POLARIZED LIGHT AT THE SURFACE OF A UNIAXAL CRYSTAL. 
601 
Fig. 1. 
Let us suppose that B R is the ordinary refracted wave corresponding to an 
incident wave B I. Then A B V =cf>' and X B V=\-f -(/>'; also B Y=~ — 0'; and 
from the right angle and triangle X B Y we have 
cos X B V=tan B V cot B X 
cot d'=tan jB cos (X+<£'). 
. ... (7) 
But we have from (5) 
* 
cot d=cot 6' sec (<£ — <f>') 
Therefore 
cot d=tan J3 cos (\+<//) sec (<£—<£') . 
. ... (8) 
If B B represent the extraordinary wave 
BY= 0", A B Y=f, 
and we have 
tan 6"=taxi /3 cos (A+c/P).. . (9) 
so that from (6) we obtain 
tan d=tan ft cos cos (<£—<£") + sin 3 <f>" tan q/ cos 6" sin (<j> + <f>") . (10) 
It remains now to find q> the angle between the extraordinary ray and the wave 
normal. 
Fig. 2. 
Let the figure (fig. 2) represent a section of the surface of wave slowness passing 
through the optic axis, 0 A, and the extraordinary wave normal, O P. Let O Y be 
4 H 2 
