62 Mr. Mac Cullagh on the Laws of 



sin2<i=sin2«'2+ SsinV^tans'. (64) 



But from (36) we have, In general, 



sm*4tan e'=(a^—b'') sin w' cos w' sin't,, (65) 



and in the present instance it is evident that 



a,'=90°-X-4, 

 vF^ere \ denotes, as before, the angle vphich the axis of the crystal makes with its 

 surface. Substituting these values in (64), and multiplying all the terms by 

 tan/j, we get 



sin^t'jzn sin t,cos 4,tan i'^ — (a^—¥) sin (\-{- 4) cos (X+t's) tan t'^sin^t,. 

 Again, from (37) we have 



sinV,=&^sin^«.4-(a*-60cos^(^+''2)sin%; HoQ) 



and by equating these two expressions for sinVj, we find 



Then if this value of tan ^ be substituted in equation {&Q), after all its terms 

 have been divided by cos^^t'^' we shall obtain the simple and rigorous formula 



. „ 1 — a'^cos^X— i^sin^X . , .^„v /^ . 



sm\ = iZTtfl? =sinV„ (68) U^i/UL 



for determining the polarising angle ■a^, when the axis of the crystal lies in the 

 plane of incidence. It is manifest, from the nature of the formula, that this 

 angle is the same, whether the azimuth is or 180° ; that is, whether the light is 

 incident at the right or left side of the perpendicular to the surface of the 

 crystal. 



This formula might be deduced more briefly by recollecting what we have 

 already proved, that the corresponding masses m^ and m!^ are proportional to the 

 ordinates y of the points where the incident ray and the extraordinary refracted 

 ray meet their respective wave surfaces ; whence it follows, that these ordinates 

 must be equal at the polarising angle ; and thus the question is reduced at once 

 to a geometrical problem. For as both rays are in the plane of incidence, the 

 axis of X will be intersected in one and the same point by right lines touching 

 the wave surfaces, or their sections, at the extremities of the ordinates. Now the 



