320 Mr. J. C. M‘Connel’s Notes on 
Since tan XQ and tan NQ occur in small terms, we may 
reject squares in finding their values. 
In NQ let us take a point Qo such that sin NPp>=p sin NQp. 
Then, since sin(NP)+ 8)=p sin (NQ) + QQo), 
8 cos NPp>=pQQ) cos NQo, 
and 
deg te Pdi epee... Vn eee 
tan XQ tan XQ, sin? XQ, tan XQ, pcos NQ sin?XQo 
1 il QQ, 1 B cos NP» 
tan NQ ~ tan NQ, sin? N Q, tanNQ, pcos NQ sin?NQ,” 
Now in the expansion of the tangent of a small angle the 
squares of the angle do not appear. So we have by (3), (2), 
and (4), 3 
ee a 1 a 1 
Xan ptan XQ p tan NQ’ 
while 
ake 1 B 
tan NP tan NPy _ sin? NP, 
So by (5), 
1 i 1 
oS ome (ar, + ptanXQo ptan = 
1 cos NP 1 Ih 
ae { 7 sin2 NP) 2 cos NOQ, Ge xO; ee } (6) 
=la—map say. 
In Nicol’s prism, as it is usually cut, 
x = ao 
Wey = AES, 
iS aut 
NQ, = 14°. 
Substituting these values, we obtain 
32) an — 154, 
y= d2a— 84a8. 82 1 ae 
In the foregoing we have assumed that there is no rotation 
of the plane of polarization on refraction out of the spar, or, 
in other words, that the angle between the plane of polariza- 
tion and the plane of incidence remains unchanged. There 
is of course a change, but it is merely due to the disturbing 
effect of reflection and is very small. If we treat the spar as 
