2

W1, W, the weights which sink the hydrometer in two fluids whose specific gravities are σ,, σ, respectively.

Then

19 2

2

Let now W be the weight necessary to sink the instrument to the same depth, in a fluid compounded of volumes V1, V2 of the above fluids. Then, observing that the specific gravity of the mixture will be

therefore

1 1

2

W

+ V1

(3);

1849.

(A). Describe Smeaton's Air-pump, and find the density of air in the receiver after any number of ascents of the piston.

(B). If instead of the receiver we use a cylindrical vessel of ten times the capacity of the barrel, and cover the upper extremity with a diaphragm capable of sustaining only half the pressure of the atmosphere, find after how many ascents of the piston the diaphragm will burst. Given

Generally, if A, B be the capacities of the receiver and barrel. respectively, p the density of atmospheric air, p, the density of air in the pump after n ascents of the piston,

Here A 10B; and if n is the number of ascents of the piston just before the diaphragm bursts, p = p, since it is only capable of sustaining half the pressure of the atmosphere;

Hence the diaphragm will burst during the 8th ascent.

62

1849.

OPTICS.

(A). When rays diverging from a point are incident on a plane mirror, prove that the reflected rays diverge accurately from a point.

(B). Within what space must the eye be situated to see a given point by reflection at the mirror; and within what space must a point be situated to be seen by the eye in a given position?

Let AB be the mirror; P (fig. 52) a given point. Its image will by (A), be a fixed point p equally distant from the mirror on the opposite side. Join BA, pB, and produce these lines to Q, R. Then will QpR be the reflected pencil, and consequently QABR the space within which an eye must be situated in order to see p.

Again, let E (fig. 53) be the eye in a given position. Draw the lines EAq, EBr. Then the image of the point must lie in the space qABr; and therefore, drawing AQ, BR at the same inclination to AB as Aq, Br respectively, the point, in order to be seen by the eye at E, must lie within the space QABR.

1851.

(A). A luminous point is placed between two parallel plane mirrors, find the position of the successive images.

(B). When the luminous point moves uniformly in a straight line, shew that all the images will move uniformly in two sets of parallel straight lines which are equally inclined to the mirrors.

2

1

Let Q (fig. 54) be the luminous point. Then if we consider the rays which fall first upon the mirror A, an image Q, is formed, then an image Q, of Q, by the mirror B, then an image Q of Q by the mirror A, and so on. And a similar set of images will be formed by the rays which fall first on the mirror B.

Now if Q move along a line QP1, Q, will by the nature of reflection move along Q,P. Let QP, produced meet the mirror B in P; then Q2 will move along Q,P2. So Q3 will move along Q,P; and so on.

2

Thus the alternate images will move along two sets of lines respectively parallel to QP, QP; lines which are equally inclined to either mirror.

It is clear that the same holds also for the images formed by rays falling first upon the mirror B.

1851.

(A). Two rays are incident at any point of a spherical mirror whose centre is E, the one parallel to the axis of the mirror, the other proceeding from a point Q in the axis, and the reflected rays cut the axis in G and spectively; shew that GQ.Gq= GE".

q re

(B). If the axis AE of the spherical mirror meet the surface produced in R, shew that a ray proceeding from R and making an angle of 30° with the axis, will be reflected to the principal focus of the mirror.

Let the ray RP (fig. 55) incident at P make an angle of 30° with RA. Join PA; and let Pq be the reflected ray.

It is clear that the triangle APE is equiangular and therefore equilateral. Hence a ray incident at P parallel to RA will be reflected in PA. Applying therefore the formula of (4), Aq.AR = AE2 = AP2 ;

which shews that Pq is perpendicular to Aq;

1851.

(4). If parallel rays be incident directly upon a spherical refracting surface, the distance of the geometrical focus of refracted rays from the surface is to its distance from the centre as the index of refraction to unity.

(B). A pencil of parallel rays is incident directly upon a spherical refracting surface, and after refraction converges to a point at a distance from the surface equal to three times the radius; find the index of refraction, (1) when the surface is concave, (2) when it is convex.

By (4), if 0 (figs. 56, 57) be the centre of the refractor, F the geometrical focus of rays parallel to the axis,

and, whether the refractor be concave or convex,

OF AF OA = 20A;

therefore, in both cases,

AF
OF

1849. (A). When divergent rays are incident nearly perpendicularly upon a spherical refracting surface, the distance of the focus of incident rays from the principal focus of rays coming in a contrary direction, is to its distance. from the centre of the refractor as its distance from the surface to its distance from the geometrical focus of refracted rays.

(B). If the conjugate foci are each at a distance from the surface equal to twice the radius, what is the index of refraction?

58) are on opposite Let O be its centre,

In (B), since the conjugate foci (Q, q) (fig. sides of the surface, the refractor is convex. F the principal focus of rays coming in a contrary direction.