a Diffraction Problem. 319 



and (2). The first case is now 



where 



n\\ (>=-nX, c^= - y -^-, . . (5a) 

 1 — n 2 B 2 ^ _ B 2 ^ B 2 "^ 



or LV±^V±,V^, V±. , n 



V 2 b* 2 ~ B.? 2 + 3y» + d* 2 . . . . w 



Thus, for example, the first of (2) is 



1 3« BY n BY _, 



TT ^ ~ ^r— = ^ r ^— , or a — ni. 

 VBf B* V B* 



The polarization corresponds with that for (3 a) and the 

 electric vector in the incident wave is 



(ZY — mX)/ \/l — n% or A- y/l — n 2 sin & ( Yt 4- fo? + nip -f ?*~) . 



The second case is 



By B* i-n 2 c^ 



X=— n&, Y=na, Z = ^. . (5 A) 



Here X=— n, J-, and when -Jr or ^7 is made to vanish 

 0# Of 



for all points of the barrier, X and Z also vanish. In 

 the incident wave X=Hwsin&( ), Y = &m?isin£( ), 

 Z= — Z-(l — rc 2 ) sin k( ) ; the magnetic vector is perpen- 

 dicular to the axis of z, the electric vector has direction- 

 cosines 



{In, mn, ~{l-n 2 )}/ \/T^n~\ 



and is of amount k V 1 — n 2 sin k(Yt + lx + my-\-nz). 



It is clear that (6) corresponds to the solution given in 

 the paper, and that the same modification is applicable to 

 other plane problems to meet the case where the incident 

 wave has motion in a third dimension. 



