28 
Dr. Young’s Lecture 
be as m : n, and as m 1 :n'; then the ratio sought will be that of 
m n l : m 1 n. For instance, let the three mediums be glass, water, 
and air ; then m — 3, n = 2, m 1 = 4, n' = 3, mn l = g, and 
m 1 n = 8. If the ratios be 4 : 3, and 13 : 14, we have mn' :m' n 
: • 39 • 56; and, dividing by 56 — 39, we obtain 2 . 3 and 3 . 3 
for m and m + 1, in Schol. 1, Prop. I. 
Proposition III. Problem. (Plate II. Fig. 1.) 
At the vertex of a given triangle (CBA), to place a given re- 
fracting surface (B), so that the incident and refracted rays may 
coincide with the sides of the triangle (AB and BC.) 
Let the sides be called d and e ; then in the base take, next to 
d(or AB), a portion (AE) equalto or (AD =) 
draw a line (EB, or DB) to the vertex, and the surface must be 
perpendicular to this line, whenever the problem is physically 
possible. When e becomes infinite, and parallel to the base, take 
— or — next to d, for the intersection of the radius of curvature. 
m n 
Proposition IV. Theorem. (Fig. 2.) 
In oblique refractions at spherical surfaces, the line (A I, KL,j 
joining the conjugate foci ( A, I ; K, L;) passes through the point 
(G), where a perpendicular from the centre (H) falls on the 
line (EF), bisecting the chords (BC, BD,) cut off from the in- 
cident and refracted rays. 
Corollary 1. Let t and 11 be the cosines of incidence and re- 
fraction, the radius being i,and d and e the respective distances 
of the foci of incident and refracted rays ; then <?= — md .l ~ — 
J mdu —ndt —ntt 
Corollary 2. For a plane surface,*? = 
