122 On the Laws of Crystalline 



tor of (63), according as the refracted ray or its wave normal 

 makes the smaller angle with a perpendicular to the face of 

 the crystal. 



To find the polarizing angle, we have only to make m\ = / 2 , 

 for then r x 3 will vanish by (62) ; and therefore, if common light 

 be incident, the whole reflected pencil will be polarized in the 

 plane of incidence. Supposing the crystal to be a negative one, 

 let us conceive the refracted ray to lie within the acute angle 

 made by the axis of the crystal with a perpendicular to its 

 surface. We shall then have to take the positive sign in the 

 numerator of (63), and the polarizing angle will be given by 

 the condition 



sin 2h = sin 2/ 2 + 2 sinY 2 tan t'. (64) 



But from (36) we have, in general, 



sinY 2 tan / = ( 2 - b z ) sin a/ cos a/ sin 2 *!, (65) 



and in the present instance it is evident that 

 (t/ = 90 A t' 2 , 



where \ denotes, as before, the angle which the axis of the 

 crystal makes with its surface. Substituting these values in 

 (64), and multiplying all the terms by tan ' 2 , we get 



sinY 2 = sin ^ cos /i tan i' 2 - (# 2 - b 2 ) sin (A + / 2 ) cos (X + 1 2 ) tan 



Again, from (37) we have 



sinY 2 = b* sin 2 *, + (a 2 - b z ) cos 2 (A + ' 8 ) sin 2 /! ; (66) 

 and by equating these two expressions for sinY 2 , we find 



, a 2 cos 2 A + b 2 sin 2 A 



tant 2 = - - 7-5 : T - r-. (o/j 



cotan 1 1 + (a* - b~) sin A cos A 



Then, if this value of tan t' 2 be substituted in equation (66), after 

 all its terms have been divided by cosY 2 , we shall obtain the 

 simple and rigorous formula 



l- 2 cos 2 A- 2 sin 2 A 



1 = " ~ 2 -- = sm " OT| ' (68) 



