356 Mr. R. T. Glazebrook on 



A, ON, and in it draw Q at right angles to N ; Q is 

 the direction of displacement which travels along N, and 

 Q is not parallel to the plane POM. 



We can resolve the displacement along Q into two, in and 

 perpendicular to the plane POM. The light then will be 

 most nearly plane-polarized when the average intensity of the 

 vibrations normal to this plane is least ; and it remains to find 

 the condition for this. 



In fig. 3 let QR be perpendicular to AP. Let AM=«, 

 NM=/3, AN=0, AMN=</>. Let p be the amplitude of the 

 displacement along OQ. The displacement normal to the 

 plane PAM is p sin QR, and the intensity of the wave is pro- 

 portional to p 2 sin 2 QR. 



We are to consider a hollow conical pencil with OM as axis. 

 An element of such a pencil at N will be sin /3d<j) ; and the 

 total energy in the pencil, so far as it depends on the dis- 

 placement normal to the plane PAM, is 



Now 



j/ si 



sin /3 sin 2 QR dcf>. 



sin QK= sin AQ sin RAQ 



= cos sin NAM, 



sin0 



.*. sin QR= cot 0sin</>sin/3 (1) 



Also cos = cos a cos /S + sin a sin /3 cos <£. . . (2) 



Substituting in the value of sin QR, we find for the energy 

 required the expression 



2 / o 2 sin 3 yS 



r^sn^^cosacos/S-f sin a sin /3 cos (f>) 2 d<f> ,„x 



J 1 — (cos a cos /3+ sin a sin /3 cos (/>) 2 



And we require to evaluate this integral. 

 Let 



cos a cos /3= a, sin a sin /3= b 

 Then 



sin 2 <f> (cos a cos /3 + sin a sin /3 cos <j>) 2 d<f> 

 1 — (cos a cos /3 + sin a sin /3 cos (p) 2 

 _ C * sin 2 ft(g + b cos (f>fd(f> 

 J. 1 



1' 



(a + b cos <pf 



