An Elementary Treatise on Optics/Chapter 6

by Henry Coddington

CHAP. VI.

OF IMAGES PRODUCED BY REFLEXION.

36. When a sufficient quantity of light falls on an object which is neither transparent nor specular, (that is, polished sufficiently to reflect properly,) it is dispersed from every point in all directions, and makes the object visible to a spectator placed on any side of it. If this light meet with a specular surface, either plane or curved, it will be reflected regularly according to the laws we have investigated in the preceding chapters, and since for every pencil of rays falling any how on a plane mirror, or nearly perpendicularly on a curved one, there is a focus of reflected rays, it follows, that the principal part of the reflected light will as it were proceed from the various foci of reflected rays, and the effect will be the same as if the same quantity of light came from an actual substance, of which each point should correspond or coincide with one of those foci; this is technically expressed by saying, that they proceed from the image of the object from which the light originally fell on the mirror.

37. There is a distinction, though not a very important one, between some optical images and others. In most cases in which the reflexion is made by a concave mirror, the rays proceeding from the various points of an object are made to converge, and actually meet in points constituting the image, and afterwards diverge from those points, as they would from a real surface placed there, so as to make it visible. In other cases the reflexion only changes the foci from which the rays diverge, and it is only by being produced backwards, that the rays can be said to meet in certain points.

38. There are some cases in which an optical image does in fact present the appearance of something real. If a candle be held before a concave mirror in a dark room, there will often be seen an appearance of an inverted flame in the air, owing to the reflected light converging to foci in front of the mirror, and there illuminating strongly the particles of dust floating about in the room. The image may generally be made more evident by holding a piece of paper at the place to be illuminated by the reflected light.

39. In the case of a plane mirror we have seen, that when any pencil of rays falls on it, the reflected light proceeds from a point at the same distance behind the mirror, as the focus of incident rays is before it. It will therefore be easily seen, (Fig. 34.) that the image of an object placed before such a mirror must be exactly similar to the object itself in point of form, all the points in the surface of the object having points corresponding to them which are similarly situated with respect to each other. Thus when a man looks at his reflexion in a plane mirror or common looking-glass, he sees a figure exactly like himself in form and dimensions, only that what is the right hand of the figure answers to his left, and vice-versâ.^[1] This reversion of the figure of course takes place in all such reflexions.

40. When an object is placed between two plane mirrors, in the first place it is reflected at each of them, which produces two images; these are again reflected at the mirrors to which they are opposite, and thus there is formed an infinite number of images growing more and more distant and more indistinct on account of the light which is lost at each reflexion.

To make this plainer, let O (Fig. 36.) be an object considered as a point placed between the two mirrors XY, ZV. AB a line through O perpendicular to the two mirrors.

Then there is first a reflexion at A,	which gives an image O′,
then a reflexion of this image at B,	which gives another image O″,
then a reflextion„ of O″ at A,	which gives another image„ O‴

and so on.

Again, there is a reflexion at B, giving an image O′,
a reflexion of this at A,	giving another image O″,
a reflexion„ of O″ at B,	giving another image„ O‴,

and so on.

An eye placed any where between the mirrors as at E, will see all these images in the directions EO′, EO″, …, Eo_′, Eo_″, ….

The reader may perhaps find some difficulty in understanding how the image O′, for instance, can be reflected at B, when it is behind the mirror XY, so that no light could come from O′ to B; but he has only to remember, that the light never goes from between the mirrors; O′, O″, &c. are merely imaginary points, where the rays intersect the line AB.^[2]

The distances OO′, OO″, &c. are easily calculated. If we put a for OA, b for OB, and c for AB or a+b,

OO′=2OA=2a,

OO″=2O′B−O′A=O′B+OB=2AB=2c,

OO‴=2A′O″−OO″=AO″+AO=2OO″+2AO=2c+2a,

⋮:

OO_′=2OB=b,

OO_″=2AO_′−OO_′=AO_′+AO=2AB=2c,

OO_‴=2BO_″−OO_″=BO_″+BO=OO_″+2BO=2c+2b.

⋮:

The angular distances between the object and the images, that is, the angles OEO′, OEO″, &c. may be calculated by means of their tangents; thus, if EN be perpendicular to AB,

OEO′	=	NEO′−NEO=tan−1NO′/EN−tan−1NO/EN,
OEO″	=	NEO″+NEO=tan−1NO″/EN+tan−1NO/EN,
OEO‴	=	NEO‴−NEO=tan−1NO‴/EN−tan−1NO/EN, &c.

Thus supposing the distance AB is 5 inches,

that AO	=	2,
BO	=	3,
EN	=	6,
NO	=	1.

Then OO′=4;	OO″=10;	OO‴=14, &c.
OO′=b;	OO″=10;	OO‴=16, &c.

NO/EN=1/6=.1666…	NEO=tan−1.1666…=9°27′1/2 nearly,
NO′/EN=5/6=.8333…	NEO′=tan−1.8333…=39°48′1/2;
∴ OEO′=30°21′,
NO″/EN=9/6=1.5	NEO″=tan−11.5=56°19′1/2;
∴ OEO″=65°46′.

The distances EO′, EO″, &c. are of course the secants of these angles to the radius EN.

41. Suppose now that the mirrors instead of being parallel are inclined to each other, as HI, IK. (Fig. 37.)

In this case, the images of an object O will no longer be in the same straight line, but it will easily be seen, that they will be all at the same distance from the intersection of the mirrors; for if IO, IO′, for example, be joined, the two right-angled triangles IHO, IHO′ are exactly equal in all respects.

There are of course here as before, two series of images,

O′, O″, O‴, &c.O_′, O_″, O_‴, &c.

their number will not, however, be unlimited, as we shall see.

In order to determine their places, we must find the values of the angles OIO′, OIO″, &c. or of the arcs OO′, OO″, which measure them in the circle round I, containing all the images.

Let HIO or HO	=θ, the radius begin taken as the unit,
OIK or OK	=θ′,
HIK or HK	=ι.

Then OO′	=	2HO=2θ,
OO″	=	2KO′−OO′=KO′+KO=2KH=2ι,
OO‴	=	2HO″−OO″=HO″+HO=OO″+2HO
		2HO =2ι+2θ,
OOiv.	=	2KO‴−OO‴=KO‴+KO=OO‴+2KO
		2HO =2ι+2θ+2θ′=4ι,
&c.&c.
OO′	=	2KO=2θ′,
OO″	=	2HO′−OO′=HO′+HO=2HK=2ι,
OO‴	=	2KO″−OO″=KO″+KO=OO″+2KO
		2HO =2ι+2θ′,
&c.&c.

The number of images is not unbounded in this case, as in that of two parallel mirrors, for when any one of the images, as O⁽ⁿ⁾, or O_(n) gets between the lines HI, KI produced, no further reflexion can take place, as no rays proceeding from such a point could fall on the face of either of the mirrors.

In order to express this condition algebraically, we must observe, that of the first series of images, O′, O‴, O^v., &c. lie on one side of the mirrors, and O″, O^iv., O^vi., &c. on the other, and that if O⁽²ⁿ⁺¹⁾, for instance, be the last, the distance KO⁽²ⁿ⁺¹⁾, or 2nι+2θ+θ′, that is, 2nι+ι+θ′, must be the first that is greater than Kk or π,

that is, (2n+1)ι+θ>π,

or 2n+1>π−θ/ι.

If O⁽²ⁿ⁾ be the last image, we must have HO⁽²ⁿ⁾<π;

that is, 2nι+θ>π, or 2n>π−θ/ι,

the same expression as before, 2n+1 being the number of images in the one case, and 2n in the other.

In like manner we should find, that the number of images in the other series is the least whole number greater than π−θ′/ι.

If ι be a measure of π, since θ/ι and θ′/ι are proper fractions, the number of images in each series must be π/ι, and therefore the whole number of images 2π/ι. In this case, however, two images of the different series will coincide; for if ${\frac {\pi }{\iota }}$ be an even number, some one of the distances $OO''$ or $OO_{''}\ (2\iota ),\ OO^{iv}$ or $OO_{iv}\ (4\iota ),$ &c. will become equal to $\pi$, so that $O^{(2n)}$ and $O_{(2n)}$ meet at the opposite point of the circle from $O$; and if ${\frac {\pi }{\iota }}$ be an odd number, we must have $(2n+1)\iota =\pi ,$

that is, $4n\iota +2\iota =2\pi ,$

that is, $2n\iota +2\theta +2n\iota +2\theta '=2\pi ,$

that is, $OO^{(2n+1)}+OO_{(2n+1)}=2\pi .$

It appears then upon the whole, that taking in the object $O$, the whole number of points visible will be ${\frac {2\pi }{\iota }}$.

42.Let us now consider the images produced by spherical reflectors. We must of course find the focus of reflected rays corresponding to each point in the object, and consider the figure which all such foci compose by their aggregation.

As a first instance, suppose there be presented to a spherical mirror, a portion of a sphere concentric with itself, (Fig. 38.). All the points of this are equidistant from the centre of the sphere, and if they be considered as foci of incident rays, the foci of reflected rays will all be at some other distance from the centre, on the same radii with them, so that the image will be a portion of a sphere like the object.

43.Prop.Let a plane object be placed in front of a spherical mirror, so as to be perpendicular to its axis; required the form of the image.

Fig. 39, represents a section of the object and mirror, (a concave one) through the axis. $PQ$ is a part of the section of the object having for its image $pq,\ q$ being the focus of reflected rays answering to $Q,$ and consequently lying on the same diameter with it.

In order to determine the form of the image $pq,$ we will consider it as a spiral curve referred to the centre $E$ as a pole, and $EA$ as an axis.

Then supposing the object $PQ$ to be beyond $E$ the centre of the mirror,

Let $r$ represent the radius $EA,$

$\theta$. . . . . . . . . . . . angle $PEQ,$

$c$. . . . . . . . . . . . . . line $EP,$

$q{\Bigl (}={\frac {c}{\cos \theta }}{\Bigr )}$. . . . . . $EQ,$

$q'$ . . . . . . . . . . . . . . . .Eq^{[errata 1]}.

Then, referring to page 8, we find

${\frac {1}{q'}}={\frac {2}{r}}+{\frac {1}{q}}={\frac {2}{r}}+{\frac {1}{c}}\cos \theta .$

Now the polar equation to an ellipse referred to the near focus is

${\frac {1}{\rho }}={\frac {1}{a(1-e^{2})}}+{\frac {e}{a(1-e^{2})}}\cos \theta .$

which coincides with the above, provided $a(1-e^{2})={\frac {r}{2}},$ and

${\frac {a(1-e^{2})}{e}}=c;\quad \mathrm {that~is,~} e={\frac {r}{2c}},$

and $a={\frac {r}{2(1-e^{2})}}={\frac {r}{2{\Bigl (}1-{\frac {r^{2}}{4c^{2}}}{\Bigr )}}}={\frac {2c^{2}r}{4c^{2}-r^{2}}}.$

It is also necessary, that $e$ be less than unity, that is, $c>{\frac {r}{2}}$.

44.There are some things to be remarked in this elliptic image. We found the quantity $a\ (1-e^{2})$ to be constantly equal to the half radius. Now $a\ (1-e^{2})$ is half the latus rectum of the ellipse, which is also the radius of curvature at the vertex. It appears then, that although the place of the vertex $p,$ and the magnitude of the ellipse, depend on the place of the point $P,$ yet the curvature of the image at the vertex is invariable, as is the latus rectum of the ellipse, which passes through $E$, and is always equal to the radius.

In order to account for this geometrically, we must observe, that rays proceeding from a point at an infinite distance from a mirror, are reflected to the principal focus, which is at the middle of the radius containing that point in its prolongation. Now $PQ$ being perpendicular to $AP$, a line drawn from $E$ to a point in $PQ$ infinitely distant from $P$ or $E$ must be perpendicular to $AE$, and the focus for rays proceeding from such a point will necessarily be $f',$ the middle point of the radius $EN'$ which is perpendicular to $AE$

It appears, that supposing the line $PQ$ to be infinitely extended both ways from $P$, and to be placed in $AE$ produced, at a distance from $E$ greater than half the radius, the image is a portion of an ellipse,^[3] extending from the extremity of the axis major to those of the latus rectum; we shall see hereafter how the ellipse may be supposed to be completed.

It is, however, necessary that we examine what change takes place in the image when $P$ is brought within the limit assigned above, namely, when $EP$ is not less than half the radius, or further when $P$ is placed on the other side of $E$.

In the first place, when $EP$ is half the radius, we have

$c={\frac {r}{2}};\quad \therefore e={\frac {r}{2c}}=1;\quad a={\frac {r}{2(1-1)}}=\infty .$

In this case then the ellipse changes to a parabola, (Fig. 40.)

Suppose now $EP$ be less than half the radius,

$c<{\frac {r}{2}};\quad \therefore e={\frac {r}{2c}}>1;\quad a=-{\frac {r}{2(e^{2}-1)}}=-{\frac {2c^{2}r}{r^{2}-4c^{2})}}.$

Here we have then a portion of an hyperbola.(Fig. 41.)

When $P$ is at the centre, $c=0,\ e=\infty .$ The hyperbola becomes a straight line coincident with $PQ.$

When $P$ is between $A$ and $E$, $EP$ or $c$ is negative; our formula then becomes

${\frac {1}{q'}}={\frac {2}{r}}-{\frac {1}{c}}\cos \theta ,$

which answers to

${\frac {1}{\rho }}={\frac {1}{a(1-e^{2})}}-{\frac {e}{a(1-e^{2})}}\cos \theta ,$

or ${\frac {1}{\rho }}=-{\frac {1}{a(e^{2}-1)}}+{\frac {e}{a(e^{2}-1)}}\cos \theta ,$

the former of which is the equation to an ellipse, when the angle $\theta$ is measured from the farther vertex, and is consequently the supplement of that used in the former cases; the latter is the equation to a pair of hyperbolas.

This latter case we will now examine, as it comes first in order.

Let then $P,$ (Fig. 42.) be between $E$ and $F,$ (the principal focus). In the first place, we know that if $k$ be that point in $PQ$ for which $Ek=EF,$ the image of $K$ must be infinitely distant, so that the line $Ek$ must be parallel to one of the asymptotes.

Again, if $g$ be the point where $PQ$ cuts the circle, its image coincides with it. Every point between $k$ and $g$ has its image without the circle, the distances of these images diminishing gradually from infinity to the radius.

It appears then that the image in this case consists of an hyperbola, and its conjugate wanting the part between the vertex and the extremities of the latus rectum.

When $P$ coincides with $F,$ (Fig. 43.) the image is a parabola, wanting the part about the vertex extending to the extremities of the latus rectum.

The equation in this case takes the form

${\frac {1}{q'}}={\frac {2}{r}}-{\frac {2}{r}}\cos \theta ={\frac {2}{r}}(1-\cos \theta ),$

or $q'={\frac {{\frac {1}{2}}r}{1-\cos \theta }},$

which we know to be that of the parabola, in which the angle $\theta$ is measured from the axis, not beginning at the vertex.

When $P$ is between $F$ and $A$, (Fig. 44.) $c>{\frac {r}{2}};\ e<1,$ and the image is part of an ellipse, namely, all but that part, which we found in the first case.

If we suppose $P$ to go outside of the circle beyond $A$, (Fig. 46.) we shall be led to the case of a convex mirror.

Our equation will still be

${\frac {1}{q'}}={\frac {2}{r}}-{\frac {1}{c}}\cos \theta$

and the image will in all cases be a part of an ellipse, turning its convexity towards $P$.

When $P$ is on the circle at $A$, (Fig. 45.) the image extends from that point both ways to $f,\ f'$ the bisection of the radii $EN,\ EN',$ which are parallel to $PQ$.

When $P$ is at an infinite distance from $A$, the image is a semi-circle with centre $P$, and radius $EF$.

45.We may now show how the curve of the image, which we have in different cases found to be a part of a conic section, may be supposed to be completed.

Supposing in all cases the line to be infinite in extent each way.

In the first place, when $P$ is at an infinite distance, (Fig. 47.) the semi-circle $NAN'$ representing a concave mirror gives a semi-circular image $fFf';$ and the convex mirror represented by $NA'N'$ gives the image $fF'f'$, which completes the circle.

When $P$ is at a finite distance outside the circle, the concave and convex parts give together a complete ellipse $pfp'f'$, (Fig. 48.).

When $P$ is on the circle at $A'$, the ellipse is such as represented in Fig. 49, where $Ep$ is two-thirds of $EF.$

When $P$ is between $A'$ and $F'$, the ellipse cuts the circle, (Fig. 50.)

When $P$ is at $F'$ the middle point of $EA'$, (Fig. 51.) the two parts of the circle divided by the line, unite to produce a complete parabola.

When $P$ is between $E$ and $F',$, the reflexion of the whole circle gives two complete hyperbolas in Fig. 52.

In the first place, the semi-circle $NAN'$ gives the portion of hyperbola $fpf'$. The part $gA'g'$ gives the infinite branches $gh,\ g'h,$ and the conjugate hyperbola $mp'm';$ and the former hyperbola is completed by the reflexions at $Ng,\ N'g',$ considered as convex mirrors. The part $kk'$ of the object has for its images the hyperbola $mp'm',$ and part of $fpf'$ namely, $\kappa p\kappa ';$ the infinite parts of the line outside the circle are represented by $fg,\ f'g',$ and by $f\gamma ,f'\gamma ';$ the remaining parts $kg,\ k'g',$ have for images only $\gamma \kappa ,$ and $\gamma '\kappa '.$

46.In all that has preceded, we have confined our attention to sections of the mirror, object, and image; but of course the reader will not find the smallest difficulty in inferring that the image of a plane object, made by a spherical mirror, is, according to circumstances, a portion of a sphere, a spheroid, a parabolic or hyperbolic conoid, or a plane.

47.By referring to the figures, it will readily be seen that when the mirror is concave, the image is, in most cases,^[4] inverted with respect to the object: a convex mirror always gives an erect image.

48.It will also be seen, that when the image is inverted, it is what is called a real image: when erect, it is imaginary.

49.Let the object presented to a concave mirror be a portion of its own sphere, (Fig. 53.)

Since rays proceeding from $P,$ the extremity of the diameter, are reflected to $p,$ making $Ep$ two-thirds of $EF,$ and that all points of the object are equally distant from the centre, it will readily be seen that the image of the portion of sphere represented by $PQ,$ is a corresponding portion of sphere, $pq,$ having its radius one-third of that of the mirror.

50.Suppose now the object be a portion of any other sphere.

Let $AEOP,$ (Fig. 54,) be the line joining the centres, which we will consider as the axis of the mirror. $PQ$ a part of the object; $pq$ its image.

Let $OP=b;\quad EO=c,\quad EQ=q,\quad Eq=q',\quad QEP,\mathrm {or~} pEq=\theta ,\quad EF=f,\quad Ep=p.$

Then we know that $q=c.\cos \theta +{\sqrt {b^{2}+c^{2}\sin \theta ^{2}}},$ and if to simplify the problem, we suppose $PQ$ and $\theta$ to be very small, we may put for this

$q=c{\Bigl (}1-{\frac {\theta ^{2}}{2}}{\Bigr )}+b-{\frac {c^{2}}{2b}}\theta ^{2}=(b+c).{\Bigl (}1-{\frac {c\theta ^{2}}{2b}}{\Bigr )}.$

Then ${\frac {1}{q'}}={\frac {1}{f}}+{\frac {1}{q}}={\frac {1}{f}}+{\frac {1}{b+c}}\left\{1+{\frac {c\theta ^{2}}{2b}}\right\}$

${\frac {1}{p}}={\frac {1}{f}}+{\frac {1}{b+c}}.$

It is not easy to determine any thing about the form of the curve from this, but we may deduce one rather remarkable conclusion.

The diameter of curvature of $pq$, is equal to ${\frac {pq^{2}}{qr}}$ if $pr$ be a tangent at $p.$ Now this is equal to ${\frac {p^{2}\theta ^{2}}{p\sec \theta -q'}}.$

Hence, calling the radius of curvature $\rho$, we have

${\frac {1}{2\rho }}={\frac {p\sec \theta -q'}{p^{2}\theta ^{2}}}={\frac {p-q'}{p^{2}\theta ^{2}}}+{\frac {p.{\frac {1}{2}}\theta ^{2}}{p^{2}\theta ^{2}}}$

$={\frac {p-q'}{pq'\theta ^{2}}}+{\frac {1}{2p}},$ nearly

$={\Bigl (}{\frac {1}{q'}}-{\frac {1}{p}}{\Bigr )}{\frac {1}{\theta ^{2}}}+{\frac {1}{2p}}$

$={\frac {c}{2b(b+c)}}+{\frac {1}{2f}}+{\frac {1}{2(b+c)}}$

$={\frac {1}{2b}}+{\frac {1}{2f'}},$

or ${\frac {1}{\rho }}={\frac {1}{b}}+{\frac {1}{f}}.$

Now $b$ is the radius of the object, and $f$ is that of the image of a straight line at the vertex; moreover the curvature of a line is aptly measured by the reciprocal of its radius of curvature. It appears then that the curvature of the image of a small portion of a sphere, is equal to that of the object, together with that of the conoidal image of a small plane object, at the same place.^[5]

51.With regard to the magnitude of the image produced by a spherical mirror, it is easy to see that as it subtends the same angle at the centre of the mirror, that the object does, if we suppose them to be plane, an hypothesis which agrees very well with ordinary cases of experiment, the linear magnitudes, that is, the lengths or breadths, of the object and image, will be in direct proportion to their distances from the centre, so that, if we put $L,\ l,$ for the lengths of the object and images, $q_{0}',\ q_{0}'$ for the distances $EP,\ Ep;$

$l=L.{\frac {q_{0}'}{q_{0}}}.$

But $q_{o}=c$, and ${\frac {1}{q_{o}'}}={\frac {2}{r}}+{\frac {1}{c}};\quad \therefore q'={\frac {cr}{2c+r}};$

$\therefore \ l=L.{\frac {r}{2c+r}}=L\div {\Bigl (}1+{\frac {2c}{r}}{\Bigr )},$

or if $f={\frac {r}{2}}$, being the principal focal distance,

$l=L.{\frac {f}{c+f}}=L\div {\Bigl (}1+{\frac {c}{f}}{\Bigr )}.$

52.It might be expected that we should treat of images produced by reflexion at surfaces not spherical, but the subject is in general too difficult for an elementary Treatise like the present. As far as common practical results are concerned, we shall find it sufficient to substitute for surfaces of revolution, portions of spheres having the same curvature, and as to others, plane sections will generally give all the information desired.

Suppose, for instance, the mirror were cylindrical, and convex, and the object a circle placed directly in front of it. It will easily be seen that that diameter of the circle, which is parallel to the axis of the cylinder, will not be altered in the reflexion, but that the diameter perpendicular to that axis with all chords parallel to it, will have for their images portions of conic sections of less breadth than themselves, so that the image will appear diminished in breadth, and distorted into a form like that of the bowl of a spoon.

---

1. We may observe, that the length or height of an upright mirror, in which a man may just see his whole person, must be just half his height, since it stands just half way between his eye and his image, Fig. 35.
2. If the mirror be inclined to the line of the body, its length may of course be easily calculated by Trigonometry. The data required are the height of the body and of the eye, and those which determine the position of the mirror.
3. When $P$ is infinitely distant, the image is a semi-circle for $e{=}0.$
4. When the object is between the mirror and the principal focus, the object is erect, otherwise not.
5. We here suppose that the curvatures of the object and mirror are opposed to each other, that is, are of the same kind; if the radii lie in the same direction, so that the one be convex, and the other concave, the curvature of the image will be less than in the case of a right line, the expression being then ${\frac {1}{\rho }}={\frac {1}{f}}-{\frac {1}{b}}$

Errata

1. Original: EQ was amended to Eq: detail