for thermal conductivity of a bad conductor.
To ensure good thermal contact, the adjoining faces of C, D and B
must be flat and clean ; those of C and B should be polished. A trace of
vaseline smeared over each face improves the contact.
When the temperatures have become steady, the heat passing from
C through D escapes from B by radiation and convection. Its rate of
escape from B is roughly proportional to the excess temperature of B
over the room (Newton's law). Thus B takes up a steady temperature
#! such that its rate of loss of heat to the outside is just equal to its
gain through D. The rate of loss of heat from the sides of D is negligible,
because their surface area is small.
This apparatus is derived from one due to Lees, and simplified for
elementary work. If we use glass or ebonite for the specimen, the
temperature 0^ is generally about 70°C; 6 2 is, of course, about 100°C.
After these temperatures have become steady, and we have measured
them, the problem is to find the rate of heat loss from B. To do this,
we take away the specimen D and heat B directly from C until its
temperature has risen by about 10°C. We then remove C, and cover
the top part of B with a thick layer of felt F (Fig. 13.5 (b)). At intervals
of a minute — or less — we measure the temperature of B, and after-
wards plot it against the time (Fig. 13.5 (c)).
While the slab B is cooling it is losing heat by radiation and con-
vection. It is doing so under the same conditions as in the first part
of the experiment, because the felt prevents heat escaping from the
top surface. Thus when the slab B passes through the temperature 6 U
it is losing heat at the same rate as in the first part of the experiment.
The heat which it loses is now drawn from its own heat content, whereas
before it was supplied from C via D ; that is why the temperature of B
is now falling, whereas before it was steady. The rate at which B loses
heat at the temperature X is given by :
heat lost/second = Mc x temperature fall/second,
where M, c are respectively the mass and specific heat capacity of the
slab.
338
ADVANCED LEVEL PHYSICS
To find the rate of fall of temperature at U we draw the tangent to
the cooling curve at that point. If, as shown in Fig. 13.5 (c), its gradient
at #! would give a fall of a deg C in b seconds, then the rate of tem-
perature fall is a/b deg C per second.
We then have, if A is the cross-sectional area of the specimen, / its
thickness, and k its conductivity,
kA 2 . l = Mc T .
I b
Thus k can be calculated.
Liquids
In finding the conductivity of a liquid, the liquid must be heated at
the top and cooled at the bottom, to prevent convection. Lees' apparatus
is therefore suitable. The liquid is held in a narrow glass ring, R,
Fig. 13.6, sandwiched between the plates (not shown) of the Lees' disc
apparatus. Let k g , k t be the conductivities of the glass and liquid
Fig. 13.6. Apparatus for finding thermal conductivity of a liquid.
respectively, and r 2 and r x the inner and outer radii of the ring. Then
the downward heat flow per second is
Q = k g n{r x 2 -r 2 2 )
#2 — 01
/
+ k l nr 2
I '
where 2 and X are the temperatures above and below the specimen
and / is its thickness. The conductivity k g need not be known ; the
heat flow through the ring may be determined in a preliminary experi-
ment with the ring, but without the liquid.
Conduction through a Tube
The conductivity of glass tubing may be measured in the laboratory
with the apparatus shown in Fig. 13.7 (a). The glass tube AB is sur-
rounded by a steam jacket J, and water flows through it from A to B
at a measured rate of m g/second. Thermometers measure the inflow and
outflow temperatures of the water, 2 and 3 , which eventually become
steady. In the steady state, the heat flowing through the walls of the
tubing is equal to the heat carried away by the water, mc w {0 2 — 2 )
joule/second.
TRANSFER OF HEAT
339
To find the conductivity, we must know the area through which the
heat flows. If r 1? r 2 , are the inner and outer radii of the tube, and L is
its length, then the areas of the inner and outer walls are 2nr x L and
2nr 2 L respectively (Fig. 13.7 (b)). If the tube is thin, we may take the
area as constant and equal to its average value.
Thus
A = 2ttzA^
At the entrance of the tube, the temperature gradient is
(Ox — 6 2 )/(r 2 — r^), where X is the temperature of the steam ; at the exit
end the gradient is (0 X — 3 )/(r 2 — rj).
Steam-
(a) (b)
Fig. 13.7. Apparatus to measure conductivity of glass in form of tubing.
If 3 and 2 differ by not more than about 10°C, we may take the
gradient as constant and equal to its average value :
temperature-gradient =
llO 1 — 2 X —
+-
A r 2-ri r 2 -r t
02 + 03
0i-
r,-r,
The conductivity k therefore is given by
k x27rlA±^x
0i-
+ 1
r-> — r,
= mc w (0 3 -d 2 )
(4)
The conductivity of rubber tubing can be found by a modification of
this method. A measured length of the tubing is submerged in a calori-
meter of water, and steam passed through for a measured time t. The
rise in temperature of the water must be corrected for cooling, as in the
measurement of the specific heat of a bad conductor (p. 203). The heat
flow through the rubber is given by the left-hand side of equation (4)
with 6 2 and 3 standing for the initial and final temperature of the
340
ADVANCED LEVEL PHYSICS
water. If m is the mass of water, and C the heat capacity of the calori-
meter, the right-hand side of the equation is
(mc w + Q(0 3 -e 2 )/t.
Comparison of Conductivities
Fig. 13.8 (a) shows an apparatus due to Ingenhousz for comparing
the conductivities of solids. Metal, wood, glass, and other rods, of
equal lengths and cross-sections, are stuck into a tank through corks.
The rods are painted with the same paint, and coated with wax over
100
(a) Apparatus
(b) Theory
Fig. 13.8. Comparison of conductivities.
their whole projecting lengths. The tank is filled with water, leaks are
stopped as well as possible, and the water is boiled. As the rods warm
up, the wax melts off them. Eventually a steady state is reached, and
the best conductor is the one from which wax has melted off the
greatest length.
If the experiment is to give a quantitative comparison of the con-
ductivities, the rods must be so long that the far end of each of them
is at room temperature. Otherwise the following argument will not
be true.
At the points where melted wax gives way to solid, each bar is at the melting-
point of wax — let us call it 50°C. The temperature distributions along any two
bars are therefore as shown in Fig. 13.8 (b); they have similar shapes, but at a
given distance from the tank, (ii) is steeper than (i). If l t is the distance along (i) to
the 50°C point, and / 2 the corresponding distance along (ii), then the curve (ii) is
the same as curve (i) except that it is horizontally contracted in the ratio ijh-
Therefore the gradient of (ii), at any distance x, is steeper than that of (i) in the
ratio ljl 2 - The temperature gradient at the tank end of each rod determines the
rate at which heat flows into it from the hot water.
Now Q oc k x temp. grad. at hot end,
where k is the conductivity of the rod. Therefore, for rods (i) and (ii)
Qx _ ki temp, grad. at end of (i)
Q 2 k 2 temp. grad. at end of (ii)
- *iv!z
X
k 2 h
(5)
TRANSFER OF HEAT 341
The heat passing into a rod at the hot end escapes by convection from its sides.
The lengths l t and l 2 respectively, of rods (i) and (ii), have the same average tem-
peratures, 75°C. Over the lengths l t and l 2 , therefore, each rod loses heat at the
same rate per unit area, since each has the same surface (p. 203). The heat lost
from either rod per second, between the 100°G and the 50°C points, is therefore
proportional to the area of the rod between those points. It is therefore propor-
tional to the distance /between them. ..■•' i-
Since the temperature curves differ only in scale, the distance to the 50°jC point
on either rod is proportional to the distance L to the point where the rod reaches
room temperature (Fig. 13.8 (b) ^Beyond this point, the bar loses no heat. There-
fore, by the above argument, the distance L is proportional to the total heat lost
per second by the bar. The distance L istherefore proportional to the total heat per
second lost by the bar, and this heat is the heat Q entering the bar at the hot end.
Therefore ** QocLacl '
0,1 L>2 '2
or
But we have seen that <L = ^1/2 . . . . (5)
Therefore
Q 2 Mi
k 2 l l '2
^2 '2
Thus the conductivity of a given rod is proportional to the square of the
distance along it to the melting-point of the wax.
The Cracking of Glass
Glass is a bad conductor of heat. Therefore, when a piece of glass is
heated in one place, the neighbouring parts of the glass do not at first
warm up with it. Consequently they resist the expansion of the heated
part, and the force set up cracks the glass (p. 262). To avoid cracking
the glass, care must be taken to warm the whole region around the
place to be made hot. Similarly, glass which has been heated must be
made to cool slowly and uniformly by playing the flame over it now
and then, for shorter and shorter times as it cools.
EXAMPLES
1. Calculate the quantity of heat conducted through 2 m 2 of a brick wall
12 cm thick in 1 hour if the temperature on one side is 8°C and on the other side
is 28°C. (Thermal conductivity of brick = 013 W m~ * K~ '.)
Temperature gradient = — — 77p2°C m" 1 .
Since 1 hour = 3600 seconds,
.'. Q = kAt x temperature gradient
= 013 x 2 x 3600 x' / ? 8 ? / 8 _ 2 joules
12x10 2J
= 156000 J.
342 ADVANCED LEVEL PHYSICS
2. Define thermal conductivity. Describe and give the theory of a method of
measuring the thermal conductivity of copper.
A sheet of rubber and a sheet of cardboard, each 2 mm thick, are pressed
together and their outer faces are maintained respectively at 0°C and 25°C. If
the thermal conductivities of rubber and cardboard^ are respectively 013 and
005 W m~ * K~ x , find the quantity of heat which flows in 1 hour across a piece
of the composite sheet of area 100 cm 2 ; (L.)
First part. The thermal conductivity of a substance is the quantity of heat per
second flowing in the steady state through opposite faces of a unit cube of the
material when a temperature difference of 1 degree is maintained across these faces.
The thermal conductivity of copper can be measured by Searle's method (p. 335).
Second part. We must first find the temperature, 0°C, of the junction of the
rubber and cardboard. The temperature gradient across the rubber = (0—0)/
2x 10" 3 ; the temperature gradient across the cardboard == (25— 0/2 x 10" 3 .
.". <2 per second per m 2 across rubber = 013 x(0— 0)/2 x 10~ 3
and Q per second per m 2 across cardboard = 0-05 x (25 — 0)/2 x 10" 3 .
But in the steady state the quantities of heat above are the same.
. 0-13(0-0) 0-05(25-0)
" 2xl0 -3 ~ 2xl0 -3
..130=125-50
125
.. = ~ = 7°C.
Now area = 100 cm 2 = 100 x 10 -4 m 2 . "
.'. Q through area in 1 hour (3600 seconds)
= 013 x 100 x 10~ 4 x 7 x 3600
2xl0~ 3
= 16380 J.
3. Define thermal conductivity and explain how you Would measure its value
for a poorly conducting solid.
In order to minimise heat losses from a glass container, the walls of the con-
tainer are made of two sheets of glass^ each 2 mm thick, placed 3 mm apart,
the intervening space being filled with a poorly conducting solid. Calculate the
ratio of the rate of conduction of heat per unit area through this composite wall
to that which would have occurred had a single sheet of the same glass been used
under the same internal and external temperature conditions. (Assume that the
thermal conductivity of glass and the poorly conducting solid = 0-63 and
0049 W m~ l K~ 1 respectively. (L.)
First part.
Second part. Let B u 4 be the respective temperatures of the outer faces of the
two glass sheets, and 2 , 3 their respective junction temperatures with the solid
between them. The thickness of glass = 2x 10 ~ 3 m, that of the solid =
0-3 x 10 -2 m. In the steady state, the quantity of heat per second per metre 2 is
the same for each. Call this Q v Then, for the first glass, since Qi = kx tem-
perature gradient,
- V1 ■ 2x10 3
„ . „ 2xl0~ 3 2 ^ ■ ...
•••^-^ 2 = ^ X ^63- = 630^ • ■ ■ »
TRANSFER OF HEAT 343
Similarly, for the solid,
For the second glass,
Adding (i), (ii), (iii) to eliminate 9 2 and 9 3 ,
»•-««- fi. x2 ^r-«|£. • • • <«>
For a single sheet of glass and the same internal and external temperatures
6 t and 6 4 respectively, the quantity of heat per second per metre 2 , Q 2 say, is
given, from (i), by
•■-««- &"S5 ■ • • ■ «)
Hence, from (1), g^+l) = Q 2 x A
Q 14
Simplifying, .-. -i = — = 005 (approx.).
RADIATION
Radiation
All heat comes to. us, directly or indirectly, from the sun. The heat
which comes directly travels through 150 million km of space, mostly
empty, and travels in straight lines, as does the light : the shade of a
tree coincides with its shadow. Both heat and light travel with the
same speed because they are cut off at the same instant in an eclipse.
Since light is propagated by waves of some kind we conclude that the
heat from the sun is propagated by similar waves, and we say it is
'radiated'.
As we show later, radiation is more copious from a dull black body
than from a transparent or polished one. Black bodies are also better
absorbers of radiation than polished or transparent ones, which either
allow radiation to pass through themselves, or reflect it away from
themselves. If we hold a piece of white card, with a patch of black
drawing ink on it, in front of the fire, the black patch soon comes to
feel warmer than its white surround.
Reflection and Refraction
If, with either a convex lens or a concave mirror, we focus the sun's
light on our skin, we feel heat at the focal spot. The heat from the sun
has therefore been re-
flected or refracted in the
same way as the light.
If we wish to show the
reflection of heat un-
accompanied by light, we
may use two search-
light mirrors, set up as in
Fig. 13.9. At the focus of Fig. 13.9. Reflection of radiant heat.
344 ADVANCED LEVEL PHYStCS
one, F t , we put an iron ball heated to just below redness. At the focus
of the other, F 2 , we put the bulb of a thermometer, which has been
blackened with soot to make it a good absorber (p. 349). The mercury
rises in the stem of the thermometer.- If we move either the bulb or
the ball away from the focus, the mercury falls back; the bulb has
therefore been receiving heat from the ball, by reflection at the two
mirrors. We can show that the foci of the mirrors are the same for heat
as for light if we replace the ball and thermometer by a lamp and
screen. (In practice we do this first, to save time in finding the foci for
the main experiment.)
To show the refraction of heat apart from the refraction of light is
more difficult. It was first done by the astronomer Herschel in 1800.
Herschel passed a beam of sunlight through a prism, as shown dia-
grammatically in Fig. 13.10, and explored the spectrum with a sensitive
/ ^ — * Infra red
t- ^^ZT" — Red
^C" Violet
^ Ultra violet
Fig. 13.10. Infra-red and ultra-violet (diagrammatic).
thermometer, whose bulb he had blackened. He found that in the
visible part of the spectrum the mercury rose, showing that the light
energy which it absorbed was converted into heat But the mercury
also rose when he carried the bulb into the darkened portion a little
beyond the red of the visible spectrum ; the sun's rays therefore carried
energy which was not light.
Ultra-violet and Infra-red
The radiant energy which Herschel found beyond the red is now
called infra-red radiation, because it is less refracted than the red.
Radiant energy is also found beyond the violet and it is called ultra-
violet radiation, because it is refracted more than the violet.
Ultra-violet radiation is absorbed by the human skin and causes
sun-burn ; more importantly, it stimulates the formation of vitamin D,
which is necessary for the assimilation of calcium and the prevention
of rickets. It is also absorbed by green plants ; in them it enables water
to combine with carbon dioxide to form carbohydrates. This process
is called photo-synthesis; we have already, on p. 196, discussed its
importance to animals and man. Ultra-violet radiation causes the
emission of electrons from metals, as in photo-electric cells; and it
excites a latent image on a photographic emulsion. It is harmful to
the eyes.
Ultra-violet radiation is strongly absorbed by glass — spectacle-
wearers do not sunburn round the eyes — but enough of it gets through
TRANSFER OF HEAT 345
to affect a photographic film. It is transmitted with little absorption
by quartz.
Infra-red radiation is transmitted by quartz, and rock-salt, but most
of it is absorbed by glass. A little, that which lies near the visible red,
passes fairly easily through glass — if it did not, Herschel would not
have discovered it. When infra-red radiation falls on the skin, it gives
the sensation of warmth. It is what we usually have in mind when we
speak of heat radiation, and it is the main component of the radiation
from a hot body ; but it is in no essential way different from the other
components, visible and ultra-violet radiation, as we shall now see.
Wavelengths of Radiation
In books on Optics, it is shown how the wavelength of light can be
measured with a diffraction grating — a series of fine close lines ruled
on glass. The wavelength ranges from 4000 x 10~ 10 m for the violet,
to 7500 xlO -10 m for the red. The first accurate measurements of
wavelength were published in 1868 by Angstrom, and in his honour a
distance of 10 ~ 10 m is called an Angstrom unit (A.U.). The wavelengths
of infra-red radiation can be measured with a grating made from fine
wires stretched between two screws of close pitch. They range from
7500 A.U. to about 1 000000 A. U. Often they are expressed in a longer
unit than the Angstrom : this unit is the micron (/mi), which is 1/1000 mm.
Thus
l//m = 10 ~ 6 m = 10 4 A.U.
We denote wavelength by the symbol A; its value for visible light
ranges from 0-4 jum to 0-75 fim, and for infra-red radiation from 0-75 //m
to about 100 //m.
We now consider that X-rays and radio waves also have the same
nature as light, and that so do the y-rays from radio-active substances.
X10 10 10" 8 10" 6 10 4 10' 2 1 10* 10 4 cm
y rays X-rays Ultra-violet
/\
Violet 4x1
Infra-red (heat) Radio waves
oV / I \ 7 — " 5 -
•5x10 red
Visible
Fig. 13.11. The electromagnetic spectrum.
For reasons which we cannot here discuss, we consider all these waves
to be due to oscillating electric and magnetic fields. Fig. 13.1 1 shows the
range of their wavelengths : it is called a diagram of the electromagnetic
spectrum.
Detection of Heat Radiation
A thermometer with a blackened bulb is a sluggish and insensitive
detector of radiant heat. More satisfactory detectors, however, are
346
ADVANCED LEVEL PHYSICS
Fig. 13.12. Bolometer strip.
less direct; they are of two main kinds, both
electrical. One kind consists of a long thin
strip of blackened platinum foil, arranged in
a compact zigzag (Fig. 13.12). On this the
radiation falls. The foil is connected in a
Wheatstone bridge, to measure its electrical
resistance. When the strip is heated by the
radiation* its resistance increases, and the
increase is measured on the bridge. The instru-
ment was devised by Langley in 1881 ; it is called a bolometer, bole
being Greek for a ray.
The other, commoner, type of radiation detector is called a thermo-
pile (Nobili and Melloni, c. 1830). Its action depends on the electro-
motive force, which appears between the junctions of two different
metals, when one junction is
hot and the other cold. The
modern thermopile is due to
Coblenz (1913). It consists of
many junctions between fine
wires, as shown diagramma-
tically in Fig. 13.13; the wires
are of silver and bismuth, 01
mm or less in diameter. Their
junctions are attached to thin
discs of tin, about 0-2 mm
thick, and about 1 mm square.
One set of discs is blackened
and mounted behind a slit,
through which radiation can
fall on them; the junctions
attached to them become the
hot junctions of the thermopile.
The other, cold, junctions are
shielded from the radiation to
be measured; the discs attached to them help to keep them cool, by
increasing their surface area.
Older types of thermopile are made from bars of metal about a
millimetre thick. They are slow to warm up when radiation falls upon
them, but are more rugged than the modern type.
When radiation falls on the blackened discs of a thermopile, it
warms the junctions attached to them, and sets up an e.m.f. This e.m.f.
can be measured with a potentiometer, or, for less accurate work, it
can be used to deflect a galvanometer, G, connected directly to the
ends of the thermopile (Fig. 13.13).
Reflection and Refraction observed with Thermopile: Inverse Square
Law
With a thermopile and galvanometer, we can repeat Herschel's
experiment more strikingly than with a thermometer. And with the
Slit
Fig. 13.13. Thermopile.
TRANSFER OF HEAT
347
simple apparatus of Fig. 13.13 we can show that, when heat is reflected,
the angle of reflection is equal to the angle of incidence. We can also
show the first law of reflection ; that the incident and reflected rays
are in the same plane as the normal to the reflector at the point of
incidence.
Polished metal plate
Thermopile
Fig." 13.14. Demonstration of reflection.
If heat is radiant energy, its intensity should fall off as the inverse
square of the distance from a point source. We can check that it does
so by setting up an electric lamp, with a compact filament, in a dark
room preferably with black walls. When we put a thermopile at dif-
ferent distances from the lamp, the deflection of the galvanometer is
found to be inversely proportional to the square of the distance.
If we wish to do this experiment with radiation that includes no
visible light, we must modify it. Instead of the lamp, we use a large
blackened tank of boiling water, A, and we fit the thermopile, B, with
a conical mouthpiece, blackened on the inside. The blackening prevents
any radiation from reaching the pile by reflection at the walls of the
mouthpiece. We now find that the deflection of the galvanometer, G,
does not vary with the distance of the pile from the tank, provided that
the tank occupies the whole field of view of the cone (Fig. 13.15). The area
S of the tank from which radiation can reach the thermopile is then
proportional to the square of the distance d. And since the deflection is
unchanged when the distance is altered, the total radiation from each
element of S must therefore fall off as the inverse square of the distance d.
Fig. 13.15. Proof of inverse square law.
348
ADVANCED LEVEL PHYSICS
Fig. 13.16. Infra-red spectrometer.
The Infra-red Spectrometer
Infra-red spectra are important in the study of molecular structure.
They are observed with an infra-red spectrometer, whose principle is
shown in Fig. 13.16. Since glass is opaque to the infra-red, the radiation
* s is focused by concave mirrors
\ instead of lenses ; the mirrors
\ are plated with copper or gold
on their front surfaces. The
source of light is a Nernst
filament, a metal filament
coated with alkaline-earth
oxides, and heated electrically.
The radiation from such a
filament is rich in infra-red. A
carbon arc, or a gas-mantle,
may be used, however.
The slit S of the spectrometer
is at the focus of one mirror
which acts as a collimator.
After passing through the rock-
salt prism, A, the radiation is
focused on to the thermopile P
by the mirror M 2 , which re-
places the telescope of an optical spectrometer. Rotating the prism
brings different wavelengths on to the slit; the position of the prism is
calibrated in wavelengths with the help of a grating.
To a fair approximation, the deflection of the galvanometer is
proportional to the radiant power carried in the narrow band of wave-
lengths which fall on the thermopile. If an absorbing body, such as a
solution of an organic compound, is placed between the source and the
slit, it weakens the radiation passing through the spectrometer, in
the wavelengths which it absorbs. These wavelengths are therefore
shown by a fall in the galvanometer deflection.
Reflection, Transmission, Absorption
Measurements whose description is outside our scope give the amount of
radiant energy approaching the earth from the sun. At the upper limit of our
atmosphere, it is about 80 J cm -2 min~ *.
At the surface of the earth it is always less than this because of absorption in the
atmosphere. Even on a cloudless day it is less, because the ozone in the upper
atmosphere absorbs much of the ultra-violet.
In Fig. 13.17, XY represents a body on which radiant energy is falling. The
symbol / denotes the latter 's intensity : to fix our ideas we may take
/ = 40 joule per cm 2 per minute
= —joule per cm 2 per second
= 0-067 watt per cm 2
TRANSFER OF HEAT 349
A=aT
absorbed
X
Fig. 13.17. Reflection, transmission, and absorption.
Some of this energy is reflected by the glass (R), some is absorbed (4), and
some is transmitted (T). The total energy transmitted, absorbed and reflected per
cm 2 per second is equal to the energy falling on the body over the same area
and in the same time :
T+A+R = I.
If we denote by t, a, and r, the fractions of energy which are respectively trans-
mitted, absorbed, and reflected by the body, then
tl+al + rl = I
or t+a+r =1 (6)
This equation expresses common knowledge: if a body is transparent (t-*l),
it is not opaque, and it is not a good reflector (a-+0, r-+0). But also, if the body is
a good absorber of radiation (a-*l), it is not transparent, and its surface is dull
(t->0, r-»0). And if it is a good reflector (r->lX it is neither transparent nor a good
absorber (t— +0^ a-*0). The term opaque, as commonly used, simply means not
transparent ; we see that it does not necessarily mean absorbent.
Equation (6), as we have written it above, is over-simplified. For a body may
transmit some wavelengths (colours, if visible) and absorb or reflect others. If
we now let / denote the intensity of radiation of a particular wavelength A, then
by repeating the argument we get
tx+ax+rx =1 (7)
where the coefficients t x , etc., all refer to the wavelength A.
The truth of equation (7) is well shown by the metal gold, which reflects yellow
light better than other colours. In thin films, gold is partly transparent, and the
light which it transmits is green. Green is the colour complementary to yellow ;
gold removes the yellow from white light by reflection, and passes on the rest by
transmission.
Radiation and Absorption
We have already pointed out that black surfaces are good absorbers
and radiators of heat, and that polished surfaces are bad absorbers
and radiators. This can be demonstrated by the apparatus in Fig. 13.18,
in which is a cubical metal tank whose sides have a variety of finishes :
dull black, dull white, highly polished. It contains boiling water, and,
therefore, has a constant temperature. Facing it is a thermopile, P,
350
ADVANCED LEVEL PHYSICS
Constanta n
Fig. 13.18. Comparing radiators.
which is fitted with the blackened conical mouth-piece described on
p. 347.
Provided that the face of the cube occupies the whole field of view
of the cone, its distance from the thermopile does not matter (p. 347).
The galvanometer deflection is greatest when the thermopile is facing
the dull black surface of the cube, and least when it is facing the highly
polished surface. The highly polished surface is therefore the worst
radiator of all, and the dull black is the best.
This experiment was first done by Leslie in 1804. There were no
thermopiles in those days, and Leslie detected the radiant heat with an
instrument depending on the expansion of air, which we shall not
describe. The tank with different surfaces is called Leslie's cube.
Leslie's cube can also be used in an experiment to compare the
absorbing properties of surfaces, due to Ritchie (1833). A modern
version of it is shown in Fig. 13.19. The cube C, full of boiling water, is
placed between two copper plates,
A, B, of which A is blackened and
B is polished. The temperature
difference between A and B is
measured by making each of them
one element in a thermo junction :
they are joined by a constantan
wire, XY, and connected to a gal-
vanometer, by copper wires, AE,
DB. If A is hotter than B, the
junction, X, is hotter than the
junction, Y, and a current flows
through the galvanometer in one
direction. If B is hotter than A, the
current is reversed.
The most suitable type of Leslie's
cube is one which has two opposite
faces similar — say grey — and the
other two opposite faces very dis-
EKL^ D similar — one black, one polished.
Fig. 13.19. Comparing absorbers. At first the plates A, B are set
TRANSFER OF HEAT
351
Vacuum
Silver
opposite similar faces. The blackened plate, A, then becomes the hotter,
showing that it is the better absorber.
The cube is now turned so that the blackened plate, A, is opposite
the polished face of the cube, while the polished plate, B, is opposite
the blackened face of the cube. The galvanometer then shows no
deflection ; the plates thus reach the same temperature. It follows that
the good radiating property of the blackened face of the cube, and the
bad absorbing property of the polished plate, are just compensated
by the good absorbing property of the blackened plate, and the bad
radiating property of the polished face of the cube.
The Thermos Flask
A thermos flask — sometimes called a Dewar flask after its inventor
(c. 1894) — is a device for reducing Cork
the transfer of heat to a minimum.
It consists of a double walled glass
vessel, as shown in Fig. 13.20; the
space between the walls is exhausted
to as high a vacuum as possible, and
the insides of the walls are silvered.
Silvered surfaces are good reflectors,
but bad absorbers and radiators.
Heat therefore passes very slowly
from the outer wall to the inner by
radiation. If the vacuum is good,
convection is almost inhibited — the
goodness of the vacuum determines
the goodness of the flask. Conduc-
tion through the glass is slight,
because the conduction paths are long. In a good flask, the main
cause of heat loss is conduction through the cork.
The Black Body
The experiments described before lead us to the idea of a perfectly
black body ; one which absorbs all the radiation that falls upon it, and
reflects and transmits none. The experiments also lead us to suppose
that such a body would be the best possible radiator.
A perfectly black body can be very nearly realized — a good one can
be made in half a minute, simply by punching a small hole in the lid
of an empty tin. The hole looks almost
black, although the shining tin is a
good reflector. The hole looks black
because the light which enters through
it is reflected many times round, the
walls of the tin, before it meets the hole
again (Fig. 13.21). At each reflection,
about 80 per cent of the light energy
is reflected, and 20 per cent is absorbed. fig. 13.21 . Multiple reflections
After two reflections, 64 per cent of the make a black body.
Fig. 13.20. A thermos flask.
352 ADVANCED LEVEL PHYSICS
original light goes on to be reflected a third time ; 36 per cent has been
absorbed. After ten reflections, the fraction of the original energy which
has been absorbed is 0-8 10 , or 01.
Any space which is almost wholly enclosed
□ approximates to a black body. And, since a
good absorber is also a good radiator, an almost
closed space is the best radiator we can find.
A form of black-body which is used in
radiation measurements is shown in Fig. 13.22.
It consists of a porcelain sphere, S, with a small
hole in it. The inside is blackened with soot
to make it as good a radiator and as bad a
reflector as possible. (The effect of multiple
Fig. 13.22. A black body. re fl ect i on s is then to convert the body from
nearly black to very nearly black indeed.)
The sphere is surrounded by a high-temperature bath of, for example,
molten salt (the melting-point of common salt is 801 °C).
The deepest recesses of a coal or wood fire are black bodies. Anyone
who has looked into a fire knows that the deepest parts of it look
brightest — they are radiating most power. Anyone who has looked
into a fire also knows that, in the hottest part, no detail of the coals
or wood can be seen. That is to say, the radiation from an almost
enclosed space is uniform ; its character does not vary with the nature
of the surfaces of the space. This is so because the radiation coming
out from any area is made up partly of the radiation emitted by that
area, and partly of the radiation from other areas, reflected at the area
in question. If the surface of the area is a good radiator, it is a bad
reflector, and vice-versa. And if the hole in the body is small, the
radiations from every area inside it are well mixed by reflection before
they can escape; the intensity and quality of the radiation escaping
thus does not depend on the particular surface from which it escapes.
When we speak of the quality of radiation we mean the relative
intensities of the different wavelengths that it comprises ; the proportion
of red to blue, for example. The quality of the radiation from a perfectly
black body depends only on its temperature. When the body is made
hotter, its radiation becomes not only more intense, but also more
nearly white ; the proportion of blue to red in it increases. Because its
quality is determined only by its temperature, black-body radiation is
sometimes called 'temperature radiation'.
Properties of Temperature Radiation
The quality of the radiation from a black body was examined by
Lummer and Pringsheim in 1899. They used a black body represented
by B in Fig. 13.23 and measured its temperature with a thermocouple;
they took it to 2000°C. To measure the intensities of the various wave-
lengths, Lummer and Pringsheim used an infra-red spectrometer and
a bolometer (p. 346) consisting of a single platinum strip.
The results of experiments such as these are shown in Fig. 13.24 (a).
Each curve gives the relative intensities of the different wavelengths,
TRANSFER OF HEAT
353
for a given temperature
of the body. The curves
show that, as the tem-
perature rises, the inten-
sity of every wavelength
increases, but the inten-
sities of the shorter wave
lengths increase more
rapidly. Thus the radia-
tion becomes, as we
have already observed,
less red, that is to say,
more nearly white. The
curve for sunlight has
its peak at about 5000
A.U., in the visible
green ; from the position
of this peak we con-
clude that the surface
temperature of the sun is
about 6000 K. Stars
which are hotter than the sun, such as Sirius and Vega, look blue, not
white.
| l 22x10" 10
Bolometer strip
Fig. 13.23. Lummer and Pringsheim's apparatus for
study of black body radiation (diagrammatic).
3
<
£
a
60X10 3
! i
Violet- I I -Red
Visible
X, A.U.
Fig. 13.24(a). Distribution of intensity in black-body radiation.
354
ADVANCED LEVEL PHYSICS
because the peaks of their radiation curves lie further towards the
visible blue than does the peak of sunlight.
The actual intensities of the radiations are shown on the right of the
graph in Fig. 13.24 (a). To speak of the intensity of a single wavelength
is meaningless because there is an infinite number of wavelengths, but
the total intensity of the radiation is finite. The slit of the spectrometer
always gathers a band of wavelengths — the narrower the slit the
narrower the band — and we always speak of the intensity of a given
band. We express it as follows (V represents 'second'):
energy radiated m~ 2 s~ 1 , in band X to X + 5X = E X 5X. (8)
The quantity E x is called the emissive power of a black body for the
wavelength X and at the given temperature ; its definition follows from
equation (8) :
_ energy radiated m ~ 2 s ~ \ in band X to X + SX
x ~ bandwidth, bX
The expression 'energy per second' can be replaced by the word
'power', whose unit is the watt. Thus
p _ power radiated m~ 2 in band X, X + SX
In the figure £ ;
Fig. 13.24 (b). Definition of E ,
k , and E x .
is expressed in watt per m 2 per Angstrom unit.
SI units may be 'watt per metre 2
per nanometre (10 -9 m)\
The quantity E x dX in equation
(8) is the area beneath the radia-
tion curve between the wave-
lengths X and X + SX (Fig. 13.24
(b)). Thus the energy radiated
per cm 2 per second between
those wavelengths is proportional
to that area. Similarly the total
radiation emitted per cm 2 per
second over all wavelengths is
proportional to the area under
the whole curve.
Laws of Black Body Radiation
The curves of Fig. 13.24 (a) can be explained only by the quantum
theory of radiation, which is outside our scope. Both theory and
experiment lead to three generalizations, which together describe well
the properties of black-body radiation :
(i) If X m is the wavelength of the peak of the curve for T K, then
X m T = constant .... (9)
The value of the constant is 2-9 x 10" 3 m K. In Fig. 13.24 (a) the dotted
line is the locus of the peaks of the curves for different temperatures.
TRANSFER OF HEAT
355
(ii) If E Xm is the height of the peak of the curve for the temperature
TK,then "*
£ Am ocT 5 .... (10)
The relationships (10) and (9) are particular cases of a general law
given by Wien in 1894 ; (9) is sometimes called WierCs displacement law.
(iii) If E is the total energy radiated per metre 2 per second at a tem-
perature T, represented by the area under the curve, then
E = trT 4 ,
where cr is a constant. This result is called Stefan's law, and the constant
a is called Stefan's constant. Its value is
<r = 5-7xl0 _8 W
m
-2 w--4
K
Prevost's Theory of Exchanges
In 1792 Prevost applied the idea of dynamic equilibrium to radiation.
He asserted that a body radiates heat at a rate which depends only
on its surface and its temperature, and that it absorbs heat at a rate
depending on its surface and the temperature of its surroundings.
When the temperature of a body is constant, the body is losing heat
by radiation, and gaining it by absorption, at equal rates.
It is easy to think of experiments which seem to support Prevost's
theory, and the reader will certainly grasp the general idea of it if he
imagines hot pies and cold ice-creams put into the same cupboard.
But in such experiments it is difficult to get rid of the possibility of
convection. Let us rather take an old-fashioned, high vacuum, electric
lamp, and put it in a can of water (Fig. 13.25 (a)). We can find the
temperature of the lamp's filament by measuring its resistance. We
find that, whatever the temperature of the water, the filament comes
to that temperature, if we leave it long enough. When the water is
cooler than the filament, the filament cools down ; when the water is
hotter, the filament warms up.
(a) (b)
Fig. 13.25. Illustrating Prevost's theory of exchanges.
356
ADVANCED LEVEL PHYSICS
In the abstract language of theoretical physics, Prevost's theory is
easy enough to discuss. If a hot body A (Fig. 13.25 (b)) is placed in an
evacuated enclosure B, at a lower temperature than A, then A cools
until it reaches the temperature of B. If a body C, cooler than B, is
put in B, then C warms up to the temperature of B. We conclude that
radiation from B falls on C, and therefore also on A, even though A
(a) (b)
Fig. 13.26. Illustrating Prevost's theory.
is at a higher temperature. Thus A and C each come to equilibrium
at the temperature of B when each is absorbing and emitting radiation
at equal rates.
Now let us suppose that, after it has reached equilibrium with B,
one of the bodies, say C, is transferred from B to a cooler evacuated
enclosure D (Fig. 13.26 (a)). It loses heat and cools to the temperature
of D. Therefore it is radiating heat. But if C is transferred from B to
a warmer enclosure F, then C gains heat and warms up to the tempera-
ture of F (Fig. 13.26 (b)). It seems unreasonable to suppose that C
stops radiating when it is transferred to F ; it is more reasonable to
suppose that it goes on radiating but, while it is cooler than F, it absorbs
more than it radiates.
Emissivity
Let us consider a body B, in equilibrium with an enclosure A, at a
temperature T K (Fig. 13.27). If the body is perfectly black, it emits
radiation characteristics of the temperature T ; let us write the total
intensity of this radiation over all wavelengths as E watts/m 2 . Since
the body is in equilibrium with the enclosure, it is absorbing as much
as it radiates. And since it absorbs all the radiation that falls upon it,
the energy falling on it per cm 2 per second must be equal to E. This
conclusion need not surprise us, since the enclosure A is full of black
body radiation characteristic of its temperature T.
Now let us consider, in the same enclosure, a body C which is not black. On
each square metre of the body's surface, E watts of radiation fall (Fig. 13.27). Of
this, let us suppose that the body absorbs a fraction a, that is to say, it absorbs aE
watts per m 2 . We may call a the total absorption factor of the body C, 'total'
because it refers to the total radiation. The radiation which the body does not
absorb, (1 — a)E, it reflects or transmits.
TRANSFER OF HEAT
357
/ =0-a)E
eE=aE
aE absorbed '
Fig. 13.27. Equilibria in an enclosure.
Thus:
aE.
power reflected or I _ p_
transmitted/m 2 j —
For equilibrium, the total power leaving the body per m 2 must be equal to the
total power falling upon it, E W/m 2 . The power emitted by the body, which must
be added to that reflected and transmitted, is therefore :
total power radiated/m 2 = aE
(12)
The ratio of the total power radiated per m 2 by a given body, to that emitted
by a black body at the same temperature, is called the total emissivity of the given
body. Hence, by equation (12),
aE
e = — = a.
E
We have therefore shown that the total emissivity of a body is equal to its total
absorption factor.
This is a formal expression of the results of Ritchie's experiment (p. 350). If
we combine it with Stefan's law, we find that the total energy £ radiated per m 2
per second by a body of emissivity e at a temperature T K is
E = eE = eaT A .
Spectral Emissivity; Kirchoff's Law
Most bodies are coloured; they transmit or reflect some wavelengths better
than others. We have already seen that they must absorb these wavelengths
weakly; we now see that, because they absorb them weakly, they must also
Fig. 13.28. Photographs
showing how a piece of
incandescent decorated
crockery appears (a) by
reflected light and (b) by
its own emitted light.
358
ADVANCED LEVEL PHYSICS
radiate them weakly. To show this, we have only to repeat the foregoing argument,
but restricting it to a narrow band of wavelengths between A and X+5X. The
energy falling per m 2 per second on the body, in this band, is E x dX where E x is the
emissive power of a black-body in the neighbourhood of A, at the temperature of
the enclosure. If the body C absorbs a fraction a x of this, we call a x the spectral
absorption factor of the body, for the wavelength A. In equilibrium, the body
emits as much radiation in the neighbourhood of A as it absorbs ; thus :
energy radiated = a x E x dX watts per m 2 .
We define the spectral emissivity of the body e x , by the equation
energy radiated by body in range X,X+SX
x energy radiated in same range, by black body at same temperature
_ energy radiated by body in range A, X+dX
JzjTx
_ <*xE x SX
Thus
E X SX
e x = a x
(13)
f- Black body
Equation (13) expresses a
law due to Kirchhoff :
The spectral emissivity of
a body, for a given wave-
length, is equal to its spec-
tral absorption factor for
the same wavelength.
Kirchhoff' s law is not easy to
demonstrate by experiment.
One reads that a plate, which
when cold shows a red pattern
on a blue ground, glows blue
on a red ground when heated
in a furnace. But not all such
plates do this, because the
spectral emissivities of many
coloured pigments vary with
their temperature. However,
Fig. 13.28 shows two photo-
graphs of a piece of pottery,
one taken by reflected light at
room temperature (left), the
other by its own light when
heated (right).
Fig. 13.29 illustrates Kirch-
hoff 's law, by showing how the
spectral emissivity and absorp-
tion factor of a coloured body may vary with wavelength, and how its
emissive power E x does likewise.. It is assumed that e x rises to unity
at the wavelength A t {which is not likely), and that it does not vary
Fig. 13.29. Illustrating Kirchhoff 's law
of radiation.
TRANSFER OF HEAT
359
with the temperature. A body for which e x is the same for all wavelengths,
but is less than unity, is said to be 'grey'.
Absorption by Gases
An experiment which shows that, if a body radiates a given wave-
length strongly, it also absorbs that wavelength strongly, can be made
with sodium vapour. A sodium vapour lamp runs at about 220°C;
compared with the sun, or even an arc-lamp, it is cool. The experiment
consists of passing sunlight or arc-light through a spectroscope, and
observing its continuous spectrum. The sodium lamp is then placed in
the path of the light, and a black line appears in the yellow. If the white
light is now cut off, the line which looked black comes up brightly — it
is the sodium yellow line.
The process of absorption by sodium vapour — or any other gas — is
not, however, the same as the process of absorption by a solid. When a
solid absorbs radiation, it turns it into heat — into the random kinetic
energy of its molecules. It then re-
radiates it in all wavelengths, but
mostly in very long ones, because the
solid is cool. When a vapour absorbs
light of its characteristic wavelength,
however, its atoms are excited; they
then re-radiate the absorbed energy,
in the same wavelength (5893 A.U.
for sodium). But they re-radiate it in
all directions, and therefore less of it
passes on in the original direction
than before (Fig. 13.30). Thus the
yellow component of the original
beam is weakened, but the yellow
light radiated sideways by the sodium is strengthened. The sideways
strengthening is hard to detect, but it was shown by R. W. Wood in
1906. He used mercury vapour instead of sodium. The phenomenon is
called optical resonance, by analogy with resonance in sound.
\
White
/
— gu
5893 A.U.
/
\
White* weakened
in 5893 A.U.
Fig. 13.30. Absorption by
sodium vapour.
CONVECTION
Liquids — except mercury, which is a molten metal — are bad con-
ductors of heat. If we hold a test-tube full of water by the bottom, we
can boil the water near the top in a Bunsen flame, without any dis-
comfort. But if we hold the tube at the top, and heat it at the bottom,
then the top becomes unbearably hot long before the water boils.
The heat is brought to the top by convection ; the warm water at the
bottom expands, becomes less dense, and rises; the cold water sinks
to take its place. If we heat a beaker of water at one side, and drop in a
crystal of potassium permanganate, we can see the currents of hot
water rising, and cold descending. Central-heating systems rely on
convection to bring hot water from the boiler, in the basement, to the
so-called radiators, and to take the cooler water back to the boiler.
360 ADVANCED LEVEL PHYSICS
The radiators of a central-heating system are wrongly named; they
are con vectors. They warm the air around them, which rises, and gives
way to cooler air from cooler parts of the room. Gases are even worse
conductors of heat than liquids, and for most practical purposes we
can neglect conduction through them altogether. Woollen clothes keep
us warm because they contain pockets of air, which hardly conduct at
all, and cannot convect because they cannot move. The wool fibres
themselves are much better conductors than the air they imprison.
Convection by air is important in ventilation : the fire in a room
maintains a draught of hot air up the chimney, and cool fresh air
from outside comes in under the door. The draught also helps to keep
the fire supplied with oxygen; factory chimneys are made tall to
stimulate convection and increase the draught.
Forced and Free (Natural) Convection
A gas or a liquid may carry away heat from a hot body by convection.
If the flow of liquid or gas is simply due to its being heated by the body,
and hence rising, the convection is said to be free, or natural. But if the
gas or liquid is flowing in a stream maintained by some other means,
then the convection is said to be forced. Thus cooling one's porridge in
the obvious way is an example of forced convection ; it causes a more
rapid loss of heat than does natural convection.
Critical Diameter of Pipes
Hot- water and steam pipes are often lagged with asbestos to reduce
the loss of heat from them. The temperature drop across the lagging
makes the outside cooler than the pipe, and so, by Newton's law, tends
to reduce the rate at which heat escapes from it. However, the lagging
increases the outside diameter of the whole, and so increases its area of
contact with the atmosphere. The increase in area tends to make
convection more vigorous, by enabling the pipe to heat a greater mass
of air. If the diameter of the pipe is small, the increase in area may more
than offset the reduction temperature of the outside, and so increase
the rate of heat loss. Thus there is a critical diameter of pipe ; if the
diameter is less than the critical value, the pipe should not be lagged.
The critical diameter depends on many factors, but is commonly of the
order of 1 cm.
The Greenhouse
A greenhouse keeps plants warm by inhibiting convection. The
glass allows radiant heat to reach the plants from the sun, but prevents
the warm air in the greenhouse from escaping. In winter, when there is
little sunshine, the heat is provided by hot water pipes. In summer the
temperature is regulated by opening or closing the roof and windows,
and so adjusting the loss of heat by convection.
TRANSFER OF HEAT 361
EXERCISES 13
Conduction
1. Define the thermal conductivity of a substance. Describe how the thermal
conductivity of a metal may be measured, pointing out the sources of error in
the experiment.
A large hot-water tank has four steel legs in the form of cylindrical rods 2-5 cm
in diameter and 15 cm long. The lower ends of the legs are in good thermal
contact with the floor, which is at 20°C, and their upper ends can be taken to be
at the temperature of the water in the tank. The tank and the legs are well lagged
so that the only heat loss is through the legs. It is found that 22 watts are needed
to maintain the tank at 60°C. What is the thermal conductivity of steel? When a
sheet of asbestos 1-5 mm thick is placed between the lower end of each leg and the
floor only 5 watts are needed to maintain the tank at 60°C. What is the thermal
conductivity of asbestos? (O. & C.)
2. Define thermal conductivity and state a unit in which it is expressed.
Explain why, in an experiment to determine the thermal conductivity of copper
using a Searle's arrangement, it is necessary (a) that the bar should be thick, of
uniform cross-section and have its sides well lagged, (b) that the temperatures used
in the calculation should be the steady values finally registered by the thermo-
meters.
Straight metal bars X and Y of circular section and equal in length are joined
end to end. The thermal conductivity of the material of X is twice that of the
material of Y, and the uniform diameter of X is twice that of Y. The exposed
ends of X and Y are maintained at 100°C and P C, respectively and the sides of
the bars are ideally lagged. Ignoring the distortion of the heat flow at the junction,
sketch a graph to illustrate how the temperature varies between the ends of the
composite bar when conditions are steady. Explain the features of the graph and
calculate the steady temperature of the junction. (N.)
3. Give a critical account of an experiment to determine the thermal con-
ductivity of a material of low thermal conductivity such as cork. Why is it that
most cellular materials, such as cotton wool, felt, etc., all have approximately the
same thermal conductivity?
One face of a sheet of cork, 3 mm thick, is placed in contact with one face of a
sheet of glass 5 mm thick, both sheets being 20 cm square. The outer faces of this
square composite sheet are maintained at 100°C and 20°C, the glass being at
the higher mean temperature. Find (a) the temperature of the glass-cork interface,
and (b) the rate at which heat is conducted across the sheet, neglecting edge effects.
[Thermal conductivity of cork = 6-3 x lO -2 W m _1 K _1 , thermal con-
ductivity of glass = 7-14 x KT 1 W m' 1 K -1 .] (O. & C.)
4. Define coefficient of thermal conductivity. Describe a method of measuring
this coefficient for a metal.
Assuming that the thermal insulation provided by a woollen glove is equivalent
to a layer of quiescent air 3 mm thick, determine the heat loss per minute from a
man's hand, surface area 200 cm 2 on a winter's day when the atmospheric air
temperature is — 3°C. The skin temperature is to be taken as 34°C and the
thermal conductivity of air as 24 x 10~ 3 W m~ l K~ 1 . (L.)
5. Define thermal conductivity. Describe in detail a method of determining the
thermal conductivity of cork in the form of a thin sheet.
The base and the vertical walls of an open thin-walled metal tank, filled with
water maintained at 35°C, are lagged with a layer of cork of superficial area 2-00 m 2
382
ADVANCED LEVEL PHYSICS
and 1-00 cm thick and the water surface is exposed. Heat is supplied electrically to
the water at the rate of 250 watts. Find the mass of water that will evaporate per
day, if the outside surface of the cork is at 15°C. [Assume that the thermal con-
ductivity of cork is 50 x 10" 2 Wm _1 K" 1 and that the latent heat of vaporiza-
tion of water at 35°C is 2520 J g -1 .] (L.)
6. Explain what is meant by the coefficient of thermal conductivity of a metal
One end of a long uniform metal bar is heated in a steam chest and the other is
kept cool by a current of water. Draw sketch graphs to show the variation of
temperature along the bar when the steady state has been attained (a) when the
bar is lagged so that no heat escapes from the sides, (b) when the bar is exposed
to the air. Explain the shape of the graph in each case.
The surface temperatures of the glass in a window are 20°C for the side facing
the room and 5°C for the outside. Compare the rate of flow of heat through (i) a
window consisting of a single sheet of glass 50 mm thick, and (ii) a double-glazed
window of the same area consisting of two sheets of glass each 2-5 mm thick
separated by a layer of still air 50 mm thick. It may be assumed that the steady
state has been attained.
[Use the following values of coefficient of thermal conductivity: glass: 10
War 1 K _1 ; air: 2-5 x KT 2 Wm" 1 K -1 .] (C.)
7. Describe the construction of a Dewar (vacuum) vessel and explain the
physical features which result in a reduction to a minimum of the heat exchange
between the interior and exterior. Explain why such a vessel is equally suitable for
thermally isolating a cold or a hot body.
A copper sphere of radius 0-5 cm is suspended in an evacuated enclosure by a
copper wire of diameter 0-01 cm and length 3 cm. An insulated electrical heating
coil in good thermal contact with the sphere is connected through the wall of
the enclosure by two copper leads of negligible resistance each of diameter 0-02 cm
and length 5 cm. What rate of heating in the coil is required to maintain the
sphere at a temperature 50°C above that of the surroundings assuming that heat
is lost only by conduction along the supports and along the electrical leads?
When a steady state has been reached, the coil is disconnected from the electrical
supply and the initial rate of fall of temperature of the sphere is found to be
0O13°C per second. Calculate the specific heat of copper, assuming that the
electrical leads are still kept at the temperature of the surroundings at the points
where they pass through the wall of the enclosure, that the temperature gradient
in the sphere is negligible and given that the thermal capacity of the heating coil
is equal to that of 1-5 g of copper. [Density of copper = 9-0 g cm -3 . Thermal
conductivity of copper = 380 Wm -1 K~VJ (0. & C.)
8. Define thermal conductivity.
A 'cold probe', i.e. an instrument to produce a low temperature at its extremity,
consists of a solid copper rod 1 mm in diameter attached axially to a well-lagged
1mm
Fig. 13a.
TRANSFER OF HEAT 363
copper reservoir which holds boiling liquid nitrogen, temperature 78 K (see
Fig. 13a). The distance from the bottom of the reservoir to the tip of the probe is
20 cm. The curved surface of the rod is coated with a non-conducting material.
Assuming that no heat can reach the copper except through its flat tip, calculate
the maximum rate at which heat can be accepted there if the temperature is not
lo rise above — 10°C.
(Thermal conductivity of copper = 385 W m _1 K _1 .)
If you were given such a probe and told to use it to determine the mean con-
ductivity of copper over this temperature range, how would you proceed?
Describe the kind of apparatus you would use and specify (a) the quantities you
would need to measure, (b) any data you would need to know. (O. & C.)
Radiation
9. A hot body, such as a wire heated by an electric current, can lose energy to
its surroundings by various processes. Outline the nature of each of these processes.
A black body of temperature t is situated in a blackened enclosure maintained
at a temperature of 10°C. When t = 30°C the net rate of loss of energy from the
body is equal to 10 watts. What will the rate become when t = 50°C if the energy
exchange takes place solely by the process of radiation? What percentage error is
there in the answer obtained by basing the solution on Newton's law of cooling?
(C)
10. Explain what is meant by a black body. How do the total energy radiated
by a black body and its distribution among the wavelengths in the spectrum
depend upon the temperature of the radiator?
Describe the structure of an optical pyrometer and explain how it is used to
measure the temperature of a furnace. (L.)
11. Explain what is meant by black body radiation and how it can be obtained
in practice.
Give an account of Prevost's theory of exchanges and show how it can be
used in conjunction with Stefan's law to obtain an expression for the net rate of
loss of heat by a black body cooling in an evacuated enclosure.
Sketch the curves relating intensity of radiation and wavelength of radiation
from a black body, for three different temperatures. (L.)
12. Explain what is meant by Stefan's constant. Defining any symbols used.
A sphere of radius 200 cm with a black surface is cooled and then suspended
in a large evacuated enclosure the black walls of which are maintained at 27°C.
If the rate of change of thermal energy of the sphere is 1-848 J s -1 when its
temperature is — 73°C, calculate a value for Stefan's constant. (N.)
13. What is Prevost's Theory of Exchanges? Describe some phenomenon of
theoretical or practical importance to which it applies.
A metal sphere of 1 cm diameter, whose surface acts as a black body, is placed
at the focus of a concave mirror with aperture of diameter 60 cm directed towards
the sun. If the solar radiation falling normally on the earth is at the rate of 014
watt cm -2 , Stefan's constant is taken as-6x 10~ 8 Wm -2 K -4 and the mean
temperature of the surroundings is 27°C, calculate the maximum temperature
which the sphere could theoretically attain, stating any assumptions you make.
(O.&C.)
14. State Newton's law of cooling and Stefan '« fourth power law. Describe an
experiment to test the validity of one of these laws.
A sphere of copper cools at the rate of 10 deg C min" * when at a temperature
of 70°C in an enclosure at 20°C. Calculate its rate of cooling when its temperature
364 ADVANCED LEVEL PHYSICS
is raised to (a) 100°C, (b)700°C, assuming the validity of Newton's law. Repeat the
calculations assuming the validity of Stefan's law. Comment on your answers. (L.)
15. Explain what is meant by (a) a black body, (b) black body radiation.
State Stefan's law and draw a diagram to show how the energy is distributed
against wavelength in the spectrum of a black body for two different temperatures.
Indicate which temperature is the higher.
A roof measures 20 m x 50 m and is blackened. If the temperature of the
sun's surface is 6000 K, Stefan's constant = 5-72 x 10" 8 Wm~ 2 K~ 4 , the
radius of the sun is 7-5 x 10 8 m and the distance of the sun from the earth is
1-5 x 10 11 m, calculate how much solar energy is incident on the roof per minute,
assuming that half is lost in passing through the earth's atmosphere, the roof
being normal to the sun's rays. (0. & C.)
16. What is black body radiationl
Using the same axes sketch graphs, one in each instance, to illustrate the
distribution of energy in the spectrum of radiation emanating from (a) a black
body at 1000 K, (b) a black body at 2000 K and (c) a source other than a black
body at 1000 K. Point out any special features of the graphs.
Indicate briefly how the relative intensities needed to draw one of these graphs
could be determined. (N.)
17. How can the temperature of a furnace be determined from observations
on the radiation emitted?
Calculate the apparent temperature of the sun from the following information :
Sun's radius : 704 x 10 5 km.
Distance from earth : 14-72 x 10 7 km.
Solar constant : ; 14 watt per cm 2 .
Stefan's constant: 5-7 x 10 -8 Wm -2 K -4 . (AT.)
18. Describe one experiment to show that a polished metal surface is a poor
absorber of heat, and one experiment to show that such a surface reflects a high
proportion of a beam of light falling on it. Briefly compare heat and light radiations
from the standpoint of (a) velocity, (b) effect at a distance, (c) simple refraction,
(d) transmission through material substances. (N.)
19. A block of metal is heated and (a) exposed to ordinary atmospheric con-
ditions, or (b) placed in a high vacuum. State concisely the factors that govern the
rate at which its temperature falls under conditions (a) and (b). Energy is supplied
at the rate of 165 watts to a closed cylindrical canister 5 cm in radius and 15 cm
high, filled with water and exposed to the air of the room, which is at 15 CT C. It is
found that the temperature of the water remains steady at 80°C. Find the rate of
heat loss per unit area of the vessel per deg C excess temperature. Estimate also
the fall of temperature in a minute, when the energy supply is shut off. Neglect
the weight of the canister itself. (L.)
20. Describe an experiment to show (a) that the spectrum of an incandescent
solid includes both visible and invisible radiations, (b) how the fraction of incident
radiation transmitted by glass depends on the temperature of the source of the
radiation.
The sun's rays are focussed by a concave mirror of diameter 12 cm fixed with its
axis towards the sun on to a copper calorimeter, where they are absorbed. If the
thermal capacity of the calorimeter and its contents is 247-8 joules per deg C and
the temperature rises 8°C in 2 min, calculate the heat received in 1 min by a square
metre of the earth's surface when the rays are incident normally. (N.)
21. Give an account of Stefan's law of radiation, explaining the character of
the radiating body to which it applies and how such a body can be experimenatlly
realized.
TRANSFER OF HEAT 365
If each square cm of the sun's surface radiates energy at the rate of 6-3 x 10 3 J
s~ 1 cm -2 and Stefan's constant is 5-7 x 10~ 8 W m~ 2 K -4 , calculate the tempera-
ture of the sun's surface in degrees centigrade, assuming Stefan's law applies
to the radiation. (L.)
22. Explain what is meant by black body radiation and how it can be obtained
in practice.
Give an account of Prevost's theory of exchanges and show how it can be used
in conjunction with Stefan's law to obtain an expression for the net rate of loss of
heat by a black body cooling in an evacuated enclosure.
Sketch the curves relating intensity of radiation and wavelength of radiation
from a black body, for three different temperatures. (L.)
chapter fourteen
Thermometry and Pyrometry
REALIZATION OF TEMPERATURE SCALE
In Chapter 8 we discussed the general idea of a temperature scale.
To establish such a scale we need :
(i) some physical property of a substance — such as the volume of a
particular liquid — which increases continuously with increasing
hotness, but is constant at constant hotness;
(ii) two standard degrees of hotness — the fixed points (ice and
steam) — which can be accurately reproduced.
The Fixed Points and Fundamental Interval
The temperature of melting ice and saturated steam are chosen as
the fixed points on the Celsius temperature scale. We have seen that
the atmospheric pressure must be specified in defining the steam point,
but this is not necessary in defining the ice point, because the melting-
point of ice changes very little with pressure (p. 296). On the other hand
impurities in water do not affect the temperature of saturated steam,
but impurities in ice do affect its melting-point (p. 221); the ice used in
realizing the lower fixed point must therefore be prepared from pure
water.
Then, if P is the chosen temperature-measuring quantity, its values
P at the ice point, and P 100 ,at the steam point determine the funda-
mental interval of the scale: P i0 o~ ?o- And the temperature 8 p on
the P-scale, which corresponds to a value P of P is, by definition,
e p= p P °~ P p * 1QQ -
• r 100~^0
On the thermodynamic scale, the triple point of water (p. 319) is
chosen as one fixed point and is defined as 273- 16 K. The other fixed
point is the absolute zero (see p. 190).
The Thermometric Substance and Property
Most thermometers are of the liquid-in-glass type, because it is
simple and cheap ; they contain either mercury or alcohol.
The mercury and alcohol scales agree fairly well with one another,
and with either of the gas scales. The gas scales depend on the change
of volume at constant pressure and of pressure at constant volume
(p. 225). In practice the constant volume scale is always used, because
a change of pressure is easier to measure accurately than a change of
volume.
The mercury scale agrees better with the gas scales than does the
366
THERMOMETRY AND PYROMETRY
367
alcohol scale. However, even the best mercury thermometers disagree
slightly amongst themselves. The discrepancies may arise because the
bores of the tubes are not uniform, or the mercury is impure, or the
glass is not homogeneous.
In most accurate work, therefore, temperatures are measured by the
changes in pressure of a gas at constant volume. At pressures of the
order of one atmosphere, different gases give slightly different tempera-
ture scales, because none of them obeys the gas laws perfectly. But as
the pressure is reduced, the gases approach closely to the ideal, and
their temperature scales come together. By observing the departure
of a gas from Boyle's law at moderate pressures it is possible to allow
for its departure from the ideal ; temperatures measured with the gas
in a constant volume thermometer can then be converted to the values
which would be given by the same thermometer if the gas were ideal.
The Constant- Volume Gas Thermometer
Fig. 14.1 shows a constant volume hydrogen thermometer, due to
Chappius (1884). B is a bulb of platinum-iridium, holding the gas.
cm
Fig. 14.1. Constant volume hydrogen thermometer (not to scale).
The volume is defined by the level of the index I in the glass tube A.
The pressure is adjusted by raising or lowering the mercury reservoir R.
A barometer CD is fitted directly into the pressure-measuring system ;
if H x is its height, and h the difference in level between the mercury
surfaces in A and C, then the pressure H of the hydrogen, in mm
mercury is
H = H 1 + h.
H is measured with a cathetometer.
The glass tubes A, C, D, all have the same diameter to prevent errors
due to surface tension ; and A and D are optically worked to prevent
errors due to refraction (as in looking through common window-glass).
368 ADVANCED LEVEL PHYSICS
Observations made with a constant-volume gas thermometer must
be corrected for the following errors :
(i) the expansion of the bulb B ;
(ii) the temperature of the gas in the tube E and A, which lies
between the temperature of B and the temperature of the room ;
(iii) the temperature of the mercury in the barometer and mano-
meter.
The expansion of the bulb can be estimated from its coefficient of
cubical expansion, by using the temperature shown by the gas thermo-
meter. Since the expansion appears only as a small correction to the
observed temperature, the uncorrected value of the temperature may
be used in estimating it. The tube E is called the 'dead-space' of the
thermometer. Its diameter is made small, about 0-7 mm, so that it
contains only a small fraction of the total mass of gas. Its volume is
known, and the temperatures at various points in it are measured with
mercury thermometers. The effect of the gas in it is then allowed for
in a calculation similar to that used to calculate the pressure of a gas
in two bulbs at different temperatures (p. 262). Mercury thermometers
may be used to measure the temperatures because the error due to
the dead-space is small ; any error in allowing for it is of the second
order of small quantities. For the same reason, mercury thermometers
may be used to measure the temperature of the manometer and baro-
meter.
A gas thermometer is a cumbersome instrument, demanding much
skill and time, and useless for measuring changing temperatures. In
practice, gas thermometers are used only for calibrating electrical
thermometers — resistance thermometers and thermocouples. The read-
ings of these, when they are used to measure unknown temperatures
can then be converted into temperatures on the ideal gas scale.
The International Temperature Scale
Because of the wide use of electrical thermometers, a scale of tempera-
ture based on them is used throughout the laboratories of the world.
It is called the international scale of temperature, and is defined so that
temperatures expressed on it agree, within the limits of experimental
accuracy, with the same temperatures expressed on the ideal gas scale.
For the purpose of calibrating electrical thermometers, subsidiary
fixed points, in addition to the fundamental fixed points of ice and steam,
have been determined with the constant-volume gas thermometer.
They are all measured at an atmospheric pressure of 760 mm mercury.
Their values are given, with those of the fundamental fixed points,
in the following table.
FIXED POINTS OF THE INTERNATIONAL TEMPERATURE SCALE
(a) Boiling point of liquid oxygen — 182-970°C.
(b) Ice point (fundamental)
(c) Steam point (fundamental)
(d) Boiling-point of sulphur
(e) Freezing-point of silver
(/) Freezing-point of gold .
0000°C.
100000°C.
444-600°C.
960-800°C.
1063000°C.
THERMOMETRY AND PYROMETRY
369
The methods of interpolating between these fixed points will be
described below.
Electric Thermometers
Electrical thermometers have great advantages over other types.
They are more accurate than any except gas thermometers, and are
quicker in action and less cumbersome than those.
The measuring element of a thermo-electric thermometer is the
welded junction of two fine wires. It is very small in size, and can there-
fore measure the temperature almost at a point. It causes very little
disturbance wherever it is placed* because the wires leading from it
are so thin that the heat loss along them is usually negligible. It has
a very small heat capacity, and can therefore follow a rapidly changing
temperature. To measure such a temperature, however, the e.m.f. of
the junction must be measured with a galvanometer, instead of a
potentiometer, and some accuracy is then lost.
The measuring element of a resistance thermometer is a spiral of fine-
wire. It has a greater size and heat capacity than a thermojunction,
and cannot therefore measure a local or rapidly changing temperature.
But, over the range from about room temperature to a few hundred
degrees Centigrade, it is more accurate.
Resistance Thermometers
Resistance thermometers are usually made of platinum. The wire
is wound on two strips of mica, arranged crosswise as shown in Fig.
(a) Construction
-»~A
(b) Connection in bridge
Fig. 14.2. Platinum resistance thermometer (P
and S = R).
Q, so that B compensates A ;
370 ADVANCED LEVEL PHYSICS
14.2 (a). The ends of the coil are attached to a pair of leads A, for con-
necting them to a Wheatstone bridge. A similar pair of leads B is near
to the leads from the coil, and connected in the adjacent arm of the
bridge (Fig. 14.2 (b)). At the end near the coil, the pair of leads B is
short-circuited. If the two pairs of leads are identical, their resistances
are equal, whatever their temperature. Thus if P = Q the dummy pair,
B, just compensates for the pair A going to the coil ; and the bridge
measures the resistance of the coil alone.
The platinum resistance thermometer is used to measure tempera-
tures on the international scale between the boiling-point of oxygen
and 630°C (630°C is approximately the freezing-point of antimony,
but it is not a fixed point on the scale). The platinum used in the coil
must be of high purity. Its purity is judged by the increase in its resistance
from the ice point to the steam point. Thus if JR and R l00 are the
resistances of the coil at these points, then the coil is fit to reproduce the
international temperature scale if
^120 > 1.3910.
From the boiling-point of oxygen (— 182-970°C) to the ice-point,
the temperature 6, on the international scale, is given by the equation
R e = R o [l + Ae + B6 2 + C(0- 100)03] . . (i)
Here R e is the resistance of the coil, and A, B, C are constants. The
constants A and B are determined in a way which we shall describe
shortly. When they are known the constant C can be determined from
the value of R at the boiling-point of oxygen.
From the ice-point to 630°C the temperature 9 is given by
R = R O (1+AO + B0 2 ).
The constants A and B are the same as A and B in equation (1) ; they
can be determined by measuring R e at the steam point and the sulphur
point (444-600°C).
At temperatures below the boiling-point of oxygen the resistance of
platinum changes rather slowly with temperature. The resistance of
lead changes more rapidly, and resistance thermometers of lead wire
have been used.
Thermocouples
Between 630°C and the gold point (10630°C) the international
temperature scale is expressed in terms of the electromotive force of a
thermocouple. The wires of the thermocouple are platinum, and
platinum-rhodium alloy (90 per cent Pt. : 10 per cent Rh.). Since the
e.m.f. is to be measured on a potentiometer, care must be taken that
thermal e.m.f.'s are not set up at the junctions of the thermocouple wires
and the copper leads to the potentiometer. To do this three junctions
are made, as shown in Fig. 14.3 (a). The junctions of the copper leads to
the thermocouple wires are both placed in melting ice. The electro-
THERMOMETRY AMD PYROMETRY
371
motive force of the whole system is then equal to the e.m.f. of two
platinum/platmum-rhodium junctions, one in ice and the other at the
unknown temperature (Fig. 14.3 '(b)).
Unknown
tempi, Q — I
OX e
(b) Electrical equivalent
(a) Actual arrangement
Fig. 14.3. Use of thermocouples.
The international temperature corresponding to an e.m.f. £ is
given by
E = a+b0 + c0 2 ,
where a, b and c are constants. The values of the constants are deter-
mined by measurements at the gold point (10630°C), the silver poult
(960-8°C), and the temperature of freezing antimony (about 630-3°C).
Since the freezing-point of antimony is not a fixed point on the inter-
national scale, its value in a given experiment is directly measured with
a resistance thermometer. This temperature therefore serves to link
the resistance and thermo-electric regions of the temperature scale.
Other Thermocouples
Because of their convenience, thermocouples are Used to measure
temperatures outside their range on the international scale, when the
highest accuracy is not required. The arrangement of three junctions
and potentiometer may be used, but for less accurate work the potentio-
meter may be replaced by a galvanometer G, in the simpler arrange-
ment of Fig. 14.4 (a). The galvanometer scale may be calibrated to
read directly in temperatures, the known melting-points of metals like
tin and lead being used as subsidiary fixed points. For rough work,
particularly at high temperatures, the cold junction may be omitted
(Fig. 14.4 (b)). An uncertainty of a few degrees in a thousand is often of
no importance.
372
ADVANCED LEVEL PHYSICS
water
Fig. 14.4. Simple thermojunction thermometer.
E.M.F.'S OF THERMOJUNCTIONS
(In millivolts, cold junction at 0°C)
Temp, of hot
Pt./PL-
Chromel 1 /
Copper/
Iron/
junction °C
10% Rh.
Alumel 2
Constantan 3
Constantan
100
0-64
41
4
5
200
1-44
84
9
11
300
2-32
12-2
15
16
400
3-25
16-4
21 5
22
500
4-22
20-6
27
600
5-22
24-9
33
700
6-26
291
39
800
7-33
33-3
45
900
8-43
37-4
1000
9-57
41-3
1200
11-92 4
48-8
1400
14-31
55-8
1600
16-67
1. Cr-Ni alloy.
2. Al-Ni alloy.
3. 60 per cent Cu, 40 per cent Ni (sometimes called Eureka).
4. Liable to a change of fundamental interval if heated above 1,100°C.
5. Cu starts to oxidize.
Intermediate Metals
Fig. 14.5 (a) shows two metals of a thermocouple, A, B, separated by a third
metal, C. The metal C may be, for example, a film of the solder used to join A and
B. At a given temperature 6, the e.m.f. E of the couple ACB is found, by measure-
ment, to be equal to that of a simple couple AB, formed by twisting or welding the
the wires together (Fig. 14.5 (ft)). This is true provided that the junctions of A to
C, and C to B, are both at the temperature 6. An intermediate metal, at a uniform
temperature, does not therefore affect the e.m.f. of a thermojunction.
Expanding-liquid Thermometers: Mercury-in-glass
Mercury freezes at — 39°C and boils, under atmospheric pressure, at
357°C. Mercury thermometers can be made to read up to about 550°C,
THERMOMETRY AND PYROMETRY
-E-
373
(a) ^-' (b)
Fig. 14.5. Intermediate metal in a thermojunction.
however, by filling the space above the liquid with nitrogen, which is
compressed as the mercury expands, and raises its boiling-point.
A mercury thermometer has a much greater heat capacity than a
thermocouple, and cannot follow a rapidly changing temperature. Also
glass, when it has been warmed and then cooled, does not immediately
contract back to its original volume at the lower temperature. If a low
temperature is measured immediately after a high one, the value given
by a mercury thermometer tends to be too low. With modern hard
glass (Jena glass) this effect is small ; but with a cheap thermometer the
ice point may be as much as half a degree low if taken immediately
after the steam point has been checked.
A mercury thermometer is filled by warming and dipping, as a weight
thermometer (p. 283). The mercury in its bulb is then boiled, to drive
out air. It is then allowed to cool and draw in the requisite amount of
mercury. Finally it is warmed a little above the highest temperature it is
to measure, sealed off, and left for about a year — to age. During the
ageing period the glass slowly contracts after its strong heating, and at
the end of the period the thermometer is calibrated.
Clinical Thermometers
A clinical thermometer has a fine stem, divided into fifths or tenths of
a degree, and calibrated over only a small range : 95°-l 10°F or 35°-45°C
(Fig. 14.6). The stem is thickened on the side remote from the gradua-
95 98-4
i*i I I I ' I I l I
110'F J
Fig. 14.6. Clinical thermometer.
tions so that it acts as a lens, to magnify the fine mercury thread. At
room temperature the mercury retreats right into the bulb. Between the
bulb and the graduations is a fine kink K. When the bulb is warmed the
mercury is forced through the kink into the stem. But when the bulb is
cooled, the mercury does not flow back past the kink — it stays in the
stem, so that the temperature can be read at leisure. It is then shaken
back into the bulb.
374 ADVANCED LEVEL PHYSICS
Mercury-in-steel Thermometers
For industrial purposes mercury-in-steel thermometers are used.
They consist of a steel bulb B, connected by a long steel capillary S to a
coiled steel tube C (Fig. 14.7). The whole is filled with mercury, and when
<(< frrrmrr 7>
S
Fig. 14.7. Mercury-in-steel thermometer (diagrammatic).
the bulb is warmed the expansion of the mercury makes the coil unwind.
The unwinding of the coil actuates a pointer, and indicates the tempera-
ture of the bulb. The distance between the bulb and the indicating dial
may be many feet.
Alcohol Thermometers
Ethyl alcohol boils at 78°C, and freezes at - 1 15°C. Alcohol thermo-
meters are therefore used in polar regions. Alcohol is also used in some
of the maximum and minimum thermometers which we are about to
describe.
Maximum and Minimum Thermometers
Meteorologists observe the highest and lowest temperatures reached
by the air day and night. They use a maximum thermometer which is a
mercury thermometer containing a small glass index I (Fig. 14.8 (a)).
The thermometer is laid horizontally, in a louvred screen. When the
temperature rises the mercury pushes the index along, but when the
temperature falls the mercury leaves the index behind. The maximum
temperature is therefore shown by the end of the index nearer the
mercury. After each observation, the index is brought back to the
mercury by tilting the thermometer.
(a) Maximum
////*/ WtWHt^ , , 3)
(b) Minimum
Fig. 14.8. Meteorological thermometers.
THERMOMETRY AND PYROMETRY
375
A minimum thermometer is similar to a maximum, except that it
contains alcohol instead of mercury. When the alcohol expands it
flows past the index, but when it contracts it drags the index back,
because of its surface tension (Fig. 14.8 (b)). The end of the index
nearer the meniscus therefore shows the minimum temperature. The
index is re-set by tilting.
A combined maximum and minimum thermometer was invented by
Six in 1782. Its construction is shown in Fig. 14.9. The bulb A, and
the part B of the stem, contain
alcohol; so does the part D of the
stem, and the lower part of the
bulb E. The part C of the stem
contains mercury, and the upper
part of the bulb E contains air
and saturated alcohol vapour. The
indices G and H are made of iron,
and are fitted with springs pressing
against the walls of the stem. When
the temperature rises, the expansion
of the large volume of alcohol in A
forces the mercury round, and com-
presses the gases in E. The mercury
pushes the index G up the tube.
When the temperature falls, the
alcohol in A contracts, and the
mercury retreats. But the spring
holds the index G, and the alcohol
flows past it. Thus the bottom of G
shows the maximum temperature.
Similarly, the bottom of H shows
the minimum. The indices are re-set
by dragging them along from the outside with a magnet.
Bimetal Strip Thermometers
If a bimetal strip is wound into a spiral, with the more expansible
metal on the inside, then the spiral will uncoil as the temperature rises.
The movement of the spiral can be made to turn a pointer, and to act
as a thermometer. Instruments of this kind do not hold their calibration
as well as liquid-in-glass thermometers.
Fig. 14.9. Combined maximum and
minimum thermometer.
PYROMETERS
High temperatures are usually measured by observing the radiation
from the hot body, and the name Pyrometry is given to this measure-
ment. Before describing pyrometers, however, we may mention some
other, rough, methods which are sometimes used. One method is to
insert in the furnace a number of ceramic cones, of slightly different
compositions; their melting-points increase from one to the next by
about 20°C. The temperature of the furnace lies between the melting-
376
ADVANCED LEVEL PHYSICS
points of adjacent cones, one of which softens and collapses, and the
other of which does not.
The temperature of steel, when it is below red heat, can be judged by
its colour, which depends on the thickness of the oxide film upon it-
Temperatures below red heat can also be estimated by the use of paints,
which change colour at known temperatures.
Radiation pyrometers can be used only above red-heat (about
600°C). They fall into two classes :
(i) total radiation pyrometers, which respond to the total radiation
from the hot body, heat and light ;
(ii) optical pyrometers, which respond only to the visible light.
Optical Pyrometers
Fig. 14.10 illustrates the principle of the commonest type of optical
pyrometer, called a disappearing filament pyrometer. It consists essen-
Fig. 14.10. Optical radiation pyrometer (not to scale).
tially of a low power telescope, OE, and a tungsten filament lamp L.
The eyepiece E is focused upon the filament F. The hot body A whose
temperature is to be found is then focused by the lens O so that its
image lies in the plane of F. The light from both the filament and the
hot body passes through a filter of red glass G before reaching the eye.
If the body is brighter than the filament, the filament appears dark on a
bright ground. If the filament is brighter than the body, it appears
bright on a dark ground. The temperature of the filament is adjusted, by
adjusting the current through it, until it merges as nearly as possible
into its background. It is then as bright as the body. The rheostat R
which adjusts the current is mounted on the body of the pyrometer,
and so is the ammeter A which measures the current. The ammeter is
calibrated directly in degrees Centigrade or Fahrenheit. A pyrometer
of this type can be adjusted to within about 5°C at 1000°C; more
elaborate types can be adjusted more closely.
The range of an optical pyrometer can be extended by introducing a
filter of green glass between the objective O and the lamp L ; this reduces
the brightness of the red light. A second scale on the ammeter is provided
for use when the filter is inserted.
The scale of a radiation pyrometer is calibrated by assuming that
the radiation is black-body radiation (p. 352). If — as usual — the hot
body is not black, then it will be radiating less intensely than a black
THERMOMETRY AND PYROMETRY
377
body at the same temperature. Conversely, a black body which radiates
with the same intensity as the actual body will be cooler than the actual
body. Thus the temperature indicated by the pyrometer will be lower
than the true temperature of the actual, not black, body. A correction
must be applied to the pyrometer reading, which depends on the
spectral emissivity of the body for red light. The wavelength X for which
the spectral emissivity e x must be known is the average wavelength of
light transmitted by the red filter — usually about 6500 A.U.
The following tables give e x for various substances, and the correc-
tions to be added for various values of e x .
SPECTRAL EMISSIVITIES, e x
(At650nm)
Substance
Solid
Liquid
Substance
Solid
Liquid
Carbon
0-85
—
Nickel
0-35
—
Copper
01
015
Nickel oxidized .
0-9
—
Copper oxidized .
0-7
—
Platinum .
0-35
0-35
Gold.
015
0-2
Silver
01
—
Iron .
0-35
0-35
Slag .
—
0-65
Iron oxidized
0-95
—
Tungsten .
045
—
Nichrome .
0-9
—
OPTICAL PYROMETER CORRECTIONS
(A = 650 nm)
[To be added to observed temperature]
Observed
temp., °C
0-3
0-4
0-5
0-6
0-7
0-8
0-9
600
44
34
26
18
13
8
4
800
67
50
37
27
19
12
6
1000
95
71
53
39
27
17
8
1200
129
96
71
52
36
22
10
1400
169
125
93
67
46
28
13
1600
214
159
117
85
58
35
17
1800
265
196
145
105
72
44
20
2000
322
238
176
127
87
53
25
2500
495
362
266
190
131
78
38
3000
713
516
377
269
183
110
53
Total Radiation Pyrometers
Total radiation pyrometers are less common than optical pyrometers.
As we shall see, they can only be used when the source of radiation is of
considerable size — such as the open door of a furnace — whereas an
optical pyrometer. can be used on a very small body such as a lamp-
filament.
Fig. 14. 1 1 illustrates the principle of a Fery total radiation pyrometer.
The blackened tube A is open at the end B ; at the other end C it carries
an eye-piece E. D is a thermocouple attached to a small blackened
378 ADVANCED LEVEL PHYSICS
A
E
^37
B
(a) Construction
©
(b) Out of focus (c) In focus
Fig. 14.11. Fery total radiation pyrometer.
disc of copper, which faces the end C of the tube and is shielded from
direct radiation. M is a gold-plated mirror, pierced at the centre to
allow light to reach the eye-piece, and moveable by a rack and pinion P.
In use, the eyepiece is first focused upon the disc D. The mirror M is
then adjusted until the furnace is also focused upon D. Since a body
which is black or nearly so shows no detail, focusing it upon D by
simply looking at the image would be almost impossible. To make the
focusing easier, two small plane mirrors m', m are fitted in front of D.
They are inclined with their normals at about 5° to the axis of the tube,
and are pierced with semi-circular holes to allow radiation from M to
reach the disc. The diameter of the resulting circular hole is less than
that of the disc. When the source of heat is not focused on the disc, the
two mirrors appear as at (b) in the figure ; when the focusing is correct,
they appear as at (c). The source must be of such a size that its image
completely fills the hole.
The radiation from the source warms the junction and sets up an
electromotive force. A galvanometer G connected to the junction is
then deflected, and can be calibrated to read directly the temperature of
the source.
The calibration gives the correct temperature if the source is a black
body. If the source is not black, its total radiation is equal to that of a
black body at some lower temperature; the pyrometer therefore reads
too low. If the total emissivity of the source is known, a correction can
be made for it. This correction is greater than it would be if an optical
pyrometer were used, that is to say, departure from perfect blackness
causes less error in an optical pyrometer than in a total radiation one.
The Foster Pyrometer
Another type of total radiation pyrometer is the Foster fixed-focus
instrument (Fig. 14.12); it also uses a thermojunction with blackened
disc. A is an open diaphragm, so placed that it and the thermojunction
D are at conjugate foci of the mirror M. The thermojunction then
collects all the radiation entering through A but, since it is much smaller
than A, it is raised to a higher temperature than if its disc were the
size of A. The radiation entering through A is limited to that within
THERMOMETRY AND PYROMETRY
B ___
379
C —
Fig. 14.12. Fixed focus total radiation pyrometer.
the cone ABC ; as long as the whole of this cone is intersected by the
hot source, the amount of radiation reaching the thermocouple is
independent of the distance to the source (compare the use of a cone
with a thermopile, p. 350).
Comparison of Pyrometers: the International Scale
Total radiation and optical pyrometers agree within the limits of
experimental error— 1° at 1750°C, about 4° at 2800°C. The choice
between them is decided solely by convenience. The international
temperature scale above the gold-point (1063-0°C) is defined in terms
of an optical pyrometer.
Extension of Range by Sectored Disc
The range of a radiation pyrometer can be extended by cutting
down the radiation admitted to it. A disc from which an angle
radians has been cut out is
Pyrometer
rotated in front of the pyrometer,
as shown in Fig. 14.13, so that the
radiation entering is cut down in
the ratio 6/2iz. The pyrometer
then indicates a temperature T x ,
which is less than the true tem-
perature T 2 of the source. The
temperatures T are expressed in
K to simplify the calculation
which follows.
If the pyrometer is of the total
radiation type, then we can use Stefan's law. The radiation from a
body at T 2 K is proportional to T 2 4 . The pyrometer receives radiation
represented by the temperature T lt and therefore proportional to 7\ 4 .
Fig. 14.13. Sectored disc with radiation
pyrometer.
Therefore
whence
_ v
2ti ~ T, 4 '
T 2 = T,
27t\*
In this way the surface temperature of the sun has been estimated. The
value found agrees with that estimated from the wavelength of the
sun's most intense radiation (p. 353) ; it is about 6000 K.
A sectored disc can be used to extend the range of an optical pyro-
meter, but the calculation is more difficult than for a total radiation
pyrometer.
380 ADVANCED LEVEL PHYSICS
EXAMPLES
1. How is Celsius temperature defined (a) on the scale of a constant-pressure
gas thermometer, (b) on the scale of a platinum resistance thermometer? A
constant mass of gas maintained at constant pressure has a volume of 2000 cm 3
at the temperature of melting ice, 273-2 cm 3 at the temperature of water boiling
under standard pressure, and 525- 1 cm 3 at the normal boiling-point of sulphur.
A platinum wire has resistances of 2-000, 2-778 and 5-280 ohms at the same tem-
peratures. Calculate the values of the boiling-point of sulphur given by the two
sets of observations, and comment on the results. (N.)
First part (a). The temperature 9 on the gas thermometer scale is given by
v — v
0=K^ Xl00 '
r 100 K o
where V , V Q , F 100 are the respective volumes of the gas at constant pressure at
the temperature concerned, the temperature of melting ice, and the temperature
of steam at 76 cm mercury pressure, (b) The temperature 6 p on the platinum
resistance thermometer scale is given by
R„—R n
x 100,
where R e , R ,R 100 are the respective resistances of the platinum at the temperature
concerned, the temperature of melting ice, and the temperature of steam at
76 cm mercury.
Second part. On the gas thermometer scale, the boiling-point of sulphur is
given by
e = 5251-2000 xl0Q
273-2 -200-0
= 444-1 °C.
On the platinum resistance thermometer scale, the boiling-point is given by
n _ 5-280-2-000 inft
Up ~ 2-778 - 2-000 X1UU
= 421 -6°C.
The temperatures recorded on the thermometers are therefore different. This is
due to the fact that the variation of gas pressure with temperature at constant
volume is different from the variation of the electrical resistance of platinum
with temperature.
2. Explain how a Celsius temperature scale is denned, illustrating your answer
by reference to a platinum resistance thermometer.
The resistance R t of a platinum wire at temperature t°C, measured on the gas
scale, is given by R t = R Q (l + at+bt 2 ), where a = 3-800 x 10" 3 and b = -5-6 x
10" 7 . What temperature will the platinum thermometer indicate when the
temperature on the gas scale is 200°C? (O. & C.)
First part. The temperature 6 p in °C on a resistance thermometer scale is
given by the relation
where Rg, R , R 100 are the respective resistances at the temperature concerned,
at 0°C, and at 100°C.
THERMOMETRY AND PYROMETRY 381
Second part. R t = R (l +at+ bt 2 )
■'■ K200 = ^o(l + 200a + 200 2 b)
and Kioo = (^o(l + 100a + 100 2 6).
■•■0„ = ^ 2OO ~^ o xlOO
p
*M00 — ^0
R o (l + 2O0a+2O0 2 b)-R o
xlOO
R (l + lOOa+100 2 b)~R
200a+200 2 fr _ 2QO{a+2O0b)
a+1006 ~ a+1006
(3-8xlO- 3 -ll-2xlO -5 )
= 200
3-8xl(T 3 -5-6xl(r 5
200x0003688
0003744
197°C.
EXERCISES 14
1. How is a scale of temperature denned? What is meant by a temperature
of 15°C?
On what evidence do you accept the statement that there is an absolute zero
of temperature at about -273°C?
In a special type of thermometer a fixed mass of gas has a volume of 1000 cm 3
and a pressure of 81-6 cm of mercury at the ice point, and volume 124-0 units,
with pressure 90-0 units at the steam point. What is the temperature when its
volume is 1200 units and pressure 850 units, and what value does the scale of
this thermometer give for absolute zero? Explain the principle of your calculation.
(O.&C.)
2. Describe the structure of (a) a platinum resistance thermometer, (b) an
optical pyrometer. Explain, giving details of the auxiliary electrical circuits
required, how you would use each type of thermometer to measure a temperature,
assuming that the instrument has already been calibrated over the required
range. (L.)
3. Explain what is meant by a change in temperature of 1 deg C on the scale
of a platinum resistance thermometer.
Draw and label a diagram of a platinum resistance thermometer together with
a circuit in which it is used.
Give two advantages of this thermometer and explain why, in its normal form,
it is unsuited for measurement of varying temperatures.
The resistance R t of platinum varies with the temperature t°C as measured by
a constant volume gas thermometer according to the equation
R t = R (l + 8,000a£-at 2 )
where a is a constant. Calculate the temperature on the platinum scale correspond-
ing to 400°C on this gas scale. (N.)
4. Give the essential steps involved in setting up a scale of temperature. Explain
why scales based on different properties do not necessarily agree at all tempera-
tures. At what temperature or temperatures do these different scales agree?
382 ADVANCED LEVEL PHYSICS
The volume of some air at constant pressure, and also the length of an iron rod,
are measured at 0°C and again at 100°C with the following results :
0°C 100°C
Volume of air (cm 3 ) 28-5 38-9
Length of rod (cm) 10000 100-20
Calculate (a) the absolute zero of this air thermometer scale, and (b) the length
of the iron rod at this temperature if its expansion is uniform according to the
air scale. (C.)
5. Describe the structure of a constant volume gas thermometer. Describe also
the method of calibrating it and using it to determine the boiling point of a salt
solution.
Compare this thermometer as a means of measuring temperature with (a) a
mercury-in-glass thermometer, (b) a thermoelectric thermometer. (L.)
6. State what is meant by a temperature on the centigrade (Celsius) scale of a
platinum resistance thermometer.
Point out the relative merits of (a) a platinum resistance thermometer, and (b) a
thermoelectric thermometer for measuring (i) the rise in temperature of the water
flowing through a continuous flow calorimeter, and (ii) the temperature of a
small crystal as it is being heated rapidly. (N.)
7. Three types of thermometer in common use are based on (a) the expansion
of a fluid, (b) the production of an electromotive force, (c) the variation of electrical
resistance. Describe briefly one example of each and the way in which it is used.
In each case, state how a value of the temperature on a centigrade scale is deduced
from the quantities actually measured.
If all three of the thermometers you have described were used to measure the
temperature of the same object, would they give the same result? Give reasons
for your answer. (O. & C.)
8. Explain the principle underlying the establishment of a centigrade tem-
perature scale in terms of some suitable physical property.
What type of thermometer would you choose for use in experiments involving
(a) the plotting of a cooling curve for naphthalene in the region of its melting point,
(b) finding the boiling point of oxygen, (c) the measurement of the thermal con-
ductivity of a small crystal? In each instance give reasons for your choice.
If the resistance R t of the element of a resistance thermometer at a temperature
of t°C on the ideal gas scale is given by R t = R (l+At+Bt 2 ), where R is the
resistance at 0°C and A and B are constants such that A = —6-50 x 10 3 B, what
will be the temperature on the scale of the resistance thermometer when t =
500°C? (L.)
9. Explain briefly what is meant by a temperature of 0°C on (a) the mercury-in-
glass scale, (b) the constant-pressure hydrogen scale, (c) the platinum resistance
scale.
For what range of temperatures could (i) a platinum resistance thermometer,
(ii) an optical pyrometer be employed? Describe the structure and the method of
use of one of these instruments. (L.)
10. Explain the principle of a constant volume gas thermometer and describe
a simple instrument suitable for measurements in the range 0° to 100°C. What
factors determine (a) the sensitivity and (ft) the accuracy of the instrument you
describe?
A certain gas thermometer has a bulb of volume 50 cm 3 connected by a capillary
tube of negligible volume to a pressure gauge of volume 50 cm 3 . When the bulb is
immersed in a mixture of ice and water at 0°C, with the pressure gauge at room
temperature (17°C), the gas pressure is 700 mm Hg. What will be the pressure
THERMOMETRY AND PYROMETRY 383
when the bulb is raised to a temperature of 50°C if the gauge is maintained at
room temperature? You may assume that the gas is ideal and that the expansion
of the bulb can be neglected.
11. Give a brief account of the principles underlying the establishment of a
scale of temperature and explain precisely what is meant by the statements that
the temperature of a certain body is (a) r°C on the constant volume air scale,
(b) t P °C on the platinum resistance scale, and (c) t T °C on the Cu-Fe thermocouple
scale. Why are these three temperatures usually different?
Describe an optical pyrometer and explain how it is used to measure the
temperature of a furnace. (N.)
12. What is the general method of calibrating any type of thermometer?
Describe briefly the method of using three of the following temperature measuring
devices, and give the temperature range in which they are most usefully employed :
(i) platinum resistance thermometer, (ii) mercury-in-glass thermometer, (iii) helium
gas thermometer, (iv) optical pyrometer. (L.)
13. Explain the precautions taken in verifying the position of the fixed points
on the stem of a mercury-in-glass thermometer.
Describe a constant volume air thermometer. How could such a thermometer
be used to determine the melting-point of a solid such as naphthalene? What
advantage has the gas thermometer and for what purpose is it used? (L.)
14. Tabulate various physical properties used for measuring temperature.
Indicate the temperature range for which each is suitable.
Discuss the fact that the numerical value of a temperature expressed on the
scale of the platinum resistance thermometer is not the same as its value on the
gas scale except at the fixed points.
If the resistance of a platinum thermometer is 1-500 ohms at 0°C, 2-060 ohms
at 100°C and 1-788 ohms at 50°C on the gas scale, what is the difference between
the numerical values of the latter temperature on the two scales? (N.)
15. Explain how a centigrade temperature scale is defined, illustrating your
answer by reference to a platinum resistance thermometer.
The resistance R t of a platinum wire at temperature t°C, measured on the gas
scale, is given by R t = R (l + at+bt 2 ), where a '= 4-000 x 10" 3 and b = -60 x
10 -7 . What temperature will the platinum thermometer indicate when the
temperature on the gas scale is 300°C?
16. Describe how you would use either (i) a constant- volume or (ii) a constant-
pressure air thermometer to calibrate a mercurial thermometer. If the difference
of mercury level in a constant-volume air thermometer is —2 cm when the
temperature of the bulb is 10°C and -I- 22 cm when the bulb is^t 100°C, what is
the height of the barometer? (L.) v
PART THREE
Optics and Sound
OPTICS
chapter fifteen
Introduction
If you wear spectacles you will appreciate particularly that the science
of Light, or Optics as it is often called, has benefited people all over the
world. The illumination engineer has developed the branch of Light
dealing with light energy, and has shown how to obtain suitable lighting
conditions in the home and the factory, which is an important factor in
maintaining our health. Microscopes, used by medical research workers
in their fight against disease; telescopes, used by seamen and astrono-
mers; and a variety oh optical instruments which incorporate lenses,
mirrors, or glass prisms, such as cameras, driving mirrors, and binoculars,
all testify to the scientist's service to the community.
In mentioning the technical achievements in Light, however, it must
not be forgotten that the science of Light evolved gradually over the
past centuries; and that the technical advances were developed from the
experiments and theory on the fundamental principles of the subject
made by scientists such as Newton, Huygens, and Fresnel.
Light Travels in Straight Lines. Eclipses and Shadows
When sunlight is streaming through an open window into a room,
observation shows that the edges of the beam of light, where the shadow
begins, are straight. This suggests that light travels in straight lines, and
on this assumption the sharpness of shadows and the formation of -
eclipses can be explained. Fig. 15.1 (i) illustrates the eclipse of the sun, S,
o
0) ^ .
Fig. 15.1. Eclipse of Sun {not to scale).
C\
(ii)
our natural source of light, when the moon, M, passes between the sun
and the earth, E. The moon is a non-luminous object which does not
allow light to pass through it, and hence the boundaries of the shadows
formed by M on the earth are obtained by drawing lines from S which
387
388 ADVANCED LEVEL PHYSICS
touch the edge of M. Consequently there is a total eclipse of the sun at
c on the earth, a partial eclipse at b, and no eclipse at a. Fig. 15.1 (ii)
illustrates the appearance of the sun in each case.
Light Rays and Beams
Light is a form of energy. We know this is the case because plants and
vegetables grow when they absorb sunlight. Further, electrons are ejected
from certain metals when light is incident on them, showing that there
was some energy in the light; this phenomenon is the basis of the
photoelectric cell (p. 1077). Substances like wood or brick which allow
no light to pass through them are called "opaque" substances; unless
an opaque object is perfectly black, some of the light falling on it is
reflected (p. 391). A "transparent" substance, like glass, is one which
allows some of the light energy incident on it to pass through, the
remainder of the energy being absorbed and (or) reflected.
A ray of light is the direction along which the light energy travels;
and though rays are represented in diagrams by straight lines, in practice
a ray has a finite width. A beam of light is a collection of rays. A search-
light emits a. parallel beam of light, and the rays from a point on a very
distant object like the sun are substantially parallel, Fig. 15.2 (i). A lamp
Parallel beam Divergent beam Convergent beam
(i) (ii) (iii)
Fig. 15.2. Beams of light.
emits a divergent beam of light; while a source of light behind a lens, as
in a projection lantern, can provide a convergent beam, Fig. 15.2 (ii), (iii).
The Velocity of Light
The velocity of light is constant for a given medium, such as air, water,
or -glass, and has its greatest magnitude, about 3-0 x 10 8 metres per
second, in a Vacuum. The velocity of light in air differs only slightly
from its velocity in a vacuum, so that the velocity in air is also about
3-0 x 10 8 metres per second. The velocity of light in glass is about
2-0 x 10 8 metres per second; in water it is about 2-3 x 10 8 metres per
second. On account of the difference in velocity in air and glass, light
changes its direction on entering glass from air (see Refraction, p. 679).
Experiments to determine the velocity of light are discussed later, p. 551 .
INTRODUCTION 389
Hie Human Eye
When an object is seen, light energy passes from the object to the
observer's eyes and sets up the sensation of vision. The eye is thus
sensitive to light (or luminous) energy. The eye contains a crystalline
lens, L, made of a gelatinous transparent substance, which normally
throws an image of the object viewed on to a sensitive "screen" R at
the back of the eye-ball, called the retina, Fig. 15.3. Nerves on the retina
Fto. 15.3. The eyeball.
are joined to the optic nerve, O, which carries the sensation produced by
the image to the brain. The iris, I, is a diaphragm with a circular hole
in the middle called the pupil, P, which contracts when the light received
by the eye is excessive and painful to the eye. The colour of a person's
eyes is the colour of the iris; the pupil is always black because no light
returns from the interior of the eye-ball. A weak salt solution, called
the aqueous humour, is present on the left of the lens L, and between L
and the retina is a gelatinous substance called the vitreous humour. The
transparent spherical bulge D in front of L is made of tough material,
and is known as the cornea.
The ciliary muscles, C, enable the eye to see clearly objects at different
distances, a property of the eye known as its "power of accommodation".
The ciliary muscles are attached to the lens L, and when they contract,
the lens' surfaces are pulled out so that they bulge more; in this way
a near object can be focused clearly on the retina and thus seen distinctly.
When a very distant object is observed the ciliary muscles are relaxed,
and the lens' surfaces are flattened.
The use of two eyes gives a three-dimensional aspect of the object or
scene observed, as two slightly different views are imposed on the
retinae; this gives a sense of distance not enjoyed by a one-eyed person.
Direction of Image seen by Eye
When a fish is observed in water, rays of light coming from a point
such as O on it pass from water into air, Fig. 1 5.4 (i). At the boundary of
390
ADVANCED LEVEL PHYSICS
the water and air, the rays OA, OC proceed along new directions AB,
CD respectively and enter the eye. Similarly, a ray OC from an object O
observed in a mirror is reflected along a new direction CD and enters
the eye, Fig. 15.4 (ii). These phenomena are studied more fully later, but
(i) ('«)
Fig. 15.4. Images observed by eye.
the reader should take careful note that the eye sees an object in the
direction in which the rays enter the eye. In Fig. 15.4 (i), for example, the
object O is seen in the water at I, which lies on BA and DC produced
slightly on the right of O; in Fig. 15.4 (ii), is seen behind the mirror at I,
which lies on DC produced. In either case, all rays from O which enter
the eye appear to come from I, which is called the image of O.
Reversibility of Light
If a ray of light is directed along DC towards a mirror, experiment
shows that the ray is reflected along the path CO, Fig. 15.4 (ii). If the ray
is incident along OC, it is reflected along CD, as shown. Thus if a light
ray is reversed it always travels along its original path, and this is known
as the principle of the reversibility of light. In Fig. 15.4 (i), a ray BA in air
is refracted into the water along the path AO, since it follows the reverse
path to OAB. We shall have occasion to use the principle of the rever-
sibility of light later in the book.
chapter sixteen
Reflection at plane surfaces
Highly-polished metal surfaces reflect about 80 to 90 per cent of
the light incident on them; mirrors in everyday use are therefore usually
made by depositing silver on the back of glass. In special cases the front
of the glass is coated with the metal; for example, the largest reflector
in the world is a curved mirror nearly 5 metres across, the front of which
is coated with aluminium (p. 544). Glass by itself will also reflect light,
but the percentage reflected is small compared with the case of a silvered
surface; it is about 5 per cent for an air-glass surface.
Laws of Reflection
If a ray of light, AO, is incident on a plane mirror XY at O, the angle
AON made with the normal ON to
the mirror is called the "angle of
incidence", i, Fig. 16.1. The angle
BON made by the reflected ray OB
with the normal is called the
"angle of reflection", r; and ex-
periments with a ray-box and a
plane mirror, for example, show
that:
(1) The reflected ray, the incident
ray, and the normal to the mirror at the point of incidence all lie in the
same plane.
(2) The angle of incidence = the angle of reflection.
These are called the two laws of reflection, and they were known to
Plato, who lived about 400 b.c.
Silvering
Fig. 16.1. Plane mirror.
Regular and Diffuse Reflection
In the case of a plane mirror or glass surface, it follows from the laws
of reflection that a ray incident at a given angle on the surface is reflected
in a definite direction. Thus a parallel beam of light incident on a plane
mirror in the direction AO is reflected as a parallel beam in the direction
OB, and this is known as a case of regular reflection, Fig. 162 (i). On the
other hand, if a parallel beam of light is incident on a sheet of paper in
a direction AO, the light is reflected in all different directions from the
paper: this is an example of diffuse reflection, Fig. 16.2 (ii). Objects in
everyday life, such as flowers, books, people, are seen by light diffusely
reflected from them. The explanation of the diffusion of light is that the
391
392
ADVANCED LEVEL PHYSICS
surface of paper, for example, is not perfectly smooth like a mirrored
surface; the "roughness" in a paper surface can be seen with a micro-
Paper
(ii)
Fig. 16.2 (i). Regular reflection.
Fig. 16.2 (ii). Diffuse reflection.
scope. At each point on the paper the laws of reflection are obeyed,
but the angle of incidence varies, unlike the case of a mirror.
Deviation of Light at Plane Mirror Surface
Besides other purposes, plane mirrors are used in the sextant (p. 395),
in simple periscopes, and in signalling at sea. These instruments
utilise the property of a plane mirror to deviate light from one direction
to another.
Consider a ray AO incident at O on
a plane mirror XY, Fig. 16.3 (i). The
angle AOX made by AO with XY
is known as the glancing angle, g,
with the mirror; and since the angle
of reflection is equal to the angle of
incidence, the glancing angle BOY
made by the reflected ray OB with
the mirror is also equal to g.
The light has been deviated from a direction AO to a direction OB.
Thus if OC is the extension of AO, the angle of deviation, d, is angle
COB. Since angle COY = vertically opposite angle XOA, it follows
that
d=2g . . . . (1);
so that, in general, the angle of deviation of a ray by a plane surface is
twice the glancing angle.
(i)
Fig. 16.3 (i).
Deviation of light at at plane mirror.
Deviation of Reflected Ray by Rotated Mirror
Consider a ray AO incident at O on a plane mirror M x , a being the
glancing angle with M l5 Fig. 16.3 (ii). If OB is the reflected ray, then, as
shown above, the angle of deviation COB — Ig = 2a.
REFLECTION AT PLANE SURFACES
393
Suppose the mirror is rotated
through an angle 6 to a position
M 2 , the direction of the incident
ray AO being constant. The ray is
now reflected from M 2 , in a direc-
tion OP, and the glancing angle
with M 2 is (a + 6). Hence the
new angle of deviation COP =
2g = 2 (a + 6). The reflected ray
M 2
$77777777777? M,
(ii)
Fig. 16.3 (ii>. Rotation of reflected ray.
has thus been rotated through an angle BOP when the mirror rotated
through an angle 0; and since
then
ZBOP = ZCOP -
ZBOP = 2 (a + 6)
ZCOB,
- 2a = 26.
Thus, if the direction of an incident ray is constant, the angle of rotation oj
the reflected ray is twice the angle of rotation of the mirror. If the mirror
rotates through 4°, the reflected ray turns through 8°, the direction of the
incident ray being kept unaltered.
Optical Lever in Mirror Galvanometer
In a number of instruments a beam of light is used as a "pointer",
which thus has a negligible weight and is sensitive to deflections of the
moving system inside the instrument. In a mirror galvanometer, used
for measuring- very small electric currents, a small mirror M x is rigidly
attached to a system which rotates when a current flows in it, and a
beam of light from a fixed lamp L shines on the mirror, Fig. 16.4. If the
light is incident normally on the mirror at A, the beam is reflected
directly back, and a spot of light is obtained at O on a graduated scale S
Fig. 16.4. Optical lever principle.
placed just above L. Suppose that the moving system, to which the
mirror is attached, undergoes a rotation 6. The mirror is then rotated
through this angle to a position M 2 , and the spot of light is deflected
through a distance x, say to a position P on the scale.
Since the direction OA of the incident light is constant, the rotation
of the reflected ray is twice the angle of rotation of the mirror (p. 393).
Thus angle OAP = 26. Now tan 26 — x}d, where d is the distance OA.
Thus 26 can be calculated from a knowledge of x and d, and hence 6
394
ADVANCED LEVEL PHYSICS
is obtained. If 26 is small, then tan 20 is approximately equal to 26 in
radians, and in this case 6 is equal to x\2d radians.
In conjunction with a mirror, a beam of light used as a "pointer" is
known as an "optical lever". Besides a negligible weight, it has the
advantage of magnifying by two the rotation of the system to which
the mirror is attached, as the angle of rotation of the reflected light is
twice the angle of rotation of the mirror. An optical lever can be used
for measuring small increases of length due to the expansion or con-
traction of a solid.
Deviation by Successive Reflections at Two Inclined Mirrors
Before we can deal with the principle of the sextant, the deviation of
light by successive reflection at two inclined mirrors must be discussed.
Consider two mirrors, XO, XB, inclined at an angle 6, and suppose
AO is a ray incident on the mirror XO at a glancing angle a, Fig. 16.5 (i).
The reflected ray OB then also makes a glancing angle a with OX, and
from our result on p. 392, the angle of deviation produced by XO in a
clockwise direction (angle LOB) = 2a.
Fig. 16.5. Successive reflection at two plane mirrors.
Suppose OB is incident at a glancing angle jB on the second mirror
XB. Then, if the reflected ray is BC, the angle of deviation produced by
this mirror (angle EBC) .= 2 £, in an anti-clockwise direction. Thus the
net deviation D of the incident ray AO produced by both mirrors
= 2£ — 2a, in an anti-clockwise direction.
Now from triangle OBX,
angle PBO = angle BOX -f angle BXO,
i.e., £ = a -f- 6
Thus 6 = - a, and hence
D = 2$ - 2a = 26.
But 6 is a constant when the two mirrors are inclined at a given angle.
Thus, no matter what angle the incident ray makes with the first mirror,
the deviation D after two successive reflections is constant and equal to
twice the angle between the mirrors.
REFLECTION AT PLANE SURFACES
395
Fig. 16.5 (ii) illustrates the case when the ray BC reflected at the second
mirror travels in an opposite direction to the incident ray AO, unlike the
case in Fig. 16.5 (i). In Fig. 16.5 (ii), the net deviation, D, after two suc-
cessive reflections in a clockwise direction is 2a + 2/S. But a + = 180 p
— 0. Hence D = 2a + 2/3 = 360° — 20. Thus the deviation, D, in
an anti-clockwise direction is 20, the same result as obtained above.
Principle of the Sextant
The sextant is an instrument used in navigation for measuring the
angle of elevation of the sun or stars. It consists essentially of a fixed
glass B, silvered on a vertical half, and a silvered mirror O which can be
rotated about a horizontal axis. A small fixed telescope T is directed
towards B, Fig. 16.6.
Fig. 16.6. Sextant principle.
Suppose that the angle of elevation of the sun, S, is required. Looking
through T, the mirror O is turned until the view H' of the horizon seen
directly through the unsilvered half of B, and also the view of it, H, seen
by successive reflection at O and the silvered half of B, are coincident.
The mirror O is then parallel to B in the position M lt and the ray HO
is reflected along OB and BD to enter the telescope T. The mirror
O is now rotated to a position M 2 until the image of the sun S, seen by
successive reflections at O and B, is on the horizon H', and the angle
of rotation, 0, of the mirror is noted, Fig. 16.6.
The ray SO from the sun is now reflected in turn from O and B so
that it travels along BD, the direction of the horizon, and the angle of
deviation of the ray is thus angle SOH. But the angle between the
mirrors M 2 and B is 0. Thus, from our result for successive reflections
at two inclined mirrors, angle SOH = 26. Now the angle of elevation
of the sun, S, is angle SOH. Hence the angle of elevation is twice the angle
396
ADVANCED LEVEL PHYSICS
of rotation of the mirror O, and can thus be easily measured from a
scale (not shown) which measures the rotation of O.
Since the angle of deviation after two successive reflections is in-
dependent of the angle of incidence on the first mirror (p. 394), the image
of the sun S through T will continue to be seen on the horizon once O
is adjusted, no matter how the ship pitches or rolls. This is an advantage
of the sextant.
Images in Plane Mirrors
So far we have discussed the deviation of light by a plane mirror. We
have now to consider the images in plane mirrors.
Suppose that a point object A is placed in front of a mirror M,
Fig. 16.7. A ray AO from A, incident on M, is reflected along OB in
such a way that angle AON = angle BON, where ON is the normal at
O to the mirror. A ray AD incident normally on the mirror at D is
reflected back along DA. Thus the rays reflected from M appear to
come from a point I behind the
mirror, where I is the point of
intersection of BO and AD pro-
duced. As we shall prove
shortly, any ray from A, such
as AP, is also reflected as if it
comes from I, and hence an
observer at E sees the image of
A at I.
Since angle AON = alternate
angle DAO, and angle BON =
corresponding angle DIO, it
follows that angle DAO = angle
DIO. Thus in the triangles
ODA, ODI, angle DAO = angle
DIO, OD is common, and angle ADO = 90° = angle IDO. The two
triangles are hence congruent, and therefore AD = ID. For a given
position of the object, A and D are fixed points. Consequently, since
AD = ID, the point I is a fixed point; and hence any ray reflected from
the mirror must pass through I, as was stated above.
We have just shown that the object and image in a plane mirror are at
equal perpendicular distances from the mirror. It should also be noted
that AO = OI in Fig. 16.7, and hence the object and image are at equal
distances from any point on the mirror.
Image in plane mirror.
Image of Finite-sized Object. Perversion
If a right-handed batsman observes his stance in a plane mirror, he
appears to be left-handed. Again, the words on a piece of blotting-
paper become legible when the paper is viewed in a mirror. This
REFLECTION AT PLANE SURFACES
397
Fig. 16.8. Perverted (laterally inverted) image.
phenomenon can be explained by considering an E-shaped object placed
in front of a mirror M, Fig. 16.8. The image of a point a on the object
is at a' at an equal distance behind the mirror, and the image of a point
b on the left of a is at b', which is on the right of a'. The left-hand side
of the image thus corresponds to the right-hand side of the object, and
vice-versa, and the object is said to be perverted, or laterally inverted
to an observer.
Virtual and Real Images
As was shown on p. 396, an object O in front of a mirror has an image
I behind the mirror. The rays reflected from the mirror do not actually
pass through I, but only appear to do so, and the image cannot be
received on a screen because the image is behind the mirror, Fig. 16.9
(i). This type of image is therefore called an unreal or virtual image.
0(Real object)
I(Real
.image)
WWAW, M
I ^Virtual image
(i)
Fig. 16.9. Virtual and real image in plane mirror.
\ \ / /
\\ //
V
O Virtual object
(ii)
It must not be thought, however, that only virtual images are obtained
with a plane mirror. If a convergent beam is incident on a plane mirror
M, the reflected rays pass through a point I in front o/M, Fig. 16.9 (ii).
If the incident beam converges to the point O, the latter is termed a
"virtual" object; I is called a real image because it can be received on a
398 ADVANCED LEVEL PHYSICS
screen. Fig. 16.9 (i) and (ii) should now be compared. In the former, a real
object (divergent beam) gives rise to a virtual image ; in the latter, a virtual
object (convergent beam) gives rise to a real image. In each case the
image and object are at equal distances from the mirror.
Location of Images by No Parallax Method
A virtual image can be located by the method of no parallax, which we
shall now describe.
Suppose O is a pin placed in front of a plane mirror M, giving rise to
a virtual image I behind M, Fig. 16.10. A pin P behind the mirror is then
A
Jl
m
r _j — n
l I
I I
Fig. 16.10. Location of image by no parallax method.
moved towards or away from M, each time moving the head from side
to side so as to detect any relative motion between I and P. When the
latter appear to move together they are coincident in position, and
hence P is at the place of the image I, which is thus located. When
P and I do not coincide, they appear to move relative to one another
when the observer's head is moved; this relative movement is called
"parallax". It is useful to note that the nearer object moves in the
opposite direction to the observer.
The method of no parallax can be used, as we shall see later, to locate
the positions of real images, as well as virtual images, obtained with
lenses and curved mirrors.
Images in Inclined Mirrors
A kaleidoscope, produced as a toy under the name of "mirrorscope",
consists of two inclined pieces of plane glass with some coloured tinsel
between them. On looking into the kaleidoscope a beautiful series of
coloured images of the tinsel can often be seen, and the instrument is
used by designers to obtain ideas on colouring fashions.
Suppose OA, OB are two mirrors inclined at an angle 6, and P is a
point object between them, Fig. 16. 1 1 (i). The image of P in the mirror OB
is Bi, and OP = OBi (see p. 396). B t then forms an image B 2 in the
mirror OA, with OB 2 = OB 1} B 2 forms an image B 3 in OB, and so on.
All the images thus lie on a circle of centre O and radius OP. Another set
REFLECTION AT PLANE SURFACES
399
of images, A lf A 2 , A 3 . . ., have their origin in the image A x formed by P
in the mirror OA. When the observer looks into the mirror OB he sees
CO
(ii)
Fig. 16.11. Images in inclined mirrors.
the series of images B 1} A 2 , B 3 . . .; when he looks into the mirror OA
he sees the series of images A l5 B 2 , A 3 . . . A finite series of images is seen
in either mirror, the last image (not shown) being the one formed on the
arc A3', because it is then behind the silvering of the next mirror.
When the mirrors are inclined at an angle of 60°, the final images of P,
A 3 , B 3 , of each series coincide, Fig. 16. 1 1 (ii). The total number of images
is now 5, as the reader can verify. Fig. 16.11 (ii) illustrates the cone of light
received by the pupil of the eye when the image B 2 is observed, reflection
occurring successively at the mirrors. The drawing is started by joining
B 2 to the boundary of the eye, then using B lt and finally using P.
EXAMPLE
State the laws of reflection of light and describe how you would verify
these laws.
A man 2 m tall, whose eye level is 1-84 m above the ground, looks at
his image in a vertical mirror. What is the minimum vertical length of the
mirror if the man is to be able to see the whole of himself? Indicate its position
accurately in a diagram.
First part. See text. A ray-box can be used to verify the laws.
Second part. Suppose the man is represented by HF, where H is his head
and F is his feet; suppose that E represents his eyes, Fig. 16.12. Since the man
400
ADVANCED LEVEL PHYSICS
Fig. 16.12. Example.
sees his head H, a ray HA from H to the top A of the mirror is reflected to E.
Thus A lies on the perpendicular bisector of HE, and hence AL == \ HE = 0*08
m, where L is the point on the mirror at the same level as E. Since the man
sees his feet F, a ray FB from F to the bottom B of the mirror is also reflected
to E; Thus the perpendicular bisector of EF passes through B, and hence BL
= iFE = i x 1-84 m =0-92m.
-..*. length of mirror = AL + LB = 0-08 m + 0-92 m = 1 m.
EXERCISES 16
1. Prove the relation between the angle of rotation of a mirror and the
angle of deflection of a reflected ray, when the direction of the incident ray
is constant.
2. Two plane mirrors are inclined at an angle of 35°. A ray of light is in-
cident on one mirror at 60°, and undergoes two successive reflections at the
mirrors. Show by accurate drawing that the angle of deviation produced is 70°.
Repeat with an angle of incidence of 45°, instead of 60°, and state the law
concerning the angle of deviation.
3. Two plane mirrors are inclined to each other at a fixed angle. If a
ray travelling in a plane perpendicular to both mirrors is reflected first from
one and then from the other, show that the angle through which it is deflected
does not depend on the angle at which it strikes the first mirror.
•Describe and explain the action of either a sextant or a rear reflector on a
bicycle. (L.)
4. State the laws of reflection of light. Two plane mirrors are parallel and
face each other. They are a cm apart and a small luminous object is placed
b cm from one of them. Find the distance from the object of an image produced
by four reflections. Deduce the corresponding distance for an image produced
by 2n reflections. (L.)
REFLECTION AT PLANE SURFACES 401
5. Two vertical plane mirrors A andB are inclined to one another at an
angle a. A ray of light, travelling horizontally, is reflected first from A and
then from B. Find the resultant deviation and show it is independent of the
original direction of the ray. Describe an optical instrument that depends on
the above proposition. (N.)
6. State the laws of reflection for a parallel beam of light incident upon a
plane mirror.
Indicate clearly by means of diagrams (a) how the position and size of the
image of an extended object may be determined by geometrical construction,
in the case of reflection in a plane; (Jb) how the positions of the images of a
small lamp, placed unsymmetrically between parallel reflecting planes, may
be graphically determined. (JF.)
7. Describe the construction of the sextant and the periscope. Illustrate your
answer by clear diagrams and indicate the optical principles involved. (L.)
chapter seventeen
Reflection at curved mirrors
Curved mirrors are reputed to have been used thousands of years ago.
Today motor-cars and other vehicles are equipped with driving mirrors
which are curved, searchlights have curved mirrors inside them, and
the largest telescope in the world utilises a huge curved mirror (p. 544).
Convex and Concave Mirrors. Definitions
In the theory of Light we are mainly concerned with curved mirrors
which are parts of spherical surfaces. In Fig. 17.1 {a), the mirror APB is
part of a sphere whose centre C is in front of the reflecting surface; in
Fig. 17. 1 (b), the mirror KPL is part of a sphere whose centre C is behind
its reflecting surface. To a person in front of it APB curves inwards
and is known as a concave mirror, while KPL bulges outwards and is
known as a convex mirror.
Incident light
Concave mirror
Convex mirror
(i) 00
Fig. 17.1. Concave (converging) and convex (diverging) mirrors.
The mid-point, P, of the mirror is called its pole; C, the centre of the
sphere of which the mirror is part, is known as the centre of curvature',
and AB is called the aperture of the mirror. The line PC is known as the
principal axis, and plays an important part in the drawing of images in
the mirrors; lines parallel to PC are called secondary axes.
Narrow and Wide Beams. The Caustic
When a very narrow beam of rays, parallel to the principal axis and
close to it, is incident on a concave mirror, experiment shows that all
the reflected rays converge to a point F on the principal axis, which is
therefore known as the principal focus of the mirror, Fig. 17.2 (i). On
this account a concave mirror is better described as a "converging"
402
REFLECTION AT CURVED MIRRORS
403
mirror. An image of the sun, whose rays on the earth are parallel, can
hence be received on a screen at F, and thus a concave mirror has a
real focus.
(i) (ii)
Fig. 17.2. Foci of concave and convex mirrors.
If a narrow beam of parallel rays is incident on a convex mirror,
experiment shows that the reflected rays form a divergent beam which
appear to come from a point F behind the mirror, Fig. 17.2 (ii). A convex
mirror has thus a virtual focus, and the image of the sun cannot be
received on a screen using this type of mirror. To express its action
on & parallel beam of light, a convex mirror is often called a "diverging"
mirror.
When a wide beam of light, parallel to
the principal axis, is incident on a con-
cave spherical mirror, experiment shows
that reflected rays do not pass through
a single point, as was the case with a
narrow beam. The reflected rays appear
to touch a surface known as a caustic
surface, S, which has an apex, or cusp,
at F, the principal focus, Fig. 17.3. Simi-
larly, if a wide beam of parallel light is
incident on a convex mirror, the reflected
rays do not appear to diverge from a
single point, as was the case with a
narrow beam.
s - 'rt
\y vr
C * ^*P^
F^
^^^"
/*\S<£
s' x \p
Fig. 17.3. Caustic surface.
Parabolic Mirrors
If a small lamp is placed at the focus, F, of a concave mirror, it
follows from the principle of the reversibility of light (p. 390) that rays
striking the mirror round a small area about the pole are reflected
parallel. See Fig. 17.2 (i). But those rays from the lamp which strike the
mirror at points well away from P will be reflected in different directions,
because a wide parallel beam is not brought to a focus at F, as shown
in Fig. 17.3. The beam of light reflected from the mirror thus diminishes
404
ADVANCED LEVEL PHYSICS
'■ <> <■ n |, i, ,,
in intensity as its distance from
the mirror increases, and a concave
spherical mirror is hence useless as a
searchlight mirror.
A mirror whose section is the
shape of a parabola (the path of a
cricket-ball thrown forward into the
air) is used in searchlights. A para-
bolic mirror has the property of
reflecting the wide beam of light from
a lamp at its focus F as a perfectly
parallel beam, in which case the
intensity of the reflected beam is practically undiminished as the dis-
tance from the mirror increases, Fig. 17.4.
Fig. 17.4. Parabolic mirror.
Focal Length (/) and Radius of Curvature (r)
From now onwards we shall be concerned with curved spherical
mirrors of small aperture, so that a parallel incident beam will pass
through the focus after reflection. The diagrams which follow are
exaggerated for purposes of clarity.
The distance PC from the pole to the centre of curvature is known as
the radius of curvature (r) of a mirror; the distance PF from the pole
to the focus is known as the focal length (f) of the mirror. As we shall
now prove, there is a simple relation between /and r.
Consider a ray AX parallel to the principal axis of either a concave
or a convex mirror, Fig. 17.5 (i), (ii). The normal to the mirror at X is
_^C^J§<
Convex
Fig. 17.5. Relation between /and r.
CX, because the radius of the spherical surface is perpendicular to the
surface, and hence the reflected ray makes an angle, 0, with CX equal
to the incident angle 6. Taking the case of the concave mirror, angle
AXC = angle XCP, alternate angles, Fig. 17.5 (i). Thus triangle FXC is
isosceles, and FX = FC. As X is a point very close to P we assume to
a very good approximation that FX = FP.
/. FP = FC, orFP = £CP.
/=
(1)
REFLECTION AT CURVED MIRRORS.
405
This relation between /and r is the same for the case of the convex
mirror, Fig. 17.5 (ii), as the reader can easily verify.
Images in Concave Mirrors
Concave mirrors produce images of different sizes; sometimes they
are inverted and real, and on other occasions they are erect (the same
way up as the object) and virtual. As we shall see, the nature of the
image formed depends on the distance of the object from the mirror.
Consider an object of finite size OH placed at O perpendicular to the
principal axis of the mirror, Fig. 17.6 (i). The image, R, of the top point
Image
Fig. 17.6. Images in concave mirrors.
H can be located by the intersection of two reflected rays coming
initially from H, and the rays usually chosen are two of the following:
(1) The ray HT parallel to the principal axis, which is reflected to pass
through the focus, F, (2) the ray HC passing through the centre of
curvature, C, which is reflected back along its own path because it is a
normal to the mirror, (3) the ray HF passing through the focus, F,
which is reflected parallel to the principal axis. Since the mirror has a
small aperture, and we are considering a narrow beam of light, the
mirror must be represented in accurate image drawings by a straight
line. Thus PT in Fig. 17.6 (i) represents a perfect mirror.
When the object is a very long distance away (at infinity), the image
is small and is formed inverted at the focus (p. 402). As the object
approaches the centre of curvature, C, the image remains real and
inverted, and is formed in front of the object, Fig. 17.6 (i). When the
object is between C and F, the image is real, inverted, and larger than the
object; it is now further from the mirror than the object, Fig. 17.6 (ii).
As the object approaches the focus, the image recedes further from
the mirror, and when the object is at the focus, the image is at infinity.
When the object is nearer to the mirror than the focus the image IR
becomes erect ancl virtual, as shown in Fig. 17.7 (i). In this case the image
406
ADVANCED LEVEL PHYSICS
is magnified, and the concave mirror can thus be used as a shaving
mirror.
^-''''" i Magnified
]- erect ~. . „
image Object-
C
Image — , ,
(') (ii)
Fig. 17.7. Images in concave mirrors.
A special case occurs when the object is at the centre of curvature, C.
The image is then real, inverted, and the same size of the object, and it
is also situated at C, Fig. 17.7 (ii). This case provides a simple method
of locating the centre of curvature of a concave mirror (p. 413).
Images in Convex Mirrors
Experiment shows that the image of an object in a convex mirror is
erect, virtual, and diminished in size, no matter where the object is
situated. Suppose an object OH is placed in front of a convex mirror,
Fig. 17.8 (i). A ray HM parallel to the principal axis is reflected as if it
1 F C
(i) I E (ii)
Fig. 17.8. Images in convex mirrors.
appeared to come from the virtual focus, F, and a ray HN incident
towards the centre of curvature, C, is reflected back along its path. The
two reflected rays intersect behind the mirror at R, and IR is a virtual
and erect image.
Objects well outside the principal axis of a convex mirror, such as
A, B in Fig. 17.8 (ii), can be seen by an observer at E, whose field of view
is that between HT and RS, where T, S are the edges of the mirror.
Thus in addition to providing an erect image the convex mirror has a
wide field of view, and is hence used as a driving mirror.
REFLECTION AT CURVED MIRRORS
407
Formula for Mirrors. Sign Convention
Many of the advances in the uses of curved mirrors and lenses have
resulted from the use of optical formula ,and we have now to consider
the relation which holds between the object and image distances in
mirrors and their focal length. In order to obtain a formula which holds
for both concave and convex mirrors, a sign rule or convention must be
obeyed, and we shall adopt the following:
A real object or image distance is a positive distance.
A virtual object or image distance is a negative distance.
In brief, "real is positive, virtual is negative". The focal length of a
concave mirror is thus a positive distance; the focal length of a convex
mirror is a negative distance.
Concave Mirror
Consider a point object O on the principal axis of a concave mirror.
A ray OX from O is reflected in the direction XI making an equal
angle 6 with the normal CX; a ray OP from O, incident at P, is reflected
back along PO, since CP is the normal at P. The point of intersection,
I, of the two rays is the image of O. Fig. 17.9.
Fig. 17.9. Mirror formula.
Suppose a, p, y are the angles made by OX, CX, IX respectively
with the axis. Since we are considering a mirror of small aperture these
angles are small in practice, Fig. 17.9 being exaggerated. As /? is the
exterior angle of triangle CXO, we have p = a + 8.
.'. 6 = j8 - a . . . . (i)
Since y is the exterior angle of triangle IXC, we have y = p -f- 9.
,'.e=y-P . . (ii)
From (i) and (ii), it follows that
f$-a = y- fi
•\a+y = 2jS . . . . (iii)
408 ADVANCED LEVEL PHYSICS
We can now substitute for a, j8, y in terms of h, the height of X above
the axis, and the distances OP, CP, IP. In so doing (a) we assume N is
practically coincident with P, as X is very close to P in practice, (b) the
appropriate sign, -f or — , must precede all the numerical values of
the distances concerned. Also, as a = tan a in radians when a is very
small, we have
XN XN h
a =
ON + OP + OP'
where OP is the distance of the real object O from the mirror in centi-
metres, say, and XN = h. Similarly,
XN _ XN _ h
"~CN ~+CP~~+CP'
as CP, the radius of curvature of the concave mirror, is real.
XN XN h
Also, y = _ =__=___,
where IP is the distance of the real image I from the mirror. Substituting
for a, jS,yin(iii),
h h „ h
= 2
(+ip)'( + op) ( + cpy
Dividing by h, TP + OP = CP'
If we let v represent the image distance from the mirror, u the object
distance from the mirror, and r the radius of curvature, we have
11 2
7 + u = r • • • • (2)
Further, since/= r/2, then - = -? .
r f
••7+u=7 • • . . (3)
The relations (2), (3) are general formulae for curved spherical mirrors;
and when they are used the appropriate sign for v, i/,/, or r must always
precede the corresponding numerical value.
Convex Mirror
We now obtain a relation for object distance (u), image distance (v),
and focal length (J) of a convex mirror. In this case the incident rays
OX, OP are reflected as if they appear to come from the point I behind
the mirror, which is therefore a virtual image, and hence the image
REFLECTION AT CURVED MIRRORS 409
distance IP is negative, Fig. 17.10. Further, CP is negative, as the centre
of curvature of X, a convex mirror, is behind the mirror.
Fig. 17.10. Mirror formula.
Since 6 is the exterior angle of triangle COX, 9 = a + j8. As y is
the exterior angle of triangle CLX, y = 6 + jS, or 6 = y — /?.
.*. y - /? = a + £
.'. y-a = 2j8 . . . , (i)
h h T . . , /i A . _.
Now y = jh = ( _ Ip) > as I is virtual; a =— = ^ + Qp) > as O is real,
h h
' f$ = -^- = _ , as C is virtual. Substituting in (i),
A /z 2A
•' (-IP) ( + OP) (-CP)
' 1+-*— *-
" IP^OP CP
,.i+I-2
v m r
1,1 1
and .*. — — = -7. •
v u f
Thus, using the sign convention, the same formula holds for concave
and convex mirrors.
Formula for Magnification
The lateral magnification, m, produced by a mirror is defined by
height of image
m =
height of object
410
ADVANCED LEVEL PHYSICS
Suppose IR is the image of an object OH in a concave or convex
mirror. Fig. 17.11 (i), (ii). Then a ray HP from the top point H of the
Convex
(ii)
Fig. 17.11. Magnification formula.
object passes through the top point R of the image after reflection
from the mirror. Now the normal to the mirror at P is the principal axis,
OP. Thus angle OPH = angle IPR, from the law of reflection.
i.e. f
tan OPH = tan IPR
OH _ IR
OP ~~ IP
IR = IP
•"' OH "" OP
But IP = image distance = v, and OP = object distance = u
IR _ v
•"* OH ~~ u
v
m =
u
(4)
Thus if the image distance is half the object distance, the image is half
the length of the object.
Since
multiplying throughout by v.
1+1=1
V V _ V
1 +--7
•' 1 + «— 7
m-j-l
REFLECTION AT CURVED MIRRORS 411
Some Applications of Mirror Formulae
The following examples will assist the reader to understand how to
111 v
apply the formulae — |- - = - and m = - correctly:
v u f u
1. An object is placed 10 cm in front of a concave mirror of focal length
15 cm. Find the image position and the magnification.
Since the mirror is concave,/ = + 15 cm. The object is real, and hence
u = -f 10 cm. Substituting in 1 — —p,
v u f
■1.1 1
v ' (+ 10) (+ 15)
1 = 1_J. = — L
*• v~15 10 30
.-. v = - 30
Since v is negative in sign the image is virtual, and it is 30 cm from the
lirror. See Fig. 17.7 (i). The magnificati
image is three times as high as the object.
v 30
mirror. See Fig. 17.7 (i). The magnification, m = - = — = 3, so that the
u 10
2. The image of an object in a convex mirror is 4 cm from the mirror. If
the mirror has a radius of curvature of 24 cm, find the object position and
the magnification.
The image in a convex mirror is always virtual (p. 406). Hence v = — 4
cm. The focal length of the mirror = £ r = 12 cm; and since the mirror is
convex, / = — 12 cm. Substituting in 1 — =-j
V I* J
1.1 1 *
( - 4) ' u ( - 12)
1 __ 1_ 1 = 1
'*• u~ 12 + 4~6
.*. u = 6
Since u is positive in sign the object is real, and it is 6 cm from the mirror.
v 4
he magnification, m = - = 2
u o
as the object. See Fig. 17.8 (i).
v 4 2
The magnification, m = - = 2 = o> an< * hence the image is two-thirds as high
u 3
3. An erect image, three times the size of the object, is obtained with a
concave mirror of radius of curvature 36 cm. What is the position of the object ?
If x cm is the numerical value of the distance of the object from the mirror,
412 ADVANCED LEVEL PHYSICS
image distance
the image distance must be 3x cm, since the magnification m = -rr-. — — jt—
object distance
= 3. Now an erect image is obtained with a concave mirror only when the
image is virtual (p. 406).
image distance, v = — 3x
Also, object distance, u = + x
and focal length, f = \r = -\- 1 8 cm.
c ,_ • • • 1 , 1 1
Substituting in - -f- - = -^,
v u f
1.11
••(-3*) ' (+*) (+18)
.•.--L + 1-- L
3x^ x 18
— =i_
*• 3x _ 18
/. x = 12
Thus the object is 12 cm from the mirror.
Virtual Object and Convex Mirror
We have already seen that a convex mirror produces a virtual image
of an object in front of it, which is a real object. A convex mirror may
sometimes produce a real image of a virtual object.
As an illustration, consider an incident beam of light bounded by
AB, DE, converging to a point O behind the mirror, Fig. 17.12. O is
^__---' (Virtu
(Virtual)
Fig. 17.12. Real image in convex mirror.
regarded as a virtual object, and if its distance from the mirror is 10 cm,
then the object distance, u = — 10. Suppose the convex mirror has a
focal length of 15 cm, i.e.,/ = — 15.
1 , 1 1
Since — -\ — -x,
v u f
REFLECTION AT CURVED MIRRORS 413
v ' (-10) (-15)
v 15 ^ 10 ^30
.-. v = + 30
The point image, I, is thus 30 cm from the mirror, and is real. The
beam reflected from the mirror is hence a convergent beam, Fig. 17.12;
a similar case with a plane mirror is shown in Fig. 16.9 (ii).
Object at Centre of Curvature of Concave Mirror
Suppose an object is placed at the centre of curvature of a concave
mirror. Then u = + r, where r is the numerical value of the radius of
1 1 2
curvature. Substituting in — + - = - to find the image distance v,
v u r
-4+ l
v ( + r) ( + r)
1-1-1=1
v ~ r r r
v = r
The image is therefore also formed at the centre of curvature. The
v r
magnification in this case is given by m =— = — = 1, and hence the
object and image are the same size. This case is illustrated in Fig. 17.7
(ii), to which the reader should now refer.
SOME METHODS OF DETERMINING FOCAL LENGTH AND
RADIUS OF CURVATURE OF MIRRORS
Concave Mirror
Method 1. A pin O is placed above a concave mirror M so that an in-
verted image of the pin can be seen, Fig. 1 7. 1 3. If the pin is moved up and
down with its point on the axis of the mirror, and an observer E moves
his eye perpendicularly to the pin at the same time, a position of
O is reached when the image I remains perfectly in line with O as E
moves; i.e., there is no parallax (no relative displacement) between
414
ADVANCED LEVEL PHYSICS
A E
,0:
^o^ M
pin and image. The pin is now at exactly the
same place as its image I, Fig. 17.13. Since an
object and image coincide in position at the
centre of curvature of a concave mirror (p.
413), the distance from the point of the pin to
the mirror is equal to its radius of curvature,
r. The focal length,/, which is — ,is then easily
obtained. 2
If an illuminated object is available
instead of a pin, the object is moved to or
Fig. 17.13. Centre of curvature 1™™** TT ^ a ^. ima 8 e is ° b "
of concave mirror. tamed beside the object. The distance of the
object from the mirror is then equal to r.
In general the method of no parallax, using a pin, gives a higher
degree of accuracy in locating an image.
Method 2. By using the method
of no parallax, or employing an
illuminated object, several, say six,
values of the image distance, v, can
be obtained with the concave mirror,
corresponding to six different values
of the object distance «. Sub-
stituting for u, v in the formula
- + -=.?, six values of / can be
V u f J
calculated, and the average value
taken.
A better method of procedure,
however, is to plot the magnitudes of
- against - ; a straight line B A can be drawn through the points thus
obtained, Fig. 17.14. Now 1 + 1 = 1. Hence, when - = 0, i = \.
v u f v u f
But OB = iwhen- =
u v
•••<»- 7. i-e.,/=5j».
Thus the focal length can be determined from the reciprocal value of
the intercept OB on the axis of- .
From - + - = I, - = - wheni = 0. Thus OA = \, Fig. 17.14, and
v u f v f u /
-hence/ = -^-r • It can thus be seen that (i)/can also be calculated from
Fig. 17.14. Graph of - against -.
v u
REFLECTION AT CURVED MIRRORS 415
the reciprocal of the intercept OA on the axis of-, (ii) OAB is an isos-
celes triangle if the same scale is employed for - and — .
Convex Mirror
Method 1. By using a convex lens, L, a real image of an object O can
be formed at a point C on the other side of L, Fig. 17.15. The convex
mirror, MN, is then placed between L and C with its reflecting face
facing the lens, so that a convergent beam of rays is incident on the
mirror. When the latter is moved along the axis OC a position will be
reached when the beam is incident normally on the mirror, in which
case the rays are reflected back along the incident path. A real inverted
image is then formed at O.
Fig. 17.15. Convex mirror measurement.
Since the rays incident on the mirror, for example at N or M, are
normal to the mirror surface, they will, if produced, pass through the
centre of curvature, C, of the mirror. Thus the distance PC = r, the
radius of curvature. Since PC = LC — LP, this distance can be obtained
from measurement of LP and LC, the latter being the image distance
from the lens when the mirror is taken away.
Method 2. A more difficult method than the above consists of
positioning a pin O in front of the convex mirror, when a virtual image,
I, is formed, Fig. 17.16. A small plane mirror M is then moved between
O and P until the image I' of the lower part of O in M coincides in
position with the upper part of the image of O in the convex mirror.
The distances OP, MP are then measured.
♦ Virtual
M K I
i
Fig. 17.16. Convex mirror measurement.
416 ADVANCED LEVEL PHYSICS
Since M is a plane mirror, the image I' of O in it is such that OM =
MI'. Thus PI = MI' - MP = OM - MP, and hence PI can be cal-
culated. But PI = v, the image distance of O in the convex mirror, and
OP = u, the object distance. Substituting for the virtual distance v, and
u, in — | — = -/., the focal length of the convex mirror can be found,
v u f
EXAMPLES
1. Show that a concave spherical mirror can produce a focused image of
an object when certain conditions are observed, and prove the usual relation
between the object and image distances. A linear object, 10 cm long, lies
along the axis of a concave mirror whose radius of curvature is 30 cm, the
near end of the object lying 18 cm from the mirror. Find the magnification
of the image. (W.)
First part. The condition for a focused image is that the light from the
object must be incident as a narrow beam round the pole of the mirror. This
implies that the object must be small (p. 402). The usual relation between the
object and image distances is proved on p. 407.
Second part. Suppose Q is the near end of the object, which is 1 8 cm from
the mirror. The distance of the image of Q is given by
1.1 1
v ' ( + 18) ( + 15)
• f r 30 i«
since/ =x = ~z = 15 cm
J 2 2
j._J 1_
*' v~15 18
from which v = 90 cm
The other end P of the object is (10 -f- 18) cm from the mirror, or 28 cm
The image of P is given by
i i
v ' ( + 28) (+15)
• I=JL__L
" v 15 28
from which v = 32-3 cm
.*. length of image = 90 - 32-3 = 57-7 cm
57-7
.'. magnification of image = -r^r- = 5-8
2. PBCA is the axis of a concave spherical mirror, A being a point object,
B its image, C the centre of curvature of the mirror and P the pole. Find a
relation between PA, PB, and PC, supposing the aperture of the mirror to
be small. A concave mirror forms, on a screen, a real image of twice the linear
REFLECTION AT CURVED MIRRORS 417
dimensions of the object. Object and screen are then moved until the image is
three times the size of the object. If the shift of the screen is 25 cm determine
the shift of the object and the focal length of the mirror. (N.)
First part. See text.
Second part. Suppose v is the distance of the screen from the mirror when
the image is twice the length of the object. Since the magnification, m, is
given by
m=- f - 1 . . . . (i)
where /is the focal length of the mirror (see p. 410),
.'.2=j-l (ii)
When m = 3, the image distance is (v + 25) cm. Substituting in (i),
. , v + 25 ,
.'.3 = —^ 1 .... (in)
Subtracting (ii) from (iii), we have
-7
.*. /= 25 cm.
v
From (ii), 2 = — — 1, or v = 75 cm. The object distance, it, is thus given
by — -f - = — , from which u — 31\ cm. From (iii), v = 100 cm. The object
distance, u, is then given by jzrz + — = — , from which u = 33 J cm. Thus the
shift of the object = 37| - 33 £ = 4$ cm.
EXERCISES 17
1. An object is placed (i) 10 cm, (ii) 4 cm from a concave mirror of radius
curvature 12 cm. Calculate the image position in each case, and the respective
magnifications.
2. Repeat Q. 1 by accurate drawings to scale. (Note. — The mirror must
be represented by a straight line.)
3. An object is placed 15 cm from a convex mirror of focal length 10 cm.
Calculate the image distance and the magnification produced. Draw an
accurate diagram to scale, and verify your drawing from the calculated results.
4. Explain with the aid of diagrams why a curved mirror can be used (i)
as a driving mirror, (ii) in a searchlight, (iii) as a shaving mirror. Why is a
special form of mirror required in the searchlight?
418 ADVANCED LEVEL PHYSICS
5. Describe and explain a method of finding the focal length of (a) a concave
mirror, (b) a convex mirror.
6. A pole 4 m long is laid along the principal axis of a convex mirror of
focal length 1 m The end of the pole nearer the mirror is 2 m from it. Find
the length of the image of the pole.
7. Deduce a formula connecting u, v and r, the distances of object, image
and centre of curvature from a spherical mirror.
A mirror forms an erect image 30 cm from the object and twice its height.
Where must the mirror be situated ? What is its radius of curvature ? Assuming
the object to be real, determine whether the mirror is convex or concave. (L.)
8. Establish the formula - + - = -j. for a concave mirror.
v u f
In an experiment with a concave mirror the magnification m of the image
is measured for a series of values of v, and a curve is plotted between m and v.
What curve would you expect to obtain, and how would you use it to deduce
the focal length of the mirror ? (C)
9. Derive an approximate relation connecting the distances of an object
and its image from the surface of a convex spherical mirror.
A small object is placed at right angles to the axis of a concave mirror so
as to form (a) a real, (6) a virtual image, twice as long as the object. If the
radius of curvature of the mirror is R what is the distance between the two
images? (L.)
10. Deduce a formula connecting the distances of object and image from
a spherical mirror. What are the advantages of a concave mirror over a lens
for use in an astronomical telescope?
A driving mirror consists of a cylindrical mirror of radius 10 cm and length
(over the curved surface) of 10 cm. If the eye of the driver be assumed at a
great distance from the mirror, find the angle of view. (O. & C.)
11. Find the relation connecting the focal length of a convex spherical
mirror with the distances from the mirror of a small object and the image
formed by the mirror.
A convex mirror, radius of curvature 30 cm, forms a real image 20 cm from
its surface. Explain how this is possible and find whether the image is erect
or inverted. (L.)
12. What conditions must be satisfied for an optical system to form an image
of an object? Show how these conditions are satisfied for a convex spherical
mirror when a small object is placed on its axis and derive a relationship
showing how the position of the image depends on the position of the object
and the radius of curvature of the mirror.
A millimetre scale is placed at right angles to the axis of a convex mirror
of radius of curvature 12 cm. This scale is 18 cm away from the pole of the
mirror. Find the position of the image of the scale. What is the size of the
divisions of the image? What is the ratio of the angle subtended by the image
to that subtended by the object at a point on the axis 25 cm away from the
object on the side remote from the mirror? (0. & C.)
REFLECTION AT CURVED MIRRORS 419
13. Describe an experiment to determine the radius of curvature of a convex
mirror by an optical method. Illustrate your answer with a ray diagram and
explain how the result is derived from the observations.
A small convex mirror is placed 60 cm from the pole and on the axis of a
large concave mirror, radius of curvature 200 cm. The position of the convex
mirror is such that a real image of a distant object is formed in the plane of a
hole drilled through the concave mirror at its pole. Calculate (a) the radius
of curvature of the convex mirror, (b) the height of the real image if the distant
object subtends an angle of 0-50° at the pole of the concave mirror. Draw a
ray diagram to illustrate the action of the convex mirror in producing the
image of a non-axial point of the object and suggest a practical application
of this arrangement of mirrors. (N.)
chapter eighteen
Refraction at plane surfaces
Laws of Refraction
When a ray of light AO is incident at O on the plane surface of a glass
medium, observation shows that some of the light is reflected from the
surface along OC in accordance with the laws of reflection, while the
rest of the light travels along a new direction, OB, in the glass, Fig. 18.1.
On account of the change in
direction the light is said to be
"refracted" on entering the glass ;
and the angle of refraction, r, is
the angle made by the refracted
ray OB with the normal at O.
Historical records reveal that
the astronomer Ptolemy, who
lived about a.d. 140, measured
numerous values of the angle of
incidence, i, and the angle of re-
fraction, r, for glass as the angle
of incidence was varied. How-
ever, he was unable to discover
any relation between i and r.
Later scientists were equally unsuccessful, until centuries later Snell,
a Dutch professor, discovered in 1620 that the sines of the angles bear a
constant ratio to each other. The laws of refraction are :
1. The incident and refracted rays, and the normal at the point of
incidence, all lie in the same plane.
- _ . ,. sin i .
2. For two given media, —. — is a constant, where i is the angle of
incidence andr is the angle of refraction (Snell's law).
[Refracted
B
Fig. 18.1. Refraction at plane surface.
Refractive Index.
The constant ratio
sm i
sin r
is known as the refractive index for the two
given media; and as the magnitude of the constant depends on the
colour of the light, it is usually specified as that obtained for yellow
light. If the medium containing the incident ray is denoted by 1, and
that containing the refracted ray by 2, the refractive index can be denoted
by !« 2 .
Scientists have drawn up tables of refractive indices when the incident
420
REFRACTION AT PLANE SURFACES 421
ray is travelling in vacuo and is then refracted into the medium con-
cerned, for example, glass or water. The values thus obtained are known
as the absolute refractive indices of the media; and as a vacuum is
always the first medium, the subscripts for the absolute refractive index,
symbol «, can be dropped. The magnitude of n for glass is about 1-5, n
for water is about 1-33, and n for air at normal pressure is about 1 '00028.
As the magnitude of the refractive index of a medium is only very slightly
altered when the incident light is in ah* instead of a vacuum, experiments
to determine the absolute refractive index w^are usually performed with
the light incident from air on to the medium; thus we can take airflgiass as
equal to V acuum«giass for most practical purposes.
We have already mentioned that light is refracted because it has
different velocities in different media. The Wave Theory of Light, dis-
cussed on p. 679, shows that the refractive index l n 2 for two given media
1 and 2 is given by
_ velocity of light in medium 1
1 2 velocity of light in medium! ' ' • w
and this is a definition of refractive index which can be used instead of
sin i
the ratio — — . An alternative definition of the absolute refractive index,
sinr '
n, of a medium is thus
__ velocity of light in a vacuum
~" velocity of light in medium " * • w
In practice the velocity of light in air can replace the velocity in
vacuo in this definition.
Relations Between Refractive Indices
(1 ) Consider a ray of light, AO, re-
fracted fromglass to air along the direc-
tion OB; observation then shows that
the refracted ray OB is bent away from
the normal, Fig. 18.2. The refractive
index from glass to air, g w a > is given
by sin xjsin y, by definition, where x
is the angle of incidence in the glass and
y is the angle of refraction in the air.
From the principle of the reversi-
bility of light (p 390), it foUows that a Fl(J lg 2 Refraction from
ray travelling along BO in air is re- glass to air.
fracted along OA in the glass. The
refractive index from air to glass, a/-g, is then given by sin yjsin x, by
, _ . . _ sin x .
definition. But gn a = — , from the previous paragraph.
1
.*. gHa = — (3)
tjtg
422
ADVANCED LEVEL PHYSICS
If artg is 1-5, then % n& = -rr — 0*67. Similarly, if the refractive index
from air to water is 4/3, the refractive index from water to air is 3/4.
(2) Consider a ray AO incident in air on a plane glass boundary, then
refracted from the glass into a water medium, and finally emerging
along a direction CD into air. If the boundaries of the media are parallel,
experiment shows that the emergent ray CD is parallel to the incident ray
AO, although there is a relative displacement. Fig. 18.3. Thus the angles
made with the normals by AO, CD are equal, and we shall denote
them by i a .
Fig. 18.3. Refraction at parallel plane surfaces.
Suppose i e , i w are the angles made with the normals by the respective
rays in the glass and water media. Then, by definition, g « w = -r-^.
sini w
But
and
sm ig sin ig sin i* a
sin i w sin i a sin i w '
sin ig sin i a
= g« a , and ^ . = sjt w
sinia
• g»w
1
sm i w
g» a X n«w •
(i)
Further, as g n& = , we can write
a«g
g«w —
a/lw
a«g
1-33
Since ^w = 1-33 and ^g = 1-5, it follows that g n w = -j-^ =0-89.
From (i) above, it follows that in general
l"3 =l"2 X 2 «3 (4)
The order of the suffixes enables this formula to be easily memorised.
REFRACTION AT PLANE SURFACES 423
General Relation Between n and Sin i
FromFig.l8.3,sin j a /sin z g = a« g
.*. sin /a = a«g sin i e (i)
Also, sin i w /sin U = w«a = l/an w
.*. sin U = a«w sin i w (ii)
Hence, from (i) and (ii),
sin ia = ajtg sin i g — a«w sin i w
If the equations are re-written in terms of the absolute refractive indices
of air (« a ), glass (« g ), and water (n w ), we have
n& sin i'a = rig sin / g = « w sin i w
since n& = 1. This relation shows that when a ray is refracted from one
medium to another, the boundaries being parallel)
n sin i = a constant (5)
where n is the absolute refractive index of a medium and / is the angle
made by the ray with the normal in that medium.
Fig. 18.4. Refraction from water to glass.
This relation applies also to the case of light passing directly from
one medium to another. As an illustration of its use, suppose a ray is
incident on a water-glass boundary at an angle of 60°, Fig. 18.4. Then,
applying "« sin / is a constant", we have
1-33 sin 60° = 1-5 sin r, . . . . (iii)
where r is the angle of refraction in the glass, and 1-33,1 -5 are the respec-
tive values of n w and « g . Thus sin r = 1-33 sin 60°/l-5 = 0-7679, from
which r = 50 1°.
Multiple Images in Mirrors
If a candle or other object is held in front of a plane mirror, a series
of faint or "ghost" images are observed in addition to one bright image.
424
ADVANCED LEVEL PHYSICS
Suppose O is an object placed in front of a
mirror with silvering on the back surface M,
as shown in Fig. 1 8 . 5 . A ray O A from O is then
reflected from the front (glass) surface along
AD and gives rise to a faint image I ls while
the remainder of the light energy is refracted
at A along AB. Reflection then takes place
at the silvered surface, and after refraction
into the air along CH a bright image is ob-
served at I 2 . A small percentage of the light
is reflected at C, however, and re-enters the
glass again, thus forming a faint image at
I 3 . Other faint images are formed in the same
way. Thus a series of multiple images is ob-
tained, the brightest being I 2 . The images lie on the normal from O to
the mirror, the distances depending on the thickness of the glass and its
refractive index and the angle of incidence.
Fig. 18.5. Multiple images.
Drawing the Refracted Ray by Geometrical Construction
Since sin //sin r = n, the direction of the refracted ray can be calculated
when a ray is incident in air at a known angle i on a medium of given
refractive index n. The direction of the refracted ray can also be obtained
by means of a geometrical construction. Thus suppose AO is a ray
incident in air at a given angle / on a medium of refractive index n,
Fig. 18.6 (i). With O as centre, two circles, a, b, are drawn whose radii
are in the ratio 1 : n, and AO is produced to cut circle a at P. PN is then
drawn parallel to the normal at O to intersect circle b at Q. OQ is then
the direction of the refracted ray.
To prove the construction is correct, we note that angle OPN = /,
angle OQN = r. Thus sin //sin r = ON/OP -^ ON/OQ = OQ/OP.
But OQ/OP = radius of circle 6/radius of circle a = n, from our drawing
Glass / >
n-VSl I
\J
>s^3\
Air \ V
/7=1 \ N
\ I
< j) (ii)
Fig. 18.6. Drawing of refracted rays.
REFRACTION AT PLANE SURFACES
425
of the circles. Hence sin z'/sin r = «. Thus OQ must be the refracted
ray. Although we have taken the case of the incident ray in air, the
same construction will enable the refracted ray to be drawn when the
incident ray is in any other medium. The radii of the circles are then in
the ratio of the absolute refractive indices.
Fig. 18.6 (ii) illustrates the drawing in the case of a ray AO refracted
from a dense medium such as glass (« = 1-5) into a less dense medium
such as air (n = 1). Circles a, b are again drawn concentric with O,
their radii being in the ratio 1 : «. The incident ray AO, however, is
produced to cut the larger circle b this time, at P, and a line PN is then
drawn parallel to the normal at O to intersect the circle a at Q. OQ is
then the direction of the refracted ray. The proof for the construction
follows similar lines to that given for Fig. 18.6 (i), and it is left as an
exercise for the reader.
The direction of the incident ray AO in Fig. 18.6 (ii) may be such that
the line PN does not intersect the circle a. In this case, which is im-
portant and is discussed shortly, no refracted ray can be drawn. See
Total Internal Reflection, p. 430.
Refractive Index of a Liquid by Using a Concave Mirror
We are now in a position to utilise our formulae in refraction, and we
shall first consider a simple method of determining roughly the refrac-
tive index, n, of a small quantity of transparent liquid.
If a small drop of the liquid is placed on a concave mirror S, a position
H can be located by the no parallax method where the image of a pin
held over the mirror coincides in position with the pin itself, Fig. 18.7.
The rays from the pin must now be striking the mirror normally, in
which case the rays are reflected back along the incident path and form
an image at the same place as the object. A ray HN close to the axis
HP is refracted at N along ND in the liquid, strikes the mirror normally
at D, and is reflected back along the path DNH. Thus if DNis produced
it passes through the centre of curvature, C, of the concave mirror.
Fig. 18.7. (Depth of liquid exaggerated.)
426
ADVANCED LEVEL PHYSICS
Let ANB be the normal to the liquid surface at N. Then angle ANH
— angle NHM = i, the angle of incidence, and angle BND = angle
ANC = angle NCM = r, the angle of refraction in the liquid. From
triangles HNM, CNM respectively, sin i = NM/HN and sin r =
NM/CN. The refractive index, «, of the liquid is thus given by
_ fiPJ _ NM/HN _ CN
n ~~ sin r ~ NM/CN ~~ HN *
Now since HN is a ray very close to the principal axis CP, HN = HM
CM
and CN = CM, to a very good approximation. Thus n — jjt-j. Further,
if the depth MP of the liquid is very small compared with HM and CM,
CM = CP and HM = HP approximately. Hence, approximately,
n =
CP
HP
HP can be measured directly. CP is the radius of curvature of the mirror,
which can be obtained by the method shown on page 414. The refractive
index, «, of the liquid can thus be calculated.
Apparent Depth
Swimmers in particular are aware that the bottom of a pool of water
appears nearer the surface than is actually the case; the phenomenon
is due to the refraction of light.
Consider an object O at a distance below the surface of a medium
such as water or glass, which has a
refractive index n, Fig. 18.8. A ray
OM from O perpendicular to the
surface passes straight through into
the air along MS. A ray ON very
close to OM is refracted at N into
the air away from the normal, in a
direction NT; and an observer view-
ing O directly overhead sees it in
the position I, which is the point
of intersection of SM and TN pro-
duced. Though we have only
considered two rays in the air, a
cone of rays, with SM as the
axis, actually enters the observer's
eye.
Suppose the angle of incidence in the glass is i, and the angle of
refraction in the air is r. Then, since "« sin i" is. a constant (p, 423), we
have
n sin i = 1 x sin r (i)
where n is the refractive index of glass; the refractive index pf air is 1.
Fig. 18.8. (Inclination of ON
to OM exaggerated.)
REFRACTION AT PLANE SURFACES
427
Since i = angle NOM, and r — MIN, sin i =? MN/ON and sin r
MN/IN. From (i), it follows that
MN MN
" ON ~ IN
ON
n =
IN
(ii)
Since we are dealing with the case of an observer directly above O,
the rays ON, IN are very close to the normal OM. Hence to a very good
approximation, ON = OM and IN = IM. From (ii),
ON _ OM
•"' n ~ IN ~~ IM *
Since the real depth of the object O = OM, and its apparent depth =
IM,
real depth
.... (6)
n =
apparent depth
If the real depth, OM, = /, the apparent depth = — , from (6). The
n
displacement, OI, of the object, which we shall denote by d, is thus
* ■
given by/ --,i.e.,
-(-9
(7)
If an object is 6 cm below water of refractive index, n — 1J, it
appears to be displaced upward to an observer in air by an amount, d
6(,-l) = .i
cm.
Object Below Parallel-sided
Glass Block
Consider an object O placed
some distance in air below a
parallel-sided glass block of thick-
ness t, Fig, 18.9. The ray OMS
normal to the surface emerges
along MS, while the ray OO t close
to the normal is refracted along
O x N in the glass and emerges in
air along NT in a direction parallel
to OOj (see p. 422). An observer
(not shown) above the glass thus
sees the object at I, the point of
intersection of TN and SM.
Suppose the normal at Oj inter-
sects IN at I x . Then, since O^ is
parallel to OI and IT is parallel to OO^ OI^Ox is a parallelogram. Thus
OI = Ojli. But OI is the displacement of the object O. Hence 0^ is
Fig. 18.9. Object below glass block.
428
ADVANCED LEVEL PHYSICS
equal to the displacement. Since the apparent position of an object
at O t is at Ix (compare Fig. 18.8), we conclude that the displacement OI
of O is independent of the position of O below the glass, and is given by
OI
-(-:>
see p. 427.
Fig. 18.10. Refractive index
by apparent depth.
Measurement of Refractive Index by
Apparent Depth Method
The formula for the refractive index of a
medium in terms of the real and apparent
depths is the basis of a very accurate method
of measuring refractive index. A travelling
microscope, S (a microscope which can travel
in a vertical direction and which has a fixed
graduated scale T beside it) is focused on
lycopodium particles at O on a sheet of white
paper, and the reading on T is noted, Fig.
18.10. Suppose it is c cm. If the refractive index
of glass is required, a glass block A is placed on
the paper, and the microscope is raised until
the particles are refocused at I. Suppose the reading on T is b cm.
Some lycopodium particles are then sprinkled at M on the top of the
glass block, and the microscope is raised until they are focused, when
the reading on T is noted. Suppose it is a cm.
Then real depth of O = OM = (a — c) cm.
and apparent depth = IM = (a — b) cm.
real depth _ a — c
~ apparent depth a — b
The high accuracy of this determination of n lies mainly in the fact
that the objective of the microscope collects only those rays near to its
axis, so that the object O, and its apparent position I, are seen by rays
very close to the normal OM. The experiment thus fulfils the theoretical
real depth
conditions assumed in the proof of the formula n = r— ; — -r
r apparent depth
p. 427.
The refractive index of water can also be obtained by an apparent
depth method. The block A is replaced by a dish, and the microscope
is focused first on the bottom of the dish and then on lycopodium
powder sprinkled on the surface of water poured into the dish. The
apparent position of the bottom of the dish is also noted, and the
refractive index of the water « w is calculated from the relation
_ real depth of water
w ~ apparent depth of water
General formula for real and apparent depth. So far we have considered the
rays refracted from a medium like glass into air. As a more general case,
suppose an object O is in a medium of refractive index n x and the rays from it
are refracted at M, N into a medium of refractive index w 2 > Fig. 18.11. The
image of O to an observer in the latter medium is then at I.
REFRACTION AT PLANE SURFACES
s
429
Fig. 18.11. Relation between real and apparent depth.
Suppose i'i is the angle of incidence at N, and i 2 is the angle of refraction as
shown. Then, since "/* sin i" is a constant,
«i sin /'x = « 2 sin / 2 .
Now angle MON = i lt and angle MIN = i 2 .
MN MN
.. Hi
ON
*i _
«2
7I 2
IN
" ON IN ' * (i)
If we consider rays very close to the normal, then IN = IM = v, say, and
ON = OM = k, say. Substituting in (i),
nx — n 2
" T 7 (u)
This formula can easily be remembered, as the refractive index of a medium
is divided by the corresponding distance of the object (or image) in that
medium.
Total Internal Reflection. Critical Angle
If a ray AO in glass is incident at a small angle a on a glass-air plane
boundary, observation shows that part of the incident light is reflected
along OE in the glass, while the remainder of the light is refracted away
from the normal at an angle jS into the air. The reflected ray OE is
weak, but the refracted ray OL is bright, Fig. 18.12 (i). This means that
most of the incident light energy is transmitted, and a little is reflected.
When the angle of incidence, o, in the glass is increased, the angle of
emergence, j9, is increased at the same time; and at some angle of in-
cidence c in the glass the refracted ray OL travels along the glass-air
boundary, making the angle of refraction of 90°, Fig. 18.12 (ii). The re-
flected ray OE is still weak in intensity, but as the angle of incidence in
the glass is increased slightly the reflected ray suddenly becomes bright,
and no refracted ray is then observed, Fig. 18.12 (iii). Since all the
430
(Bright)L
ADVANCED LEVEL PHYSICS
(«)
OgCP L
(Bright)
Critical angle = c
(ii)
Total reflection
(iii)
Fig. 18.12. Total internal reflection.
incident light energy is now reflected, total reflection is said to take place
in the glass at O.
When the angle of refraction in air is 90°, a critical stage is reached
at the point of incidence O, and the angle of incidence in the glass is
accordingly known as the critical angle for glass and air, Fig. 18.12 (ii).
Since "« sin /" is a constant (p. 423), we have
n sin c = 1 x sin 90%
where n is the refractive index of the glass. As sin 90° = 1, then
n sin c = 1,
or,
1
sinc = -
n
(8)
Crown glass has a refractive index of about 1-51 for yellow light, and
thus the critical angle for glass to air is given by sin c = 1/1*51 = 0-667.
Consequently c — 41-5°. Thus if the incident angle in the glass is
greater than c, for example 45°,
total reflection occurs, Fig. 18.12
(iii). The critical angle between two
media for blue light is less than for
red light, since the refractive index
for blue light is greater than that for
red light (see p. 458).
The phenomenon of total reflec-
tion may occur when light in glass
(rig = 1-51, say) is incident on a
boundary with water (« w = 1-33).
Applying "/i sin i is a constant" to
the critical case, Fig. 18.13, we have
n 8 sin c = « w sin 90°,
where c is the critical angle. As sin 90° = 1
n g sin c =n w
ZL Water -
Fig. 18.13. Critical angle for
water and glass.
sm c = — =
"8
/. c = 63°.
1-33
1-51
= 0-889
REFRACTION AT PLANE SURFACES 431
Thus if the angle of incidence in the glass exceeds 63°, total internal
reflection occurs.
It should be carefully noted that the phenomenon of total internal
reflection can occur only when light travels from one medium to another
which has a smaller refractive index, i.e., which is optically less dense.
The phenomenon cannot occur when light travels from one medium
to another optically denser, for example from air to glass, or from water
to glass, as a refracted ray is then always obtained.
SOME APPLICATIONS OF TOTAL INTERNAL REFLECTION
1. Reflecting prisms are pieces of glass of a special shape which are
used in prism binoculars and in certain accurate ranging instruments
such as submarine periscopes. These prisms, discussed on p. 449, act
as reflectors of light by total internal reflection.
2. The mirage is a phenomenon due to total reflection. In the desert
the air is progressively hotter towards the sand, and hence the density
of the air decreases in the direction bed, Fig. 18.14 (i). A downward ray
OA from a tree or the sky is thus refracted more and more away from
the normal; but at some layer of air c, a critical angle is reached, and
the ray begins to travel in an upward direction along eg. A distant
observer P thus sees the object O at I, and hence an image of a palm
tree, for example, is seen below the actual position of the tree. As an
image of part of the sky is also formed by total reflection round the
image of the tree, the whole appearance is similar to that of a pool of
water in which the tree is reflected.
O
Appleton
layer r
(ii)
Fig. 18.14. Examples of total reflection.
3. Total reflection of radio waves. A radio wave is an example of an
electromagnetic wave because it comprises electric and magnetic forces.
Light waves also are electromagnetic waves (p. 983). Light waves and
radio waves are therefore the same in nature, and a close analogy can
be made between the refraction of light and the refraction of a radio
wave when the latter enters a medium containing electric particles.
In particular, the phenomenon of total reflection occurs when radio
waves travel from one place, S, on the earth, for example England, to
another place, R, on the other side of the earth, for example America,
Fig. 18.14 (ii). A layer of considerable density of electrons exists many
432
ADVANCED LEVEL PHYSICS
miles above the earth (at night this is the Appleton layer), and when a
radio wave SA from a transmitter is sent skyward it is refracted away
from the normal on entering the electron layer. At some height, corres-
ponding to O, a critical angle is reached, and the wave then begins to
be refracted downward. After emerging from the electron layer it returns
to R on the earth, where its presence can be detected by a radio receiver.
Measurement of Refractive Index of a Liquid by an Air-cell Method
The phenomenon of total internal reflection is utilised in many
methods of measuring refractive index. Fig. 18.15 (i) illustrates how the
c~zr~zzrjrji^. — Liquid
0)
Fig. 18.15 (i). Air-cell method.
refractive index of a liquid can be determined. Two thin plane-parallel
glass plates, such as microscope slides, are cemented together so as to
contain a thin film of air of constant thickness between them, thus
forming an air-cell, X. The liquid whose refractive index is required is
placed in a glass vessel V having thin plane-parallel sides, and X is placed
in the liquid. A bright source of light, S,. provides rays which are in-
cident on one side of X in a constant direction SO, and the light through
X is observed by a person on the other side at E.
When the light is incident normally on the sides of X, the light passes
straight through X to E. When X is rotated slightly about a vertical
axis, light is still observed ; but as X is rotated farther, the light is suddenly
cut off from E, and hence no light now passes through X, Fig. 18.15 (i).
Fig. 18.15 (ii) illustrates the behaviour of the light when this happens.
The ray SO is refracted along OB in the glass, but at B total internal
reflection begins. Suppose i x is the angle of incidence in the liquid, i,
is the angle of incidence in the glass, and «, « g are the corresponding
refractive indices. Since the boundaries of the media are parallel we
can apply the relation "n sin i is a constant'*, and hence
n sin ix — «g sin i a = 1 x sin 90°, (i)
the last product corresponding to the case of refraction in the air-film.
.*. Msini'i = 1 X sin 90° = 1 X 1 = 1
.-. n =-^- r (9)
sin i x w
REFRACTION AT PLANE SURFACES
433
It should now be carefully noted that i x is the angle of incidence in
the liquid medium, and is thus determined by measuring the rotation
of X from its position when normal to SO to the position when the light
*>
Liquid
Glass Air
(ii)
Fig. 18.15 (ii). Air-cell theory.
is cut off. In practice, it is better to rotate X in opposite directions and
determine the angle 6 between the two positions for the extinction of the
light. The angle d is then half the angle 0, and hence n — 1/sin^.
From equations (i) and (9), it will be noted that i x is the critical
angle between the liquid and air, and i z is the critical angle between the
glass and air. We cannot measure i a , however, as we can i lt and hence
the method provides the refractive index of the liquid.
The source of light, S, in the experiment should be a monochromatic
source, i.e., it should provide light of one colour, for example yellow light.
The extinction of the light is then sharp. If white light is used, the colours
in its spectrum are cut off at slightly different angles of incidence, since
refractive index depends upon the colour of the light (p. 458). The ex-
tinction of the light is then gradual and ill-defined.
Pulfrich Refractometer
A refractometer is an instrument which measures refractive index by
making use of total internal reflection. Pulfrich designed a refracto-
meter enabling the refractive index of a liquid to be easily obtained,
which consists of a block of glass G with a polished and vertical face.
On top of G is cemented a circular glass tube V, Fig. 18.16. The liquid L
is placed in V, and a convergent beam of monochromatic light is directed
so that the liquid-glass interface is illuminated. On observing the light
refracted through G by a telescope T, a light and dark field of view are
seen.
434
ADVANCED LEVEL PHYSICS
Beam
Fig. 18.16. Pulfrich refractometer
The boundary between the light and dark fields corresponds to the
ray which is incident just horizontally on the liquid-glass boundary, as
shown in Fig. 18.16. If c is the angle of refraction in the glass, it follows
that
n sin 90° = n g sin c (i)
where n, n g are the refractive indices of the liquid and glass respectively.
For refraction at B,
n g sin r = sin i (ii)
Also, c + r = 90° (iii)
From (i), sin c = n\n g . Now from (iii), sin r =■ sin (90° — c) = cos c.
Substituting in (ii), we have « g cos c = sin i, or cos c = sin */n g .
But sin 2 c + cos 2 c = 1
m 2 sin 2 i
2 ' 2~ 1
«g 2 «g 2
n 2 + sin 2 / == n g a
n = V He 2 —
sm^z
Thus if i is measured and n g = 1-51, n can be calculated. In practice,
tables are supplied giving the refractive index in terms of i, and another
block is used in place of G for liquids of higher n than 1-51.
Abbe refractometer. Abbe designed a refractometer for measuring the
refractive index of liquids whose principle is illustrated in Fig. 18.17. Two
similar prisms X, Y are placed on a table A, the prism X being hinged at H
so that it could be swung away from Y. A drop of the liquid is placed on
the surface a, which is matt, and the prisms are placed together so that the
liquid is squeezed into a thin film between them. Light from a suitable source
is directed towards the prisms by means of a mirror M, where it strikes the
surface a and is scattered by the matt surface into the liquid film. The emergent
rays are collected in a telescope T directed towards the prisms, and the field
of view is divided into a dark and bright portion. The table A is then turned
until the dividing line between the dark and bright fields is on the crosswires
of the telescope, which is fixed. The reading on the scale S, which is attached
to, and moves with, the table, gives the refractive index of the liquid directly,
as explained below.
REFRACTION AT PLANE SURFACES
435
Liquid
film
Fig. 18.17. Abbe refractometer principle.
Theory. The dividing line, BQ, between the bright and dark fields corres-
ponds to the case of the ray DA, incident in the liquid L at grazing incidence
on the prism Y, Fig. 18.18. The refracted ray AB in the prism then makes the
critical angle c with the normal at A, where n sin 90° = « g sin c, n and n g being
the respective refractive indices of the liquid and the glass.
n = « g sin c (i)
(c-45)
Dj/Z- y"
Fig. 18.18. Theory of refractometer.
For simplicity, suppose that Y is a right-angled isosceles prism, so that
the angle P of the prism is 45°. The angle of incidence at B in the glass is then
(c - 45°), by considering the geometry of triangle PAB, and hence for refrac-
tion at B we have
« g sin (c — 45°) = sin (ii)
where 9 is the angle with the normal at B made by the emerging ray BQ. By
eliminating c from (i) and (ii), we obtain finally
n = sin 45° (n g 2 - sin 2 0*) + cos 45° sin 6
=^[(»g 2 -sin^* + sin^]
since sin 45° = 1/V2 = cos 45°. Thus knowing n g and 9, the refractive index
of the liquid, n, can be evaluated. The scale S, Fig. 18.17, which gives 6, can
thus be calibrated in terms of n.
436
ADVANCED LEVEL PHYSICS
EXAMPLES
1. Describe a method, based on grazing incidence or total internal r6flection,
for finding the refractive index of water for the yellow light emitted by a
sodium flame.
The refractive index of carbon bisulphide for red light is 1-634 and the
difference between the critical angles for red and blue light at a carbon bisul-
phide-air interface is 0° 56'. What is the refractive index of carbon bisulphide
for blue light? (W.)
First part. See air-cell method, p. 432.
Second part. Suppose n r and c r are the refractive index and critical angle
of carbon bisulphide for red light.
Then sin c r = - = t-4t, = 0-6119
/. c r = 37°
1-634
44'
The critical angle, c b , for blue light is less than that for red light.
/. ^ = 37° 44' - 0° 56' = 36° 48'
1
The refractive index for blue light, « to , =
1
fib
sm Cb
= 1-669
II
11
4-5
Water
Liquid
Glass
sin 36° 48'
2. Find an expression for the distance through which an object appears
to be displaced towards the eye when a
plate of glass of thickness t and refractive
index n is interposed.
A tank contains a slab of glass 8 cm
thick and of refractive index 1-6. Above
this is a depth of 4-5 cm of a liquid of
refractive index 1-5 and upon this floats 6
cm of water (n = 4/3). To an observer
looking from above, what is the apparent
position of a mark on the bottom of the
tank? (0. & C.)
First part. See text.
Second part. Suppose O is the mark at
the bottom of the tank, Fig. 18.19. Then
since the boundaries of the media are parallel, the total displacement of O
is the sum of the displacements due to each of the media.
Forglass, displacement, d =± / ( 1 J = 8 ( 1 — r-, J = 3 cm
For liquid,
cm
For water,
d=6
(-id-
1-5 cm
.*. total displacement = 3 + 1-5 + 1-5 = 6 cm
/. apparent position of O is 6 cm from bottom.
3. A small object is placed on the principal axis of a concave spherical
mirror of radius 20 cm at a distance of 30 cm. By how much will the position
REFRACTION AT PLANE SURFACES
437
and size of the image alter when a parallel-sided slab of glass, of thickness
6 cm and refractive index 1-5, is introduced between the centre of curvature
and the object? The parallel sides are perpendicular to the principal axis.
Prove any formula used. (N.)
Suppose O is the position of the object before the glass is placed in position,
112
Fig. 18.20. The image position is given by — + - = - ,
Mirror
Fig. 18.20.
I+1 = A
v ^ 30 20
Solving,
v = 15 cm.
v IS
The magnification, m = - = — = 05.
u 30
When the glass slab G of thickness, t, 6 cm is inserted, the rays from O
appear to come from a point O' whose displacement from O is f ( 1 — -J,
where n is the glass refractive index. See p. 428. The displacement is thus
6 ( ! ~ j^ ) = 2 cm. The distance of O' from the mirror is therefore (30 - 2),
112
or 28 cm. Applying the equation- + - =-, we find v = + 154 cm. The
v u r •»
image position changes by (15f - 15) or 4 cm. The magnification becomes
15| -f- 28, or 0-52.
EXERCISES 18
1. A ray of light is incident at 60° in air on an air-glass plane surface. Find
the angle of refraction in the glass by calculation and by drawing (n for glass
= 1-5).
2. A ray of light is incident in water at an angle of 30° on a water-air plane
surface. Find the angle of refraction in the air by calculation and by drawing
(» for water = 4/3).
3. A ray of light is incident in water at an angle of (i) 30°, (ii) 70° on a water-
glass plane surface. Calculate the angle of refraction in the glass in each case
(an g = 1-5, a/*w = 1-33).
438 ADVANCED LEVEL PHYSICS
4. (a) Describe the apparent depth method of finding the refractive index
of glass, and prove the formula used, (b) What is the apparent position of an
object below a rectangular block of glass 6 cm thick if a layer of water 4 cm
thick is on top of the glass (refractive index of glass and water = 1 J and 1$
respectively)?
5. Describe and explain a method of measuring approximately the refractive
index of a small quantity of liquid.
6. Calculate the critical angle for (i) an air-glass surface, (ii) an air-water
surface, (iii) a water-glass surface; draw diagrams in each case illustrating
the total reflection of a ray incident on the surface ( a «g = 15, a «w = 1*33).
7. Explain what happens in general when a ray of light strikes the surface
separating transparent media such as water and glass. Explain the circum-
stances in which total reflection occurs and show how the critical angle is
related to the refractive index.
Describe a method for determining the refractive index of a medium by
means of critical reflection. (L.)
8. Define refractive index of one medium with respect to another and show
how it is related to the values of the velocity of light in the two media.
Describe a method of finding the refractive index of water for sodium light,
deducing any formula required in the reduction of the observations. (JV.)
9. Explain carefully why the apparent depth of the water in a tank changes
with the position of the observer.
A microscope is focused on a scratch on the bottom of a beaker. Turpentine
is poured into the beaker to a depth of 4 cm, and it is found necessary to
raise the microscope through a vertical distance of 1-28 cm to bring the
scratch again into focus. Find the refractive index of the turpentine. (C.)
10. What is meant by total reflection and critical angle! Describe two
methods of measuring the refractive index of a material by determining the
critical angle, one of which is suitable for a solid substance and the other
for a liquid. (L.)
11. (a) State the conditions under which total reflection occurs. Show that
the phenomenon will occur in the case of light entering normally one face
of an isosceles right-angle prism of glass, but not in the case when light enters
similarly a similar hollow prism full of water, (b) A concave mirror of small
aperture and focal length 8 cm lies on a bench and a pin is moved vertically
above it. At what point will image and object coincide if the mirror is filled
with water of refractive index 4/3 ? (JV.)
12. State the laws of refraction, and define refractive index.
Describe an accurate method of determining the refractive index of a
transparent liquid for sodium light. Give the theory of the method, and
derive any formula you require. Discuss the effect of substituting white light
for sodium light in your experiment. ( W.)
13. Describe an experiment for finding the refractive index of a liquid by
measuring its apparent depth.
A vessel of depth 2d cm is half filled with a liquid of refractive index m,
and the upper half is occupied by a liquid of refractive index n 2 . Show that
the apparent depth of the vessel, viewed perpendicularly, is d I 1 — ) . (JL.)
REFRACTION AT PLANE SURFACES 439
14. The base of a cube of glass of refractive index m is in contact with the
surface of a liquid of refractive index n%. Light incident on one vertical face
of the cube is reflected internally from the base and emerges again from the
opposite vertical face in a direction making an angle 6 with its normal. Assum-
ing that m > n it show th at the light has just been totally reflected internally
ifn 2 = V(/i 1 2 - sin*0).
Describe how the above principle may be used to measure the refractive
index of a small quantity of liquid. (N.)
15. Explain the meaning of critical angle and total internal reflection.
Describe fully (a) one natural phenomenon due to total internal reflection,
(b) one practical application of it. Light from a luminous point on the
lower face of a rectangular glass slab, 2-0 cm thick, strikes the upper face and
the totally reflected rays outline a circle of 3-2 cm radius on the lower face.
What is the refractive index of the glass ? (N.)
16. Describe one experiment in each case to determine the refractive index
for sodium light of (a) a sample of glass which could be supplied to any shape
and size which you specify, (b) a liquid of which only a very small quantity is
available. Show how the result is calculated in each case. (You are not expected
to derive standard formulae.)
How would you modify the experiment in (a) to find how the refractive
index varies with the wavelength of the light used? What general result would
you expect? (O. & C).
17. Explain the meaning of critical angle, and describe how you would
measure the critical angle for a water-air boundary.
ABCD is the plan of a glass cube. A horizontal beam of light enters the
face AB at grazing incidence. Show that the angle 6 which any rays emerging
from BC would make with the normal to BC is given by sin 6 = cot a, where
a is the critical angle. What is the greatest value that the refractive index of
glass may have if any of the light is to emerge from BC? (JV.)
18. State the laws of refraction of light. Explain how you would measure
the refractive index of a transparent liquid available only in small quantity,
i.e., less than 0-5 cm 3 .
A ray of light is refracted through a sphere, whose material has a refractive
index «, in such a way that it passes through the extremities of two radii which
make an angle a with each other. Prove that if y is the deviation of the ray
caused by its passage through the sphere
cos | (a — y) = n cos \a. (L.)
19. Explain what is meant by the terms critical angle and total reflection.
Describe an accurate method of determining the critical angle for a liquid,
indicating how you would calculate the refractive index from your measure-
ments.
A man stands at the edge of the deep end of a swimming bath, the floor
of which is covered with square tiles. If the water is clear and undisturbed,
explain carefully how the floor of the bath appears to him. (O. & C.)
20. Summarize the various effects that may occur when a parallel beam of
light strikes a plane interface between two transparent media.
440 ADVANCED LEVEL PHYSICS
Explain why, when looking at the windows of a railway carriage from
inside, one sees by day the country outside and by night the reflection of the
inside of the carriage.
An observer looks normally through a thick window of thickness d and
refractive index n at an object at a distance e behind the farther surface.
Where does the object appear to be, and how can this apparent position be
found experimentally? (0. & C.)
chapter nineteen
Refraction through prisms
In Light, a prism is a transparent object usually made of glass which
has two plane surfaces, XDEY, XDFZ, inclined to each other, Fig. 19.1
Z
Fig. 19.1. Prism.
Prisms are used in many optical instruments, for example prism bino-
culars, and they are also utilised for separating the colours of the light
emitted by glowing objects, which affords an accurate knowledge of
their chemical composition. A prism of glass enables the refractive
index of this material to be measured very accurately.
The angle between the inclined plane surfaces XDFZ, XDEY is
known as the angle of the prism, or the refracting angle, the line of
intersection XD of the planes is known as the refracting edge, and any
plane in the prism perpendicular to XD, such as PQR, is known as a
principal section of the prism. A ray of light ab, incident on the prism
at b in a direction perpendicular to XD, is refracted towards the normal
along be when it enters the prism, and is refracted away from the normal
along cd when it emerges into the air. From the law of refraction (p. 420),
the rays ab, be, cd all lie in the same plane, which is PQR in this case.
If the incident ray is directed towards the refracting angle, as in Fig.
19.1, the light is always deviated by the prism towards its base.
Refraction Through a Prism
Consider a ray HM incident in air on a prism of refracting angle A,
and suppose the ray lies in the principal section PQR, Fig. 19.2. Then, if
i ly r x and i 2 , r 2 are the angles of incidence and refraction at M, N as
shown, and n is the prism refractive index,
sin i x = n sin r x (i)
sin i 2 = n sin r 2 (ii)
441
442
Fig. 19.2. Refraction through prism.
Further, as MS and NS are normals to PM and PN respectively, angle
MPN + angle MSN = 180°, considering the quadrilateral PMSN. But
angle NST + angle MSN = 180°.
.'. angle NST = angle MPN = A
:. A = r x + r % . . . . (in)
as angle NST is the exterior angle of triangle MSN.
In the following sections, we shall see that the angle of deviation, d,
of the light, caused by the prism, is utilised considerably. The angle of
deviation at M = angle OMN = i x — r x \ the angle of deviation at N =
angle MNO = i 2 — r 2 . Since the deviations at M, N are in the same
direction, the total deviation, d (angle BOK), is given by
d = {i x - r x ) + (/ 2 - ra) . . . . (iv)
Equations (i) — (iv) are the general relations which hold for refraction
through a prism. In deriving them, it should be noted that the geo-
metrical form of the prism base plays no part.
Minimum Deviation
The angle of deviation, d, of the incident ray HM is the angle BOK
in Fig. 19.2. The variation of d with the angle of incidence, i, can be
obtained experimentally by placing the prism on paper on a drawing
board and using a ray AO from a ray-box (or two pins) as the incident
ray, Fig. 19.3 (i). When the direction AO is kept constant and the draw-
ing board is turned so that the ray is always incident at O on the prism, the
angle of incidence i is varied; the corresponding emergent rays CE, HK,
LM, NP can be traced on the paper. Experiment shows that as the angle
of incidence i is increased from zero, the deviation d begins to decrease
continuously to some value D, and then increases to a maximum as Ms
increased further to 90°. A minimum deviation, corresponding to the
emergent ray NP, is thus obtained. A graph of </ plotted against i has the
appearance of the curve X, which has a minimum value at R, Fig. 19.3 (ii).
REFRACTION THROUGH PRISMS
T
443
Maximum
deviation .
Fig. 19.3. Minimum deviation.
Experiment and theory show that the minimum deviation, D, of the
light occurs when the ray passes symmetrically through the prism. Suppose
this corresponds to the case of the ray AONP in Fig. 19.3 (i). Then the
corresponding incident angle, i, is equal to the angle of emergence, i lt
into the air at N for this special case. See also Fig. 19.5 and Fig. 19.9.
A proof of symmetrical passage of ray at minimum deviation. Experiment
shows that minimum deviation is obtained at one particular angle of incidence.
On this assumption it is possible to prove by a reductio ad absurdum method
that the angle of incidence is equal to the angle of emergence in this case. Thus
Fig. 19.4. Minimum deviation proof.
suppose that minimum deviation is obtained with a ray PMNR when these
angles are not equal, so that angle PMB is not equal to angle RNC, Fig. 19.4.
It then follows that a ray YX, incident on AC at an angle CXY equal to angle
PMB, will emerge along TS, where angle BTS = angle CNR; and from the
principle of the reversibility of light, a ray incident along ST on the prism
emerges along XY. We therefore have two cases of minimum deviation,
corresponding to two different angles of incidence. But, from experiment,
this is impossible. Consequently our initial assumption must be wrong, and
hence the angle of emergence does equal the angle of incidence. Thus the ray
passes symmetrically through the prism in the minimum deviation case.
444 ADVANCED LEVEL PHYSICS
Relation Between A, D, and n
A very convenient formula for refractive index, n, can be obtained in
the minimum deviation case. The ray PQRS then passes symmetrically
through the prism, and the angles made with the normal in the air and
in the glass at Q, R respectively are equal, Fig. 19.5. Suppose the angles
are denoted by i, r, as shown. Then, as explained on p. 442,
Fig. 19.5. Formula for n of prism.
i — r + i — r = D
.
and r + r = A
•
From (ii), r ~ ~2
Substituting for r in (i), 2/ = A + D
. A + D
••' I== 2
sin i
. A + D
sin— 2~
smr
. A
sin 2
(i)
(ii)
(1)
The Spectrometer
The spectrometer is an optical instrument which is mainly used to
study the light from different sources. As we shall see later, it can be
used to measure accurately the refractive index of glass in the form of a
prism. The instrument consists essentially of a collimator, C, a telescope,
T, and a table, R, on which a prism B can be placed. The lenses in C, T
are achromatic lenses (p. 515). The collimator is fixed, but the table
and the telescope can be rotated round a circular scale graduated in
half-degrees (not shown) which has a common vertical axis with the
table, Fig. 19.6. A vernier is also provided for this scale. The source of
light, S, used in the experiment is placed in front of a narrow slit at one
end of the collimator, so that the prism is illuminated by light from S.
REFRACTION THROUGH PRISMS 445
R
Fig. 19.6. Spectrometer.
Before the spectrometer can be used, however, three adjustments
must be made: (1) The collimator C must be adjusted so that parallel
light emerges from it; (2) the telescope T must be adjusted so that
parallel rays entering it are brought to a focus at cross-wires near its
eye-piece; (3) the refracting edge of the prism must be parallel to the
axis of rotation of the telescope, i.e., the table must be "levelled".
Adjustments of Spectrometer
The telescope adjustment is made by first moving its eye-piece until
the cross-wires are distinctly seen, and then sighting the telescope on to
a distant object through an open window. The length of the telescope is
now altered by a screw arrangement until the object is clearly seen at
the same place as the cross-wires, so that parallel rays now entering
the telescope are brought to a focus at the cross-wires.
The collimator adjustment. With the prism removed from the table,
the telescope is now turned to face the collimator, C, and the slit in C is
illuminated by a sodium flame which provides yellow light. The edges
of the slit are usually blurred, showing that the light emerging from the
lens of C is not a parallel beam. The position of the slit is now adjusted
by moving the tube in C, to which the slit is attached, until the edges of
the latter are sharp.
"Levelling" the table. If the rectangular slit is not in the centre of the
field of view when the prism is placed on the table, the refracting edge of
the prism is not parallel to the axis of rotation of the telescope. The
table must then be adjusted, or "levelled", by means of the screws a,
b, c beneath it. One method of procedure consists of placing the prism
on the table with one face MN approximately perpendicular to the line
joining two screws a, b, as shown in Fig. 19.6. The table is turned until
MN is illuminated by the light from C, and the telescope T is then moved
to receive the light reflected from MN. The screw b is then adjusted
until the slit appears in the centre of the field of view. With C and T
fixed, the table is now rotated until the slit is seen by reflection at the
face NP of the prism, and the screw c is then adjusted until the slit is
again in the middle of the field of view. The screw c moves MN in its
own plane, and hence the movement of c will not upset the adjustment of
MN in the perpendicular plane.
446
ADVANCED LEVEL PHYSICS
Measurement of the Angle, A, of a Prism
The angle of a prism can be measured very accurately by a spectro-
meter. The refracting edge, P, of the prism is turned so as to face the
collimator lens, which then illuminates the two surfaces containing
Collimator
Fig. 19.7. Measurement of angle of prism.
the refracting angle A with parallel light, Fig. 19.7 (i). An image of the
collimator slit is hence observed with the telescope in positions T^ T 2 ,
corresponding to reflection of light at the respective faces of the prism.
It is shown below that the angle of rotation of the telescope from T t to
T 2 is equal to 2A, and hence the angle of the prism, A, can be obtained.
Proof. Suppose the incident ray MN makes a glancing angle a with one
face of the prism, and a parallel ray at K makes a glancing angle £ with the
other face, Fig. 19.7 (ii). The reflected ray NQ then makes a glancing angle a
with the prism surface, and hence the deviation of MN is 2a (see p. 392). Simi-
larly, the deviation by reflection at K is 2 p. Thus the reflected rays QN, LK are
inclined at an angle equal to 2a -j- 2/3, corresponding to the angle of rotation
of the telescope from Ti to T 2 . But the angle, A, of the prism = a + £, as
can be seen by drawing a line through P parallel to MN, and using alternate
angles. Hence the rotation of the telescope = 2a + 2j8 = 2A.
Measurement of the Minimum Deviation, D
In order to measure the minimum deviation, 2>, caused by refraction
through the prism, the latter is placed with its refracting angle A pointing
away from the collimator, as shown in Fig. 19.8 (i). The telescope is then
turned until an image of the slit is obtained on the cross-wires, corres-
ponding to the position T v The table is now slowly rotated so that the
angle of incidence on the left side of the prism decreases, and the image
of the slit is kept on the cross-wires by moving the telescope at the same
time. The image of the slit, and the telescope, then slowly approach the
REFRACTION THROUGH PRISMS
447
fixed line XY. But at one position, corresponding to T 2 , the image of
the slit begins to move away from XY. If the table is now turned in the
opposite direction the image of the slit again moves back when the
telescope reaches the position T 2 . The angle between the emergent ray
CH and the line XY is hence the smallest angle of deviation caused
by the prism, and is thus equal to D.
The minimum deviation is obtained by finding the angle between the
positions of the telescope (i) at T 2 , (ii) at T; the prism is removed in the
latter case so as to view the slit directly. Alternatively, the experiment to
find the minimum deviation is repeated with the refracting angle pointing
(0 (ii)
Fig. 19.8. Measurement of minimum deviation.
the opposite way, the prism being represented by dotted lines in this
case, Fig. 19.8 (ii). If the position of the telescope for minimum deviation
is now O, it can be seen that the angle between the position O and the
other niinimum deviation position T 2 is 2D. The value of D is thus
easily calculated.
The Refractive Index of the Prism Material
The refractive index, n, of the material of the prism can be easily
calculated once A and D have been determined, since, from p. 444.
A + D I . A
— / Sm r
In an experiment of this nature, the angle, A, of a glass prism was found
to be 59° 52', and the minimum deviation, D, was 40° 30'. Thus
n= sin
n= sin
59° 52' + 40° 30'
/
sin
59° 52'
2 / "" 2
= sin 50° 11'/ sin 29° 56'
= 1-539
The spectrometer prism method of measuring refractive index is
capable of providing an accuracy of one part in a thousand. The refrac-
tive index of a liquid can also be found by this method, using a hollow
glass prism made from thin parallel-sided glass strips.
448 ADVANCED LEVEL PHYSICS
Grazing Incidence for a Prism
We shall now leave any further considerations of minimum deviation,
and shall consider briefly other special cases of refraction through a
prism.
When the surface PQ of a prism is illuminated by a source of yellow
light placed near Q, the field of view seen through the other surface
PR is divided into a bright and dark portion, Fig. 19.9. If NM is the
P
R
Fig. 19.9. Grazing incidence.
emergent ray corresponding to the incident ray QH which just grazes
the prism surface, the dark portion lies above NM and the bright portion
exists below NM. The boundary of the light and dark portions is thus
NM.
Since the angle of incidence of QH is 90°, angle NHL is equal to c, the
critical angle for the glass of the prism. From Fig. 19.9, it follows that
A = c + r, and hence
r = A - c . . . . (i)
Further, for refraction at N,
sin 6 = n sin r.
.*. sin 6 = n sin {A — c) = n (sin A cos c — cos A sin c) . (ii)
Butsin c = -, i.e., cos c = Vl — sin 2 c= / 1 =■ = - Vn* - 1.
Substituting in (ii) and simplifying, we obtain finally
/ , , /cos A + sin 0\ a
n = J 1 + ( sinA ) *
Thus if A and 6 are measured, the refractive index of the prism material
can be calculated.
It should be noted that maximum deviation by a prism is obtained (i)
at grazing incidence, / = 90°, (ii) at an angle of incidence 6 (Fig. 19.9),
corresponding to grazing emergence.
Grazing Incidence and Grazing Emergence
If a ray BM is at grazing incidence on the face of a prism, and the
angle A of the prism is increased, a calculation shows that the refracted
REFRACTION THROUGH PRISMS
449
ray MN in the glass will make a bigger and bigger angle of incidence on
the other face PR, Fig. 19.10. This is left as an exercise for the reader. At
Fig. 19.10. Maximum angle of prism.
a certain value of A, MN will make the critical angle, c, with the normal
at N, and the emergent ray NR will then graze the surface PR, as shown
in Fig. 19.10. As A is increased further, the rays in the glass strike PR
at angles of incidence greater than c, and hence no emergent rays are ob-
tained. Thus Fig. 19.10 illustrates the largest angle of a prism for which
emergent rays are obtained, and this is known as the limiting angle of the
prism. It can be seen from the simple geometry of Fig. 19.10 that A =
c 4- c in this special case, and hence the limiting angle of a prism is
twice the critical angle. For crown glass of n = 1-51 the critical angle c
is 41 ° 30', and hence transmission of light through a prism of crown glass
is impossible if the angle of the prism exceeds 83°.
Total Reflecting Prisms
When a plane mirror silvered on the back is used as a reflector,
multiple images are obtained (p. 424). This disadvantage is overcome by
using right-angled isosceles prisms as reflectors of light in optical
instruments such as submarine periscopes (see p. 540).
Consider a ray OQ incident normally on the face AC of such a prism.
Fig. 19. 1 1 (i). The ray is undeviated, and is therefore incident at P in the
(0
(ii)
Fig. 19.11. Images in prisms.
450 ADVANCED LEVEL PHYSICS
glass at an angle of 45° to the normal at P. If the prism is made of crown
glass its critical angle is 41° 30'. Hence the incident angle, 45°, in the
glass is greater than the critical angle, and consequently the light is
totally reflected in the glass at P. A bright beam of light thus emerges
from the prism along RT, and since the angle of reflection at P is equal
to the incident angle, RT is perpendicular to OQ. The prism thus deviates
the light through 90°. If the prism is positioned as shown in Fig. 19.11
(ii), an inverted bright virtual image I of the object O is seen by total
reflection at the two surfaces of the prism.
There is no loss of brightness when total internal reflection occurs at a
surface, whereas the loss may be as much as 10 per cent or more in
reflection at a silver surface.
EXAMPLES
1. Describe a good method of measuring the refractive index of a substance
such as glass and give the theory of the method. A glass prism of angle 72°
and index of refraction 1-66 is immersed in a liquid of refractive index 1-33.
What is the angle of minimum deviation for a parallel beam of light passing
through the prism? (L.)
First part. Spectrometer can be used, p. 444.
Second part.
sin
n =
(^)
. A
sin —
2
where n is the relative refractive index of glass with respect to the liquid.
1-66
But
* = F33
1-66 _ Sm (,— 2— ) _ Sm \—2~)
1-33 . 72° sin 36°
sm — -
2
. . f 72° + D \ 1-66 . '
. . sm ( — J = r-^r sin 36 = 0-7335
72° 4- D
/z -r^ = 47 o n ,
2
.*. D = 22° 22'
2. How would you measure the angle of minimum deviation of a prism?
(a) Show that the ray of light which enters the first face of a prism at grazing
incidence is least likely to suffer total internal reflection at the other face.
(b) Find the least value of the refracting angle of a prism made of glass of
refractive index 7/4 so that no rays incident on one of the faces containing this
angle can emerge from the other. (N.)
First part. See text.
REFRACTION THROUGH PRISMS 451
Second part, (a) Suppose PM is a ray which enters the first face of the
prism at grazing incidence, i.e., at an angle of incidence of 90°, Fig. 19.12. The
refracted ray MQ then makes an angle of refraction c, where c is the critical
angle. Suppose QN is the normal at Q on the other face of the prism. Then
since angle BNQ = A, where A is the angle of the prism, the angle of refraction
MQN at Q = A — c, from the exterior angle property of triangle MQN.
Similarly, if RM is a ray at an angle of incidence i at the first face less than
90°, the angle of refraction at S at the second face — A — r, where r is the
angle of refraction BMS.
Fig. 19.12. Example
Now c is the maximum angle of refraction in the prism.
Hence MQ makes the minimum angle of incidence on the second face, and
thus is least likely to suffer total internal reflection.
Q>) The least value of the refracting angle of the prism corresponds to a
ray at grazing incidence and grazing emergence, as shown in Fig. 19.10.
Thus minimum angle = 2c
where c is the critical angle (p. 449).
1 4
But sin c = - = - = 0-5714
n 7
.-. c = 34°51'
.*. minimum angle = 1c = 69° 42'
EXERCISES 19
1. A ray of fight is refracted through a prism of angle 70°. If the angle of
refraction in the glass at the first face is 28°, what is the angle of incidence in
the glass at the second face?
2. (i) The angle of a glass prism is 60°, and the minimum deviation of
light through the prism is 39°. Calculate the refractive index of the glass,
(ii) The refractive index of a glass prism is 1-66, and the angle of the prism
is 60°. Find the minimum deviation.
452 ADVANCED LEVEL PHYSICS
3. By means of a labelled diagram show the paths of rays from a mono-
chromatic source to the eye through a correctly adjusted prism spectrometer.
Obtain an expression relating the deviation of the beam by the prism to the
refracting angle and the angles of incidence and emergence.
A certain prism is found to produce a minimum deviation of 51° 0', while
it produces a deviation of 62° 48' for two values of the angle of incidence,
namely 40° 6' and 82° 42' respectively. Determine the refracting angle of the
prism, the angle of incidence at minimum deviation and the refractive index
of the material of the prism. (L).
4. A ray of light passing symmetrically through a glass prism of refracting
angle A is deviated through an angle D. Derive an expression for the refractive
index of the glass.
A prism of refracting angle about 60° is mounted on a spectrometer table
and all the preliminary adjustments are made to the instrument. Describe
and explain how you would then proceed to measure the angles A and D.
PQR represents a right-angled isosceles prism of glass of refractive index
1-50. A ray of light enters the prism through the hypotenuse QR at an angle
of incidence /, and is reflected at the critical angle from PQ to PR. Calculate
and draw a diagram showing the path of the ray through the prism. (Only
rays in the plane of PQR need be considered.) (N.)
5. Give a labelled diagram showing the essential optical parts of a prism
spectrometer. Describe the method of adjusting a spectrometer and using it
to measure the angle of a prism.
A is the vertex of a triangular glass prism, the angle at A being 30°. A ray
of light OP is incident at P on one of the faces enclosing the angle A, in a
direction such that the angle OPA = 40°. Show that, if the refractive index
of the glass is 1-50, the ray cannot emerge from the second face. (L.)
6. Define refractive index and derive an expression relating the relative
refractive index n AB for light travelling out of medium A into medium B with
the velocities of light v^ and v B respectively in those media.
Draw a diagram showing how a parallel beam of monochromatic light is
deviated by its passage through a triangular glass prism. Given that the angle
of deviation is a minimum when the angles of incidence and emergence are
equal show that the refractive index n of the glass is related to the refracting
angle a of the prism and the minimum deviation 6 by the equation
n = sin |( a + <5)/sin £a.
Describe how you would apply this result to measure the dispersive power
of the glass of a given triangular prism. You may assume the availability of
sources of light of standard wavelengths. (O. & C.)
7. Explain how you would adjust the telescope of a spectrometer before
making measurements.
Draw and label a diagram of the optical parts of a prism spectrometer after
the adjustments have been completed. Indicate the position of the crosswires
and show the paths through the instrument of two rays from a monochromatic
source when the setting for minimum deviation has been obtained.
The refracting angle of a prism is 620° and the refractive index of the glass
for yellow light is 1-65°. What is the smallest possible angle of incidence of a
ray of this yellow light which is transmitted without total internal reflection?
Explain what happens if white light is used instead, and the angle of incidence
is varied in the neighbourhood of this minimum. (N.)
REFRACTION THROUGH PRISMS 453
8. Explain the meaning of the term critical angle. Describe and give the
theory of a critical angle method for determining the refractive index of
water.
A right-angled prism ABC has angle BAC = angle ACB = 45°, and is
made of glass of refractive index 1-60. A ray of light is incident upon the
hypotenuse face AC so that after refraction it strikes face AB and emerges at
minimum deviation. What is the angle of incidence upon AC?
What is the smallest angle of incidence upon AC for which the ray can still
emerge at AB? If the angle of incidence upon AC is made zero, what will be
the whole deviation of the ray? (L.)
9. Draw a labelled diagram of a spectrometer set up for studying the
deviation of light through a triangular prism. Describe how you would adjust
the instrument and use it to find the refractive index of the prism material.
Indicate briefly how you would show that the radiation from an arc* lamp
is not confined to the visible spectrum. (L.)
10. How would you investigate the way in which the deviation of a ray of
light by a triangular glass prism varies with the angle of incidence on the first
face of the prism? What result would you expect to obtain?
The deviation of a ray of light incident on the first face of a 60° glass prism
at an angle of 45° is 40°. Find the angle which the emergent ray makes with
the normal to the second face of the prism and determine, preferably by
graphical construction, the refractive index of the glass of the prism. (L.)
11. A prism has angles of 45°, 45°, and 90° and all three faces polished.
Trace the path of a ray entering one of the smaller faces in a direction parallel
to the larger face and perpendicular to the prism edges. Assume 1-5 for the
refractive index.
If you had two such prisms how would you determine by a simple pin or
ray-box method the refractive index of a liquid available only in small
quantity? (£.)
12. Under what circumstances does total internal reflection occur? Show
that a ray of light incident in a principal section of an equilateral glass prism
of refractive index 1-5, can only be transmitted after two refractions at
adjacent faces if the angle of incidence on the prism exceeds a certain value.
Find this limiting angle of incidence. (W.)
13. Draw a graph showing, in a general way, how the deviation of a ray of
light when passed through a triangular prism depends on the angle of
incidence.
You are required to measure the refractive index of glass in the form of a
prism by means of a spectrometer provided with a vertical slit. Explain how
you would level the spectrometer table and derive the formula from which
you would calculate the refractive index. (You are not required to explain
any other adjustment of the apparatus nor to explain how you would find
the refracting angle of the prism.) (L.)
chapter twenty
Dispersion. Spectra
Spectrum of White Light
In 1666, Newton made a great scientific discovery. He found that
sunlight, or white light, was made up of different colours, consisting of
red, orange, yellow, green, blue, indigo, violet. Newton made a small
hole in a shutter in a darkened room, and received a white circular patch
of sunlight on a screen S in the path of the light, Fig. 20.1 (i). But on
interposing a glass prism between the hole and the screen he observed
Prism
(i)
Fig. 20.1. Impure spectrum.
Coloured
images
(impure
spectrum)
(«')
a series of overlapping coloured patches in place of the white patch, the
total length of the coloured images being several times their width,
Fig. 20.1 (ii). By separating one colour from the rest, Newton demon-
strated that the colours themselves could not be changed by refraction
through a prism, and he concluded that the colours were not introduced
by the prism, but were components of the white light. The spectrum
(colours) of white lightconsists of red, orange, yellow, green, blue, indigo,
and violet, and the separaion of the colours by the prism is known as
dispersion.
The red rays are the least deviated by the prism, and the violet rays are
the most deviated, as shown in the exaggerated sketch of Fig. 20.1 (i).
Since the angle of incidence at O in the air is the same for the red and
violet rays, and the angle of refraction made by the red ray OB in the
glass is greater than that made by the violet ray OC, it follows from
sin i/sin r that the refractive index of the prism material for red light is
less than for violet light. Similarly, the refractive index for yellow light
lies between the refractive index values for red and violet light (see also
p. 458).
454
DISPERSION. SPECTRA
455
Production of Pure Spectrum
Newton's spectrum of sunlight is an impure spectrum because the
different coloured images overlap, Fig. 20.1 (ii). A pure spectrum is one
in which the different coloured images contain light of one colour only,
i.e., they are monochromatic images. In order to obtain a pure spectrum
(i) the white light must be admitted through a very narrow opening, so
as to assist in the reduction of the overlapping of the images, (ii) the
beams of coloured rays emerging from the prism must be parallel, so
that each beam can be brought to a separate focus.
The spectrometer can be used to provide a pure spectrum. The
collimator slit is made very narrow, and the collimator C and the
telescope T are both adjusted for parallel light, Fig. 20.2. A bright source
White
Violet
Fig. 20.2. Pure spectrum.
of white light, S, is placed near the slit, and the prism P is usually set
in the minimum deviation position for yellow light, although this is not
essential. The rays refracted through P are now separated into a number
of different coloured parallel beams of light, each travelling in slightly
different directions, and the telescope brings each coloured beam to a
separate focus. A pure spectrum can now be seen through T, consisting
of a series of monochromatic images of the slit.
If only one lens, L, is available, the prism P must be placed in the
minimum deviation position for yellow light in order to obtain a fairly
pure spectrum, Fig. 20.3. The prism is then also approximately in the
Spectrum
Redl
Violet/
Fig. 20.3. Fairly pure spectrum.
minimum deviation position for the various colours in the incident
convergent beam, and hence the rays of one colour are approximately
deviated by the same amount by the prism, thus forming an image of
the slit S at roughly the same place.
Infra-red and ultra-violet rays. In 1800 Herschel discovered the existence
456 ADVANCED LEVEL PHYSICS
of infra-red rays, invisible rays bevond the red end of the spectrum. Funda-
mentally, they are of the same nature as rays in the visible spectrum but
having longer wavelengths than the red, and produce a sensation of heat
(see p. 344). Their existence may be demonstrated in the laboratory by means
of an arc light in place of S in Fig. 20.3, a rocksalt lens at L and a rocksalt
prism at P. A phototransistor, such as Mullard OCP71, connected to an ampli-
fier and galvanometer is very sensitive to infra-red light. When this detector is
moved into the dark part beyond the red end of the spectrum, a deflection is
obtained in the galvanometer. Since they are not scattered by fine particles
as much as the rays in the visible spectrum, infra-red rays can penetrate fog
and mist. Clear pictures have been taken in mist by using infra-red filters and
photographic plates.
About 1801 Ritter discovered the existence of invisible rays beyond the
violet end of the visible spectra. Ultra-violet rays, as they are known, affect
photographic plates and cause certain minerals to fluoresce. They can also
eject electrons from metal plates (see Photoelectric effect, p. 1077). Ultra-
violet rays can be detected in the laboratory by using an arc light in the place
of S in Fig. 20.3, a quartz lens at L, and a quartz prism at P. A sensitive detector
is a photoelectric cell connected to a galvanometer and battery. When the cell
is moved beyond the violet into the dark part of the spectrum a deflection is
observed in the galvanometer.
Deviation Produced by Small-angle Prism for Small Angles of
Incidence
Before discussing in detail the colour effect produced when white light
is incident on a prism, we must derive an expression for the deviation
produced by a small-angle prism.
Consider a ray PM of monochromatic light incident almost normally
on the face TM of a prism of small angle A, so that the angle of incidence,
i 1} is small, Fig. 20.4. Then sin i t /sin r\ = n, where r t is the angle of refrac-
tion in the prism, and n is the refractive index for the colour of the light.
As r x is less than ii, r t also is a small angle. Now the sine of a small angle
is practically equal to the angle measured in radians. Thus i^r x = n, or
*i = Wi (i)
From the geometry of Fig. 20.4, the angle of incidence r 2 on the face
TN of the prism is given by r 2 = A — r t \ and since A and r t are both
Fig. 20.4. Deviation through small-angle prism.
DISPERSION. SPECTRA 457
small, it follows that r 2 is a small angle. The angle of emergence i 2 is thus
also small, and since sin i 2 /sin r 2 = n we may state that f 2 /r 2 = n, or
h = nr 2 (ii)
The deviation, </, of the ray on passing through the prism is given by
d = (*i — rj) + (i a — r 2 ). Substituting for i x and /, from (i) and (ii),
.\ d=nr t - r ± + «r 2 — r 2 = «(r x + rj - (r 2 + r»)
/. ^=(n-l)(r 1 + r 2 )
But r x + r 2 = ^4
/. d = («-l)A (1)
This is the magnitude of the deviation produced by a sma//-angle
prism for small angles of incidence. If A is expressed in radians, then d
is in radians; if A is expressed in degrees, then dis in degrees. If A = 6°
and n = 1-6 for yellow light, the deviation </ of that colour for small
angles of incidence is given by d = (1-6 - 1) 6° = 3-6°. It will be nSted
that the deviation is independent of the magnitude of the small angle
of incidence on the prism.
Dispersion by Small-angle Prism
We have already seen from Newton's experiment that the colours
in a beam of white light are separated by a glass prism into red, orange,
yellow, green, blue, indigo, violet, so that the emergent light is no
longer white but coloured. The separation of the colours by the prism
is known generally as the phenomenon of dispersion, and the angular
dispersion between the red and blue emergent rays, for example, is
defined as the angle between these two rays. Thus, in Fig. 20.5, is the
angular dispersion between the red and blue rays. Of course, the angular
dispersion is also equal to the difference in deviation of the two colours
produced by the prism; and since we have already derived the
expression d = (n - 1) A for the deviation of monochromatic light by
a small-angle prism we can obtain the angular dispersion between any
two colours.
Red
Yellow
Blue
Fig. 20.5. Dispersion.
458
ADVANCED LEVEL PHYSICS
Suppose dn, d T are the respective deviations of the blue and red light
when a ray of white light is incident at a small angle on a prism of small
angle A, Fig. 20.5. Then, if n b , n T are the refractive indices of the prism
material for blue and red light respectively,
4 = ("b - 1) A,
and d T = (ra r - 1) A.
.'. angular dispersion, d h — d T = (h„ — 1) A — (n T
:. d h - d T = (n b - /* r ) A
I) A
(2)
For a particular crown glass, n b = 1-521, n T = 1-510. Thus if A = 8°,
the angular dispersion between the blue and red colours =
d h -d T = (n h -n t )A = (1-521 - 1-510) 8° = 0-09°
The mean deviation of the white light by the prism is commonly
chosen as the deviation of the yellow light, since this is the colour
approximately in the middle of the spectrum; the mean refractive
index of a material is also specified as that for yellow light. Now the
deviation, d, of monochromatic light is given by d — (n — 1) A, from
equation (1), and unless otherwise stated, the magnitudes of d and n
will be understood to be those for yellow light when these symbols
contain no suffixes. If w b = 1-521 and n r = 1-510, then approximately
the refractive index, n, for yellow light is the average of n b and n T , or
1(1-521 + l-510);thusn = 1-515. Hence if the prism has an angle of 8°,
the mean deviation, d, = (« - 1) A = (1-515 - 1) 8° = 4-1°.
Dispersive Power
The dispersive power, a, of the material of a small-angle prism for blue
and red rays may be defined as the ratio
angular dispersion between blue and red rays
<«> =
mean deviation
(3)
The dispersive power depends on the material of the prism. As an
illustration, suppose that a prism of angle 8° is made of glass of a type
X, say, and another prism of angle 8° is made of glass of a type Y.
"b
«r
n
Crown glass, X
1-521
1-510
1-515
Flint glass, Y
1-665
1-645
1-655
Further, suppose the refractive indices of the two materials for blue
red, and yellow light are those shown in the above table.
For a small angle of incidence on the prism of glass X, the angular
dispersion
DISPERSION. SPECTRA 459
= d b — d T = (n h — I) A — (n T — 1) A
= (n b - n r ) A = (1-521 - 1-510) 8° = 0-09° . (i)
The mean deviation, d = (n-l)A = (1-515 - 1) 8° = 4-1° . (ii)
009 nM1
.*. dispersive power, w, = -^-r- = 0-02 1 .... (iii)
Similarly, for the prism of glass Y,
angular dispersion = (« b - n T ) A = (1-665 - 1-645) 8 = 0-16°
and mean deviation = (n — 1) A = 1-655 — 1) 8 = 5-24°
016
.'. dispersive power = -^r = 003 .... (iv)
From (iii) and (iv), it follows that the dispersive power of glass Y
is about 1-5 times as great as that of glass X.
General Formula for Dispersive Power
We can now derive a general formula for dispersive power, w, which
is independent of angles. From equation 3, it follows that
d h — d T
<8= — —
as d\, d Xi d denote the deviations of blue, red, and yellow light respec-
tively.
But 4 — d T = (« b - 1) A - (n T - 1) A = (« b - n T ) A
and d=(n - I) A.
dh — d t (n h — « r ) A
CD =
ct>
d (n -\)A
« b — n t
n - 1
(4)
From this formula, it can be seen that (i) «■> depends only on the
material of the prism and is independent of its angle, (ii) w is a number
and has therefore no units. In contrast to "dispersive power'*, it should
be noted that "dispersion" is an angle, and that its magnitude depends
on the angle A of the prism and the two colours concerned, for
d b - d T = dispersion = (« b - n T ) A.
Achromatic Prisms
We have seen that a prism separates the colours in white light. If a
prism is required to deviate white light without dispersing it into colours,
two prisms of different material must be used to eliminate the dispersion,
as shown in Fig. 20.6. The prism P is made of crown glass, and causes
dispersion between the red and blue in the incident white light. The
prism Q is inverted with respect to P, and with a suitable choice of its
angle A' (discussed fully later), the red and blue rays incident on it can
be made to emerge in parallel directions. If the rays are viewed the eye-
lens brings them to a focus at the same place on the retina, and hence
460 ADVANCED LEVEL PHYSICS
the colour effect due to red and blue rays is eliminated. The dispersion
of the other colours in white light still remains, but most of the colour
effect is eliminated as the red and blue rays are the "outside" (extreme)
rays in the spectrum of white light.
Flint
Crown
Fig. 20.6. Achromatic prisms.
Prisms which eliminate dispersion between two colours, blue and red
say, are said to be achromatic prisms for those colours. Suppose w b , « r
are the refractive indices of crown glass for blue and red light, and A
is the angle of the crown glass prism P. Then, from p. 458,
dispersion = (« b — n T ) A . . . . (i)
If w' b , n' T are the refractive indices of flint glass for blue and red light
and A' is the angle of the flint glass prism Q, then similarly,
dispersion = («' b — n't) A' . . . . (ii)
Now prism P produces its dispersion in a "downward" direction since
a prism bends rays towards its base, Fig. 20.6, and prism Q produces
its dispersion in an "upward" direction. For achromatic prisms, there-
fore, the dispersions produced by P and Q must be equal.
:.(n h -nr)A = (n\-n'r)A' .... (5)
Suppose P has an angle of 6°. Then, using the refractive indices for
n b , n t , «' b , n' T in the table on p. 458, it follows from (5) that the angle A'
is given by
(1-521 - 1-510) 6° = (1-665 - 1-645) A'
Thus A'=^^ x 6° = 3-3°
Deviation Produced by Achromatic Prisms
Although the colour effects between the red and blue rays are
eliminated by the use of achromatic prisms, it should be carefully
noted that the incident light beam, as a whole, has been deviated.
This angle of deviation, d, is shown in Fig. 20.6, and is the angle between
the incident and emergent beams. The deviation of the mean or yellow
light by prism P is given by (« — 1) A, and is in a "downward" direction.
DISPERSION. SPECTRA 461
Since the deviation of the yellow light by the prism Q is in an opposite
direction, and is given by (n' - 1) A', the net deviation, d, is given by
</=(« - I) A -(»' - 1)A'.
Using the angles 6° and 3-3° obtained above, with n = 1-515 and «' =
1-655,
d = (1-515 - 1) 6° - (1-655 - 1) 3-3° = 0-93°.
Direct-vision Spectroscope
The direct-vision spectroscope is a simple instrument used for examin-
ing the different colours in the spectrum obtained from a glowing gas
in a flame or in a discharge tube. It contains several crown and flint
prisms cemented together, and contained in a straight tube having
lenses which constitute an eye-piece. The tube is pointed at the source
of light examined, when various colours are seen on account of the
dispersion produced by the prisms, Fig. 20.7.
In practice, the direct-vision spectroscope contains several crown
and flint glass prisms, but for convenience suppose we consider two
such prisms, as in Fig. 20.7. For "direct vision", the net deviation of the
mean (yellow) ray produced by the prisms must be zero. Thus the
mean deviation caused by the crown glass prism in one direction must
be equal to that caused by the flint glass prism in the opposite direction.
Hence, with the notation already used, we must have
(n- l)A = (n' - \)A'.
Suppose A = 6°, n = 1-515, n' = 1-655. Then A' is given by
(1-515 -1)6° = (1-655 - \)A'
*. 0-515 ,
A ~ 6*55 x6° = 4-7"
The net dispersion of the blue and red rays is given by
(n h -n T ) A -(«; - n' t )A'
= (1-521 - 1-510)6° - (1-665 - 1-645)4-7°
= 0066 - 0094 = - 0-028°.
The minus indicates that the net dispersion is produced in a "blue-
upward" direction, as the dispersion of the flint glass prism is greater
than that of the crown glass prism.
^s^jhrj^!^. M ean ray V.
"r T r \T ^r"
Inciden t
light
>r y Flint
Fig. 20.7. Dispersion; but no deviation of mean ray.
462 ADVANCED LEVEL PHYSICS
SPECTRA
The Importance of the Study of Spectra
The study of the wavelengths of the radiation from a hot body comes
under the general heading of Spectra. The number of spectra of elements
and compounds which have been recorded runs easily into millions, and
it is worth while stating at the outset the main reasons for the interest
in the phenomenon.
It is now considered that an atom consists of a nucleus of positive
electricity surrounded by electrons moving in various orbits, and that
a particular electron in an orbit has a definite amount of energy. In
certain circumstances the electron may jump from this orbit to another,
where it has a smaller amount of energy. When this occurs radiation
is emitted, and the energy in the radiation is equal to the difference in
energy of the atom between its initial and final states. The displacement
of an electron from one orbit to another occurs when a substance is
raised to a high temperature, in which case the atoms present collide
with each other very violently. Light of a definite wavelength will then
be emitted, and will be characteristic of the electron energy changes in
the atom. There is usually more than one wavelength in the light from a
hot body (iron has more than 4,000 different wavelengths in its spectrum),
and each wavelength corresponds to a change in energy between two
orbits. A study of spectra should therefore reveal much important
information concerning the structure and properties of atoms.
Every element has a unique spectrum. Consequently a study of the
spectrum of a substance enables its composition to be readily deter-
mined. Spectroscopy is the name given to the exact analysis of mixtures
or compounds by a study of their spectra, and the science has developed
to such an extent that the presence in a substance of less than a millionth
of a milligram of sodium can be detected.
Types of emission spectra. There are three different types of spectra, which
are easily recognised. They are known as (a) line spectra, (b) band spectra,
(c) continuous spectra.
(a) Line spectra. When the light emitted by the atoms of a glowing substance
(such as vaporised sodium or helium gas) is examined by a prism and spectro-
meter, lines of various wavelengths are obtained. These lines, it should be
noted, are images of the narrow slit of the spectrometer on which the light
is incident. The spectra of hydrogen, Fig. 20.8, and helium are line spectra, and
it is generally true that line spectra are obtained from atoms.
6563 48614340 4102 ^ (in10~ 8 cm)
Fig. 20.8. Visible line spectra of hydrogen.
(6) Band spectra. Band spectra are obtained from molecules, and consist
of a series of bands each sharp at one end but "fading" at the other end, Fig.
DISPERSION. SPECTRA 463
20.9. Hie term "fluting" is often used to describe the way in which the bands
are spaced. Careful examination reveals that the bands are made up of
numerous fine lines very close to each other. Two examples of band spectra
are those usually obtained from nitrogen and oxygen.
Fig. 20.9. Diagrammatic representation of band spectra.
(c) Continuous spectra. The spectrum of the sun is an example of a con-
tinuous spectrum, and, in general, the latter are obtained from solids and
liquids. In these states of matter the atoms and molecules are close together,
and electron orbital changes in a particular atom are influenced by neigh-
bouring atoms to such an extent that radiations of all different wavelengths
are emitted. In a gas the atoms are comparatively far apart, and each atom
is uninfluenced by any other. The gas therefore emits radiations of wave-
lengths which result from orbital changes in the atom due solely to the high
temperature of the gas, and a line spectrum is obtained. When the temperature
of a gas is decreased and pressure applied so that the liquid state is approached,
the line spectrum of the gas is observed to broaden out considerably.
Production of spectra. In order to produce its spectrum the substance under
examination must be heated to a high temperature. There are four main
methods of excitation, as the process is called, and spectra are classified under
the method of their production.
(a) Flame spectra. The temperature of a Bunsen flame is high enough to
vaporise certain solids. Thus if a piece of platinum wire is dipped into a
sodium salt and then placed in the flame, a vivid yellow colour is obtained
which is characteristic of the element sodium. This method of excitation can
only be used for a limited number of metals, the main class being the alkali
and alkaline earth metals such as sodium, potassium, lithium, calcium, and
barium. The line spectra produced in each case consist of lines of different
colours, but some lines have a greater intensity than others. Thus sodium is
characterised by two prominent yellow lines barely distinguishable in a small
spectroscope, and lithium by a prominent green line.
(6) Spark spectra. If metal electrodes are connected to the secondary of
an induction coil and placed a few millimetres apart, a spark can be obtained
which bridges the gap. It was discovered that a much more intense and violent
spark could be obtained by placing a capacitor in parallel with the gap. This
spark is known as a condensed spark. The solid under investigation forms one
of the electrodes, and is vaporised at the high temperature obtained.
(c) Arc spectra. This is the method most used in industry. If two metal rods
connected to a d.c. voltage supply are placed in contact with each other and
then drawn a few millimetres apart, a continuous spark, known as an arc,
is obtained across the gap. The arc is a source of very high temperature, and
therefore vaporises substances very readily. In practice the two rods are placed
in a vertical position, and a small amount of the substance investigated is
placed on the lower rod.
(d) Discharge-tube spectra. If a gas is contained at low pressure inside a
tube having two aluminium electrodes and a high a.c. or d.c. voltage is
applied to the gas, a "discharge" occurs between the electrodes and the gas
464 ADVANCED LEVEL PHYSICS
becomes luminous. This is the most convenient method of examining the
spectra of gases. The luminous neon gas in a discharge tube has a reddish
colour, while mercury vapour is greenish-blue.
Absorption Spectra. Kirchhoff's Law
The spectra just discussed are classified as emission spectra. There is
another class of spectra known as absorption spectra, which we shall
now briefly consider.
If light from a source having a continuous spectrum is examined
after it has passed through a sodium flame, the spectrum is found to be
crossed by a dark line; this dark line is in the position corresponding to
the bright line emission spectrum obtained with the sodium flame alone.
The continuous spectrum with the dark line is naturally characteristic
of the absorbing substance, in this case sodium, and it is known as an
absorption spectrum. An absorption spectrum is obtained when red
glass is placed in front of sunlight, as it allows only a narrow band of
red rays to be transmitted.
Kirchhoff's investigations on absorption spectra in 1855 led him
to formulate a simple law concerning the emission and absorption of
light by a substance. This states : A substance which emits light of a certain
wavelength at a given temperature can also absorb light of the same
wavelength at that temperature. In other words, a good emitter of a certain
wavelength is also a good absorber of that wavelength. From Kirch-
hoff's law it follows that if the radiation from a hot source emitting a
continuous spectrum is passed through a vapour, the absorption
spectrum obtained is deficient in those wavelengths which the vapour
would emit if it were raised to the same high temperature. Thus if a
sodium flame is observed through a spectrometer in a darkened room,
a bright yellow line is seen; if a strong white arc light, richer in yellow
light than the sodium flame, is placed behind the flame, a dark line is
observed in the place of the yellow line. The sodium absorbs yellow light
from the white light, and re-radiates it in all directions. Consequently
there is less yellow light in front of the sodium flame than if it were
removed, and a dark line is thus observed.
Fraunhofer Lines
In 1814 Fraunhofer noticed that the sun's spectrum was crossed
by many hundreds of dark lines. These Fraunhofer lines, as they are
called, were mapped out by him on a chart of wavelengths, and the more
prominent were labelled by the letters of the alphabet. Thus the dark
line in the blue part of the spectrum was known as the Fline, the dark
line in the yellow part as the D line, and the dark line in the red part as
the C line.
The Fraunhofer lines indicate the presence in the sun's atmosphere
of certain elements in a vaporised form. The vapours are cooler than
the central hot portion of the sun, and they absorb their own character-
istic wavelengths from the sun's continuous spectrum. Now every element
DISPERSION. SPECTRA 465
has a characteristic spectrum of wavelengths. Accordingly, it became
possible to identify the elements round the sun from a study of the
wavelengths of the Fraunhofer (dark) lines in the sun's spectrum, and
it was then found that hydrogen and helium were present. This was how
helium was first discovered. The D line is the yellow sodium line.
The incandescent gases round the sun can be seen as flames many
miles high during a total eclipse of the sun, when the central portion of
the sun is cut off from the observer. If the spectrum of the sun is observed
just before an eclipse takes place, a continuous spectrum with Fraun-
hofer lines is obtained, as already stated. At the instant when the eclipse
becomes total, however, bright emission lines are seen in exactly the
same position as those previously occupied by the Fraunhofer lines, and
they correspond to the emission spectra of the vapours alone. This is
an illustration of Kirchhoff 's law, p. 464.
Measurement of Wavelengths by Spectrometer
As we shall discuss later (p. 690) the light waves produced by different
colours are characterised by different wavelengths. Besides measuring
refractive index, the spectrometer can be adapted for measuring un-
known wavelengths, corresponding to the lines in the spectrum of a
glowing gas in a discharge tube, for example.
A prism is first placed on the spectrometer table in the minimum
deviation position for yellow (sodium) light, thus providing a reference
position for the prism in relation to incident light from the collimator.
The source of yellow light is now replaced by a helium discharge tube,
which contains helium at a very low pressure, glowing as a result of the
high voltage placed across the tube. Several bright lines of various
colours can now be observed through the telescope (they are differently
coloured images of the slit), and the deviation, 9, of each of the lines is
obtained by rotating the telescope until the image is on the cross-wires,
a and then noting the corresponding
reading on the circular graduated
scale. Since the wavelengths, A,
of the various lines in the helium
spectrum are known very accur-
ately from tables, a graph can
now be plotted between and A.
s,^^ The helium discharge tube can
^ N *" , » then be replaced by a hydrogen
or mercury discharge tube, and
the deviations due to other lines
=»^X °f known wavelength obtained.
Fig. 20.10. Measurement of In this wa ? a calibration curve for
wavelength by spectrometer. the spectrometer can be obtained,
Fig. 20.10.
The wavelength due to a line Q in the spectrum of an unknown
glowing gas can now be easily derived. With the prism still in the mini-
mum deviation position for yellow light, the deviation, 6, of Q is
466 ADVANCED LEVEL PHYSICS
measured. If this angle corresponds to C in Fig. 20.10, the wavelength
A is OA.
EXAMPLES
1. Show that when a ray of light passes nearly normally through a prism
of small angle a and refractive index n, the deviation 8 is given by 8 = (n — 1) a.
A parallel beam of light falls normally upon the first face of a prism of small
angle. The portion of the beam which is refracted at the second surface is
deviated through an angle of 1° 35', and the portion which is reflected at the
second surface and emerges again at the first surface makes an angle of 8° 9'
with the incident beam. Calculate the angle of the prism and the refractive
index of the glass. (C)
First part. See text.
Second part. Let = angle of prism, n = the refractive index, and RH the
ray incident normally on the face AN, striking the second face at K, Fig. 20. 1 1 .
35' B
Fio. 20.11. Example.
Then the angle of incidence at K = 0, and angle HKN = 20. By drawing the
normal NS at N, which is parallel to HK, it can be seen that angle KNS = 20.
The angle of emergence from the prism = 8° 9' since the incident beam was
normal to AN.
The angle of deviation, 8, of the beam by the prism is given by
8= (« - 1)
:. 1° 35' = ( B - 1) . . . . (i)
„ f .. 4XT sin 8° 9'
For refraction at N, n = —. — r-*
sin 20
Since the angles concerned are small,
8° 9'
" 20
where is in degrees.
(ii)
DISPERSION. SPECTRA
467
From (ii),
8° 9'
2n
; substituting in (i),
1° 35' = («-!)
8° 9'
In
n- 1
21
489
190«
In
489« - 489
/. k = 1-63
8° 29'
=
8° 29'
= 2° 30'
2n 3-26
2. Define dispersive power. The following table gives the refractive indices
of crown and flint glass for three lines of the spectrum.
C
D
F
Crown
Flint
1-514
1-644
1-517
1-650
1-523
1-664
Calculate the refracting angle of a flint glass prism which, when combined
with a crown glass prism of refracting angle 5°, produces a combination that
does not deviate the light corresponding to the D line. What separation of the
rays corresponding to the C and F lines will such a compound prism pro-
duce? (JL.)
For definition, see text.
The D line corresponds to the mean, or yellow, ray, the F and C lines to
the blue and red rays respectively. Let n, n — the refractive indices for crown
and flint glass respectively, A\ A — the corresponding angles of the prisms.
For no deviation («' D - 1) A' - (n D - 1) A = 0,
.'. (1-517 - 1) 5° - (1-650 - 1) A =
. 0-517
0-650
X 5 = 3-99°
The separation of the F and C lines
= (« F —no) A — (/i' F — n'c) A'
= (1-664 — 1-644) 3-99° - (1-523 - 1-514) 5°
= 0-0798° - 0-045° = 00348°
3. Prove that for a prism of small angle A the deviation of a ray of light is
(n — 1) A, provided that the angle of incidence also is small. A crown glass
prism of refracting angle 6° is to be achromatised for red and blue light with a
flint glass prism. Using the data below and the formula above find (a) the
angle of the flint glass prism, (b) the mean deviation.
Crown glass Flint glass
nred 1-513 1-645
n blue 1-523 1-665
468 ADVANCED LEVEL PHYSICS
First part. See text.
Second part. Let A = the angle of the flint prism, n, n = the refractive
indices of the crown and flint glass respectively. For achromatism,
(«'b - «'r) 6° = (n b - «r) A
.'. (1-523 - 1-513) 6° = 1-665 - 1-645) A
. A _ 0010 V *° V
The mean refractive index, n, for crown glass = i = 1-518
and mean refractive mdex n, for flint glass = ^ = 1-655
.*. deviation of mean ray = (jn — 1) 6° — (n — 1) 3°
= (1-518 - 1) 6° - (1-655 - 1) 3° = 1043°
EXERCISES 20
1. Write down the formula for the deviation of a ray of light through a
prism of small angle A which has a refractive index n for the colour con-
cerned. Using the following table, calculate the deviation of (i) red light,
(ii) blue light, (iii) yellow light through a flint glass prism of refracting angle
4°, and through a crown glass prism of refracting angle 6°.
Crown gloss
Flint glass
wred
1-512
1-646
n blue
1-524
1-666
2. Using the above data, calculate the dispersive powers of crown glass and
flint glass.
3. Explain how it is possible with two prisms to produce dispersion with-
out mean deviation. A prism of crown glass with refracting angle of 5° and
mean refractive index 1-51 is combined with one flint glass of refractive
index 1-65 to produce no mean deviation. Find the angle of the flint glass
prism. The difference in the refractive indices of the red and blue rays in
crown glass is 00085 and in flint glass 001 62. Find the inclination between
the red and blue rays which emerge from the composite prism. (L.)
4. Draw a ray diagram showing the passage of light of two different
wavelengths through a prism spectrometer. Why is it that such a spectro-
meter is almost invariably used with (a) a very narrow entrance slit, (b)
parallel light passing through the prism, (c) the prism set at, or near, minimum
deviation?
A spectrometer is used with a small angle prism made from glass which
has a refractive index of 1-649 for the blue mercury line and 1-631 for the
green mercury line. The collimator lens and the objective of the spectrometer
both have a focal length of 30 cm. If the angle of the prism is 01 radian what
is the spacing of the centres of the blue and green mercury lines in the focal
plane of the objective, and what maximum slit width may be used without
the lines overlapping? The effect of diffraction need not be considered.
(O. & C.)
DISPERSION. SPECTRA 469
5. A glass prism of refracting angle 60° and of material of refractive
index 1-50 is held with its refracting angle downwards alongside another
prism of angle 40° which has its refracting angle pointing upwards. A narrow
parallel beam of yellow light is incident nearly normally on the first prism,
passes through both prisms, and is observed to emerge parallel to its original
direction. Calculate the refractive index of the material of the second prism.
If white light were used and the glasses of the two prisms were very different
in their power to disperse light, describe very briefly what would be seen on a
white screen placed at right angles to the emergent light. (C.)
6. A ray of monochromatic light is incident at an angle i on one face of a
prism of refracting angle A of glass of refractive index n and is transmitted.
The deviation of the ray is D.
Considering only rays incident on the side of the normal away from the
refracting angle, sketch graphs on the same set of axes showing how D varies
with / when (a) A is about 60°, (b) A is very small.
From first principles derive an expression for D when i and A are both very
small angles. (N.)
7. Distinguish between emission spectra and absorption spectra. Describe
the spectrum of the light emitted by (i) the sun, (ii) a car headlamp fitted
with yellow glass, (iii) a sodium vapour street lamp.
What are the approximate wavelength limits of the visible spectrum? How
would you demonstrate the existence of radiations whose wavelengths lie
just outside these limits? (0. & C.)
8. State what is meant by dispersion and describe, with diagrams, the
principle of (i) an achromatic and (ii) a direct-vision prism.
Derive an expression for the refractive index of the glass of a narrow angle
prism in terms of the angle of minimum deviation and the angle of the prism.
If the refractive index of the glass of refracting angle 8° is 1-532 and 1-514
for blue and red light respectively, determine the angular dispersion produced
by the prism. (L.)
9. Describe the processes which lead to the formation of numerous dark
lines (Fraunhofer lines) in the solar spectrum. Explain why the positions of
these lines in the spectrum differ very slightly when the light is received from
opposite ends of an equatorial diameter of the sun. (N.)
10. Describe with the aid of diagrams what is meant by dispersion and
deviation by a glass prism. Derive a formula for the deviation D produced by
a glass prism of small refracting angle A for small angles of incidence. Sketch
the graph showing how the deviation varies with angle of incidence for a
beam of light striking such a prism, and on the same axes indicate what
would happen with a prism of much larger refracting angle but of material
of the same index of refraction. (C.)
11. Describe the optical system of a simple prism spectrometer. Illustrate
your answer with a diagram showing the paths through the spectrometer of
the pencils of rays which form the red and blue ends of the spectrum of a
source of white light. (Assume in your diagram that the lenses are achro-
matic.)
The prism of a spectrometer has a refracting angle of 60° and is made of
glass whose refractive indices for red and violet are respectively 1-514 and
1-530. A white source is used and the instrument is set to give minimum
deviation for red light. Determine (a) the angle of incidence of the light on
the prism, (b) the angle of emergence of the violet light, (c) the angular
width of the spectrum. (N.)
470 ADVANCED LEVEL PHYSICS
12. Calculate the angle of a crown glass prism which makes an achro-
matic combination for red and blue light with a flint glass prism of refracting
angle 4°. What is the mean deviation of the light by this combination? Use
the data given in question 1.
13. Describe and give a diagram of the optical system of a spectrometer.
What procedure would you adopt when using the instrument to measure
the refractive index of the glass of a prism for sodium light? What additional
observations would be necessary in order to determine the dispersive power
of the glass?
The refractive index of the glass of a prism for red light is 1-514 and for
blue light 1-523. Calculate the difference in the velocities of the red and blue
light in the prism if the velocity of light in vacuo is 3 x 10 5 kilometres per
second. (JV.)
14. Explain, with diagrams, how a 'pure' spectrum is produced by means
of a spectrometer. What source of light may be used and what readings must
be taken in order to find the dispersive power of the material of which the
prism is made? (£,.)
15. (a) Explain, giving a carefully drawn, labelled diagram, the function
of the various parts of a spectrometer. How is it adjusted for normal laboratory
use? (b) Distinguish between a continuous spectrum, an absorption spectrum,
a band spectrum, and a line spectrum. State briefly how you would obtain
each type with a spectrometer. (W.)
16. Describe a prism spectrometer and the adjustment of it necessary for
the precise observation of the spectrum of light by a gaseous source.
Compare and contrast briefly the spectrum of sunlight and of light emitted
by hydrogen at low pressure contained in a tube through which an electric
discharge is passing. (L.)
chapter twenty-one
Refraction through lenses
A lens is a piece of glass bounded by one or two spherical surfaces.
When a lens is thicker in the middle than at the edges it is called a convex
or converging lens, Fig. 21.1 (i); when it is thinner in the middle than at
the edges it is known as a concave or diverging lens, Fig. 21.1 (ii). Fig. 21 .9,
on p. 478, illustrates other types of converging and diverging lenses.
Lenses were no doubt made soon after
the art of glass-making was discovered;
and as the sun's rays could be con-
centrated by these curved pieces of glass
they were called "burning glasses".
Aristophanes, in 424 B.C., mentions a
burning glass. To-day, lenses are used
in spectacles, cameras, microscopes, and
Bi- convex Bi- concave telescopes, as well as in many other
(i) (ii) optical instruments, and they afford yet
another example of the many ways in
Converging F anddidrging lenses. w Wch Science is used to benefit our
everyday lives.
Since a lens has a curved spherical surface, a thorough study of a
lens should be preceded by a discussion of the refraction of light through
a curved surface. We shall therefore proceed to consider what happens
in this case, and defer a discussion of lenses until later, p. 478.
REFRACTION AT CURVED SPHERICAL SURFACE
Relation Between Object and Image Distances
Consider a curved spherical surface NP, bounding media of refractive
indices n lt n 2 respectively, Fig. 21.2. The medium of refractive index n x
might be air, for example, and the other of refractive index n 2 might be
glass. The centre, C, of the sphere of which NP is part is the centre of
curvature of the surface, and hence CP is the radius of curvature, r.
The line joining C to the mid-point P of the surface is known as its
principal axis. P is known as the pole.
Suppose a point object O is situated on the axis PC in the medium of
refractive index n x . The image of O by refraction at the curved surface
can be obtained by taking two rays from O. A ray OP passes straight
through along PC into the medium of refractive index n 2 , since OP is
normal to the surface, while a ray ON very close to the axis is refracted
471
472 ADVANCED LEVEL PHYSICS
at N along NI towards the normal CN, if we assume n 2 is greater than
n x . Thus at the point of intersection, I, of OP and NI is the image O,
and we have here the ease of a real image.
Fig. 21.2. Refraction at curved surface.
Suppose 1*1, i 2 are the angles made by ON, NI respectively with the nor-
mal, CN, at N, Fig. 21.2. Then, applying "n sin *" is a constant (p. 423),
n x sin i x = n 2 sin i 2 (i)
But if we deal with rays from O very close to the axis OP, i x is small;
and hence sin i x — i x in radians. Similarly, sin i 2 = i 2 ui radians. From
(i), it follows that
n x i x = n 2 i 2 (ii)
If a, ft y are the angles with the axis made by ON, CN, IN respec-
tively, we have
/j = a + j8, from the geometry of triangle ONC,
and i 2 = p — y, from the geometry of triangle CNI.
Substituting for i x , i 2 in (ii), we have
"i(a+0) = « 2 O3-y)
.*. n x a + n 2 y = (n 2 - n^ /3 . . . (in)
If h is the height of N above the axis, and N is so close to P that NP
is perpendicular to OP,
a ~op ,y ~pr ^~pc*
From (iii), using our sign convention on p. 407, we have
(n 2 - «j)
Kh^ + hSi) = *
PC *
since O is a real object and I is a real image.
"i , ?H _ n % — n x
" OP^PI PC *
If the object distance, OP, from P = w, the image distance, IP, from
P = v, and PC = r, then
Ul + U*. = n * ~ ** (i)
REFRACTION THROUGH LENSES 473
Sign Convention for Radius of Curvature
Equation (1) is the general relation between the object and image
distances, m, v, from the middle or pole of the refracting surface, its
radius of curvature r, and the refractive indices of the media, n 2 , n v
The quantity (» 2 — n^/r is known as the power of the surface. If a ray is
made to converge by a surface, as in Fig. 21 . 1 , the power will be assumed
positive in sign; if a ray is made to diverge by a surface, the power will
be assumed negative. Since refractive index is a ratio of velocities
(p. 421), n x and n 2 have no sign. (n 2 — «i) on the right side of equation
(1) will be taken always as a positive quantity, and thus denotes the
smaller refractive index subtracted from the greater refractive index.
The sign convention for the radius of curvature, r, of a spherical surface
is now as follows: if the surface is convex to the less dense medium, its
radius is positive; if it is concave to the less dense medium, its radius is
negative. We have thus to view the surface from a point in the less
dense medium. In Fig. 21.3 (i), the surface A is convex to the less dense
Air KalasSyl Air
(i) (ii)
Fig. 21.3. Sign convention for radius of curvature.
medium air, and hence its radius is positive. The surface C is concave to
the less dense medium air, and its radius is thus negative, Fig. 21.3 (ii).
The radii of the surfaces B and D are both positive. ^
Special Cases
The general formula — H — - = — can easily be remembered
u v r,
on account of its symmetry. The object distance u corresponds to the
refractive index n x of the medium in which the object is situated; while
the image distance v corresponds to the medium of refractive index « 2 in
which the image is situated.
Suppose an object O in air is x cm from a curved spherical surface,
and the image I is real and in glass of refractive index n, at a distance
of v cm from the surface, Fig. 21.4 (i). Then u — + x, v = -f y, n t = 1,
w 2 = n. If the surface is convex to the less dense medium, as shown in
Fig. 21.4 (i), the radius of curvature, a cm, is given by r = + a.
474
ADVANCED LEVEL PHYSICS
Substituting in
«1 «2 _ «2 ~**1
u • v r
i + ^
x y
n - 1
Fig. 21.4 (ii) illustrates the case of an object O in glass of refractive
index n, the surface being concave to the less dense surface air. The
radius, b cm, is then given by r = - b. If the image I is virtual,
its distance v = — m. If / is the distance of O, then u = + /.
Substituting in
u v r
U '
/ ' -m
n-l
-b
If a surface is plane, its radius of curvature, r, is infinitely large.
fl„ r+J ft.
Hence is zero, whatever different values rix and n 2 may have.
Deviation of Light by Sphere
Suppose a ray AO in air is incident on a sphere of glass or a drop of
water, Fig. 21.5 (i). The light is refracted at O, then reflected inside B,
('')
Fig. 21.5. Deviation of light by sphere.
REFRACTION THROUGH LENSES
475
and finally emerges into the air along CD. If/, r are the angles of incidence
and refraction at O, the deviation of the light at O and C is (/ - r) each
time ; it is (1 80° - 2r) at B. The total deviation, 8, in a clockwise direction
is thus given by
8 = 2 (i - r) + 180° - 2r = 180° + 2z - 4r . . (i)
It can be seen that the deviation at each reflection inside the sphere
is (180° - 2r) and that the deviation at each refraction is (/ - r). Thus
if a ray undergoes two reflections inside the sphere, and two refractions,
as shown in Fig. 21.5 (ii), the total deviation in a clockwise direction =
2 (/ - r ) + 2 (180° - 2r) = 360° +2/ - 6r. After m internal reflections,
the total deviation = 2 (i - r) + m (180° - 2r).
The Rainbow
The explanation of the colours of the rainbow was first given by
Newton about 1667. He had already shown that sunlight consisted of a
mixture of colours ranging from red to violet, and that glass could
disperse or separate the colours (p. 454). In the same way, he argued,
water droplets in the air dispersed the various colours in different
directions, so that the colours of the spectrum were seen.
The curved appearance of the rainbow was first correctly explained
about 1611. It was attributed to refraction of light at a water drop,
followed by reflection inside the drop, the ray finally emerging into
the air as shown in Fig. 21.6. The primary bow is the rainbow usually
seen, and is obtained by two refractions and one reflection at the drops,
as in Fig. 21.6. Sometimes a secondary bow is seen higher in the sky,
and it is formed by rays undergoing two refractions and two reflections
at the drop, as in Fig. 21.6.
Sun's
rays
Droplets
Secondary
bow
Primary
bow
Fig. 21.6. The Rainbow.
The total deviation 8 of the light when one reflection occurs in the drop,
Fig. 21.6, is given by d = 180° + Ii — 4r, as proved before. Now the light
476 ADVANCED LEVEL PHYSICS
emerging from the drop will be intense at those angles of incidence corres-
ponding to the minimum deviation position, since a considerable number
of rays have about the same deviation at the minimum value and thus emerge
almost parallel. Now for a minimum value, -p = 0.
di
j
Differentiating the expression for 8, we have 2 — 4 -j. = 0.
di
• ^=1
' * di ~~ 2
But sin i = n sin r,
where n is the refractive index of water.
dr n
cos i = n cos r-r. = — cos r
di 2
4 cos 2 1 = n 2 cos 2 r = n 2 — n 2 sin 2 r= n 2 — sin 2 1
= 7* 2 - (1 - cos 2 i)
3 cos 2 i = n 2 — 1
/« 2 - 1
cos/= / — ^— .... (ii)
The refractive index of water for red light is 1*331. Substituting this value
in (ii) i can be found, and thus r is obtained. The deviation 8 can then be cal-
culated, and the acute angle between the incident and emergent red rays,
which is the supplement of 8, is about 42-1°. By substituting the refractive
index of water for violet light in (ii), the acute angle between the incident and
emergent violet rays is found to be about 40-2°. Thus if a shower of drops is
illuminated by the sun's rays, an observer standing with his back to the sun
sees a brilliant red light at an angle of 42-1° with the line joining the sun to
him, and a brilliant violet light at an angle of 40-2° with mis line, Fig. 21.6.
Since the phenomenon is the same in all planes passing through the line, the
brightly coloured drops form an arc of a circle whose centre is on the line.
The secondary bow is formed by two internal reflections in the water drops,
as illustrat ed in Fig. 21 .5 (ii) and Fig. 21.6. The minimum deviation occurs when
cos i = V(/i 2 — l)/8 in this case. The acute angle between the incident and
emergent red rays is then found to be about 51-8°, and that for the violet rays
is found to be about 54-5°. Thus the secondary bow has red on the inside and
violet on the outside, whereas the primary bow colours are the reverse, Fig.
21.6.
EXAMPLES
1. Obtain a formula connecting the distances of object and image from a
spherical refracting surface. A small piece of paper is stuck on a glass sphere
of 5 cm radius and viewed through the glass from a position directly opposite.
Find the position of the image. Find also the position of the image formed,
by the sphere, of an object at infinity. (O. & C.)
First part. See text.
Second part. Suppose O is the piece of paper, Fig. 21.7 (i). The refracting
surface of the glass is at P, and u = + 10. Now
REFRACTION THROUGH LENSES
477
(0 (ii)
Fig. 21.7. Example
H2 , Bl flt^"
v u r
«i
where « x = 1*5, n % =■ 1, r = + 5 p. 472 and v is the image distance from P.
Substituting,
1 1-5 = 15-
v ~ l ~ 10 ~ 5
1 = 01
V
015 = - 005
v = — 20 cm.
Thus the image is virtual, i.e., it is 20 cm from P on the same side as O.
Third part. Suppose I is the position of the image by refraction at the first
surface, A, Fig. 21 .7 (ii). Now — + — = ^-^
v u r
«1
, where u = oo, «i = 1
« 2
1-5, r = + 5.
1-5 1-5 - 1
v 5
v = 15 cm = AI, or BI = 5 cm.
I is a virtual object for refraction at the curved surface B. Since u —
= — 5 cm, /ii = 1-5, m = 1, r — + 5, it follows from
BI
n% , «i «2 ^ »i
v u r
that
1 1-5 1-5 - 1
v ' (- 5 5
from which
v = 2-5cm = BI'.
2. An object is placed in front of a spherical refracting surface. Derive an
expression connecting the distances from the refracting surface of the object
and the image produced. The apparent thickness of a thick plano-convex lens
is measured with (a) the plane face uppermost (b) the convex face uppermost,
the values being 2 cm and 2f cm respectively. If its real thickness is 3 cm,
calculate the refractive index of the glass and the radius of curvature of the
convex face. (L.)
First part. See text.
Second part. With the plane face uppermost, the image I of the lowest
point O is obtained by considering refraction at the plane surface D, Fig.
21.8 (i). Now
478
ADVANCED LEVEL PHYSICS
D
M
Vk <^2^i/^^ <fs77.
7/X&77>^
o.
(i)
(>D
Fig. 21.8. Example
real depth
~ apparent depth
/. "==1=15
With the curved surface uppermost, the image Ii of the lowest point Oi is
obtained by considering refraction at the curved surface M, Fig. 21.8 (ii). In
this case Mix = v = apparent thickness = — 2§ cm, the image Ii being
virtual. Now u = MOi = 3 cm, n 2 = 1, n x = n = 1-5. Substituting in
»2 X «1 «2 ^ »1
v u r
we have
1 1-5 1-5-
- 2f 3 r
Simplifying,
r = 10 cm.
REFRACTION THROUGH THIN LENSES
Converging and Diverging Lenses
At the beginning of the chapter we defined a lens as an object, usually
of glass, bounded by one or two spherical surfaces. Besides the converg-
ing (convex) lens shown in Fig. 21.1 (i) on p. 471, Fig. 21.9 (i) illustrates
two other types of converging lenses, which are thicker in the middle
than at the edges. Fig. 84 (ii) illustrates two types of diverging (concave)
lenses, a diverging lens being also shown in Fig. 21.1 (ii) on p. 471.
Piano- Converging Piano- Diverging
convex meniscus concave meniscus
(') (ii)
Fig. 21.9. (i). Convex (converging) lenses, (ii). Concave (diverging) lenses.
The principal axis of a lens is the line joining the centres of curvature
of the two surfaces, and passes through the middle of the lens. Experi-
REFRACTION THROUGH LENSES
479
ments with a ray-box show that a thin convex lens brings an incident
parallel beam of rays to a. principal focus, F, on the other side of the lens
when the beam is narrow and incident close to the principal axis, Fig.
21.10 (i). On account of the convergent beam contained with it, the con-
vex lens is better described as a "converging" lens. If a similar parallel
beam is incident on the other (right) side of the lens, it converges to a
focus F', which is at the same distance from the lens as F when the lens
is thin. To distinguish F from F' the latter is called the "first principal
focus" ; F is known as the "second principal focus".
Principal
(ii)
Fig. 21.10. Focus of converging (convex) and diverging (concave) lenses.
When a narrow parallel beam, close to the principal axis, is incident
on a thin concave lens, experiment shows that a beam is obtained which
appears to diverge from a point F on the same side as the incident beam,
Fig. 21.10 (ii). F is known as the principal "focus" of the concave lens.
Since a divergent beam is obtained, the concave lens is better described
as a "diverging" lens.
Explanation of Effects of Lenses
A thin lens may be regarded as made up of a very large number of
small-angle prisms placed together, as shown in the exaggerated sketches
of Fig. 2 1 . 1 1 . If the spherical surfaces of the various truncated prisms are
Fig. 21.11. Action of converging (convex) and diverging (concave) lenses.
480 ADVANCED LEVEL PHYSICS
imagined to be produced, the angles of the prisms can be seen to increase
from zero at the middle to a small value at the edge of the lens. Now
the deviation, d, of a ray of light by a small-angle prism is given by d =
(n — 1) A, where A is the angle of the prism, see p. 457. Consequently the
truncated prism corresponding to a position farther away from the
middle of the lens deviates an incident ray more than those prisms nearer
the middle. Thus, for the case of the converging lens, the refracted rays
converge to the same point or focus F, Fig. 21.11 (i). It will be noted that
a ray AC incident on the middle, C, of the lens emerges parallel to AC,
since the middle acts like a rectangular piece of glass (p. 422). This fact is
utilised in the drawing of images in lenses (p. 485).
Since the diverging lens is made up of truncated prisms pointing the
opposite way to the converging lens, the deviation of the light is in the
opposite direction, Fig. 21.11 (ii). A divergent beam is hence obtained
when parallel rays are refracted by the lens.
The Signs of Focal Length,/
From Fig. 21.1 1 (i), it can be seen that a convex lens has a real focus;
the focal length,/, of a converging lens is thus positive in sign. Since the
focus of a diverging lens is virtual, the focal length of such a lens is
negative in sign, Fig. 21.11 (ii). The reader must memorise the sign of/
for a converging and diverging lens respectively, as this is always re-
quired in connection with lens formulae.
Relations Between Image and Object Distances for Thin Lens
We can now derive a relation between the object and image distances
when a lens is used. We shall limit ourselves to the case of a thin lens,
i.e., one whose thickness is small compared with its other dimensions,
and consider narrow beams of light incident on its central portion.
Suppose a lens of refractive index n 2 is placed in a medium of refractive
index n u and a point object O is situated on the principal axis, Fig. 21.12.
I 7 "
(Virtual object)
Fig. 21.12. Lens proof (exaggerated for clarity).
A ray from O through the middle of the lens passes straight through as
it is normal to both lens surfaces. A ray OM from O, making a small
angle with the principal axis, is refracted at the first surface in the
direction MNT, and then refracted again at N at the second surface
so that it emerges along NI.
Refraction at first surface, MP X . Suppose u is the distance of the
REFRACTION THROUGH LENSES 481
object from the lens, i.e., u = OP lt and v' is the distance of the image
I' by refraction at the first surface, MP 1? of the lens, i.e., v' = rP x . Then,
since I' is situated in the medium of refractive index n 2 (T is on the ray
MN produced), we have, if « 2 > n lt
v' ^ u r x ' ' ' ' ' K)
where r x is the radius of the spherical surface MP^ see p. 472.
Refraction at second surface, NP 2 . Since MN and PiP 2 are the incident
rays on the second surface NP 2 . it follows that I' is a virtual object for
refraction at this surface (see p. 412). Hence the object distance I'P 2 is
negative; and as we are dealing with a thin lens, I'P 2 = — v\ The
corresponding image distance, IP 2 or v, is positive since I is a real image.
Substituting in the formula for refraction at a single spherical surface,
»* 7 + >h = «*-"i . ... 00
— v v r 2
where r 2 is the radius of curvature of the surface NP 2 of the lens.
Lens equation. Adding (i) and (ii) to eliminate v', we have
V U V A Vl r 2/
and dividing throughout by n u
i+I^Afi+IY . . (iii)
Now parallel rays incident on the lens are brought to a focus. In this
case, u = oo and v =/. From (iii),
f oo \n x J \/i rj
■ !-e-oe + 3 • ■ »
Substituting —~ for the right-hand side of (iii), we obtain the important
1,1 1
equation 7 + i = ? < 3 >
This is the "lens equation", and it applies equally to converging and
diverging lenses if the sign convention is used (see also p. 407).
Focal Length of Lens. Small-angle Prism Method
The focal length /of a lens can also be found by using the deviation
formula due to a small-angle prism. Consider a ray PQ parallel to the
principal axis at a height h above it. Fig. 21.13 (i). This ray is refracted to
the principal focus, and thus undergoes a small deviation through an
angle d given by
J h
d = J ®
482
ADVANCED LEVEL PHYSICS
d C
<■"> (ii)
Fig. 21.13. Focal length by small angle prism.
This is the deviation through a prism of small angle A formed by the
tangents at Q, R to the lens surfaces, as shown. Now for a small angle of
incidence, which is the case for a thin lens and a ray close to the principal
axis, d = (n — 1) A. See p. 457.
h
From(/), -j=(n-l)A
1 t n A
(ii)
The normals at Q, R pass respectively through the centres of curvatures
Ci, C 2 of the lens surfaces. From the geometry, angle ROQ = A =
a + £, where a, 0, are the angles with the principal axis at Q,
C 2 respectively, as shown. But a = hjr ly /? = hjr 2 .
A=a+f} = *+!L
r*
(iii)
h r x r 2
Substituting in (ii),
j-o.
i)
Focal Length Values
Since v=( — -ljf- + — J,it follows that the focal length of a
lens depends on the refractive index, n 2 , of its material, the refractive
index, n v of the medium in which it is placed, and the radii of curvature,
/*!, r 2 , of the lens surfaces. The quantity — may be termed the "relative
refractive index" of the lens material; if the lens is made of glass of n 2 =
1-5, and it is placed in water of n x = 1-33, then the relative refractive
index = ^=1.13.
In practice, however, lenses are usually situated in air; in which case
n x = 1. If the glass has a refractive index, « 2 , equal to n, the relative
REFRACTION THROUGH LEVELS
483
refractive index, — = j = n. Substituting in (22), then
i-.-»G*9
(4)
Fig. 21.14 illustrates four different types of glass lenses in air, whose
refractive indices, n, are each 1-5. Fig. 21.14 (i) is a biconvex lens, whose
radii of curvature, r lf r 2 , are each 10 cm. Since a spherical surface convex
to a less dense medium has a positive sign (see p. 473), r x = + 10 and
r 2 = + 10. Substituting in (4).
/. j= (1-5 - 1) [^fJo) + ( + 10)
/. /= + 10cm.
)-
05 X *>
01
(«©)
(ii) (iii)
Fig. 21.14. Signs of radius of lens surface.
Fig. 21.14 (ii) is a biconcave lens in air. Since its surfaces are both con-
cave to the less dense medium, r t = — 10 and r 2 = — 10, assuming the
radii are both 10 cm. Substituting in (24),
••• 7 -< 1,5 - ') (irmj+r^) =°- 5 x -is - - 01
/. /= - 10cm.
In the case of a plano-convex lens, suppose the radius is 8 cm. Then
r 1 = + 8andr 2 = oo,Fig.21.14(iii).Hencei = (l-5-l)f— t- + -^
f \( + 8) co/
= 0-5 x g=^.Thus/= + 16cm.
In Fig. 21.14 (iv), suppose the radii r u r 2 are numerically 16 cm, 12 cm
respectively. Then r x = — 16, but r 2 = + 12. Hence
7= (1 - 5 - 1) (r^) + (+W) = ' 5 (-^ + ^) =+ ^
Thus/= + 96 cm, conm-ming that the lens is a converging one.
Some Applications of the Lens Equation
The following examples should assist the reader in understanding
111
how to apply correctly the lens equation — I — = -? :
1. An object is placed 12 cm from a converging lens of focal length 18 cm.
Find the position of the image.
484 ADVANCED LEVEL PHYSICS
Since the lens is converging, /= + 18 cm. The object is real, and therefore
= + 12 cm. Substituting in — +-=->,
v u f
:. 1+ >
v ' (+12) ( + 18)
• I = J__1 = _1.
" v 18 12 36
/. v = - 36
Since v is negative in sign the image is virtual, and it is 36 cm from the lens.
See Fig. 21.17 (ii).
2. A beam of light, converging to a point 10 cm behind a converging lens,
is incident on the lens. Find the position of the point image if the lens has
a focal length of 40 cm.
If the incident beam converges to the point O, then O is a virtual object.
Fig. 21.15. See p. 412. Thus u = — 10 cm. Also,/= + 40 cm since the lens is
converging. Substituting in — | — = - ,
V U J
v + (-10) ( + 40)
" v ~ 40 + 10 ~ 40
40
Since v is positive in sign the image is real, and it is 8 cm from the lens. The
image is I in Fig. 21.15.
3. An object is placed 6 cm in front of a diverging lens of focal length 12
cm. Find the image position.
I ^>0
Fig. 21.15. Virtual object.
Since the lens is concave, / = — 12 cm. The object is real, and hence u
+ 6cm. Substituting in — \--=-
v u f
v ' ( + 6) (-12)
REFRACTION THROUGH LENSES 485
I=_ J__ I ==-2
•'* v~ 12 6 12
= _12 = _ 4
Since v is negative in sign the image is virtual, and it is 4 cm from the lens.
See Fig. 21.18 (i).
4. A converging beam of light is incident on a diverging lens of focal length
15 cm. If the beam converges to a point 3 cm behind the lens, find the position
of the point image.
Fig. 21.16. Virtual object.
If the beam converges to the point O, then O is a virtual object, as in example
3, Fig. 21.15. Thus u = — 3 cm. Since the lens is diverging,/^ - 15 cm. Sub-
...1,1 1
stitutmg m - -|- - = - ,
1.1 1
•' v ' (-3) (-15)
*• v 15 + 3 15
15 41
•• v = T = 3|
Since v is positive in sign the point image, I, is real, and it is 3£ cm from the
lens, Fig. 21.16.
Images in Lenses
Converging lens, (i) When an object is a very long way from this
lens, i.e., at infinity, the rays arriving at the lens from the object are
parallel. Thus the image is formed at the focus of the lens, and is real
and inverted.
(ii) Suppose an object OP is placed at O perpendicular to the principal
axis of a thin converging lens, so that it is farther from the lens than its
principal focus, Fig. 21.17 (i). A ray PC incident on the middle, C, of the
lens is very slightly displaced by its refraction through the lens, as the op-
posite surfaces near C are parallel (see Fig. 21.11, which is an exaggerated
sketch of the passage of the ray). We therefore consider that PC passes
straight through the lens, and this is true for any ray incident on the
middle of a thin lens.
A ray PL parallel to the principal axis is refracted so that it passes
through the focus F. Thus the image, Q, of the top point P of the object
486
ADVANCED LEVEL PHYSICS
n Real
u inverted
image
CO
Fig. 21.17. Images in converging lenses.
is formed below the principal axis, and hence the whole image IQ is
real and inverted. In making accurate drawings the lens should be
represented by a straight line, as illustrated in Fig. 21.17, as we are only
concerned with thin lenses and a narrow beam incident close to the
principal axis.
(iii) The image formed by a converging lens is always real and in-
verted until the object is placed nearer the lens than its focal length, Fig.
2 1 . 1 7 (ii). In this case the rays from the top point P diverge after refraction
through the lens, and hence the image Q is virtual. The whole image,
IQ, is erect (the same way up as the object) and magnified, besides being
virtual, and hence the converging lens can be used as a simple "magnify-
ing glass" (see p. 527).
Diverging lens. In the case of a converging lens, the image is some-
times real and sometimes virtual. In a diverging lens, the image is
Virtual
erect
image^
0)
(ii)
Fig. 21.18. Images in diverging lenses.
always virtual; in addition, the image is always erect and diminished.
Fig. 21.18 (i), (ii) illustrate the formation of two images. A ray PL appears
to diverge from the focus F after refraction through the lens, a ray PC
passes straight through the middle of the lens and emerges along CN,
and hence the emergent beam from P appears to diverge from Q on the
same side of the lens as the object. The image IQ is thus virtual.
The rays entering the eye from a point on an object viewed through
a lens can easily be traced. Suppose L is a converging lens, and IQ is the
REFRACTIQN THROUGH LENSES W
image of the object OP, drawn as already explained, Fig. 21.19. If the eye
E observes the top point P of the object through the lens, the cone of
rays entering E are those bounded by the image Q of P and the pupil of
the eye. If these rays are produced back to meet the lens L, and the
Fig. 21.19. Rays entering the eyes.
points of incidence are joined to P, the rays entering E are shown shaded
in the beam. The method can be applied to trace the beam of light
entering the eye from any other point on the object; the important thing
to remember is to work back from the eye.
Another proof of - + - = -^ . We have already shown how the lens equation
v u f
1 -j — = - can be derived by considering refraction in turn at the two curved
v u f
surfaces (p. 480). A proof of the equation can also be obtained from Fig. 21.17
or Fig. 21.18, but it is not as rigid a proof as that already given on page 481.
In Fig. 21.18, triangles CQI, CPO are similar. Hence IQ/PO = CI/CO.
Since triangles FQI, FLC are similar, IQ/CL = FI/FC. Now CL = PO. Thus
the left sides of the two ratios are equal.
CI = FI
* * CO FC
But CI = - v; CO = + "J FI = FC - IC = - /- ( - v) = v - /; and
FC = - /.
. ~" v _ v ~ f
"T~H~ -f
vf= uv — uf
.'. uf-\-vf=uv
Dividing throughout by «v/and simplifying each term,
v^u f
The same result can be derived by considering similar triangles in Fig. 21.17,
a useful exercise for the student.
Lateral Magnification
The lateral or transverse or linear magnification, m, produced by a
lens is defined by
m =
height of image
height of object
(5)
488
ADVANCED LEVEL PHYSICS
IQ..
Thus m = ^ in Fig. 21.17 or Fig. 21.18. Since triangles QIC, POC are
similar in either of the diagrams,
IQ _ CI _ v
OP
CO
v
m = —
u
where v, u are the respective image and object distances from the lens.
(6)
Equation (6) provides a simple formula for the magnitude of the
magnification; there is no need to consider the signs of v and u in this
case.
Other formula for magnification. Since- + - = -, we have, by multi-
plying throughout by v,
V
•• m== r
1 .... ( 7 )
Thus if a real image is formed 25 cm from a converging lens of focal
+ 25
length 10 cm, the magnification, m, =
+ 10
1 = 1-5.
By multiplying both sides of the lens equation by u, we have
m f
Object at Distance 2/ from Converging Lens
When an object is placed at a distance of 2/ from a convex lens,
drawing shows that the real image obtained is the same size as the image
Fig. 21.20. Object and image of same size.
REFRACTION THROUGH LENSES 489
and is also formed at a distance 2/ from the lens, Fig. 21.20. This result
1,1 1.
can be accurately checked by using the lens equation— +— — y
Substituting u = + 2/, and noting that the focal length,/, of a conver-
ging lens is positive, we have
v 2/ /
1 Jl _i = J_
v / If If
v = 2f== image distance.
•o • v 2/ ,
.*. lateral magnification, m = — = Yf = *»
showing that the image is the same size as the object.
Least Possible Distance Between Object and Real Image with
Converging Lens
It is not always possible to obtain a real image on a screen, although
the object and the screen may both be at a greater distance from a con-
verging lens than its focal length. The theory below shows that the
distance between an object and a screen must be equal to, or greater
than, four times the focal length if a real image is required.
Theory. Suppose I is the real image of a point object O in a converging lens.
If the image distance = x, and the distance OI = d, the object distance =
d-x)-
H
Fig. 21.21. Minimum distance between object and image.
(d — x), Fig. 21 .21 . Thus v = + x, and u = + id — x). Substituting in the lens
equation - + -=-„, in which /is positive, we have
v u f
1
X
+rb-
1
"/
d
1
x(d- x)
/
x 2 -
-dx + df=
=
(0
490 ADVANCED LEVEL PHYSICS
For a real image, the roots of this quadratic equation for x must be real
roots. Applying to (i) the condition b* - 4 ac> for the general quadratic
ax 2 -f- bx -f c = 0, then
d*-4df>0
.'. d*>4d,
/. d>4f
Thus the distance OI between the object and screen must be greater
than 4/, otherwise no image can be formed on the screen. Hence 4/ is
the minimum distance between object and screen; the latter case is
illustrated by Fig. 21.20, in which u = 2/ and v = 2/. If it is difficult to
obtain a real image on a screen when a converging lens is used, possible
causes may be (i) the object is nearer to the lens than its focal length,
Fig. 21.17 (ii), or (ii) the distance between the screen and object is less
than four times the focal length of the lens.
Conjugate Points. Newton's Relation
Suppose that an object at a point O in front of a lens has its image
Fig. 21.22. Newton's relation.
formed at a point I. Since light rays are reversible, it follows an object
placed at I will give rise to an image at O. The points O, I are thus "inter-
changeable", and are hence called conjugate points (or conjugate foci)
with respect to the lens. Newton showed that conjugate points obey the
relation xx' = /*, where x, x' are their respective distances from the
focus on the same side of the lens.
The proof of this relation can be seen by taking the case of the conver-
ging lens in Fig. 21.22, in which OC = u = x +/, and CI = v = x' +f.
Substituting in the lens equation- + - = -
v u f*
1.1 1
.-. fix' + x + if) = (*' +/)(* + n
/. xx' =f* . . . . (g)
Since x' =P\x, it follows that x' increases as jc decreases. The image
I thus recedes from the focus F' away from the lens when the object O
approaches the lens.
The property of conjugate points stated above, namely that an
object and an image at these points are interchangeable, can also be
REFRACTION THROUGH LENSES
491
derived from the lens equation - + - = ->. Thus if u= 15 cm and v
v U J
= 10 cm satisfies this equation, so must u = 10 cm and v = 15 cm.
Displacement of Lens when Object and Screen are Fixed
Suppose that an object O, in front of a converging lens A, gives rise
to an image on a screen at I, Fig. 21.23. Since the image distance AI (v) is
O
•*4
I
Fig. 21.23. Displacement of lens.
^H
greater than the object distance AO (w), the image is larger than the
object. If the object and the screen are kept fixed at O, I respectively,
another clear image can be obtained on the screen by moving the lens
from A to a position B. This time the image is smaller than the object,
as the new image distance BI is less than the new object distance OB.
Since O and I are conjugate points with respect to the lens, it follows
that OB = IA and IB = OA. (If this is the case the lens equation will
be satisfied by
1
^OB
IB ' OB 7andby 1 l + ^ : =i.)Ifthe^foce.
merit, AB, of the lens = d, and the constant distance OI = /, then OA
+ BI = / - d. But, from above, OA = IB. Hence OA = (/ - d)\2.
Further, AI = AB + BI = OA + AB = (/ - d)\2 + d = (/ + d)\2.
But u = OA, and v = AI for the lens in the position A. Substituting
for OA and AI in - + - = -7 ,
v u f
1,1-
1
(/ + d)l2 ' (/ - d)j2
/
2 2 1
/ + d ' / - d f
4/ 1
' * / 2 - rf 2 /
/ 2 -d 2
f
4/
(9)
Thus if the displacement d of the lens, and the distance/ between the
object and the screen, are measured, the focal length/of the lens can be
found from equation (9). This provides a very useful method of mea-
492
ADVANCED LEVEL PHYSICS
suring the focal length of a lens whose surfaces are inaccessible (for
example, when the lens is in a tube), when measurements of v and u
cannot be made (see p. 445).
Magnification. When the lens is in the position A, the lateral magni-
fication rn x of the object = - = ^—- , Fig. 21.23.
h x AI
7T = AO ®
where h x is the length of the image and h is the length of the object.
When the lens is in the position B, the image is smaller than the
BI
object. The lateral magnification, m 2 = 7^ .
where h 2 is the length of the image. But, from our previous discussion,
AI = OB and OA = BI. From (i) and (ii) it follows that, by inverting
(0,
h_ = h
h x h
h* = h x h z
:. h^VhJh^ . . . (10)
The length, h, of an object can hence be found by measuring the
lengths h lt h 2 of the images for the two positions of the lens. This method
of measuring h is most useful when the object is inaccessible, for example,
when the width of a slit in a tube is required.
EXAMPLES
1. A converging lens of focal length 30 cm is 20 cm away from a diverging
lens of focal length 5 cm. An object is placed 6 metres distant from the
former lens (which is the nearer to it) and on the common axis of the system.
Determine the position, magnification, and nature of the image formed.
(O. & C.)
= +30
Fig. 21.24.
/^-5
Suppose O is the object, Fig. 21.24.
from which v = t^- = 31ft cm = LI
REFRACTION THROUGH LENSES 493
For the converging lens,
u = + 600 cm, /= + 30 cm.
Substituting in the lens equation,
•* v^( + 600) ( + 30)
600
19
PI = LI - LP = 31ft - 20 = lift cm.
For the diverging lens, I is a virtual object. Thus u = PI = — lift. Also
/= — 5. Substituting in the lens equation, we have
v + (-llii) (-5)
from which v = — 8-8 cm.
The image is thus virtual, and hence the rays diverge after refraction through
P, as shown. The image is 8-8 cm to the left of P.
The magnification, m, is given by m = m x X m 2 , where m it /n 2 are the magni-
fications produced by the converging and diverging lens respectively.
But mx = £=31ft/600
and m 2 = 8-8/1 lft
.'. m==31ft/600x8f/llft = l
2. Establish a formula connecting object-distance and image-distance for
a simple lens. A small object is placed at a distance of 30 cm from a con-
verging lens of focal length 10 cm. Determine at what distances from this lens
a second converging lens of focal length 40 cm must be placed in order to
produce (i) an erect image, (ii) an inverted image, in each case of the same
size as the object. (L.)
First part. See text.
Second part. Suppose O is the object, Fig. 21.25. The image I in the convex
lens L is formed at a distance v from L given by
A = H-10 f= + 40
Fig. 21.25.
1.1 1
v ' ( + 30) (+10)
from which v = + 15.
v 15 1
For an erect image. Since- = ~ = -, the image at I is half the object size;
also, the image is inverted, since it is real (see Fig. 21.17 (i)). If an erect image is
required, the second lens, M, must invert the image at I. Further, if the new
494
ADVANCED LEVEL PHYSICS
image, Ii, say, is to be the same size as the object at O, the magnification pro-
duced by M of the image at I must be 2. Suppose IM = x numerically; then,
since the magnification
have, from
(0-
2, Mix = 2x. As I and I x are both real, we
!+!=!
1
i l - *
( + 2*)
'(+*) ( + 40)
3 1
*' 2x "~40
x = 60 cm.
from which
Thus M must be placed 75 cm from L for an erect image of the same size
asO.
For an inverted image. Since the image at I is inverted, the image I 2 of I
in M must be erect with respect to I. The lens M must thus act like a magnifying
glass which produces a magnification of 2, and the image I 2 is virtual in this
case. Suppose IM = * numerically; then I 2 M = 2* numerically. Substituting in
1
v^u f
1
1
'• (-2*) ' ( + jc) ( + 40)
.*. x = 20 cm.
Thus M must be placed 35 cm from L for an inverted image of the same
size as O.
SOME METHODS OF MEASURING FOCAL LENGTHS OF LENSES,
AND THEIR RADII OF CURVATURE
Converging Lens
(1) Plane mirror method. In this method a plane mirror M is placed
on a table, and the lens L is placed on the mirror, Fig. 21.26. A pin O
is then moved along the axis of the lens until its image I is observed to
coincide with O when they are both viewed from above, the method
of no parallax being used. The dis-
tance from the pin O to the lens is
then the focal length,/, of the lens,
which can thus be measured.
The explanation of the method
is as follows. In general, rays from O
pass through the lens, are reflected
from the mirror M, and then pass
through the lens again to form an
image at some place. When O and
the image coincide in position, the
rays from O incident on M must
have returned along their incident path after reflection from the mirror.
Fig. 21.26. Plane mirror method.
REFRACTION THROUGH LENSES
495
This is only possible if the rays are incident normally on M. Consequently
the rays entering the lens after reflection are all parallel, and hence the
point to which they converge must be the focus, F, Fig. 21.26. It will thus
be noted that the mirror provides a simple method of obtaining parallel
rays incident on the lens.
(2) Lens formula method. In this method five or six values of u and v
are obtained by using an illuminated object and a screen, or by using
two pins and the method of no parallax. The focal length,/, can then be
calculated from the equation - + - = ■?, and the average of the values
obtained. Alternatively, the values of- can be plotted against -, and a
straight line drawn through the points. When - = 0, - = OA = -x ,
1
from the lens equation; thus/ = -pr-r, and hence can be calculated,
Fig. 21.27. Since - = -when - = 0, from the lens equation, OB = -.
u f v f
Thus/also be evaluated from ~g-.
O **
Fig. 21.27. Graph of l/« against 1/v.
(3) Displacement method. In this method, an illuminated object O
is placed in front of the lens, A, and an image I is obtained on a screen.
Keeping the object and screen fixed, the lens is then moved to a position
B so that a clear image is again obtained on the screen, Fig. 21.28. From
our discussion on p. 491, it follows that a magnified sharp image is
obtained at I when the lens is in the position A, and a diminished sharp
image when the lens is in the position B. If the displacement of the
lens is d, and the distance between the object and the screen is /, the
/2_</2
focal length,/, is given by/= , from p. 491. Thus /can be cal-
culated. The experiment can be repeated by altering the distance
between the object and the screen, and the average value of/ is then
calculated. It should be noted that the screen must be at a distance from
the object of at least four times the focal length of the lens, otherwise
an image is unobtainable on the screen (p. 490).
496
ADVANCED LEVEL PHYSICS
(
Fig. 21.28. Displacement method for focal length.
Since no measurements need be made to the surfaces of the lens (the
"displacement" is simply the distance moved by the holder of the lens),
this method can be used for finding the focal length of (i) a thick lens,
(ii) an inaccessible lens, such as that fixed inside an eye-piece or telescope
tube. Neither of the two methods previously discussed could be used
for such a lens.
Lateral Magnification Method of Measuring Focal Length
On p. 488, we showed that the lateral magnification, m, produced by
a lens is given by
m = -f— 1
(i)
where /is the focal length of the lens and v the distance of the image.
If an illuminated glass scale is set up as an
object in front of a lens, and the image is
received on a screen, the magnification, m,
can be measured directly. From (i) a
straight line graph BA is obtained when
m is plotted against the corresponding
image distance v, Fig. 21.29. Further, from
(i), v[f — 1 = when m = 0; thus v =/in
this case. Hence, by producing BA to cut
the axis of v in D, it follows that OD = /;
the focal length of the lens can thus be
found from the graph.
Fig. 21.29.
Graph of m against v.
Diverging Lens
(1) Converging lens method. By itself, a diverging lens always forms
a virtual image of a real object. A real image may be obtained, how-
ever, if a virtual object is used, and a converging lens can be used to
provide such an object, as shown in Fig. 21.30. An object S is placed at a
distance from M greater than its focal length, so that a beam converging
to a point O is obtained. O is thus a virtual object for the diverging lens
L placed as shown in Fig. 21 .30, and a real image I can now be obtained.
REFRACTION THROUGH LENSES
497
I is farther away from L and O, since the concave lens makes the in-
cident beam on it diverge more.
The image distance, v, from the diverging lens is CI and can be
measured; v is + ve in sign as I is real. The object distance, w, from this
Fig. 21.30. Focal length of diverging lens.
lens = CO = AO — AC, and AC can be measured. The length AO is
obtained by removing the lens L, leaving the converging lens, and
noting the position of the real image now formed at O by the lens Mj
Thus u (= CO) can be found; it is a - ve distance, since O is a virtual
1 ,_
j> the
object for the diverging lens. Substituting for u and v in - + -
focal length of the diverging lens can be calculated.
(2) Concave mirror method. In this method a real object is placed in
front of a diverging lens, and the position of the virtual image is located
with the aid of a concave mirror. An object O is placed in front of the
lens L, and a concave mirror M is placed behind the lens so that a
divergent beam is incident on it, Fig. 21.31. With L and M in the same
position, the object O is moved until an image is obtained coincident
with it in position, i.e., beside O. The distances CO, CM are then
measured.
As the object and image are coincident at O, the rays must be in-
cident normally on the mirror M. The rays BA, ED thus pass through
the centre of curvature of M, and this is also the position of the virtual
image I. The image distance, v, from the lens = IC = IM - CM =
Fig. 21.31. Focal length of diverging lens.
498
ADVANCED LEVEL PHYSICS
r — CM, where r is the radius of curvature; CM can be measured, while
r can be determined by means of a separate experiment, as described on
p. 413. The object distance, w, from the lens = OC, and by substituting
for u and v in the formula - + - = j, the focal length /can be cal-
culated. Of course, v is negative as I is a virtual image for the lens.
Measurement of Radii of Curvature of Lens Surfaces
Diverging lens. The radius of curvature of a lens concave surface A
can easily be measured by moving an object O in front of it until the
image by reflection at A coincides with the object. Since the rays from
Fig. 21.32. Radius of diverging lens surface.
O are now incident normally on A, its radius of curvature, r lf — OC, the
distance from O to the lens, Fig. 21.32. If the radius of curvature of the
surface B is required, the lens is turned round and the experiment is
repeated.
Converging lens: Boys* method. Since a convex surface usually gives a
virtual image, it is not an easy matter to measure the radius of curvature
of such a lens surface. C. V. Boys, however, suggested an ingenious
method which is now known by his name, and is illustrated in Fig. 21 .33.
In Boys' method, an object O is placed in front of a converging lens,
and is then moved until an image by reflection at the back surface NA
is formed beside O. To make the image brighter O should be a well-
illuminated object, and the lens can be floated with NA on top of mercury
to provide better reflection from this surface.
Fig. 21.33. Boys' method for radius of converging lens surface.
Since the image is coincident with O, the rays are incident normally
on NA. A ray OM from O would thus pass straight through the lens
REFRACTION THROUGH LENSES 499
along NP after refraction at M. Further, as PN produced passes through
I, I is a virtual image of O by refraction in the lens. On account of the
latter fact, we can apply the lens equation - + — = ■?, where v = IC =
r, the radius of curvature of NA, and u — OC. Thus knowing OC and
the focal length,/, of the lens, r can be calculated. The same method can
be used to measure the radius of curvature