7^ SOLUTIONS OF PROBLEMS. 



To find the law of the index of refraction of the medium, so that a ray 

 from A may describe the circle APB, /* must be made to vary inversely as p 

 by Lemma I. 



Let AO = r,, and let the refractive index at A=p^; then generally 



_C _ 2Cp 

 ^^"oM^ 5 



2Cp 

 but at A /*! = 



therefore P- = P L \ ~TT 3 



a +? 



The value of p at any point is therefore independent of p, the radius of 

 the given circle ; so that the same law of refractive index will cause any other 

 ray to describe another circle, for which the value of a' is the same. The 



value of OB is , which is also independent of p ; so that every ray which 

 proceeds from A must pass through B. 



Again, if we assume /A, as the value of p. when r = 0, 



therefore = . 



a result independent of r,. This shews that any point A' may be taken as 

 the origin of the ray instead of A, and that the path of the ray will still be 

 circular, and will pass through another point R on the other side of 0, such that 



OR= a 



Next, let CP be a ray from C, a point without the medium, falling at P 

 on a spherical surface whose centre is C. 



Let be the fixed point in the medium as before. Join PO, and produce 



a* 

 to Q till OQ = 7jp- Through Q draw a circle touching CP in P, and cutting 



CO in A and B ; then PBQ is the path of the ray within the medium. 



