206 ' REFLECTION OF LIGHT AT THE [378
pasted paper. The tube should be lowered over the candle until the whole of the flame projects, when it will be apparent that the illumination of the candle extends decidedly lower down than before.
In imagination we may get quit of the lateral loss by supposing the diameter of the candle to be increased without limit, the source of light being at the same time extended over the whole of the horizontal plane.
To come to a definite question, we may ask what is the proportion of light reflected when it is incident equally in all directions upon a surface of transition, such as is constituted by the candle fracture. The answer depends upon a suitable integration of Fresnel's expression for the reflection of light of the two polarisations, viz.
sni^-^0 ten* (0-8')
*~~ ' ............... (1)
where 6, 6' are the angles of incidence and refraction. We may take first the case where 0 > 6', that is, when the transition is from the less to the more refractive medium.
The element of solid angle is STrsm&dO, and the area of cross-section corresponding to unit area of the refracting surface is cos 0 ; so that we have to consider
2 ('sin 0008008*01 T2)d6, ..................... (2)
the multiplier being so chosen as to make the integral equal to unity when S2 or T2 has that value throughout. The integral could be evaluated analytically, at any rate in the case of S2, but the result would scarcely repay the trouble. An estimate by quadratures in a particular case will suffice for our purposes, and to this we shall presently return.
In (2) 0 varies from 0 to ^TT and Q' is always real. If we suppose the passage to be in the other direction, viz. from the more to the less refractive medium, $2 and T2, being symmetrical in d and 0', remain as before, and we have to integrate
2 sin 0' cos 6' (& or T^ d6'.
The integral divides itself into two parts, the first from 0 to o, where a is the critical angle corresponding to 6 — |TT. In this S2, T'2 have the values given in (1). The second part of the range from 6' = a to &' — fyrr involves " total reflection," so that S2 and T2 must -be taken equal to unity. Thus altogether we have
'••'- 2 ["sin 9' cos ff (S* or T2) de' + zf" sin B' cos B' d6', ..... ,(3)
• '''• therefore dzU/dys, is everywhere positive and accordingly that a non-viscous liquid, moving laminarly as the viscous fluid moves in any of these stages, is stable. It would appear then that no explanation is to be found in this direction.