1913] - FINE SLITS IN THIN OPAQUE SCREENS 163
This %-solution makes d%m/dn 0, d^p/dn 0 over the whole plane x 0, and over the same plane %m = 2, %p = 0.
For the supplementary solution, distinguished in like manner upon the two sides, we have
f____ r7<3 /"7\
p , C60' ............\')
where r denotes the distance of the point at which ^ is to be estimated from the element dS of the aperture, and the integration is extended over the whole of the area of aperture. "Whatever functions of position "^fm, typ may be, these values on the two sides satisfy (2), and (as is evident from symmetry) they make d-tym/dn, d-fyp/dn vanish over the wall, viz., the un-perforated part of the screen, so that the required condition over the wall for the complete solution is already satisfied. It remains to consider the further conditions that </> and d<f>/daa shall be continuous across the aperture. These conditions require that on the aperture
The second is satisfied if "ifp = - tym ; so that
r r pikr
-dS ............. (9)
making the values of tym, typ equal and opposite at all corresponding points, viz., points which are images of one another in the plane x = 0. In order further to satisfy the first condition, it suffices that over the area of aperture
and the remainder of the problem consists in so determining ^m that this shall be the case.
It should be remarked that "*? in (9) is closely connected with the normal velocity at dS. In general,
duo JJ doo
At a point (x) infinitely close to the surface, only the neighbouring elements contribute to the integral, and the factor e~ikr may be omitted. Thus
or *= , ...(12)
2vr dn ' ^ '
being the normal velocity at the point of the surface in question.
* The use of dx implies that the variation is in a fixed direction, while dn may be supposed to be drawn outwards from the screen in both cases.
112 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).