An Elementary Treatise on Optics/Chapter 5

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Chap IV.: Reflexion at Curved Surfaces in general

An Elementary Treatise on Optics
by Henry Coddington

Chapter 5

Chap. VI.: Images produced by Reflexion

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2936283An Elementary Treatise on Optics — Chapter 5Henry Coddington

CHAP. V.

OF CAUSTICS PRODUCED BY REFLEXION.

21. We have hitherto considered only the places where reflected rays intersect the axis of a spherical mirror; we will now examine their intersections with each other, and treat the question in a more general manner, that is, we will suppose a cone of rays to be reflected at any surface, any how disposed with respect to the radiant point. We will, however, simplify the question a little by taking only a plane section of the surface through the radiant point.

In Fig. 15, $Q$ represents a point from which proceed rays $QR$ , $QR'$ , $QR″$ … which are reflected at the curve $R$ … $R iv.$ into the directions $Rr$ , $R'r'$ ….

The question being merely one of Plane Geometry, it will immediately occur to the reader that these reflected rays must be very analogous to normals drawn at different points of a curve, and that as these by their mutual intersections produce a broken line which becomes a regular curve, when their number is increased without limit, so the reflected rays should give rise to a similar broken line and curve, which is, in fact, the case, the curve being what is technically termed a caustic.^[1]

22. Prop. Given a point from which a thin pencil of rays proceeding, fall on a spherical reflector, to determine their intersections after the reflexion.

Let $QR$ , $QR'$ (Fig. 16.) be two incident rays, $Rq$ , $R'q$ the reflected rays meeting in $q$ , $RE$ , $R'E$ the normals at $R$ , $R'$ meeting in $E$ , which if we suppose the distance $RR'$ to be, according to the phrase, infinitely small, will be the centre of the osculating circle.

Let $QE$	$= q$
$Eq$	$= t$	$q 2 = u 2 + r 2 -2 ru cos φ$
$QR$	$= u$	$t 2 = v 2 + r 2 -3 rv cos φ$
$Rq$	$= v$
$∠ QRE$ , or $ERq = φ$
$ER$	$= r$ .

Then since a small variation in the place of $R'$ causes an infinitely less variation in that of $q$ , we may establish the following equations by differentiating those above,

$0= udu - rdu cos φ + ru sin φdφ$

$0= vdv - rdv cos φ + rv sin φdφ$ .

Moreover, since $QR$ , $Rq$ , $QR'$ , $R'q$ make respectively equal angles with the curve as in an ellipse, of which $Q$ and $q$ would be the foci,

$QR + Rq = QR' + R'q$ or $d (u + v)=0$ ,

so that our equations now stand

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$0=- vdu + rdu cos φ + rv sin φdφ$ ,

and if we multiply the first by

v

, and the second by

u

, and subtract, we shall have

$0=2 uv -(u + v) r cos φ$ ;
and $∴ v = ur cos φ / 2 u - r cos φ$ , or $1 / u + 1 / v = 2 / r cos φ$ ,

and if we put $2 f$ for $r$ the expressions will be rather simpler,

$v = uf cos φ / u - f cos φ$ , or $1 / u + 1 / v = 1 / f cos φ$ .

23. Here we may observe, that if $u$ be infinite, that is, the incident rays parallel, we have simply $v = f cos φ$ , which referred to the geometrical figure, shows that $Aq$ is in that case one-fourth of the chord of the osculating circle so that if $Ry$ be a tangent, and $Qy$ perpendicular to it, calling $Qy$ , $p$ , we have in general

$1 / u + 1 / v = 2 dp / pdu$ , and $v = pudu / 2 udp - pdu$	$= du / 2 dp / p - du / u$
	$= du / dl p 2 / u$ .

24. Prop. Given the radiant point, and the reflecting surface, to describe the caustic.

In order to determine the nature of the caustic curve, that is, the section of the superficial caustic, we may consider it as a spiral having $Q$ for its pole.

Let $Qx$ (Fig. 18.) be drawn perpendicular on $Rq$ which is of course a tangent to the caustic at $q$ . Call $Qq$ , $u'$ ; and $Qx$ , $p'$

$u' 2$	$= u 2 + v 2 -2 uv cos2 φ$ ,
$p'$	$= u sin2 φ$ ,
$cos φ$	$= p / u$ ; $∴ cos2 φ = 2 p 2 / u 2 -1$ ; $sin2 φ = 2 p / u \sqrt 1- p 2 / u 2$ ;
$∴ u' 2$	$= u 2 + v 2 -2 uv (2 p 2 / u 2 -1)=(u + v) 2 - 4 p 2 v / u$ ,

$p' = u \cdot 2 p / u \sqrt 1- p 2 / u 2 =2 p \sqrt 1- p 2 / u 2$ .

Then if for $v$ be put its value $uf cos φ / u - f cos φ$ , or $du / dl p 2 / u$ , and for $p$ the proper function of $u$ given by the equation to the original curve, and $u$ be then eliminated, we shall have an equation in $u'$ and $p'$ , which will be that of the caustic.^[2]

25. In order to obtain an equation in rectangular co-ordinates, we may proceed as follows:

Reasoning as before, since the caustic is formed by the continual intersections of the reflected rays, two of these are necessary to determine one point of the caustic, and the point where one of them meets the caustic, is that which it has in common with the next; so that if we refer the two reflected rays to the same abscissa, their ordinates, differing in general, coincide at this point, and as far as that point is concerned, a change in the point of the reflecting curve, or in its co-ordinates, takes place without any alteration in the co-ordinates of the reflected ray (Fig. 19.)

We have therefore only to find the equation to the reflected ray belonging to an assumed point of the curve; to differentiate this, considering the co-ordinates of the curve as the only variables, and eliminate these co-ordinates between this equation, its primitive, and that of the curve.

An Example will make this more intelligible.

To diminish the length of the process, we will confine ourselves to the simple case of parallel rays, and take one of them for our principal co-ordinate axis. Let $AN$ (Fig. 20.) be this axis, $AM$ , $MP$ co-ordinates of the curve, $PN$ a normal, $QP$ , $Pv$ an incident and a reflected ray. The question is first to determine the equation of the line $Pv$ . Since this line passes through the point $P$ whose co-ordinates are $x$ , $y$ , the equation must be

$Y - y = α (X - x)$ ,

$α$ being the tangent of the angle $PvN$ .

Now,

$tan PvN =-tan vPQ =-tan2 NPQ =-tan2 PNv$ .

And since $PN$ is a normal,

$tan PNv = dx / dy$ ; $∴ tan2 PNv = 2 dx / dy / 1- dx 2 / dy 2 = 2 dxdy / dy 2 - dx 2$ .

The equation is therefore

$Y - y +2 dxdy / dy 2 - dx 2 (X - x)=0$ (1);

and we have to put for $dxdy / dy 2 - dx 2$ its value in terms of the co-ordinates given by the equation to the curve, and eliminate $x$ and $y$ between this, the equation (1), and its derivative.

26. The process is sometimes facilitated by taking for the variable a function of the angle $PNM$ , as its tangent which is equal to $dx / dy$ . The quantity we have called α is the tangent of twice this angle, and if we put $θ$ for this angle, the equation to the reflected ray is

$Y - y +2\cdot tan θ / 1-tan θ 2 (X - x)=0$ .

Example. Suppose the curve to be a common parabola, its equation is $y 2 =4 αx$ ,

$tan θ = dx / dy = y / 2 α$ ;

$∴ y =2 a tan θ$ ; $x = y 2 / 4 a = a tan θ 2$ .

Then if we put $t$ for $tan θ$ ,

$Y -2 at + 2 t / 1- t 2 (X - at 2)$	$=0$ ;
$∴ Y (1- t 2)+2(X - a) t$	$=0$ (1).

Then differentiating with respect to $t$ ,

(2);

$- ty + X - a =0$ ; $∴ t = X - a / Y$

and substituting this value of $t$ in (1), we have

$Y (1- (x - a) 2 / Y 2)+2 (x - a) 2 / Y$ , or $Y 2 +(x - a) 2 =0$ ;

the equation to a point, namely, the focus, where $X = a$ .

27. The method pursued in the following example is perhaps less elegant than that just given, but it has often the advantage of being simpler and less prolix.

Required the form of the caustic, when the reflecting curve is a common parabola, and the incident rays are perpendicular to the axis.

Let $P$ (Fig. 21.) be a point of the curve $MP$ , $Pq$ an incident and a reflected ray. Then, taking for granted that when $q$ is a point of the caustic, $Pq$ is one-fourth of the chord of the circle of curvature at $P$ , we have the following easy method of determining the co-ordinates of $q$ .

{{{1}}}

$sin φ = y' / \sqrt 1+ y' 2$ ;	$cos φ = 1 / \sqrt 1+ y' 2$ ;
$sin2 φ = 2 y' / 1+ y' 2$ ;	$cos2 φ = 1- y' 2 / 1+ y' 2$ ;

$X =$	$x + v sin2 φ = x - y' / y ″$ ,
$Y =$	$y + 1- y' 2 / 2 y ″$ .

So far all is general; in the particular example proposed,

$y 2 =4 ax$ ; $y =2 a 1 / 2 x 1 / 2$ , $y'= a 1 / 2 x - 1 / 2$ , $y ″=- 1 / 2 a 1 / 2 x - 3 / 2$ ;

$∴ X$	$=$	$x +2 x =3 x$ ,
$Y$	$=$	$y + 1- ax -1 / - a 1 / 2 x - 3 / 2$	$=$	$2 a 1 / 2 x 1 / 2 - x 3 / 2 - ax 1 / 2 / a 1 / 2$
			$=$	$3 ax 1 / 2 - x 3 / 2 / a 1 / 2$
			$=$	$9 a - X / 3 \sqrt 3 a 1 / 2 X 1 / 2$ .

From this it appears that $Y =0$ , or the curve crosses the axis, where $X =9 a$ , which answers to the point in the parabola for which $x =3 a$ ,

$dy / dx$	$=$	$1 / 3 \sqrt 3 a 1 / 2 {9 a - X / 2 X 1 / 2 - X 1 / 2$ }
	$=$	$1 / 6 \sqrt 3 a 1 / 2 \cdot 9 a -3 X / X 1 / 2$ .

When $x =0$ this is infinite, so that the caustic like the reflecting curve is perpendicular to the axis at its origin: when

$x =9 a$ , $Y =0$ ; $dY / dX = 1 / 6 \sqrt 3 a 1 / 2 \cdot -18 a / 3 a 1 / 2 = -1 / \sqrt 3$ .

The angle at which the caustic afterwards cuts the axis is therefore that having for its natural tangent $1 / \sqrt 3$ , which shows it to be one of $30°$ .

The curve extends without limit in the same directions with its generating parabola.

28. Required the form of the caustic when the reflecting curve is an ellipse, and the radiating point its centre. Fig. 22, &c.

The polar equation to an ellipse about its centre being

$p 2 = a 2 b 2 / a 2 + b 2 - u 2$ ,

we have

$p 2 / u = a 2 b 2 / u (a 2 + b 2 - u 2)$ ,

and $dl p 2 / u / du = 1 / u + 2 u / a 2 + b 2 - u 2 = 3 u 2 -(a 2 + b 2) / u (a 2 + b 2 - u 2)$ ;

$∴ v = a 2 + b 2 - u 2 / 3 u 2 -(a 2 + b 2) u$ .

Hence, when $u = a$ ,	$v = b 2 / 2 u 2 - b 2 a$ ,
and when $u = b$ ,	$v = a 2 / 2 b 2 - a 2 b$ .

The former of these values is always essentially positive, since $a$ is supposed to represent the semi-axis major, and therefore $2 a 2$ must be greater than $b 2$ ; but $2 b 2$ may be equal to, or greater than $a 2$ , so that when $u = b$ , $v$ may be infinite or negative.

When $2 b 2 > a 2$ , the form of the caustic is such as that shewn in Fig. 22.

When $b = \sqrt 3 / 2 a$ , $u = b$ gives $v =2 b$ , and the curve is that of Fig. 23.

When $2 b 2 = a 2$ , we have infinite branches asymptotic to the axis minor, as in Fig. 24.

When $2 b 2 < a 2$ , there are asymptotes inclined to that line (Fig. 25.).

29. There are some simple cases in which it is easy to determine the nature of the caustic by geometrical investigation.

Prob. To find the form of the caustic, when parallel rays are reflected by a spherical mirror. (Fig. 26.)

Taking as usual a section of the mirror, let $Pp$ be one of the reflected rays, touching the caustic in $p$ . Then, since we know that in this case $Pp$ is one quarter of the chord, if $EP$ be bisected in $O$ , and $Op$ be joined, $OpP$ will be a right angle, and if a circle be described through the points $PpO$ , $OP$ will be the diameter of it. Let $R$ be the centre, join $Rp$ . Then since $OPp = EPQ = PEA$ , if a circle be described with centre $E$ and radius $EO$ , cutting the axis $EA$ in $F$ , the principal focus, the arc $OF$ which measures the angle $PEA$ to the radius $EF$ , must be equal to the arc $Op$ , which measures twice the angle $OPp$ to the radius $OR$ , which is half of the other radius $EO$ .

It is plain therefore that the curve $CpF$ must be an epicycloid described by the revolution of a circle equal to $PpO$ , on that of which $FO$ is a part: and that there must be a similar epicycloid on the other side of the axis; moreover that if the other part of the circle, $CBc$ , represent a convex mirror, there will be a similar pair of epicycloids formed by the intersection of the reflected rays, considering them to extend behind the mirror without limit as all straight lines are supposed to do in the higher analysis.

Prob. To describe the caustic given by a spherical reflector, when the radiating point is at the extremity of the diameter.

We shall see that the section of the caustic consists again of two epicycloids.

Let $Q$ , (Fig. 28.) be the radiating point. $Pp$ a reflected ray touching the caustic in $p$ .

It appears from the equation $1 / u + 1 / v = 2 / r cos φ$ , that $Pp$ is in this case one-third of the chord, since $u =2 r cos φ$ ; and therefore if $E$ , $G$ be joined, and $PE$ be trisected in the points $R$ , $O$ , we shall have the triangle $PRp$ similar to $PEG$ , two sides of which being radii, $PR$ must be equal to $Rp$ . With centre $R$ let a circle be described through the points $P$ , $p$ , $O$ ; and with centre $E$ and radius $EO$ , another circle which will of course cut $EC$ in $V$ , the focus of rays reflected at points infinitely near $A$ , $AV$ being one third of $AQ$ . Then, since the angle $ORp$ is double of $OPp$ , it must be equal to the angle $AEP$ , which is double of $EPQ$ , and the radii $EO$ , $RO$ being equal, it follows, that the arcs $Op$ , $OV$ , must in every case be equal, and that the locus of the point $p$ is an epicycloid described by the revolution of a circle equal to $PpO$ round that of which $OV$ is a part.

30. These are the only cases in which the caustic to a spherical reflector is a known curve; we may, however, without much difficulty, make out the kind of figure that it assumes in other cases, though the investigation of the equation is very tedious.

When the radiant point is at a finite distance without the circle, the form of the caustic is naturally intermediate to those already found. It is represented in Fig. 27. The caustic touches the circle at the points $C$ , $c$ where tangents from $Q$ meet it, and it has a cusp at $V$ , the focus of the principal reflected rays. There is also an imaginary branch $Cvc$ .

When $Q$ is within the circle, and $EQ$ is greater than half the radius, the caustic takes the form represented in Fig. 30, with a cusp at $V$ , and two others at $C$ , $c$ . There are two infinite branches extending along the asymptotes $DG$ , $DG'$ .

There is another part with a cusp at $v$ , and two infinite branches having the same lines $DG$ , $DG'$ for asymptotes.

When $Q$ bisects the radius $EB$ , the caustic is such as represented in Fig. 30. $V$ bisects $EF$ , and the axis $EA$ supplies the place of the asymptotes of the last case.

When $EQ$ becomes less than half the radius, the caustic contracts to the form shown in Fig. 31. $EV$ is then less than half $EF$ .

When $Q$ comes to the centre, the caustic is reduced to that one point.

31. Prob. Supposing the form of the reflecting surface to be that generated by the revolution of a cycloid about its axis, and the incident rays to be parallel to that axis, it is required to describe the caustic. (Fig. 32.)

Let $Pp$ be a reflected ray, touching the caustic in $p$ , $PG$ the normal. Then, since $Pp$ must be one-fourth of the chord of the oscillating circle, and that in the cycloid, the normal is one-fourth of the diameter of that circle, $GpP$ must be a right angle. Let $GK$ be the diameter of the generating circle of the cycloid, when the describing point is at $P$ ; $H$ the intersection of this with $Pp$ . Then, since $HpG$ has been proved to be a right angle, a circle passing through $H$ , $p$ , $G$ , will have $HG$ for its diameter. Moreover $H$ is the centre of the circle $GPK$ , for the angle $HGP$ is equal to $GPN$ , and consequently to $GPH$ . Since then the radius of the smaller circle $HpG$ is half that of the other, and since the angle $GHp [errata 1]$ is double of the angle $GPr$ , it follows, that the arcs $Gp$ , $Gr$ are equal, that $Gp$ is equal to $GD$ , and that the locus of $p$ is a cycloid having $GpH$ for its generating circle.

The surface generated by the revolution of this cycloid about $AD$ , has of course a cusp at $D$ .

32. Enough has probably been said on this subject, which might without much difficulty be prosecuted further, but it is one rather of curiosity than utility, and more suited to the speculative Geometer, or Algebraist, than the practical Optician.

33. Some writers have investigated expressions for the density of rays, or brightness, at different parts of a caustic: it may be sufficient to observe, that it is much greatest at and near a cusp.

34. Some of the forms of caustics above described may be exhibited in an imperfect manner. If a concave reflector^[3] be placed so as to reflect directly the rays of the Sun admitted through an aperture into a dark room in which there is a good deal of dust or smoke, there will be observed a bright funnel-shaped form, such as that represented in Fig. 33. This, however, is not a simple caustic, such as that described in Fig. 26, because the solar rays would not in any case converge to one single point, but to a circular image, as we shall see in the next Chapter.

35. It is somewhat remarkable, that an infinite number of different mirrors may reflect rays proceeding from a given point, so as to produce the same caustic.

Let $Q$ , (Fig. 17.) be the radiant point, $MqN$ the caustic, $Rq$ a reflected ray touching it in $q$ . All we know about the reflecting curve $ARB$ is, that every pair of lines such as $QR$ , $Rq$ , make equal angles with it. Now it will easily be seen, that there is no limit to the number of curves that will answer this condition, any more than to the number of ellipses that may have the same two foci. If $QRq$ were part of a string fastened at $Q$ and $N$ , so as to lap round the part $Nq$ of the caustic, and extended by a pin, this pin would describe the reflecting curve $ARB$ , and of course it is only necessary to change the length of the string to get any number of different curves.

The algebraical explanation of this seems to be, that the equation $du + dv =0$ , requires for its integration a constant, which is just as arbitrary here as in the case of the ellipse in which the line represented by $v$ is always measured from the same point.

↑ Strictly speaking this curve is but a section of the caustic, which is a surface like the reflector which produces it.
↑ For instance, if the reflecting curve be a logarithmic spiral, and $Q$ its pole, its equation is of the form $p = mu$ ,

$v = du / 2 dp / p - du / u = u$ ; $p' =2 mu \sqrt 1- m 2$ ; $u' 2 =4 u 2 -4 m 2 u 2 \cdot u / u =4 u 2 (1- m 2)$ ,

whence we find $p' = mu'$ ; and the caustic is therefore another logarithmic spiral differing from the former only in position.
↑ The inside of the cover of a watch will answer the purpose extremely well.

Errata

↑ Original: GHP was amended to GHp: detail

[1] Strictly speaking this curve is but a section of the caustic, which is a surface like the reflector which produces it.

[2] For instance, if the reflecting curve be a logarithmic spiral, and $Q$ its pole, its equation is of the form $p = mu$ ,

$v = du / 2 dp / p - du / u = u$ ; $p' =2 mu \sqrt 1- m 2$ ; $u' 2 =4 u 2 -4 m 2 u 2 \cdot u / u =4 u 2 (1- m 2)$ ,

whence we find $p' = mu'$ ; and the caustic is therefore another logarithmic spiral differing from the former only in position.

[4] The inside of the cover of a watch will answer the purpose extremely well.

[3] Original: GHP was amended to GHp: detail

[1]

[2]

[errata 1]

[3]