SCATTERING OF PLANE ELECTRIC WAVES BY SPHERES. 
205 
Thus, when we regard ajr as small, we may take 
E, = r—a {I sin Q cos 0 + m sin 9 sin (f> + n cos 0), 
(7'65) •< 
and 
011 
^ = — {I sin 0 cos ^ + m sin 6 sin 0 + n cos 6). 
CP 
0R 
Now, in applying (7‘13), — occurs only in the coefficients and not in the exponential 
0T> 
index: thus we can get the first approximation by putting 0 = tt in ^; this gives 
CP 
the value 
(7-66) 
Thus, to our degree of accuracy 
aK 
= n. 
1 f dv dtv\ V ( aE div 
47r \ CP CP/ 47r \ or cp 
V I , a^o 
= - —[iKnw+ — 
47r \ oj/ 
We can now substitute for xv and — the values given by (7‘63) and (7*64): it will 
be seen that the components parallel to ij, z give zero (to this order), and that the 
component parallel to x gives 
iKn ^ _ 
27r r 
(aji—R) 
Accordingly the reflected wave is given by 
(7*67) 
X = 
IK 
‘IttX' 
ndSe' 
(an—R) 
where E is found from (7*65) and the integral extends over the positive hemisphere. 
For the purpose of integration we write 
I = sin 6' cos (f>', m = sin 0' sin <f>', n = cos O'. 
Then in (7*67) we have 
E = —r + a {(l +cos O) cos O' + sin 0 sin 0' cos (0 — 0')}. 
The integration of (7*67) with respect to cf> can be carried out at once, because 
(7*68) 
^ 27 rJo (/ca sffi Osill O'). 
Jo 
