the reflected light is much more brilliant than when it is reflected by an ordinary mirror.

Total internal reflexion may be exhibited by holding a glass of water above the level of the eye; the under surface of the water will appear very bright from the light internally reflected at it, and any object in the water will be seen by reflexion at the under surface more brilliantly than when it is reflected in a mirror. A glass prism may easily be held so that the eye may receive light through it after internal reflexion at one of its faces. That face will appear very bright. An arrangement of prisms fixed in this manner is often used to light under-ground rooms.

The subject of this article may also be illustrated by describing the appearance presented to an eye placed under the surface of still water. All external objects would appear compressed within a conical space whose axis is vertical and semi-vertical angle 48° 27' 40", the objects near the horizon being much distorted. Beyond this conical space objects within the water would be seen reflected by the surface of the water.

=

=

If we suppose that μ= μ, then o - p', and the refraction becomes a reflexion. All the subsequent theorems relating to refraction will give corresponding theorems for reflexion by making the same substitution

18. There are two other useful theorems relating to the incident and refracted rays.

The angles which the incident and refracted rays make with any plane through the normal to the refracting surface, obey the law of refraction.

Also the projections of the incident and refracted rays on any plane through the normal are connected by a law of refraction, with a refractive index depending on the inclinations of the rays to the plane.

For let AO, OB be any two refracted rays, and let the lengths of AO, OB be taken equal to μ and ', the refracting indices of the two media, respectively. Then if AM, BN be drawn from A and B perpendicular to the normal to the refracting surface, AM, BN will be equal and parallel, since they are equal to μ sin o, μ' sin o', respectively.

M

Let PO, OQ be the projections of AO, OB on any plane through the normal, P and Q being the projections of the points A, B respectively. Then the triangles APM, BQN are equal in all respects.

Let n, n' be the acute angles which the incident and refracted rays make with the plane; o, the acute angles which the projections of these rays on the plane make with the normal. Then

B

and therefore, since AP is equal to BQ,

μ sin n = μ'sin n'.

This proves the first theorem.

=

Also OPμ cos n, OQ=μ' cos n'; and therefore, since PM is equal to QN,

μ cos n sin = μ' cos n' sin p',

which proves the second theorem.

19. In any refraction, the greater the angle of incidence, the greater will be the angle of deviation.

For if o, o be the angles of incidence and refraction,

But the deviation is equal to -. If o increase, -0. and therefore also p', tan (6+) will increase, since (+) is less than 27, and therefore the deviation will

increase.

When the ray is passing into a rarer medium, we have only to suppose the ray reversed; then since the angle of refraction increases as the angle of incidence increases, the proof comes under the case just considered.

20. We shall now give a geometrical proof of this theorem *.

Let be the centre of any circle of radius r; take an external point 0, such that OC=μr, and draw any line OPQ through O to meet the circle in P, Q, and join CP, CQ. Then if we denote the angle CPQ by 4, and COP by p',

The angles and d' are therefore related like angles of incidence and refraction of a ray of light. The deviation -' will be the angle PCO, or D say. By varying the direction of the line OPQ from the position OAB, to the position OT in which it touches the circle, the angle will be made to increase from 0 to π, and during this

* This proof is due to Prof. P. G. Tait.

change D is increasing also. This proves that the deviation increases with the angle of incidence.

The angle of refraction increases from zero to the value COT; this angle represents the critical angle.

But further, as por d' increases uniformly, the deviation increases faster and faster.

For let Opq be another chord of the circle close to OPQ. Then the change in D is the angle subtended at the centre by the arc Pp. And since the angle PCQ is π-20, the increase in is represented by the arc (Qq+Pp); and therefore, by subtraction, the increase in is represented by an arc (Qq − Pp).

If we suppose 4, ' and D to have become +x, '+x' and D+8 respectively, we have

and similarly-09-1.

But as P moves from A to T, OQ becomes more and more nearly equal to OP; so that x/8 and x'/8 become smaller and smaller, which proves the proposition.

1. The angle of incidence being 60o, and the index of refraction √3; find the angle of refraction.

Ans. 30o.

2. The absolute refractive indices of two media being √5–1 and 2, respectively, find the angle of refraction of a ray travelling in the first medium and incident on the second at an angle of 30o. Ans. 180.

3. A ray of light is incident on a refracting surface at an angle whose tangent is equal to the refractive index. Prove that the angle of refraction is the complement of that of incidence.

4. The height of a cylindrical cup is 4 inches and the diameter of its base 3 inches. A person looks over its rim so that the lowest point of the opposite side visible to him is 24 inches below the top. The cup is filled with water; looking in the same direction he can just see the point of the base farthest from him. Find the refractive index of water. Ans. .

5. A ray is incident on a refracting sphere whose refractive index is, at an angle whose sine is √6. Show that if the ray be refracted into the sphere, that portion of it which emerges after having been twice internally reflected will be in the same direction as the original ray.

6. A ray of light is incident on a refracting sphere, whose refractive index is √3. It is refracted into the sphere and when it is incident on the inner surface of the sphere, part is reflected internally, and part is refracted out into vacuum. Show that if the original angle of incidence be 60o, these two parts will be at right angles.

If the part internally reflected be again incident internally and be refracted out into vacuum, its final course will be parallel to that of the ray first incident.

7. Show that when a ray of light enters a medium whose refractive index is √2, its greatest deviation is 45o.

8. A ray of light is incident on a portion of the refracting medium in the shape of a prolate spheroid, in a direction parallel to the axis of the spheroid. If the refractive index be 1/e, where e is the eccentricity of the generating ellipse, prove that after refraction the ray will pass through one focus.

9. Prove that light which has been refracted into a sphere from vacuum can never be totally internally reflected.

10. If light be incident on the curved surface of a hemisphere of a refracting medium in a direction parallel to its axis, show that there will be no total internal reflexion at the plane surface, unless the refractive index be greater than √2.

11. If and p' be the angles of incidence and refraction of a ray of light, show that sin (p −4′)/sin (p+p') increases as increases.

21. Any medium bounded by two plane faces meeting in an edge, is called a prism. The inclination of the faces to each other is called the refracting angle of the prism.