322 Mr. J. C. M‘Connel’s Notes on 
What we want to measure is the rotation of P’z round P’; 
so let eP/K=. As before, let P’/P=a, PPo>=£, and let 
AP,)=a, AP’=r. Then a, 8, a, 7 are all small. Since the 
figure A Py P P’ is small, we have to our order of approxi- 
mation 
ae =a—Pr COS ere 
B=r sin (W+ x). 
Let us now find the difference between 0 anday+y=PP’K. 
By spherical triangles, 
cos K=sin (+) cos PP’, t 
cos K=sin @ cos a; 
*, sin (r+) cos PP’=sin @ cosa, 
2 2 
sin O=sin (W+y) (ig + ) 
sin (P4x) + O—Y—x) cos (P+ A =sin'+y)(1-F +9) 
oe Bet 
2. 0pm y= 9S tan (+x) 
a* — a? 
sae tan yr, neglecting cubes. 
But : 
a* —a’=2ar cos p—71’ cos’ Wp, 
and 
v= la—maB 
=l(a—r cos +x) —mr sin ¥re(a—r cos p) 
=l{a—rcosy + lr sin (a—r cos )}—mr sin yr(a—?7 cos wr) 
=la—lr cos +(?—m)r sin W(a—r cos Wr); 
6 0=W+la—Ilr cos p+ (?—m+1)arsinw 
—(?—m-+4)r’ sin w cos w. 
Now let the plane of polarization be turned through 180° 
and the new reading of the Nicol circle be 180°+6,. Instead 
of y we must write in the last formula 180+. So 
6,=W+latlr cos — (?—m-+1)ar sin p 
| —(?—m+4)r’ sin p cos; 
AO) ig gd a oe 
5 = +la—(?—m-+ })r’ sin cos yp, 
6,—6 
2 
=lr cos y—(l?—m-+1)ar sin wp. 
