Glass under the Polariscope. 45 



When /3= ±45°, the light becomes circularly polarized, and 

 therefore the value of a becomes indefinite ; consequently in 

 equation (6) we must have cos 2<£ = 0. Putting this value of 

 cos 2</> in (7), we get the following values of <j>, <r, and j3 at 

 points where the polarization is circular: — 



0. o-. /ST. 



+ 45°, nir + 45° + /?, -45° "| 



+ 45°, nir + 135° + p, +45° t (n) 



-45°, mr+ 45°-/?, +45° | 

 -45°, mr+lS5°-p, -45°. J 



These points will be called the " circular points." All the 

 isoclinals pass through all the circular points. The sign of 6 

 indicates the direction of rotation of the aether. Where the 

 sign is positive, the direction of rotation has not been altered 

 by the passage of the light through the body ; where the sign 

 is negative, the direction of rotation has been reversed. 



To draw the isomorphals : — 



Mark the circular points by (21). Draw the circles of 

 M(/o) and M( — p) by (19) and (20). Draw one branch of 

 M(0) from = to 0=45° by (13). 



Obtain a branch of K(0) by (17). 



Obtain an oval of M(p) between the above limits by (14). 



Draw the part of a branch of M(/3), for example, M(XV.) 

 in fig. 3,. or M(XXX.) in fig. 4, which lies on one side of 

 M(0), and complete on the other side by (14). 



Draw the remaining branches of M(±/3) between the above 

 limits by (18). 



Draw the isomorphals between <j> = and 0= — 45° by (11), 

 and complete the figure by means of the symmetry about 

 UF and V V'. 



Write against the isomorphals the values of /3, taking care 

 to make the sign of /3 the same as that at the nearest circular 

 point. 



To draw the isoclinals :< — 



Equations (6) and (8) are not in a form available for cal- 

 culation ; but by solving (8) as a quadratic in sin 2<£, we 

 obtain the equation to K(±a) in the form 



sin 2/o cos 2cr sin 2cr cos 2 2a + cos 2p(l — sin 2 2cr cos 2 2a)sin 2<j> 

 = ± sin 2a (cos 2 2/? — sin 2 2<r cos 2 2a) 2, . (22) 



from which, by putting successive values for <r, we can obtain 

 corresponding value of c/>. 



By putting et=p, we obtain the equation to ~K(±p), viz. 

 sin 2/> cos 2<r— cos 2p sin 2a gin 2(p= ± sin 2<£. . (23) 



