I was doing a practice exam for maths and I got the following geometry question: A beam of light reflects against two mirrors with an adjustable angle $\alpha$ between them. For which value of $\alpha$ is the doubly reflected beam of light perpendicular to the source beam?

Illustration of the doubly reflected beam of light.

I named the initial angle of incidence $\beta$ (bottom left) and the final angle of reflection $\gamma$ (top right). The total difference between the incident beam and the reflected beam is then equal to $\theta_{total} = \beta + \pi - \alpha + \gamma$ measured anticlockwise from the initial incident beam.

Same illustration annotated with extra angle names.

If the incident and reflected beams must be perpendicular, then $\theta_{total} \in \left\{ \frac{\pi}{2}, \frac{3\pi}{2} \right\}$, but since the light beam cannot magically fly back into the mirror, the only option here is $\frac{3\pi}{2}$. Which means that $\beta + \pi - \alpha + \gamma = \frac{3\pi}{2}$. Moreover, using the sum of the interior angles of a triangle, $\gamma = \pi - \alpha - \beta$ and therefore $\theta_{total} = \beta + \pi - \alpha + \left(\pi - \alpha - \beta\right) = 2\cdot{\left(\pi - \alpha\right)}$.

Solving for $\alpha$ when $\theta_{total}=\frac{3\pi}{2}$ gives $\alpha = \frac{\pi}{4}$, but this does not work with the given illustration where it's implicitly assumed that $\alpha > \frac{\pi}{2}$ (all the multiple choice answers were also greater than $\frac{\pi}{2}$).

The correct answer is $135°$ or $\frac{3\pi}{4}$, where is the fault in my reasoning?

EDIT: Why I chose $\theta_{total} = \frac{3\pi}{2}$.

I tried to mentally connect the incident and reflected beam at a point and considered the total angular difference. I figured it would look something a bit like this, hence the $270°$ and not just $90°$.

Blue area for theta total.

EDIT 2: The image above actually made me realise I ignore the whole $180°$ traversed when going from the left side of the horizontal to the right before sweeping past $\gamma + \left(\pi - \alpha\right)$. That was the fault in my reasoning.

The correct expression would then be $\theta_{total} = 3\pi - 2\alpha$, in which case I do get the correct value of $\alpha = \frac{3\pi}{4}$.

  • 1
    $\begingroup$ " but since the light beam cannot magically fly back into the mirror, the only option here is $\frac{3\pi}2$" can you clarify this for me please? I'm noting that the correct answer occurs if $\theta$ is in fact $\frac{\pi}2$ $\endgroup$
    – Alborz
    Dec 9, 2022 at 15:53
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    $\begingroup$ It seems to me that that reasoning of "cannot fly back in to the mirror" applies to discarding the option of $\theta$ as $\frac{3\pi}2$ $\endgroup$
    – Alborz
    Dec 9, 2022 at 15:56
  • $\begingroup$ @Alborz I added a picture and a brief explanation why I thought 270° would be more suitable here. $\endgroup$ Dec 9, 2022 at 16:05
  • $\begingroup$ @Alborz oh, duh, actually drawing out the picture made me realise I ignored an entire 180° in the calculation there, oops $\endgroup$ Dec 9, 2022 at 16:06

2 Answers 2


You have to split the problem into three scenarios:

  1. $\alpha>\pi/2$
  2. $\alpha=\pi/2$ (we have parallel lines, it does not work)
  3. $\alpha<\pi/2$
  • You covered case 1. correctly with one exception. In this case there $\beta$ cases for which the light never touches the second mirror - this is if $\alpha+\beta < \pi$. So in this case the answer is conditional: $\alpha=3\pi/4$ but only if $\beta < \pi/4$.

  • The case 2 is more straightforward, if you draw the picture you get a triangle above the mirror which needs to be a right triangle, which quickly leads to the equation $$ (\pi-2\beta)+(2\beta+2\alpha-\pi) + \pi/2 = \pi $$ yielding $\alpha=\pi/4$, regardless of $\beta$.


If the incident angle is $\beta$ then the final angle must be $\frac{\pi}{2} -\beta$.

To see this, imagine a beam going to the right. Its perpendicular direction is straight up which is $\frac{\pi}{2}$ so if we rotate the beam and this direction clockwise by $\beta$ we get that the perpendicular direction goes from $\frac{\pi}{2}$ to $\frac{\pi}{2}-\beta$.

The angle $\gamma$ is the difference between $\frac{\pi}{2}-\beta$ and $\pi-\alpha$ which is $\alpha -\beta -\frac{\pi}{2}$

The angle $\gamma$ is also the incident angle on the second mirror and must be supplementary to $\alpha +\beta$. You can see this by looking at the triangle consisting of $\alpha$ and the two incident points on the mirrors.

Therefore, $$\alpha +\beta +\alpha -\beta -\frac{\pi}{2}=\pi\implies 2\alpha=\frac{3\pi}{2}\implies\alpha =\frac{3\pi}{4}$$

  • $\begingroup$ I don't understand why $\gamma = \left(\frac{\pi}{2} - \beta\right) - \pi - \alpha$, sorry... $\endgroup$ Dec 9, 2022 at 16:25
  • $\begingroup$ @JansthcirlU Look at the diagram. The final angle is $\frac{\pi}{2}-\beta$. The incline is at $\pi-\alpha$. We subtract the second from the first to get $\gamma$. $\endgroup$
    – John Douma
    Dec 9, 2022 at 16:29
  • $\begingroup$ Ooh, I see it now, thanks! The official exam solution used yet another idea where they extended the incident and reflected beams through the mirrors so that they would be perpendicular. I didn't realize there were so many ways to approach this problem. $\endgroup$ Dec 9, 2022 at 16:34
  • $\begingroup$ Guys, I believe in the case $\alpha>\pi/2$, there is a restriction on $\beta$, see my answer. Also, IMO the case $\alpha<\pi/2$ should be treated separately. $\endgroup$
    – Veliko
    Dec 9, 2022 at 17:14

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