1 1 8 On the Laws of Crystalline 



The deviation 6 3 or 0' 3 is found from the second of equations 



(38), by putting . - r for tan 0, and by substituting for 

 sin (p 2) 



cos (ti + < 2 ) the value (49) or (50) which it has at the polarizing 

 angle. The result is 



jr 



3 = 0' 3 = - 7r sin 2g sin (p + < 2 ), (56) 



since the small arc 3 may be taken for its tangent. This result 

 is easily transformed into 



3 = 0' 3 = - .ZT sin q cos tf>, (57) 



where denotes the arc -4/, or the angle which the incident ray 

 makes with the axis of the crystal ; and this last expression is 

 equivalent to the following, 



#3 = 0' 3 = - K cos X sin a (sin X cos or + cos X sin OT cos a), (58) 



which gives the deviation in terms of X and a. 



As an example of the application of our formulae, we shall 

 make some computations relative to Iceland spar. According 

 to M. Rudberg, the ordinary index of that crystal, for a ray 

 situated in the brightest part of the spectrum, at the boundary 

 of the orange and yellow, is T66 ; and the least extraordinary 

 index for the same ray is 1'487. Dividing unity by each of 

 these numbers, we get a = -6725, b = '6024 ; whence -us = 58 56' ; 

 k = -1164 - 6 40' ; K= '1587 = 9 5'. Having thus determined 

 the constants, we can readily calculate the polarizing angle and 

 the deviation, for any given values of X and a. 



First, let us see how the polarizing angle varies on different 

 faces of the crystal. 



1. When X = 90, the face of the crystal is perpendicular to 

 its axis, and OTI is independent of a. In this case the formula 

 (55) gives 



which is the maximun value of the polarizing angle. 



