196 Prof. Anderson, On the maximum [May 25, 



describe small circles with radii equal to the critical angle. These 

 circles must intersect if it be possible for a ray to get through, 

 and the point R representing the direction of the ray in the prism 

 lies on the part of the sphere common to the areas of the circles. 



Draw a small circle whose distance from N^N* is rj' the angle 

 which any ray in the prism makes with the principal plane, and 

 let it intersect the two small circles already drawn in E and F. 

 Then for positions of R on EF, G the middle point is that which 

 corresponds to minimum deviation and E, F are the points cor- 

 responding to maximum deviation 1 . Hence to find the absolute 

 maximum deviation we must examine the values of the deviation 

 for positions of R on the small circle CA. 



Let R be one of these positions. Join NjR and take NJP a 



A . A 



right angle. Join N. 2 R and produce so that sin JS T 2 Q = /j, sin N. 2 R. 

 Then P is the direction of the incident ray and Q that of the 

 emergent ray, PQ being the deviation. 



Let PQ = 8, JS T 2 R = tf>, RQ = 6; then if A be the angle of the 

 prism it is easy to shew that 



cos 8 = (u? — 1) cos A — — r — u? + 1 + a cos 9. 

 sin 9 



The deviation is therefore stationary when 



A Sm n 



cos 7 cos A — — - + cos 

 sin 9 



is stationary, which may be shewn to hold when 



cos A sin </> = tan y sin 0, 



a condition equivalent to 



1 + cos 2 J. 



cos <r> = cos 7 — r t— . 



r ' 2 cos A 



Since 1 + cos 2 A > 2 cos A, the necessary condition <f> < 7 is 

 satisfied. 



We may easily find for what values of the angle of the prism 



a stationary value of 8 is possible. Since cos is positive, A <—. 



Again, the least value of the variable <£ is A - 7 or 7 — A and 

 consequently 



1 + cos 2 A , . . 



COS ' y -^^osX <COs( ^- fy) ' 



which reduces to tan A < 2 tan 7. 



1 Proc. Camb. Phil. Soc. ix. p. 109. 



