328 Proceedings of Royal Society of Edinburgh . [sess. 
just suffers total reflection at the surface of the balsam; this is one 
of the boundaries of the field of view. 
Again, from the same point R apply the plane RST touching the 
spheroide in S, and cutting the surface OP in T ; the tangent plane 
TU will give the direction of OU the limiting extraordinary ray. 
The question is to place OR, so that the angle UO u may be 
bisected by the line OM. 
For the analytic solution of this problem we have only to go into 
detail with the investigation for extraordinary light ; since the 
insertion of (3 = a in any formula for that species of light will give 
the corresponding formula for ordinary light. 
Denote the angle LOP by /x, the required angle LOR by y, and 
the velocity of light in Canada Balsam by y: then are the co- 
ordinates of the point R 
x R = y cos v ; y R = y sin y . 
But the equation of a plane passing through R, and touching the 
spheroide, is 
± y* V(«y R + /3 2 4 - a*0$} ) 
> = o?yl + (3 2 xt 
+ + % JWyl + /3 2 x\ - a 2 /? 2 )} J 
or 
x | /3 2 cosv ± y sin v sinv 2 + /5 2 cosv 2 - 
+ y | a 2 sin v + y cos v siny 2 + /3 2 cosy 2 
a 2 /? 2 
y 2 . 
a 2 /5 2 
y y(a 2 sin y 2 + (3 2 COS y 2 
From the inspection of the figure it is clear that 
x v = sec POU. cos /x 
y T = sec POU. sin g 
which values inserted in the above equation of the tangent plam 
give, after reduction, and putting e 2 = j3 2 - a 2 , 
cos POU = 
e 2 COS fJj COS y + a 2 COS (g — y) + y sin 
in (/j.-v) 
(J8* 
' 2) ) 
y(a 2 + e 2 COS y 2 ) 
The supposition /5 = a, e = 0 in this formula will give the value 
of tbe cosine of the limiting angle for ordinary light : thus 
