# 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 $$\alpha = \frac{3\pi}{4}$$.

• " 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 $\frac{\pi}2$ Dec 9, 2022 at 15:53
• It seems to me that that reasoning of "cannot fly back in to the mirror" applies to discarding the option of $\theta$ as $\frac{3\pi}2$ Dec 9, 2022 at 15:56
• @Alborz I added a picture and a brief explanation why I thought 270° would be more suitable here. Dec 9, 2022 at 16:05
• @Alborz oh, duh, actually drawing out the picture made me realise I ignored an entire 180° in the calculation there, oops Dec 9, 2022 at 16:06

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

• I don't understand why $\gamma = \left(\frac{\pi}{2} - \beta\right) - \pi - \alpha$, sorry... Dec 9, 2022 at 16:25
• @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 $\gamma$. Dec 9, 2022 at 16:29
• 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. Dec 9, 2022 at 16:34
• 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. Dec 9, 2022 at 17:14