: the eye will see the stick as in the figure the image appearing curved. This effect may be noticed when standing above water of an uniform depth. The bottom appears to slope gradually up to the water as it recedes from the place where we are standing. 4I. We will now give an example which combines the principles of plane reflexion and plane refraction. We shall assume that the image in each refraction is at the Geometrical Focus. Example. A pencil of light passes through a triangular prism of glass, after having suffered one internal reflexion at one face of the glass. To trace its course from the origin. Let ABC be a section of the prism made by the plane of the paper. Let PQ be the object. Then (i) By refraction at AC, PQ will have an image PQ, of itself on the normals to AC, such that mP1 =μ.mP, nQ1=μ.nQ. By Art. 44 it will be so inclined as that it meets PQ produced in a on the surface AC. (ii) After entering the glass the light goes as if it came from PQ, until it falls upon the reflecting surface BC, when, by the principles of plane reflexion, it will have its image at P,Q,, on the normals to BC, where m‚P ̧=m ̧P and n1Q2 = n1Q1 (iii) After striking BC the light goes as if it came from PQ, until it emerges at AB, where by refraction it has an image PQ, such that I 3 MP1 = μ μ 3 As before, PQ, and P,Q, would meet in BA produced. It would be a very good exercise for the reader if he would trace the course of a small pencil of rays from Q an eye placed anywhere in the plane of the paper. He must remember that the Geometrical Focus is taken for the image at the refractions, and that at the refractions he must therefore only consider rays falling nearly at right angles. 42. A coin is placed at the bottom of an empty hemispherical basin of given radius, and is just not visible to an eye looking over the edge. When the basin is filled with water, the whole of the coin is just visible to an eye in the same position. Find the diameter of the coin. Fig. 35. N B Let AB be the diameter of the coin = 2x. Since at first it is just not visible to the eye, therefore the eye lies in AC produced. Since, when the basin is full it is just visible entirely, therefore the ray from B through C is just refracted to the eye. Thus = angle of incidence of BC= BCN, $= angle of emergence of BC= ACN, where μ=index of refraction from water into air; This problem was explained as early as A.D. 140 by Ptolemy, the Alexandrian astronomer. . CHAPTER VIII. Refraction at a spherical surface. 43. To find the Geometrical Focus of a pencil of rays after direct refraction at a concave spherical surface. M Fig. 36. Let Q be the luminous point, and let O be the centre. Then QOA is the axis of the pencil from Q which falls directly on the surface. Of the pencil consider any ray QR which falls on the surface at R; let Rr be its direction after refraction, nearer to the normal OR than before. Let RM be the plane touching the spherical surface at R, (the whole figure being, as usual, a section by the plane of the paper): let QM be a perpendicular to this plane. Let MQ and rR produced backwards meet in S. Since RM is a tangent plane at R, the refraction which |