250 Mr. lx. T. Glazebrook on Nicole Prism. 



and the two lines A C and B D. P Q is parallel to the trace 

 oii the incident wave on ABC D. Let the equation to this 

 wave be 



Lv + my + nz — 0. 

 Equation to trace is 



my + nz — 0. 

 Equation to B D is 



*-o. 



Hence 



tan\|r= 



1 m 



Now, when the wave-normal is parallel to AE, since the edge 

 of the rhomb is inclined at about 20° to the normal to the face 

 of incidence, 



n = Q, I— cos 20°, m — cos 70°. 



Consider now a wave-normal in a plane through this and 

 the axis of z inclined at, say, 5° to this, m will not be much 

 altered, and the value of n will be about cos 85°. From these 

 values we find 



^ = 14° about (4) 



Also, taking the value of <f> as about 20°, we have for mean 

 rays 



<// = 12° about (5) 



From this we find 



6 f = ll°W (6) 



Also 



<£-(// = 8° about. 



Thus, substituting in tan x = sin 6' tan (<£ — <//), 



X=l°39' (7) 



Thus the planes of polarization of the light in the prism and 

 the emergent light are inclined at an angle of 1° 39 / instead 

 of being coincident. 



Again, let us find the angle between the plane of polariza- 

 tion of the emergent light and that principal plane of the prism 

 which is perpendicular to BD. 



Let OV be the direction of vibration of the emergent light. 

 Then NV'V is a right angle, and LV is the angle required. 



NV = 0, LNV = <£. 

 Let LV = <r, 



cos LV = cos NV cos NL + sin NV sin NL cos LN V, 



cos <r= cos 6 cos -v/r + sin 6 sin yjr cos cj) } . . (8) 

 . a tnn 6' 



cos (<£ — <//) 



