Glass under the Poldriscope. 41 



where 



a = cos (p — <j>) cos <j — cos (p + </>) sin cr, 



b = cos (p — <f>) sin cr + cos (p + 0) cos cr, 



a' = sin (/) — c/>) cos cr — sin (p + </>) sin cr, 



V = — sin (p — c/>) sin cr— sin (p + <£>) cos cr. 



Differentiating (1) and (2) with respect to t, we have 



cos #-7- — sin 0f -77 = acos£— 6sin£, . . . (3) 

 at at 



sin 0^- + cos^^-=a / cos^— ^/sin*. . . (4) 

 dt at 



Multiplying (3) by r sin 6, and (4) by r cos 0, and subtracting, 

 we have 



r 2 — = (a sin t + b cos t)(a' cos £— If sin tf) 

 etc 



— (a'sin^ + £/cos t)(a cos £—6 sin t) ; 



whence 



^=a!b-aV 



dt 



= sin 2/o cos 2a — cos 2p sin 2cr sin 2<£. . (5) 



Again, in (1), (2), (3), (4), let t have a value which makes 



dt* 

 r a maximum or minimum ; then 6 becomes a, and -7- vanishes. 



Eliminating r from (1) and (2), we get 



(a sin a — a' cos a) sin t= — (b sin a — V cos a) cos £; 



and eliminating r -7- from (3) and (4), we get 



(b cos u + V sin a) sin t= (a cos a + a f sin a) cos t. 



Hence 



a sin a— a! cos a 6 sin a— £/cos a _ n 

 6 cos a + V sin a a cos a + a' sin a ~~ ' 



or 



that is, 



sin 2p sin 2<r + cos2p cos 2crsin 2c/> + cos 2p cos 20 tan 2a =0. (6) 



Now r 2 -7- is proportional to the area of the ellipse, as the 



period of vibration is constant ; and the axes of the ellipse are 

 proportional to sin /3 and cos /3, since the intensity of the light, 

 and consequently the sum of the squares of the axes is con- 

 stant over the whole section Hence the area of the ellipse is 

 also proportional to sin /3 . cos 3 (that is, to sin 2/3). Putting 



