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 MI = w' 2 , 

 for then T' Z 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 e = (a 2 - o 2 ) sin a/ cos a/ sin 2 ii, (65) 



and in the present instance it is evident that 

 u/ = 90 X t' 2 , 



where X 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 t' z , we get 



sinY 2 = sin ^ cos < i tan i' z - (a* b' z ) sin (X + 1' 2 ) cos (X + 1' 2 ) tan t' 2 siu^. 



Again, from (37) we have 



sinY 2 = 6 2 sinY + (a 2 - b~} cos 2 (X + t' 2 ) sin 2 /, ; (66) 

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



, a 2 cos 2 X + & 2 sin 2 X 



tant 2 = 7-5 ,TT : r 7. (o7) 



cotan ti + (a 2 - ) sin X cos X 



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 



1 - 2 cos 2 X - 1* sin 2 X . 



2 = sm " ( ) 



