:! 



136 Mr. A. W. Ward on the Use of the Biquartz in 



h 2 = c 2 cos 2 y cos 2 co + cj> — 6 + c 2 sin 2 7 sin 2 co + <£ — 6, 1 



. . (3) 



P = c 2 cos 2 7COS 2 co —0 — 6 + c 2 sin 2 7 sin 2 co — <f> — 6- s | 



or 



/i 2 == ^-{1 + cos 2 7 cos 2(a> + 0-0)1 



2 

 #» = | {1 + cos 2 7 cos 2(a>-<£-0)}. 



We have now to determine what value of makes h* = k* 

 for all values of\. 



Equating h 2 to k 2 , we get 



cos 27 sin 2$ sin 2co — = (5) 



This equation is satisfied whenever 



cos 27 = 0, (6) 



or 



sin2$ = Q, (7) 



or 



sin2(a>-0)=O (8) 



* IT 



The first of these solutions occurs when 7= j, *'• 0. when the 



elliptically-polarized light is really circularly polarized. In 

 this case the phrase plane of polarization has no meaning at 

 all, and so it need not be discussed. 



The second solution (7) gives <£= ~-. This can only be the 



case for one particular wave-length, and depends simply on 

 the thickness of the biquartz. A biquartz is usually made of 



TT 



such a thickness that <f> is ~ for the yellow light from the 



brightest part of the spectrum. We shall suppose this to be 

 the case here. 



The third solution gives 



CO = d. 



If, then, this solution does not hold for all values of X, then, 

 however the analysing Nicol be turned, both halves of the 

 biquartz can never be made of the same uniform tint. 

 Now, considering the equation 



tan 2co = tan 2a cos /?, 



we see that co = ot always if y3 = 0, that is, if the incident plane- 

 polarized light always remains so. If/8 is not equal to 0, then 



IT 



still (o = a. for all values of X, if a = or j. If a is 0, then 



