io8 On the Laivs of Crystalline 



shall have F / P / = <,, Y,T, = ( < 2) TP, = , T / Z / = v = t 2 + *, by 

 (23); whence P,,= K . 



As the transversal r 2 is perpendicular to the plane OTP, or to 

 the plane of the great circle Tf^ the cosine of the spherical 

 angle Tf,Y f is the sine of 2 ; and therefore, from the triangle 

 Tf f Yfi we have 



cos 1(2) = cos ( 2 cos + sin 2 sin sin 2? (25) 



which being substituted in (24), gives 



nh sin 2t 2 + 2 sin 2 < 2 sin 2 tan e 

 #h sin 2ti 



and comparing this result with (10), we find 



sin 2 * 2 tan 



(26) 



h = ^ ; (27) 



sm0 2 



whence, and from (20), it follows that 



sin 2 < 2 tan e 



tan K = J-T-Z - ~ 2 \ /i- 28 



2 



Draw the great circle Z^ff^ at right angles to Tf^ and meet- 

 ing it in K, ; then the plane of L,K, will be the plane of the 

 transversals, since the latter plane passes through L^ and is per- 

 pendicular to Tf r But the tangent of P,E, is equal to the 

 tangent of P^L, multiplied by the cosine of the angle P / or by 

 the sine of 2 ; therefore, denoting P,K, by 1, and recollecting 

 that P, = fc, we find, by (28), 



tan sin 2 ii - sm 2 t 2 



(29 ) 



Now we have seen that the ratio of OP to OS, or OS to OG 

 (Fig. 17), is the index of refraction ; so that shrt, is to sin 2 t 2 as 

 OP to OG. Therefore, by (29), 



tan fl OG OG 



tan OP-OG 6P' 



(-30) 



