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6. . 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.

$QR$, (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 $I$. and so on.