CHAPTER II.

REFLEXION AND REFRACTION OF RAYS OF LIGHT.

8. WHEN a ray of light travelling in one medium is incident on the surface of another medium, it is usually divided into three separate portions which behave in different manners.

(i) A portion is reflected back into the original medium, in a direction determined according to a certain law.

(ii) Another portion passes into the new medium, having its direction changed according to another law; this portion is said to be refracted into the new medium.

(iii) A third portion is said to be scattered by the surface bounding the two media; the bounding surface becomes illuminated and itself acts like a source of light sending rays in all directions.

When a ray of light is incident on a solid opaque body, the second portion does not exist, and all the light is either reflected or scattered. The quantity of light reflected depends upon the nature of the surface; the smoother and more highly polished the surface is, the more light is reflected. The scattering of light is probably due to the unevenness of the surface; the incident light is reflected by minute portions of the surface which act as mirrors distributed irregularly in all directions. It is by the scattering of light that non-luminous bodies become visible when in the presence of a bright body.

Thus if the rays of light from the sun strike a window, part of the light passes into the glass and out again into the room; for an

observer inside the room can see the sun distinctly. Another part of the sun's light is reflected back into the air; for a person stationed outside can see a picture of the sun in the window as in a lookingglass. A third portion serves to render visible the specks and marks on the window-pane; this light is said to be scattered.

9. The plane containing the incident ray and the normal* to the surface separating the two media, is called the plane of incidence, and the acute angle between the incident ray and the normal is called the angle of incidence, and the acute angle between the reflected ray and the normal, the angle of reflexion.

When the direction of a ray of light is changed by reflexions or refractions, the angle through which the original ray produced must be turned in order to bring it into the position of the final ray, is called the deviation of the ray.

The law according to which a ray of light is reflected at a surface may be thus stated.

The angles of incidence and reflexion always lie in the same plane and are equal to each other.

10. If the ray be incident on a plane surface the reflected ray may be found by a simple geometrical con

P

R

N

struction. If P be any point on the incident ray PQ, and if from P a perpendicular PN be drawn to the reflecting plane and be produced to P' so that P'N is equal to PN, then in the triangles PNQ, P'NQ, the two sides PN, NQ are equal to the two sides P'N, NQ, each to each, and the

* See note to p. 4.

angles PNQ, P'NQ are both right angles, therefore the triangles are equal in all respects, and the angle PQN is equal to the angle P'QN. Hence P'Q produced and PQ make equal angles with the plane NQ and therefore with the normal to it. Thus PQ produced will be the reflected ray.

If the surface be not plane, we may substitute the tangent plane to the surface at Q, for the plane of the mirror in the previous construction.

11. When a ray is reflected at a plane surface the incident ray and the reflected ray make equal acute angles with any line in or parallel to the reflecting plane.

For let PO, OQ be the incident and reflected rays, and let the plane of incidence meet the reflecting plane in the

line MON. Also let AOB be a line drawn through O parallel to the given line. On the lines OP, OQ measure equal lengths OP, OQ, and through P, Q draw planes perpendicular to the line MON meeting this line in the points. M, N and the line AOB in the points A, B, respectively. Then in the triangles POM, QON, the angles at M and N are right angles and the angles POM, QON are equal, by the law of reflexion, and OP is equal to OQ; therefore the triangles are equal in all respects, so that OM ON, PM - QN. Also in the triangles AOM, BON, the angles at M and N are right angles and the angle AOM is equal to the angle BON; therefore AM = BN and AO OB. Also in the triangles PMA, QNB the two sides PM, MA are equal to the two sides QN, NB each to each, and the angles at

=

=

=

=

M and N are right angles; therefore AP BQ. Finally in the triangles AOP, BOQ, the three sides of the one are equal to the three sides of the other, each to each; therefore the angles AOP, BOQ are equal. This proves the proposition.

Conversely, if two lines PO, OQ lie in a plane normal to the reflecting plane and make equal acute angles with any given line in the plane, they may be taken to represent an incident and reflected ray, respectively. The proof is similar to the preceding.

It follows from the preceding proposition that if a ray of light be reflected in any manner successively at two plane surfaces, the initial and final rays are equally inclined to the line of intersection of the plane surfaces.

N

B

12. If a ray of light be reflected at a surface, the projections of the incident and reflected rays on any plane through the normal, themselves obey the law of reflexion. For along the incident and reflected rays measure equal distances OA, OB; then AB will be bisected at right angles by the normal to the surface ON. Let PNQ be the projection of ANB on any plane through the normal, so that OP, OQ are the projections of the incident and reflected rays. Then in the triangles APN, BQN, the angles at P and Qare right angles and the angles ANP, BNQ are

equal to each other, and

AN = NB. Therefore PN = NQ.

Hence the triangles

PNO, QNO are equal in all respects, and therefore OP, OQ are equally inclined to the normal ON.

Further, it is easily seen that the triangles AOP, BOQ are equal in all respects, and therefore the angle AOP is equal to the angle BOQ. In other words, the incident and

reflected rays are equally inclined to any plane through the normal.

13. Let a ray of light be reflected successively at two plane mirrors, to find the direction of the ray after any number of reflexions.

We shall first consider the case in which the reflexions take place in a plane perpendicular to both mirrors.

Let OA, OB be the plane mirrors and let PQRST... be the ray of light which is reflected successively at Q, R, S, T...

19

Let e denote the angle between the mirrors, and let 01, 02, 0...be the acute angles formed by the ray with the reflecting surfaces at the successive incidences. Thus the angles at Q are each 0,, those at R, 2, and so on; so that from the triangle QOR we find 0, 0, +e, and similarly 01 = 02+ €, &c. These equations may be written

3

2

2

=

1

When n is even, the angles +1 and 0,, are measured from the same mirror, and therefore 0+10, is the angle between the initial and final rays; therefore the total deviation is equal to n times the inclination of the mirrors.