This equation shows that the position of the final image coincides with the position of the object. We shall next show that the final image has the same magnitude as the object. Let x, x', x", x'", x'""' be the linear magnitudes of the object and the images in succession. Then It follows by Helmholtz' theorem, or by elementary geometry, that the divergence of the emergent pencil is the same as that of the incident pencil, and therefore any ray passing through the lenses emerges in the same straight line as before incidence. 2. If an eye be supposed to consist of a sphere of fluid (radius r, refractive index 13), in which is placed at a distance r from the centre a convex lens whose axis coincides with the diameter and whose focal length and refractive index in air are, respectively, fr and; show that the distance from the centre of the sphere for clear vision is 3r. 503 3. From a cubic inch of glass, the inscribed sphere is removed, a film of glass remaining at the points of contact. The cavity is filled with water. A bright point is placed on the axis at a distance of one inch from one face of the cube. Prove that the conjugate focus is at the point of the cube nearest to the luminous point. 4. Prove that the magnifying power of a thin double-convex lens, the radius of each surface being p, when the space between the lens and an object at distance a is filled with fluid of index μ', is given by 5. If ƒ be the focal length of a double-convex lens, show that the smallest distance between an object and its image is 4f. 6. Two thin lenses of equal numerical focal length ƒ are placed on the same axis at a distance a apart, the one nearest the origin of light being concave and the other convex; show that the least distance between an object and its final image is a +4f2/a. 7. Two lenses of equal focal length ƒ are placed so as to be on a common axis at a distance 2e from one another, and midway between them is placed a glass sphere of radius r and index μ. A thin pencil diverges from a point distant c behind the first lens, and after refraction through it, through the sphere and the second lens, converges to a point at a distance c behind the second lens ; prove that μη μ-1 ec+of+fe 8. A pencil of rays is directly refracted through a series of thin lenses separated by finite intervals a1, a2... an-1, the axes being coincident. Show that if the focal lengths of the lenses (considered as concaves in the typical case) be 1/k1, 1/k2 ... 1/kn, the abscissæ of a pair of conjugate points reckoned from the first and final lens, respectively, are connected by the equation 9. If m, m', m" be the magnifying powers of a combination of any number of lenses on the same axis for objects at distances u, u', u' from the first lens, show that 10. If x be the distance between two objects and x' the distance between the corresponding images due to any system of lenses, and if m be the magnification of the first image and n that of the where μ and are the refracting indices of the initial and final media. 11. Three lenses A, B, C (of which A and C are double-concave and B is double-convex) are mounted on an axis in the order named, so that the foci of A and C coincide at the centre of B. An object beyond C is viewed through the system by an eye behind A; show that the distance through which it would have to be displaced in order that, when viewed directly, it may have the same apparent magnitude as when viewed through the system is independent of the position of both the eye and the object, if the focal lengths of the lenses are connected by the relation If 12. A thin lens has one face silvered so as to form a mirror. be the image of a point P, formed by the mirror (by two refractions and one reflexion), show that will be the same as if the lens were replaced by a spherical mirror whose radius R is given by the equation r and s being the radii of the surfaces of the lens. 13. Show that the image of an arc of a conic whose focus is at one principal point of a thick lens, is an arc of a conic whose focus is at the other. 14. A double-convex lens is formed by two equal paraboloidal surfaces cut off by planes through the focus perpendicular to the axis. Prove that for rays passing in the neighbourhood of the axis, the focal length measured from the posterior surface of the lens is 2a/(2-1), and the distance between a bright point and its image is a minimum when it is 2a/(μ+1) (μ −1), 4a being the latus rectum of either of the generating parabolas, and μ the refractive index of the glass. 15. A system of 2n thin convex lenses of equal numerical focal length, f, are placed with their axes in the same straight line, and their centres at a distance 4f apart, except the two middle ones, which are at a distance 8f apart. Show that the focal length of a lens which must be placed midway between the two middle ones in order that the image of a bright point at a distance 4f in front of the first lens may be formed at an equal distance behind the last CHAPTER V. GENERAL THEOREMS. CAUSTICS. 74. IF a ray of light pass from a point A to another point B, through any number of media, undergoing any number of reflexions and refractions, then the actual laws of reflexion and refraction are such as to make Σ (up) a minimum, where p represents the length of the path of the ray situated in the medium whose refractive index is μ. Conversely if we assume the path of light to be such as to make Zup a minimum, we are led to the actual laws of reflexion and refraction. The expression Σup is frequently called the reduced path. We shall first prove this general theorem for a single reflexion and a single refraction, and afterwards extend it to any number of reflexions and refractions. B Let APB be the path of a ray of light which travels in a homogeneous medium from a point A to a point B, undergoing one reflexion at a surface CD; then the total path between A and B is a minimum, that is, AP + PB is less along the actual path than along any consecutive path as AQB. P For a variation of P perpendicular to the plane APB, this proposition is clearly true. Let AQB be a consecutive path in the plane APB. Then the difference AQ-AP is equal to the projection of PQ on AP; and similarly the difference BP – BQ is equal to the projection of PQ on PB. But these projections are equal, because AP and PB are equally inclined to PQ. Thus, AQ+QB (AP+ PB), which proves that the increment of the total path vanishes, and therefore the total path is a minimum. = A similar theorem holds if we take the path from A to B, supposing the ray to suffer a refraction at a surface CD. Let μ, μ be the refractive indices of the two media, then μAP + 'PB is a minimum for the actual path. N ୪ B Let Draw the normal PN, and let the angles of incidence and refraction be , '; then μ sin pμ sin p'. AQB be a consecutive path; it will be sufficient to take the case when Q is in the plane APB. and μ'BP — μ'BQ = μ'PQ sin p'. Hence the whole variation, μAQ + μ'BQ — (μAP + μ'BP) |