the Use of Nicol’s Prism. 319 
Let P’x be the plane of polarization of P’ and PP’x=y. 
Then y is a function of P’P and PN. 
The exact function would be a very complicated expression, 
so we shall content ourselves with an approximation. P’P is 
always small, and P lies near some fixed point, Po say, in the 
plane XN. Let 
eek hr 
We shall reject the cubes and higher powers of « and 8 
and limit ourselves to expanding y as far as the squares. We 
know that when «@ is zero, y is zero; so our expression can 
only contain the terms @, a’, and a8. 
In a Nicol’s prism it is the extraordinary ray which emerges, 
so the plane of polarization of Q’ is perpendicular to Q’X. 
Let us assume for the present that the plane of polarization 
of P’ makes the same angle with the plane of incidence P’Q’N 
as does the plane of polarization of Q’. The plane of polari- 
zation of Q’ makes with P’Q’N anangle 90°—NQ’X. Hence 
eH NOX NEP ee a ey) 
But in the triangle NP’P, P is a right angle ; so we have 
sin P’P a 
tan (90°—NP’P)= cot NP’P= tan NP == tan NP” ° (2) 
neglecting cubes. 
It remains to find NQ’X. 
NQYX=XQY'Q-NYQ | 
~=90°—NP/P+ NQ’'Q—XQ’Q. . . . (8) 
By spherical ee Fine 
tan (90°— XQ’Q) = a. 
tan (90°—NQ'Q)= peer 
Also 
sin Q’/Q= sin Q/N sin N, 
sin P’P=sin P’N sin N.- 
But sin P’N=wpsin Q’/N, where p» is the extraordinary index 
of refraction for that wave, which we may take to be constant. 
.. sin Q’Q = : sin P’ Pas to our order of approximation. 
pe 
Hence 3 ee ee 
tan (90°— XQ’Q) = ey 
? of OrO\— oa 
tan (90° —NQ’Q) = eee 
Z2 
(4) 
