Crystalline Reflexion and Refraction. 179 



have, by Lemma I., 



cos a 2 = sin & 2 sin + sin 2 cos 2 cos c, cos j3 2 = - cos 2 cos c, 



cos y 2 = cos 2 sin * - sin 2 sin i z cos f. (35) 



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

 fracted ray makes with its wave-normal, we have 



cos a' 2 = sin *" 2 sin c' + sin 0' 2 cos i\ cos c', cos /3' 2 = - cos 0' 2 cos E ', 

 cos 7' 2 = cos i\ sin t' sin 0' 2 sin ' 2 cos c'. (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 ai, cos o'i, and attend to the relations 



T COS = S. / COS e = /, 



(37) 

 sin 4 = * sin e'i, sin i\ = s' sin i iy 



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

 sin t\, 



(TI sin 0i - r'i sin 0'i) sin t'i cos e\ = r 2 (sin 2 sin ea cos 2 



(38) 

 + sin^ 2 tan a) + r' 2 (sin ^ 2 sin i' t cos / 2 + 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 i^ we obtain 



^ (n 2 - r?) = JI,T,' + /,rV + Mr z r^ (39) 



where we have put 



/ui = cos e\, fji z = s (cos 4 + sin 2 sin h tan ), 



ju r 2 = s' (cos e' z + sin 0' 2 sin ^ 2 tan '), 



M sin i\ = sin (i z + ?*) {cos 2 cos 0' 2 + sin 2 sin 0' 2 cos ( 2 - i\) } 

 + sin ^jj sin 2 i t tan e + sin 2 sin 2 i\ tan s'. 



