1*6 Mr potter, on THE PROBLEM OF THE RAINBOW, 



Again, substituting for sin 7 its value, we have 



. ^ v^l + tan- (p {d^yf 

 '^ ^ tan (pa^y 



^ /cos- (b . ., ^ 

 whence p=.r \/ -t-j — ^ + sin- ^ 

 (u,t,y) 



/ cos' cos' d)' , • " , /c\ 



= '• ,/f-7i ^^ ^i— p+sin-<^ (3). 



a/<2 (- cos (p — eos <p'j} 



This (3) with the equations (1) and (2) would suffice to eliminate (p 

 and D, and give p in terms of 6 and constants, if the transcendental 

 relations of (p, <p' and Q did not prevent it: that expression would be 

 the polar equation of the caustic for the primary bow : we may how- 

 ever trace this curve from equation (3). 



Taking m = ^ , as is usually done for red light, we find two 

 values of (p which give p = r, 



namely, (jb = 0, and ^ = 3 • 



Again p becomes = + 00, when 



2 



- cos d> — cos <p' = 0, 



or 4 cos'^ <p = ix' — sin' (p ; 



cos 



<^ - V'^. 



which, as seen in elementary books, is the angle of minimum deviation, 

 the ray at this angle being an asymptote to the caustic. 



The deviation diminishing from (p = to (p — (p,„ the intersections of 



consecutive rays are behind the sphere, but from (p = (p„, to = — 



the deviation increases, and the intersections are in front. So that the 

 two branches of the caustic are as at Act, Bb in the Fig. 2. being 

 perpendicular to the sphere at A, and tangential to it at B, and the 



