7
Draw QA through E the centre of the surface.
We must here extend the acceptation of the second law, which was understood only with respect to plane reflected surfaces, but which is just as true for curved ones, since we may consider the point of such a surface where reflexion takes place as belonging either to it, or to its tangent-plane, and the angles of incidence and reflexion will be those made by the incident and reflected rays with a normal to the surface at the point of reflexion.
In the present case, the surface being spherical, the radius ER is a normal, and if Rq be drawn making with this an angle ERq equal to ERQ, it will represent the reflected ray.
The line QEA passing through the centre, is technically termed the axis of the reflecting surface, and the question is now to determine the point q, the intersection of the reflected ray and the axis.
We may be supposed to have given,
- the radius of the surface EA, (r),
- the distance QE, (q),
- the angle REA, (θ),
and we might calculate the angle QRE, (φ).
We will call the distance eq, q′.
| Now EREQ | =sinEQRsinERQ; that is, rq=sin(θ−φ)sinφ; |
| and EREq | =sinExrsinERx; that is, rq′=sin(θ+φ)sinφ; |
| ∴ rq′−rq | =sin(θ+φ)−sin(θ−φ)sinφ=2cosθ. |
Hence 1q′=1q+2cosθr, or q′=qrr+2qcosθ.
