502 
SOLUTIONS OF A SMITH’S PRIZE PAPER FOR 1871. 
[537 
4. Show that the caustic by refraction for ‘parallel rays of a circle, radius c, index 
of refraction y, is the same curve as the caustic by refraction for parallel rays of the 
concentric circle, radius — imd.p.nr. nf rp/frnrt.imr - 
Take as usual y > 1. Imagine the ray AP (fig. 1) parallel to the axis of x, 
incident at P on the circle radius c, and let the refracted ray after cutting the circle 
radius -, cut it again in Q, and then cut the axis in R. Take <f>, <f>' for the angles 
№ 
of incidence and refraction ; sin cf> = y sin f. 
Fig. l. 
y 
K 
x 
c 
Moreover in the triangle PQO, we have sin Q : sin P = c : - ; that is, sin Q = y sin P, 
= y sin </>', = sin (j); or zQ = <j>. And then in the triangle RQO, Z R = $ — ft, Z Q = 180° — </>, 
whence ZO = <£', that is, Z QOR = <f>'. 
Consider now a ray BQ incident at Q and refracted in the direction QR; the 
index of refraction being -, that is, the denser medium being on the outside of the 
y 
small circle. Taking 6, 6' for the angles of incidence and refraction, we have sin 6 = - sin 6'\ 
but, the refracted ray being by hypothesis QR, we have by what precedes 6' = (f>, hence 
sin 6 = - sin <f) = sin </>', that is, 0 = <//, or zBQO = Z QOR, that is, the incident ray BQ 
A 6 
is parallel to the axis of x. We have thus two pencils of rays each parallel to the 
axis, such that for any ray AP of the first pencil there is a corresponding ray BQ of 
the second pencil, the rays AP and BQ each giving rise to the same refracted ray 
PQR ; hence the two pencils have the same caustic. 
[It is proper to remark that for the ray BQ it has been assumed that the 
refraction takes place not at Q' where it first meets the small circle, but at Q; if