Sir: I do not think there is much difficulty in the rainbow business. We cannot see the reflection of the same rainbow which we behold in the sky, but we see the reflection of another invisible one within it. Suppose A and B, [Fig. 1], are two falling raindrops, and the spectator is at S, and X Y is the water surface. If R A S be a sun ray giving, we will say, the red ray in the visible rainbow, the ray, B C S, will give the same red ray, reflected from the water at C.
It is rather a long business to examine the lateral angles, and I have not time to do it; but I presume the result would be, that if a m b, [Fig. 2], be the visible rainbow, and X Y the water horizon, the reflection will be the dotted line c e d, reflecting, that is to say, the invisible bow, c n d; thus, the terminations of the arcs of the visible and reflected bows do not coincide.