268 Lord Rayleigh on the Passage of Waves 



great in comparison with r, kp is small and e~ ik § may be 

 identified with unity. Thus in the limit 



I 



-H^-^K-^*'];-^ 



and (39) becomes 



^r = 7 + logA:?'-f i*7r— log2 = y + log {^ikr), . (40) 

 in agreement with (35). 



When kr is extremely small (40) may be considered for 

 some purposes to reduce to log kr ; but the term \iir is re- 

 quired in order to represent the equality of work done in the 

 neighbourhood of the linear source and at a great distance 

 from it. 



We may now proceed to solve four problems relative to 

 narrow slits and reflecting blades analogous to the four 

 already considered in which the aperture or the reflecting 

 plate was small in both its dimensions in comparison with the 

 wave-length. 



Narroio Slit. — Boundary Condition d<f)/dn = 0. 

 As in the former problem the principal solution is 



Xm = e~ ikx + e ik % x P =0, .... (41) 

 making dxm/dn, d^p/dn vanish over the whole plane #=0 

 and over the same plane ^ TO = 2, % P = 0. The supplementary 

 solution, which represents the effect of the slit, may be 

 written 



f m =^ m D(kr)d^, ^.j'VjDCfcOrfy, . (42) 



"*P m , "Wp being certain functions of y to be determined, and 

 the integration extending over the width of the slit from 

 y= —b to y= +b. 



These additions do not disturb the condition to be satisfied 

 over the wall. On the aperture continuity requires, as in 

 (8), that 



2 + yjr m = y]r p , df m Jdx = d%jr p /dx. 



The second of these is satisfied by taking s l? p =—'*& m , so that 

 at all corresponding pairs of points -^r m = — yfr p . It remains 

 to determine "*P m so that on the aperture ^y= —1 ; and then 

 by what has been said ^r p = + 1. 



At a sufficient distance from the slit, supposed to be very 

 narrow, D(kr) may be removed from under the integral sign 

 and also be replaced by its limiting form given in (35). Thus 



*-=~{m$ t '~l*-*- ■ ■ ■ (43) 



