Required angle to deflect a beam of light 90° 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?

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.

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 $90°$.

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 $\alpha = \frac{3\pi}{4}$.
 A: You have to split the problem into three scenarios:

*

*$\alpha>\pi/2$

*$\alpha=\pi/2$ (we have parallel lines, it does not work)

*$\alpha<\pi/2$


*

*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 $\beta$.
A: 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, $$\alpha +\beta +\alpha -\beta -\frac{\pi}{2}=\pi\implies 2\alpha=\frac{3\pi}{2}\implies\alpha =\frac{3\pi}{4}$$
