Crystalline Reflexion and Refraction. 179 



have, by Lemma I., 



cos a 2 = sin 4 sin E + sin 2 cos i 2 cos E , cos /3 2 = - cos 2 cos e, 



cos 72 = cos 4 sin t - sin 2 sin 4 cos e. (35) 



In like manner, putting i for the angle which the other re- 

 fracted ray makes with its wave-normal, we have 



cos a' 2 = sin i\ sin E' + sin S f z cos e' 2 cos e', cos j3' 2 = - cos 0' 2 cos e', 

 cos 7' 2 = cos *' 2 sin E' - sin 0' 2 sin ' 2 cos E'. (36) 



If we substitute, in the first of the equations (31), the values 

 just given for cos a 2 , cos a' 2 , along with the above values of 

 cos en, cos a'i, and attend to the relations 



r cos E = s. r' cos e' = s', 



(37) 

 sin 4 = s sin 4, sin e' 2 = s' sin 4, 



we find, after multiplying all the terms of the equation by 

 sin 4, 



(ri sin 0j - T'I sin ^i) sin i^ cos \ = r 2 (sin 2 sin 4 cos i z 



(38) 

 + sin 2 e' z tan E) + r^ (sin & z sin e' 2 cos % + sin 2 " 2 tan E). 



Joining this equation to the equations (34) we have all the con- 

 ditions that are necessary for the solution of the question. 



Multiplying the first of the equations (34) by the third, also 

 the second of these equations by the equation (38), adding the 

 products together, and then dividing by sin H we obtain 



fii (r! 2 - r?) = /W + /u' 2 r? + ^fr 2 r / 2 , (39) 



where we have put 



/MI = cos e'j, /u 2 = s (cos 4 + sin 2 sin ? 2 tan E), 



fjL-i = s' (cos i\ + sin & z sin i' z tan E'), 

 M sin 4 = sin (4 + *' 2 ) {cos 2 cos 0' 2 + sin 2 sin 0' z cos (4 - 

 + sin & t sin 2 4 tan E + sin 2 sin 2 'j tan c'. 



