of the double refraction in strained glass. 483 
beams, shown in cross-section, whose thicknesses are r l5 r 2 . Let 
a be the distance from the scale to the diaphragm, b the horizontal 
distance between the two beams and e the vertical height of the 
central axis of the second beam above that of the first. 
Then if P be any point of the scale, the pencil of rays which 
ultimately form an image of P in the focal plane of the telescope 
is the broken cone bounded by the rays APA X A 2 A^, BPB X B 2 B 3 in 
Fig- 2 - ... . 
Now in investigating the relative retardations, all that we are 
concerned with is the path in the strained glass. 
Now the paths in the glass may be found by the following 
construction. 
Imagine the beam EF removed a distance (g — 1)6 to the 
left, as E'F ' (Fig. 4), and similarly the diaphragm AB removed 
(/jb — l)a to the right as A'B'. Then the paths of the actual rays 
in the glass will now lie inside a straight cone A'PB' , the paths in 
the two glass beams of any ray which emerges as PQ in the real 
case being given by the intercepts PQ X , Q 2 Q 3 of the corresponding 
ray PQ' in Fig. 4. Consider the relative retardation of the two 
oppositely polarized parts of the ray PQ'. 
i 1 
5 === ==S 
< \ 
^ ■ 
: 
K A ! 
c 
r. 
bn- 
Fig. 4. 
This can easily be shown to be 
cp cq 2 ' 
GT X cos 2 6 sec y/r dr x + CT 2 cos 2 6 sec yjr dr 2 , 
J Qi J Os' 
7T 
. — — 0 being the angle which the ray PQ' makes with the direction 
fa 
of the axes of the beams and yjr the angle PQ' makes with the 
horizontal perpendicular to the sides of the beams. 
This gives for the retardation 
C cos 2 6 sec yjr 
y v y 2 being measured vertically from the neutral axes CD , E'F' in 
the two beams. 
VOL. XI. PT. VI. 
34 
