An Elementary Treatise on Optics/Chapter 1

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Introductory Observations

An Elementary Treatise on Optics
by Henry Coddington

Chapter 1

Chap. II.: Reflexion at Spherical Surfaces

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2920004An Elementary Treatise on Optics — Chapter 1Henry Coddington

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	$φ$ ₁,
that at	$S$ ……	$φ$ ₂,

and so on.

Let also $a$ represent the angle at $I.$

The reader will find no difficulty in following these equations,

$\phi _{1}-\phi _{2}=a,$

$\phi _{2}-\phi _{3}=a,$

$\phi _{3}-\phi _{4}=a,$

:: $\quad \quad \quad \quad$

$\phi _{1}-\phi _{n}={\overline {n-1}}\ a,$

or $\phi _{n}-\phi _{1}-{\overline {n-1}}\ a.$

If now $\phi _{1}$ be any multiple of $a,$ as ${\overline {n-1}}\ a,$ we shall have somewhere $\phi _{n}=0;$ that is, some reflected ray will be perpendicular to one of the mirrors, and these of course will end the series of reflexions.

If $\phi _{1}$ be not a multiple of $a,$ some value of $n$ will make ${\overline {n-1}}\ a,$ greater than $\phi _{1},$ and then $\phi _{n}$ will become negative. The geometrical fact indicated by this is that the broken line $QRST...$ will at length be turned back upon itself, and the light after coming down the angle $HIK$ will go up again.

Let $a$ be the intersection of $QR$ and $ST,$

$b$ . . . . . . . . . . . . . . . . . . $ST$ and $UV,$

$g$ . . . . . . . . . . . . . . . . . . $QR$ and $UV.$

Then it will immediately be seen that the value of the angle at $a$ is $2\phi _{1}-2\phi _{2},$ or $2a;$ that at $b$ is the same; that at $g$ is double of these or $4a:$ so that if we represent the lines $QR,\ RS,\ ST,...$ by $q_{1},\ q_{2},\ q_{3},...$ and the angles between them by ${\overline {q_{1}q_{2}}},\ {\overline {q_{1}q_{3}}},$ &c., we shall have

${\overline {q_{1}q_{3}}}={\overline {q_{3}q_{5}}}=...{\overline {q_{n-1}q_{n+1}}}=2a,$

${\overline {q_{m}q_{n+m}}}=2na,$

(provided $m$ be an odd number.)

Let $k$ be the intersection of $QR$ and TU^{[errata 1]}.

$\angle gkR=kRT+kTR$

$={\frac {\pi }{2}}-\phi _{1}+{\frac {\pi }{2}}-\phi _{3}$

$=\pi -(\phi _{1}+\phi _{3});$

$\therefore {\overline {q_{1}q_{4}}}=\phi _{1}+\phi _{3}=2\phi _{1}-2a.$

It will be easily seen, that

${\overline {q_{1}q_{6}}}=\phi _{1}+\phi _{5}=2\phi _{1}-4a,$

${\overline {q_{1}q_{2m}}}=\phi _{1}+\phi _{2m-1}=2\phi _{1}-(2m-2)a.$

Let us resume now the first equation

\phi _{n}=\phi _{1}-{\overline {n-1}}\ a.

Suppose $a={\frac {2\pi }{n-1}};$ then $\phi _{n}=\phi _{1}-2\pi ,$ or $-\phi _{n}=2\pi -\phi _{1}.$

The first and n^th angles will then be equal.

We must observe, that there is a limit to the angle of incidence after it becomes negative, namely, the double right angle; if it becomes exactly equal to this, the last ray will be parallel to one of the mirrors; if greater, it would meet it if produced backwards.

If