ELECTRIC WAVES ROUND A PERFECTLY REFLECTING OBSTACLE. 119 



Now the distance D from the oscillator to the point (r^} is given by 



D 2 = r 2 +r 1 2 -2rr 1 cos6>, 

 whence retaining only the most important terms 



i - w - 



and therefore S t = $1. 



The same result follows when r>?v Again writing 



S 2 = -i, sin 6 cos \B 2 (n + 1) 3 R^'V^^-*-*,) ^{(2^+1) sin 0}, 



' 



the principal part of S 2 arises from the terms in the neighbourhood of the term for 

 which 



vanishes. When n + ^ is less than 2,,, this requires that 2*,, n a, is small, since is ; 

 when n + ^ is greater than 2 , it requires that TT a ^ is small, that is and ! are 

 each nearly a right angle, which means that the point is close to the oscillator, thus 

 the case of 2a u a j small need only be considered. In tin's case the principal part 

 of S 2 is equal to the principal part of 



~ sin 6 cos ^6 2 (n + |-) 2 e lC2 """-"- |+ "- ( " +1/ - >2(2 - vl - v ' )] J 1 {(2u+ 1) sin J-0}, 

 that is 



S 2 = sin cos 4-0 e' (2 *""'""" " l) (j 1 ' 2 '' 2 """ 1 "*"" 1 "" 1 " 1 '^^, (2t sin W] <1L 



K1\ Jo 



whence 



S 2 = - -^-j sin ^ cos ^Oe l( -'~'~ -'-">'> ^~ '" 1 3r/ _ r 



or 



This represents waves reflected from the neighbourhood of the points Q on the 

 sphere, which are such that O L Q makes an angle with the radius OQ, and if D! is 

 the distance C^Q, D 2 the distance PQ, where P is the point (?-, 0), the above becomes, 

 after some reduction, 



S 2 = -IK sn 



2 



the result that would have been obtained by elementary methods. Hence the eifect 

 at a point P for which 6 is small is the sum of the direct waves and the waves 

 reflected at the surface of the sphere. 



