248 Dr. Brewster on the laws of polarisation , &c. 
Sin. 1 x X Cos ? m Sin. 2 * x Cos. 1 n = Sin. 2, q> x Cos. 2 w-|- Sin. 2 9x Cos. 2 a 
and Sin 2 cc = Sin. 2 ip X Cos. 2 g + Sin. 2 9 x Cos. a &) 
Cos. 2 m x Cos. 2 n 
But Cos . 2 m x Cos . 2 n = 1. Hence 
Sin.*# = Sin. a <p x Cos . 2 tt -}- Sin . a 0 x Cos. a 6 >. 
But since Cos.'tt = Cos. 2 7 r + Sin . 2 tt and 
Cos.* w = Cos. 2 -f- Sin. a w we have 
Sin. 3 #=Sin. 3 <pxCos.2 7r -f- Sin. a x Sin.V-f Sin . a 0 xCos. 2 *>+Sin . a 0 x Sin. 2 w 
But Sin.® (?» x Sin . 2 7 r-j-Sin . 3 (5 x Sin. a w = 1; hence 
Sin. 3 # = Sin . a <p x Cos. 2 tt -j- Sin . 3 9 x Cos. 2 a -j- 1 , and 
Since Cos. 3 # = 1 — Sin. 3 # we have 
— Cos. 3 # = Sin.*<£> x Cos. 2 % + Sin . 3 9 x Cos. 2 u 
Now in the parallelogram of forces, the two forces AC, AD 
(PI. xvi. fig. 12) are Sin. 3 <?>, Sin . 3 0 when the axes are of equal 
intensity, that is AC = Sin. a (p, and AD = Sin . 3 9 . But 
Sin. 3 p : Sin . 3 0 = Sin. a w : Sin. 3 ??, and the 
Angle CAD = 2 tt -f 2 &>. Hence 
CAB = 2 tt and 
BAD — 2 u 
Consequently AF = Sin. a <p x Cos. 2 ?r and 
BF = Sin . 3 0 x Cos. 2 «, and therefore 
AF+BF = AB = Sin. a <p x Cos. 2 tt -f- Sin . 3 0 x Cos. 2 w 
which is the very same value which we have found above 
for Cos. 3 #, the measure of the tint produced by one axis. 
Hence it follows that the intensity of the tints, and conse- 
quently the form of the curves of equal tint, are the same, 
whether they are produced by one negative, or by two equal 
and rectangular positive axes. 
We may therefore conclude, 
1st. That a single negative axis may be resolved into two 
