112 



Mr. E. A. Milne 



clear from the symmetry of the ascending and descending 

 portions that neighbouring rays cannot intersect unless f 

 increases as t decreases, i. e. unless rff/tff is negative. As 

 in §6, the ray leaving at 6 = l is totally reflected at a 

 level f, given by 



W(t) + o(f) = W 1 + « 1 Bec^; 



selecting the smallest value of f satisfying this, and putting 



we have from (33) or (40) 



dz 



? = (*«) 



i 



(»(&-«.(«))* 



(41) 



To differentiate this, put z = ty> we nR d on returning to the 



variable 2 



^1 



i=(i«)*r 



b Jo 



xr 



i?»' (?)}-{«(*) 



{•<0—(*)}i 



i* 



(42) 



The sign of dg/d£ thus depends on the precise behaviour of 

 the function <o(z). We can illustrate the essentially different 

 cases by considering the example 



where, of course, to give total reflexion at all, 6 X must he 

 positive. If n = l, djjjdZ is positive ; if n = 2, dg/dg is 

 identically zero, and the locus of vertices is a vertical 

 straight line about which the whole family of rays is 

 symmetrical; if n>2, df-/d% is negative. In general, the 

 behaviour of the curve of vertices near f=0 can be judged 

 from the following table, which gives the limits of f and 

 rff/df as f->0. The values n=l, n—2, n>2 in the above 

 example illustrate the three rows respectively. 



Conditions. 



Lim |. 



Lim d%jd{. 



o)'(0)^0 



G ) , (0) = 0, o)"(0)gfc0 



a>'(0) = 0, o>"(0) = 







7r / a \* 



laV'(O) 

 3(g>"(0))I 



-co 



2WW 

 -hoc • 



