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799. The glass must be a double concave lens, whose focal length is (Wood, p. 133,) QEx Eq 12 x 3

= 4 feet.

FE =

800. Generally, required the ratio of the sine of incidence to that of refraction, when, the cylinder being the pth part full, the eye (placed so as to see the farther extremity of the vessel when empty,) shall just see the centre.

Lett be the radius of the cylinder's base, Q, of the angles of incidence and refraction from the eye into the medium, and x the distance of the centre of the base from the I at the point of incidence; then sin. Q : sin. o' :: 1 + X : 2

and r + x : 2r :: : 1

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801. When the object is between the principal focus and the centre of the lens, the image is erect. (Wood, p. 114.)

Also since the object and image are concentric and similar arcs, we have, by the question 1:3 :: QE : Eq :: QF : FE (Wood, p. 99,)

:: FE – QE : FE :: FE =


x 4 = 6 inches.

802. Let x be the distance of the centre of the base from the vertex of the segment of the sphere whose surface is the nth part of the whole surface; then (Vince's Flux. p. 110,) this partial surface is

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Again, since the extreme rays proceeding from the sphere to the eye will touch the sphere, we easily prove that the required distance from the vertex of the sphere to the eye is


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d =


803. The image recedes from the reflector as fast as the reflector from the fixed object, and in the same direction. Therefore the image recedes twice as fast from the object as the reflector.

804, The rays proceeding from the circumference of the circle to the eye form an oblique cone, which being produced, will be cut by a vertical plane, and the section thus made will be a circle when it is the subcontrary section of the cone. (See p. 18, and Fig. 15.) Hence the eye must be in such a position that the subcontrary section of the cone may be vertical.

Let a be the distance of the given locus of the eye from the centre of the given horizontal circle, whose radius call r, and I the required altitude of the eye. Also let 0, b' be the 2 subtended at the circumference, by the rays proceeding from the extremities of the diameter which meets the given locus. Then if the seco tion be vertical we must have

6 = 90 + 0
and x =
(2r + a) tan. 0 = a tan, (180 - 90 - 0)

=a. cot. 0

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805. Let r, m' be the radii of the spheres, and r + guld + d the distance between their centres; then if any point be taken in the line joining the centres, and from it tangents be drawn to both spheres, the surfaces S, S' visible at that point will be the portions cut off by planes through the extremities of the tangents I to d. Let , & be the abscissæ measured from the vertices of the spheres along d, of these portions ; then (Vince's Flux. p. 110,)

S = 2#rx, S'= 2r'x

.. 2#rr + 2 r'x' = max.

:. rdx + r'dx' = 0. Again, it is easily shewn that

glx +

.. (1) and this gives dx

dx' +

0 (r – x*) (r—r')?

(2) x' Hence, and from equatiou (1), we get

(r+d)r - po

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d =

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1 + p + d

(r+d) x – 5 ve and x' =

ototd and the distance from the required to the surface whose radius is r, is

(r'+d) r - gone

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806. Let a be the height of the person's eye, b the length of the mirror, whose top is supposed to be on a level with the eye, and x the variable distance of the person from the glass; theu since the image is equal and similar to the object, and at the same distance from the reflector's surface, if P be the length visible, we have (by similar A)

P:6:: 2x : *

.. P = 26.

807. Let m be the given ratio of the sines of incidence and refraction, r the radius of the sphere, m' the given ratio of the rays included by the sphere; also let p, p be the unknown distances of these rays from the centre of the sphere ; then it may easily be shewn that


and 2 v pole - pe

= m
2 Vp-p

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which give

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the distance of the incident ray from the centre.

808. Let a be the I distance of A from CP, (see Fig. to the Enunciation, or of B from CQ; also let 6 be the distance of C from these perpendiculars, and suppose x, a' the distances of P and Q from the I from A and B, and 0, of the of incidence at A and B; then we easily get these equations

x = a tap. A
x = a tan 6

cos. (C-6)

cos. O box

cos. (C-a) btal

cos. 6

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and by further elimination and reduction, we finally obtain

a sin.C-6(1-cos.C) ab sin. C+bäsin. C-2ab 1 - cos. C

sin. C.(1- cos. C) which being resolved gives the positions of P.


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The problem may be resolved more easily, although not so strictly, by considering the symmetry of the loci of P and Q with respect to the points A, C, B. It is evident, although not very easily demonstrable, that 0 = 0', and x = x'.

809. The caustic is the common cycloid, whose base is equal to, and coincident with, the base of the given semi-cycloid. (Wood, Prop. 100.)

Again, let dx denote the constant element of the base, and ds the corresponding variable element of the arc of the caustic; then the densities of the rays of light upon these elements will be inversely as their magnitudes; that is, since the light will be uni. form upon the base, density cc



But if y be the ordinate of the reflector parallel to the rays, by the nature of the curve, we have

dy : dx :: vers. 0 : sin. and dy + ds = 0. (Codd. p. 28.)

sin. 0 :. Density cc

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It does not therefore oc as in the enunciation,

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