SOLUTIONS OF PROBLEMS. 77 



dicular from the fixed point on the tangent to the path of the ray at any 

 point will vary inversely as the refractive index of the' medium at that point. 



We may easily prove that when a ray of light passes through a spherical 

 surface, separating a medium whose refractive index is ^ from another where 

 it is p. 2 , the plane of incidence and refraction passes through the centre of 

 the sphere, and the perpendiculars on the direction of the ray before and after 

 refraction are in the ratio of p, to p. r Since this is true of any number of 

 spherical shells of different refractive powers, it is also true when the index of 

 refraction varies continuously from one shell to another, and therefore the 

 proposition is true. 



LEMMA II. If from any fixed point in the plane of a circle, a perpen- 

 dicular be drawn to the tangent at any point of the circumference, the rectangle 

 contained by this perpendicular and the diameter of the circle is equal to the 

 square of the line joining the point of contact with the fixed point, together 

 with the rectangle contained by the segments of any chord through the fixed 

 point. 



Let APS be the circle, the fixed point ; then 



OY.PR=OP*+AO.OB. 



Produce PO to- Q, and join QR, then the triangles YP, PQR are similar ; 



therefore 



OY.PR=OP.PQ 



.: OY.PR=OP* + AO. 

 If we put in this expression AO . 0-8 = a 2 , 



P0 = r, OY=p, 

 it becomes 2pp = r^ + a\ 



