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162                           ON THE PASSAGE OF  WAVES  THROUGH                          [875
Plane waves of simple type impinge upon a parallel screen. The screen is supposed to be infinitely thin and to be perforated by some kind of aperture. Ultimately, one or both dimensions of the aperture will be regarded as small, or, at any rate, as not large, in comparison with the wavelength (A,) ; and the investigation commences by adapting to the present purpose known solutions concerning the flow of incompressible fluids.
The functions that we require may be regarded as velocity-potentials (/>, satisfying
d*<j>/dt* = VV^,  .............................. (1)
where                            V2 = dz/dx- + d2/df + d^dzz,
and V is the velocity of propagation.    If we assume that the vibration is everywhere proportional to eint, (1) becomes
(V2 + &2)<£ = 0,    .............................. (2)
where                                     k = nj V = 2ir/X, ............................... (3)
It will conduce to brevity if we suppress the factor eint. On this understanding the equation of waves travelling parallel to as in the positive direction, and accordingly incident upon the negative side of the screen situated at sc = 0, is
When the solution is complete, the factor eint is to be restored, and the imaginary part of the solution is to be rejected. The realised expression for the incident waves will therefore be
<jb = cos (nt — lets) ............................... (5)
There are two cases to be considered corresponding to two alternative boundary conditions. In the first (i) d$>jdn = 0 over the unperforated part of the screen, and in the second (ii) <j6 = 0. In case (i) dn is drawn outwards normally, and if we take the axis of z parallel to the length of the slit, </> will represent the magnetic component parallel to z, usually denoted by c, so that this case refers to vibrations for which the electric vector is perpendicular to the slit. In the second case (ii) <£ is to be identified with the component parallel to z of the electric vector R, which vanishes upon the walls, regarded as perfectly conducting. We proceed with the further consideration of case (i).
If the screen be complete, the reflected waves under condition (i) have the expression 0 = eikx. Let us divide the actual solution into two parts, •% and -V//-; the first, the solution which would obtain were the screen complete ; the second, the alteration required to take account of the aperture ; and let us distinguish by the suffixes TO and p the values applicable upon the negative. (minus), and upon the positive side of the screen. In the present case we have                ...es*, to which it would 1