So for the problem, you have a flat mirror, a light source and a point A . The problem is to move the mirror in a way in wich the light source always hits the target point A, with the light source being able to move on an axis.

So this is basically the problem I've defined to simplify this situation: The light source and the point A are both above the mirror with constant heights $h_1$ and $h_2$. The mirror is fixed on a point and cannot move on the x or y axis, but it can rotate in relation to its center. The distance $k$ between the mirror and the point A in relation to the x axis is constant, and the the distance $x$ between the mirror and the light source in relation to the x axis is variable, since the light source can move in relation to the x axis. The problem is to find the inclination os the mirror as a function of the distance $x$. I don't really know how to proceed from now on to solve the problem.

  • 3
    $\begingroup$ Please provide a sketch! $\endgroup$
    – garyp
    Sep 30, 2017 at 3:00
  • $\begingroup$ Indeed it is difficult to understand the arguments without a sketch. $\endgroup$
    – Bruce Lee
    Sep 30, 2017 at 3:52

1 Answer 1


I think this diagram describes what you are trying to do:

enter image description here

As you see from this drawing, the normal to the mirror needs to bisect the angle between the source (at x=0) and the destination A at (x, h2).

The angle is independent of the height of the light source. The angle that needs to be bisected is $$\theta=\tan^{-1}\frac{x}{h_2}$$

So the angle that the mirror makes with the horizontal is half that, $\frac12 \tan^{-1}\frac{x}{h_2}$


After re-reading your question, I think the above diagram was too simplified, and that you are actually trying to solve a slightly more complicated situation - where the source is no longer directly above the mirror, but the horizontal distance between the source and destination are a fixed distance $k$:

enter image description here

In that case, while the same principles apply we need to work a little bit harder. The angles $b$ and $c$ can be found some simple trigonometry:

$$\begin{align} c &= \tan^{-1}\frac{k-x}{h_1}\\ b &= \tan^{-1}\frac{h_2}{x} \end{align}$$

Now the angle we need to bisect is $(\frac{\pi}{2}-b+c)$, and $a$ is half that angle, minus $c$. So

$$\begin{align} a &= \frac{\pi}{4}-\frac{b}{2}+\frac{c}{2}-c \\ &= \frac{\pi}{4}-\frac{b+c}{2} \end{align}$$

Now we know that $\tan^{-1}x = \frac{\pi}{2} - \tan^{-1}\frac{1}{x}$; this allows us to simplify the expression to

$$a = \frac12\left(\tan^{-1}\frac{x}{h_2} - \tan^{-1}\frac{k-x}{h_1}\right)$$

When $k=x$ (which was the original problem), the result reduces to the same answer that I got before...

  • $\begingroup$ Is the angle θ also independent from the distance $k$ between the mirror and the source? $\endgroup$
    – Matheus C.
    Sep 30, 2017 at 15:18
  • $\begingroup$ @RickSanchez yes the light travels in a straight line so $h_1$ (distance from source to mirror) doesn't matter $\endgroup$
    – Floris
    Sep 30, 2017 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.