Reflexion and Refraction. 1 2 1 



ginning at the azimuth 0, the points t and A will lie on the same 

 side of XY until A reaches the position A', where the angle 

 ATi is equal to 127 38' ; the point t will then pass over to the 

 side opposite A y at which side it will remain until A arrives at 

 the azimuth 232 22'. Thenceforward, to the end of the revo- 

 lution, both points will be found on the same side of the circle 

 XY. 



We have seen that the deviation always vanishes when the 

 axis of the crystal lies in the plane of incidence. The reason is, 

 because the crystal is then symmetrical on opposite sides of that 

 plane. In this case the problem of reflexion offers peculiar faci- 

 lities for solution, since the uniradial directions are obviously 

 parallel and perpendicular to the plane of incidence. Let us, 

 therefore, consider the case at length. 



1. In the first place, when the only refracted ray is the 

 ordinary one, the three transversals are in the plane of incidence, 

 and the transversal of each ray is proportional to the sine of the 

 angle between the other two rays. Hence the proportions are 



Tl = Tz = T * _ (60) 



sin (ii + 1 2 ) sin 2ii sin (^ - 1 2 )' 



the same as in ordinary media. 



2. In the second place, when the sole refracted ray is the 

 extraordinary one, the three transversals are perpendicular to 

 the plane of incidence ; and, if we use accents to mark the quan- 

 tities connected with this ray, we have the equations 



T+miT, 



which give the proportions 



r / r 



r * = _L 2 _ = Tz (62) 



mi + m'z 2nii nh - m'J 

 wherein 



m\ _ sin 2/2 2 sin Y 2 tan e' ( . 



mi sin 2ii 



by (26) ; the upper or lower sign being taken, in the numera- 



