Dr. Brewster on the laws of polarisation, &c. 237 
number of axes, let us suppose that ABC, PLxvi. fi g.9, repre- 
sents the quadrant of a spherical surface, such as we have 
described in p. 2 66 ; and that the position of G, one of the re- 
sultant axes, where the tint is nothing, has been carefully 
determined by experiment, it is required to find the tint at 
any point E by the action of certain polarising forces which 
are in equilibrio at the point G. If w r e now suppose, 
1st. That the tints are produced by forces emanating from 
two negative axes whose poles are C,A, it is obvious that their 
relative intensities must be in the ratio of 1 :=r — GC re- 
presenting half the inclination of the diameters of no polarisa- 
tion. For as the tint at G produced by A is equal to the tint 
produced at the same point by C, and since the tint produced 
there by the axis A is its maximum tint, AG being 90°, then 
the maximum tint produced by C will be found by the 
analogy Sin.- GC : Rad.’= i 
2d. If we suppose that the forces emanate from two positive 
axes A, B, A being greater than B, then the relative intensities 
must be as i : e^GC' 
3d. If the forces emanate from two axes B,C, one of which 
is positive and the other negative, the intensity of B must be 
to that of C as Sin. 2 GC: Cos.* GC. 
Through E draw three great circles AEF, BE and CE, 
PI. xvi. fig. 9. and let 
T = tint required at the point E. 
9 =3 the arch between the point E and the axis C. 
<p = the arch between the points E and B. 
a — the tint produced separately at E by the greater 
axis. 
b = the tint produced separately at E by the lesser axis. 
