4
CHAP. I.
REFLEXION AT PLANE SURFACES.
6. Prop. To find the direction which a ray of light, emanating from a given point, takes after reflexion at a plane mirror in a given position.
Let QR, (Fig. 2.) represent a ray of light, proceeding from the point Q; XY, the section of the reflecting surface by a plane perpendicular to it containing the line QR; RS, the reflected ray making with XY an angle SRX equal to the angle QRY which QR makes with the same line: let QA be perpendicular to XY, and let SR meet it in q.
Then since the angle QRA is equal to SRX, that is, to qRA, the right-angled triangles QAR, qAR, having the side AR in common, are equal in all respects. Therefore qA is equal to AQ.
Any other reflected ray R′S′ will of course intersect QA in the same point q; so that if several incident rays proceed from Q, the reflected rays will all appear to proceed from q, which as we have seen is at the same distance behind the mirror as Q is before it.
7. Suppose now that a ray QR (Fig. 3.) reflected into the direction RS by a plane mirror HI, meet in S another mirror IK inclined to the former at an angle I. It will of course undergo a second reflexion, and returning to meet the first mirror be reflected again, and so on; so that the course of the light will be the broken line QRSTUVX.
Let perpendiculars be drawn to HI, IK, at the points R, S, T, V, … meeting each other successively in L, M, N, O, … Each of the angles at these points will of course be equal to the angle at I.
| Let the angle of incidence at | R be called | φ1, |
| that at | S…… | φ2, |
and so on.
