An Elementary Treatise on Optics/Chapter 2

by Henry Coddington

CHAP.II.

REFLEXION AT SPHERICAL SURFACES.

8.Prop. Rays meeting in a point being incident on a spherical reflecting surface; it is required to determine the directions of the reflected rays.

Let $RAV,$ Fig. 4, represent the spherical surface, which we will suppose concave, or rather a section of it by a diametric plane containing an incident ray $QR,\ Q$ being the point from which that and the other rays are supposed to proceed.

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 ER/EQ	=sinEQR/sinERQ; that is, r/q=sin(θ−φ)/sinφ;
and ER/Eq	=sinExr/sinERx; that is, r/q′=sin(θ+φ)/sinφ;
∴ r/q′−r/q	=sin(θ+φ)−sin(θ−φ)/sinφ=2cosθ.

Hence 1/q′=1/q+2cosθ/r, or q′=qr/r+2qcosθ.

9.It appears then, that the place of the point $q,$ or the distance $Eq$ depends on the value of $\theta$, the angle $REA$, and is therefore not the same for different rays $QR.$ It diminishes as $\cos \theta$ increases, that is, as $\theta$ diminishes, or $QR$ approaches to coincidence with $QA$. It is important to know what its final value is, which is in fact to determine the point of intersection of the reflected rays when the incident rays are nearly coincident with $QA,$ the axis of the surface, forming consequently a very small pencil.

If we suppose $\theta =0$, we shall have $\cos \theta =1,$ and

${\frac {1}{q'}}=$ 1/q^{[errata 1]}$+{\frac {2}{r}},$or $q'={\frac {qr}{r+2q}}.$

10.If moreover we suppose $q$ infinite, which is supposing the rays parallel, we shall have

${\frac {1}{q'}}={\frac {2}{r}},$^[1] or $q'={\frac {r}{2}}.$

This is what is technically termed, the principal focal distance of the reflector, the place of $q$ being then $F,$ which is called the principal focus, and if we call $AF,f,$ we shall have in general, that is, for rays nearly coincident with $QC,$

${\frac {1}{q'}}={\frac {1}{q}}+{\frac {1}{f}}.$

11.These formulæ might easily have been obtained directly by supposing QR equivalent to QA, and we will make use of this method to deduce them in another form which is often more convenient.

Since RE, (Fig. 5.) bisects the angle QRq, we have

QR/Rq=QE/Eq.

And in the extreme case

QA/Aq=QE/Eq;

that is, if we call AQ, ∆; Aq, ∆′; AE, r as before,

∆/∆′=∆−r/r−∆′;

so that in fact ∆, r, and ∆′ are in harmonic progression, and we have

1/∆+1/∆′=2/r, or ∆′=∆r/2∆−r,

and if Q be supposed infinitely distant, or 1/∆=0,

then 1/∆′=2/r, or ∆′=r/2,

which agrees with the preceding formula.

If, as before, we put f for r/2, we shall have

∆′=∆f/∆−f, and ∆′−f=f²/∆−f,
that is, Fq=FE²/FQ,

from whence it appears, that

Fq:FE::FE:FQ.

12. Upon the whole we may collect that if a small luminous body be placed before a spherical concave mirror, at some distance from it, the distance of the focus will always be something more than half the radius of the surface, which is its accurate value for the light of the Sun, the rays of which are considered as parallel; that if the luminous point be moved towards the mirror, the focus q will come forwards to meet it, and at the centre E, they will coincide, (Fig. 6.) as the formula will easily show, and as one might naturally expect, for the rays in that case being all normal to the surface, would be reflected back upon themselves. When the light is brought between the centre and the surface, or between E and A, (Fig. 7.) Q and q in a manner change places, as one might expect from observing that in the formula

1/∆+1/∆′=2/r,

∆ and ∆′ are quite similarly involved, and therefore may be commuted without altering the equation. When the light is at the middle of the line EA, namely, in the principal focus F, (Fig. 8.) the formula shows that we must have 1/∆′=0, that is, ∆′ infinite, which answers to what we found before, namely, that when the incident rays are parallel, or Q is at an infinite distance, the reflected rays meet in F.

When Q is brought between F and A, (Fig. 9.), or ∆ is made less than r/2, the formula 1/∆+1/∆′=2/r, or 1/∆′=2/r−1/∆, shows that 1/∆′, and consequently ∆′, must be negative, that is, q goes to the other side of the reflector, and the reflected rays instead of converging, diverge.

When Q comes to A, q meets it there.

As a converse to the last case but one, we may take that of rays converging to a point Q behind the reflector, and reflected to a focus q in front, (Fig. 10.) To accommodate the formula to this case, we must make ∆ negative, and we have then

−1/∆+1/∆′=2/r, or 1/∆′=2/r+1/∆,

which shows ∆′ to be essentially positive.

The student will find no difficulty in examining particular cases; the one most likely to occur, is that in which the radiant point is at the opposite point of the sphere from the centre of the mirror.

Making ∆=2r, we find here ∆′=∆r/2∆−r=2/3r.

It will occur to every one, that of the two foci Q and q, that which lies between E and A moves much more slowly than the other, when their places are changed; in fact, we have seen that by merely bringing up Q from E to F, q was sent from E to an infinite distance, and that when Q moved on from F towards A, q came back from an infinite distance on the reverse side of the reflector to meet Q at A.

13. We have hitherto considered only one species of spherical reflector, the concave; let us now take the convex, (Fig. 11.) where as before, E is the centre, Q the radiant point, QR, RS an incident and a reflected ray, making equal angles with ER the radius or normal. Let SR cut AE in q.

Then we have, keeping the same notation as before,

ER/EQ=sinEQR/sinERQ;

that is, as before, r/q=sin(π−φ)−θ/sinφ=sin(θ+φ)/sinφ,

ER/Eq=sinEqR/sinERq;

that is, r/q′=sin(π−φ)+θ/sinφ=sin(θ−φ)/sinφ;

∴ r/q′−r/q=sin(θ−φ)−sin(θ+φ)/sinφ=−2cosθ,

and finally, 1/q′=1/q−2cosθ/r,

which is the same result as before, except the sign of the 2d term; it will however immediately occur that they may be reconciled completely by supposing the radius r to be positive in the one case, and negative in the other, which is exactly true in algebraical language as applied to Geometry. This is however more clearly distinguished in the case of the other formula, which here becomes for, as before,

${\frac {1}{\Delta }}-{\frac {1}{\Delta '}}=-{\frac {2}{r}},$

for, as before, ${\frac {QA}{Aq}}={\frac {QE}{Eq}},$ that is, ${\frac {\Delta }{\Delta '}}={\frac {\Delta +r}{r-\Delta '}},$

whence $\Delta r-2\Delta \Delta '-r\Delta '=0,$ and dividing by $\Delta \Delta 'r,$

${\frac {1}{\Delta '}}-{\frac {2}{r}}-{\frac {1}{\Delta }}=0,$

or ${\frac {1}{\Delta }}-{\frac {1}{\Delta '}}=-{\frac {2}{r}}$ as above.

Here then $\Delta '$ and $r$ become both negative, which the Figure plainly shows.

Making $\Delta$ infinite, or ${\frac {1}{\Delta }}=0,$ we find ${\frac {1}{\Delta '}}={\frac {2}{r}}$, or $\Delta '={\frac {r}{2}}$ as before, so that the principal focus is here at the bisection of the radius of the reflector, and of course behind it. In fact, as long as $\Delta$ is positive, or $Q$ is in front of the mirror, so long must $q$ necessarily be behind it, and the reflected rays diverge.

If, however, the incident rays converge to a point behind the mirror, that is, $\Delta$ be negative, the formula will become

$-{\frac {1}{\Delta }}-{\frac {1}{\Delta '}}=-{\frac {2}{r}},$ or ${\frac {1}{\Delta '}}={\frac {2}{r}}-{\frac {1}{\Delta }},$

and $\Delta '$ may be positive, provided ${\frac {1}{\Delta }}>{\frac {2}{r}},$ that is, $\Delta <{\tfrac {1}{2}}r,$ so that if the focus of the incident rays lie between the principal focus and the back of the mirror, the reflected rays will converge. Of course, if the incident rays converge to the principal focus, the reflected rays are parallel.^[2]

14.It will easily be seen that the effect of a concave mirror is to give convergence to rays, that is, to increase convergence when it exists in the incident rays, to give convergence to parallel rays, and even to divergent within a certain limit, and beyond that to lessen their divergence.

Convex mirrors on the contrary give divergence to parallel rays, increase previous divergence, and make even convergent rays diverge, or at least diminish their convergence.

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1. Taking the general formula in this case, we find
   
   ${\frac {1}{q'}}={\frac {2\cos \theta }{r}};$ that is,$q'={\frac {r.\sec \theta }{2}},$
   
   

   so that, generally speaking, when the incident ray is parallel to the axis of the reflector, the reflected ray bisects the secant of the angle, which, in the more particular case of a ray nearly coincident with the axis, becomes the radius.

2. Particular examples may easily be multiplied; we will only observe, that when $Q$ is at the distance of half a radius in front of the mirror, $q$ is at one-fourth of the radius behind. When $AQ$ equals the radius, $Aq$ is one third of it. There may be some little obscurity attending the application of the particular formula deduced above for convex mirrors, owing to our having put positive symbols for $AE$ and $Aq,$ which, in algebraical strictness, are negative. The student should use only the original formula ${\frac {1}{\Delta }}+{\frac {1}{\Delta '}}-{\frac {2}{r}}=0,$ putting negative values for $\Delta$ or $r$ when necessary. Thus if the radius of a convex mirror be 6 inches, and the distance of the radiant point from it 9, we have ${\frac {1}{9}}+{\frac {1}{\Delta '}}+{\frac {2}{6}}=0,$ whence $\Delta '=-{\frac {9}{4}}$ or $2{\tfrac {1}{4}}$ inches behind the mirror.

Errata

1. Original: r was amended to q: detail