io8 On the Laws of Crystalline 



shall have F,P, = i 2 , Y,T, = ( < 2) T,P, = e, F,Z, = u = i, + K, by 

 (23) ; whence P,Z, = K . 



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

 the plane of the great circle T / P / , the cosine of the spherical 

 angle 7 7 / P / J r / is the sine of 2 ; .and therefore, from the triangle 

 F,, we have 



cos (2) = cos i 2 cos e + sin t 2 sin e sin 2 , (25) 



which being substituted in (24), gives 



m* _ sin 2t 3 + 2 sin 2 < 2 sin 2 tan 

 Wi sin 2ti 



and comparing this result with (10), we find 



(26) 



sin'ia tan e 



h = r a ; (27) 



sin 02 



whence, and from (20), it follows that 



sin 2 < 2 tan c 



tan K = -^ 2 - . . -. (28) 



(sin ii - Bin t 3 ) sin 2 



Draw the great circle L,K, at right angles to Tf,, and meet- 

 ing it in K 4 ; 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 T,P r But the tangent of P / -ff" / 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 / jff" / by EI, and recollecting 

 that P^L t = K, we find, by (28), 



tane 1= sin 2 ta 

 tane sin'ti - sin'ta* 



Now we have seen that the ratio of OP to 08, or OS to OG 

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

 OP to OG. Therefore, by (29), 



OG OG 



tan t OP-OG GP J 



