angle it subtends—which call ß; and then denoting by P, Q the angles of incidence and refraction, we have B 1 777. If F, F be the focal lengths of the simple lenses, and that of the compound lens; then (Coddington, p. 68) 1 + F F But by the question F - F, and (Coddington, p. 62) - (m ) 26m – 1) F when r = per 4(m – 1). 778. The apparerit image is the caustic formed by the intersections of the refracted rays proceeding from the eye as a radiant point to the linear surface of the water, which is in the same plane with the given straight line. Let x, y be the co-ordinates of this caustic measured along the I upon the surface from the eye, and originating in the intersection of the I and surface; also, let , o' be the inclinations of any incident and refracted ray to the I, and A the distance of the eye from the surface; then (Coddington, p. 81) x tan. A tan. 0 + y = 0 do' de 0 cos. and sin. 0 = m sin. e'. cos.' A' 23) ✓ y=(of mi m3 ma 1 the equation to the caustic, which being traced by the known rules for describing curves in general, the problem will be completely resolved. 779. Let 2t be the thickness of the lens, 26 its breadth, and , the focal length; then, since the focal length is equal to the radius (Wood, Art. 171), by the nature of the circle we have 6% = 2r t { But t being very small, t = 0 nearly :.6 r X 2t. Q. E. D. 780. If , be the distance of the focus of incident, and " m mr that of the emergent rays from the centre of the sphere; then (Coddington, p. 66) 1 1 9 mra 2(m – 1) a by the question. Hence 2gn 9(2n + 1) - nr the ratio required. mr m = 781. If A' be the real and apparent I depths of the object; then (Coddington, p. 45) A'EA X m m being the given ratio of refraction. Hence, having found the real situation of the object, let a I be let fall from the situation fired from upon the line joining the object and image, and call this I p. Also, let the distance of this I from the surface of the water be d, and ß denote the inclination of the required direction to the surface of the water ; then pid+A:: 1 : tan. O d + ma' .. tan. A р which gives the required. 782. If a circle be described passing through the two extremities of the flag-staff and touching the horizon, the 2 subtended by the flag-staff at the horizon will be the greatest possible; because all others must fall without the circumference of this circle. 783. Let a be the given distance of the object from the reflector, b that of the eye, and x, y, the co-ordinates of the required locus measured from the extremity of the rectilinear locus of the object; then, from the conditions of the problem, and the nature of reflection, it is easily seen that or the locus is an ellipse whose semi-axes are b and a. 784. Let a, b, c, be the length of the three objects, each subtending at the eye the same 20. Also, let x be the first side of the angle subtended by a, and o its opposite 2; then we have a : x :: sin. 0 : sin. which give Hence, by expanding the sines of the multiple arcs, we get cot. p = ba b ta cot. A X btc 4ac-(b+c).(a+6) 6 a (ca —6) sin. sin. 62 785. Describe a circle passing through the two extremities of the given object, and touching the circle (see p. 7, vol. 1.); the point of contact gives the maximum or minimum, according as the circles touch with their convexities or concavities. 786. The focal length is (Coddington, p. 63) . m-1 But m = 1.5. 787. The image of the ring will be equal to the object, and equally distant from the surface of the reflector. Let a, b, be the visual angles of the diameter 2r of the object and image, a the distance of the centre of the object from the reflector, and b the distance of the eye from that centre; then and we easily get 68 — pole + 4a cos, B = ✓{84 +(8a? +2r*)? +(4a? — )} Hence a and ß may be found by the table, and the apparent magnitudes of the object and image may be compared. . |