R

given circle, describe the circle BR bisecting the radii CA, CP in B, R, respectively. On PR as diameter describe another circle meeting the reflected ray in Q, and join QR. Since SP is parallel to CB, the angle SPC is equal to the angle PCB; and therefore the angle QPR is equal to the angle RCB. The angle QPR is subtended at the circumference of the circle by an arc QR; and the angle RCB is subtended at the centre of the other circle by the arc RB, and the radius of the second circle is double the radius of the first, and therefore the arc QR is equal to the arc RB; and if the circle PQR were to roll along the circle RB, the point would finally coin

B

cide with B. Now as Q begins to move, the point of contact R is for an instant fixed, so that the motion of Qis perpendicular to QR; and therefore the reflected ray PQ touches the curve described by Q. This is true whatever the position of the point P. The locus of Q is an epicycloid, and this is the caustic curve required.

80. If the incident rays diverge from a point in the circumference of the reflecting circle, the caustic curve is a cardioid, or, in other words, the caustic may be described as an epicycloid in which the rolling circle is equal to the fixed circle.

Let O be the origin of the incident rays, OCA the diameter of the reflecting circle; then the caustic curve will be symmetrical about the line OCA. Let OP be any incident ray which is reflected at P by the circle in the direction PQ. Join CP; then by the law of reflexion, CP will bisect the angle OPQ. With centre C and

radius equal to one-third of the radius of the given circle,

The

describe a circle meeting CA and CP in B and R, respectively, and on PR as diameter describe another circle cutting the reflected ray in Q; join QR. radii of the two smaller circles will be equal to each other. Now, since the triangle CPO is isosceles, the external angle PCB is double of the angle CPO, and therefore double of the angle QPR. Hence the arcs RB, QR subtend equal angles at the centres of their respective circles, and therefore these arcs are equal. If the circle PQR were to roll along the circle RB, the point Q would finally come to B. As the circle PQR begins to roll, the point of contact R is for a moment stationary, and therefore begins to move perpendicular to QR along PQ. From this it follows that the reflected ray touches the curve described by the point Q. This is true whatever the position of the point P. The locus of Q is a cardioid, and this is the caustic required.

81. To find the caustic by refraction at a straight line, for rays issuing from a point.

Let S be the bright point; draw SC perpendicular to

the line, and produce it to H, so that CHCS. Let SQ be any ray incident at Q, and QR the corresponding re

fracted ray. Describe a circle about the triangle SHQ, and let QR be produced backwards to cut the circle in P; then PQ bisects the angle SPH. Let be the angle of incidence and 'the angle of refraction at Q; then the angle POS p', and 4=▲ HSQ = 2 HPQ = 2 SPO.

Hence

=

and therefore

SO: SP = sin : sin d',

μSO = μ'SP.

But since the angle P is bisected,

and therefore

HO: HP SO SP,

=

μHP = μ'SP.

By addition, μ SH = μ' (SP+ HP).

Thus the locus of P is an ellipse whose foci are S and H and whose eccentricity is μ'/μ; and PQ is normal to the ellipse, and therefore the ellipse is an orthotomic The evolute of this ellipse is the caustic required.

curve.

If the second medium is more highly refractive than the first, it may be shown in the same way that the caustic is the evolute of a hyperbola whose foci are S and H.

H. O.

7

82. To find the length of the arc of a caustic.

The length of the arc of a caustic of any orthotomic system of rays in one plane can always be found. For the caustic is the evolute of the orthogonal curves.

Suppose a system of rays issuing from a point, or normal to a given surface, to be reflected and refracted

P

A

B

0

any number of times. For each ray, form the function Σup, and let V=Eup. Let the final medium be of refractive index μ, and let VV, be the value of the reduced path for an orthogonal curve in this medium, say the curve PQ. Let AB be any arc of the caustic, and let PA, QB be the rays touching at A, B. Then the arc AB=QB-PA, by the properties of evolutes.

83. We can now, by means of caustics, indicate more accurately the manner and position in which an object under water is seen by an eye outside.

Suppose for instance that the water had a horizontal level bottom not very deep. Let P be a point on the bottom, let us trace the pencil of rays by which an eye sees the point P. Draw the normal PM and consider rays in the plane EPM. Construct the caustic in this plane which is touched by refracted rays originally di

verging from P. We must draw the two extreme tangents

to this caustic which will meet the eye, and then these lines will bound the part of the pencil which traverses the air; if we join the points where these tangents meet the surface to P, the joining lines will bound the pencil as it passes through the water. The two tangents to the caustic meet at the point of contact of either of them, very nearly. Thus to an eye outside the point P appears to be at p.