810, The whole number of images is (Codd. p. 34,) 360° 6. 60 811. The focal length of a double convex lens is, (Codd. 812. The distance of the focus of refracted rays from the surface of a spherical refractor is (Codd. p. 56) :: A' = 3r, so that the parallel rays will just reach in their convergence the plane reflector, and be reflected back again, converging to the spherical surface. 813. Let a denote the inclination of the two mirrors, and 6 that of the first incident and second reflected ray intersecting at the luminous point; then it may be easily shewn that AF 2a. Again, let a be the distance of the given point at which the ray is always reflected from the intersection of the two mirrors, and suppose the radiant point moving in the arc whose radius is z, to be now in the middle point of that are; then it easily appears that 814. Let q be the distance of any section of the cylinder I to the coincident axis from the centre of the reflector, and q' that of the centre of the image of the circumference of this section, also let R, R' denote the radii of the section and image; then R:R'::9:9 9 2 1 ra 9 9 1-29 Let q = a (the distance of the end of the cylinder from the reflector), then = ; also when q = 00,6' : 2a Consequently the vertex of the image is in the principal focus, and the distance of any section of the image from this vertex being put = d, we have d = g + r . 7 ; r(d hence R' OC od, which is the property of the cone. The base of the cone bas a radius R' =R. 2a 815. If 8, & denote the distances of the foci of incidence and reflection of any pencil of rays from the point of incidence 8,8, their next values, and D, D' values of them for a given point of the reflector; then , being the length of the caustic between these two points of incidence we easily prove, (Wood, Prop. 97) da = 8+– (9, + 8) = d (0) + d (8) ::.= 8 + + C = d+ – (D + D). In the problem 8 - D= ordinate (y) of the circle, and D' = 0 for the beginning of the caustic, ..^= y + 816. Since the image is always at the same distance from the reflector which produces it as the object, and on the opposite side of it, that part of the distance between the eye and image, which is included between the image and reflector is equal to the distance between the object and point of incidence. The remaining parts of the paths of the object and image to the :, the two paths are equal. eye coincide. 817. The 2 between the incident and emergent rays is (Wood Prop., 93) 2 (n + 1) 6' - 20 0 and 6 being the 2 of incidence and refraction. But when the rays emerge parallel to the direction of incidence 2(n + 1) 0 - 20 = 0 1 818. The longitudinal aberration is (Coddington, p. 14) r vers. A 2 r being the radius of the reflector, and 6 = = the aperture. Hence, if a denote the lateral aberration, we have r vers. O a: :: r sin. 0 : 2 2 isar sin. 0 X vers. 0. But the diameter (d) of the least circle of aberration is half the lateral aberration nearly, and the diameter (d”) of the aperture is 2r sin. 6. Consequently we have d:d :: r sin. A x vers. A : 2r sin.e 2 : 1 :: 4mn : * -("vers. e)" the required proportion between the densities. 819. Sin. I = m sin. R, and, sin. im sin. , and the angle of deviation is (Codd. p. 48) It i R+ra. :. dI + di = 0. di and and aR dr cos, COS. I m cos. R n 820. If Q be the distance of the focus of refraction at the first surface, we have (Codd. p. 56) p - nd Q (n + 1) a (n+1) (n + 1)rd Also 1 n+1 (n + 1) X (1+ n 1 1+) -1 + (n + 1) n+1 .t nearly Q Q nd (nd - r) t t. r-2nd rd 821. This problem may be simplified by enouncing it, Given the position of a horizontal straight line, and of a vertical plane mirror, required the inclination of another mirror to the former, that the angle at a given point of the line subtended by the image of the line in the image of the former mirror may be zero. Let o be the angle required, and a the given inclination of the straight line to the surface of the given reflector; then if o denote the inclination of the second image of the line to the given mirror, we easily get Q = 0 + 0 a = 20 |