Glass under the Polariscope* 43 



The symbol K(±a) will be used to include K(«) ; K(«— 90°), 

 K( — a), and K(90° — a); its equation is 



(sin 2p sin 2a + cos 2p cos 2a sin 20) 2 = cos 2 2/? cos 2 2(f) tan 2 2a. (8) 



And the symbol M(±j8) will be used to include M(/3) and 

 M(— j8); its equation is 



(sin 2p cos 2cr— cos 2/o sin 2a sin 20) 2 = sin 2 2/3. . (9) 



Putting 90° — 0, or — 90°— 0, for in these equations 

 makes no change in the equations. Hence K(±«) and 

 M(±/3) are each symmetrical with respect to UU 7 and 



w. 



Adding (8) and (9) ? we get 



cos 2 20. cos 2 2p= cos 2 2«. cos 2 2/3, . . . (10) 



from which equation it appears that the intersections of 

 K(±a) with M(±6) all lie on the two diameters defined by 

 the equation ; and since a and (3 are interchangeable in the 

 equation, we see that K(±y) and M( + S) intersect on the 

 same two diameters at K(±8) and M(±y). 

 Let 



then K(±a) becomes 



(sin 2/3 cos 2§— cos 2p sin 2$ sin 20) 2 = cos 2 20 cos 2 2p tan 2 2«, 



and M(±/3) becomes 



(sin 2p sin 2$ 4- cos 2p cos 2d sin 20) 2 = sin 2 28. 



From these equations it appears that, in changing the sign 

 of 0, we shall change only the sign and not the magnitude of 

 £ ; so that if we have drawn part of K(±a) or M(±«) be- 

 tween the limits = and = 45°, we can draw the corre- 

 sponding part of the curve between the limits 0=0 and 

 0= —45°. If a' be the value of a corresponding to — ; 



and therefore 



< 7 + o- / =(n+ ^\ir (11) 



Putting a=0 and |3 = in (8) and (9), we get for K(0), 



sin 2p sin 2cr + cos 2p cos 2a sin 20 = ; . . (12) 



and for M(0), 



sin 2/3 cos 2a— cos 2p sin 2a sin 20 = 0. . . (13) 

 Let s be the value of a where a radius intersects a branch 



