58 BELL SYSTEM TECHNICAL JOURNAL 



a.s B = nb — fXa is the birefringence we have: 



B = 



2 2 

 Ma Mb 



Ma + M6 



By spherical trigonometry: 



{Ki ^ — Ks^) sin gi sin g2 



cos gi — cos F cos ^ + sin V sin 9 cos <f) 



cos g2 = cos V cos — sin V sin cos <^ 



where ^ is the angle the wave normal makes with Xs and 4> is the an] 

 plane containing the normal and x^ makes with xi. 



From (19.5) it follows that: 

 sin gi sin g2 = 



/(I - cos2 F cos2 d - sin2 F sin^ 6 cos^ 0)2 

 r — 4 sin F cos F sin" ^ cos" 6 cos" 



Hence if the rediagonalized K~^ is 



(19.4) 



.(19.5) 

 de the 



(19.6) 



2 2 

 MaM6 



Ma 



M6 



'(ii:r'° - i^'° + Ai - As) - (K2'° - i^'° + A2 - A3) cos' d 



- {KT'° - Kt° + Ai - A2) sin' 5 cos' </>' - 4(Ar'° - A^'° 



+ Ai - A2)(AT'' - Kt° + A2 - A3) sin' ^ cos' 6 cos' . . (19.7) 

 3 

 5 = ^0 + y (A1A3) sin gi sin g2 (3') 



For most practical purposes we may take 



22 3 



MaM6 _ M_ 

 Ma + M6 2 



where /i is some intermediate value of the refractive index.* 



2 

 * Note: It might seem that as A'71 = A'-i" + Ai gives us in = fxl - ^ Ai + f/iiA? • 



we could form the 3 principal birefringences directlj- from the m's instead of using (6a • ■ • t 



from 6i if JU2 ?^ Mi • Equation 66 is correct; the one from the /u expression is an approxi- 

 mation. 



