Reflexion and Refraction. 109 



but OG is to GP as the tangent of the angle GPT is to the tan- 

 gent of the angle GOT', and since 6 is the angle GOT, it follows 

 that fi is equal to the angle GPT or KOP. Consequently, OK 

 will meet the surface of the sphere in the point J5T,. Thus we 

 have proved our assertion, that, when there is only one refracted 

 ray, the plane of the transversals is the polar plane of that ray. 



The sign of the quantity h is always the same as that of the 

 cosine of the spherical angle T^P^Y^ But to remove all ambi- 

 guity respecting signs, we must make a few additional conven- 

 tions. Supposing, as we have hitherto done, that the refracted 

 light moves from to T, and conceiving a right line to be 

 drawn from the origin parallel to GT, and directed from G 

 towards T, let the angle 2 , which this right line makes with the 

 plane of incidence, be reckoned, like 0,, 2 , from an initial posi- 

 tion comprised between the negative directions of x and y ; and 

 let S 2 , like the angles Lt 2 , 3 , increase on the side of z positive, 

 and range from to 360. Then Sr z will always be equal either 

 to the angle P / of the spherical triangle Tf 4 F,, or to the re- 

 entrant angle, which is the difference between P 4 and 360. In 

 either case, the cosine of 3" 2 will be the same, both in magnitude 

 and sign, as the cosine of the angle Tf,Y,. Consequently, if, 

 instead of (25), we use the direct trigonometrical formula 



cos t( 2 ) = cos 't z cos e + sin i 2 sin e cos 2 , (31) 



we shall find 



sin 2 i 2 tane 

 ~stfO~~ 



showing that the sign h is always the same as the sign of cos 3v 

 Now as 2 differs from S 2 by a right angle, we will suppose 



2 = S 2 + 90, (33) 



and then we shall have sin 2 = cos S 2 , algebraically as well as 

 numerically. Thus we see that, by adopting these conventions, 

 the value of h in (27) will have the proper sign. Therefore, 

 substituting this value of h in formulae (13), we obtain 



