through Apertures in Plane Screens. 271 



theorem that ^\jr p dy = ihk. The value of h* is \irlJ 1 ; so that 



" ^-^&-*;-- ■ • • <»> 



The same algebraic expression gives y{r m} if the minus sign be 

 omitted. 



The realized solution from (53) is 



f p = - -27- V2I7- ; cos ( ~ ~ ^ • • ( 54 > 



corresponding to 



^ Hl = 2 sin n£ sin #<# (55) 



Reflecting Blade. — Boundary Condition d(f>/dn = 0. 



We have now to consider two problems which differ from 

 the last in that the opaque and transparent parts of the screen 

 are interchanged. As in the case of the circular aperture, we 

 shall find that the correspondence lies between the reflecting 

 blade under the condition d<f>/dn = Q and the transmitting 

 aperture under the condition <£ = 0, and reciprocally. 



The principal solution remains as in (30). The supple- 

 mentary solution must satisfy (31), where 



^JS^' ^=|S^^' • • (56) 



since *P m and M/p must be equal in order that the continuity 



of d<j>/da? over the aperture may be maintained. Thus TJr m 



and yjr p have opposite values at any pair of corresponding 



points. 



If we compare these conditions with those by which (53) 



was determined, we see that^ m has the same value as in that 



case, but that the sign of -^ymust be reversed. Thus in the 



present problem 



kWx / it \i 

 f m = f p = — ^— J cos ( n *-*r-iw), . (57) 



corresponding to 



%m=X P =cos (nt—kx) (58) 



Reflecting Blade. — Boundary Condition <p = 0. 

 In this case % still remains as in (30) . The general forms 

 for yjr m , yfr p are as in (42), which secure that d^r m /dcV, d^jdx 

 shall vanish on the aperture [i. e. the part of the plane a?=0 

 unoccupied by the blade). But in order that the continuity 

 of cf) may also be maintained over that area we must have 

 ^ / v». = "^ /, p. Thus yfr m , ty v have equal values at corresponding 

 points. On the blade itself \^ TO =^ p = —1. 



* Lamb's ' Hydrodynamics,' § 71. 



