Capture a laser beam

Question

Design a mirror box that can capture a laser beam, so that it will keep reflecting forever. The setup looks like in the following image:

The goal is to design a box in a way, that the light beam will never escape.

Rules:

• The walls are perfectly smooth and reflective. The beam will always reflect under the same angle it impacts the wall. There is no absorption. Also, no quantum-mechanical effects exist in this universe. One-directional mirrors are not permitted either
• If the beam escapes the box and points away to infinity, the game is over
• If the beam ever returns back to its source, it is also game over
• Neither the box or any part of it, or light source can move. The setup is completely static, except for the laser beam

Preference:

• This is not a hard rule, but the simple the solution is, the better. Ideally using only basic 2D/3D geometry. However, if you have an interesting solution using some more advanced geometric, or mathematical concepts, it is more than welcome!

Disclaimer: I do not have a solution. Let's create one!

Updates

Based on the feedback from other users, I provide additional details here:

• The shape of the box is not limited to a rectangle, it can be anything, polygon, sphere, even more exotic shapes are allowed (like fractals for examples), or shapes from non-euclidean geometry, higher number of dimensions than 3 is also allowed, although not preferred. Mirrors do not need to be straight lines, they can be curved in any way you see fit.
• Laser source and box hole sizes are not negligible.

Answer 1

Assuming that you can create curved mirrors with infinite precision, then you could catch a ray ...

between two parabolic curves, as shown in the image. . If approached at the right angle, then the ray will get more and more vertical with every reflection. In fact every bounce is about (*) half as far from the Y-axis than the previous bounce, therefor the ray will not reach the Y-axis in a finite length (nor in a finite time).

About that footnote:

(*) In the limit this will become exactly half as far with every bounce.

If curved mirrors are more expensive then you could simplify the room:

Since the figure is symmetrical, you could replace one mirror by a straight mirror on the X-axis.

A real life equivalent (albeit with a reverse light direction):

Looking into the polished bell of a trumpet (or other brass wind instrument). Notice the environment repeating in rings with increasing number of internal reflections. With the proper bell shape and under the right viewing angle, the rings can get increasingly narrow and converge into a (dark) singularity.

Answer 2

As long as we assume that the hole the laser goes in is not infinitesimal in size, and the laser can escape if it hits anywhere "near enough", then any solutions based on irrational numbers are out, and the only remaining solution is to capture the laser beam in some kind of fixed loop where it goes around forever.

To capture the beam in an infinite loop,

you'd need tools that are not available for us in this setup.

The way to see this is to build a mirror setup that captures the laser in an infinite loop, and then

reverse the laser beam.

If we do that, every reflection will still happen at exactly the same place, in exactly the same way, but now

the reversed laser will travel backwards in the same infinite loop (no matter how complex) for an infinite time, and after that, the reversed laser will somehow decide to leave the loop and hit the laser gun.

Obviously, that's not how infinite loops work, so

it's impossible to capture the laser using static mirrors alone.

NB: In the comments, @JaapScherphuis astutely points out that in the first paragraph, my argument does not rule out non-looping infinite paths for all possible mirror shapes. I do confess to thinking about "boxes" as consisting of a finite number of straight mirror segments.

Answer 3

Similar to Kris Van Bael's answer, but not only trapping a single ray of light:

Two parabolic mirrors of different size can trap a continuum of vertical rays: Credits go to Selene Routley.

Why this works:

The parabolic mirrors have the following forms: $$\begin{align}p_a (x) &= a x^2 - \frac{1}{4a}, & x \in [-\frac{1}{2a}, \frac{1}{2a}] \\ p_b (x) &= -b x^2 + \frac{1}{4b}, & x \in [-\frac{1}{2b}, \frac{1}{2b}]\end{align}$$ Their vertical offsets are chosen such that both have their focal points at $(0, 0)$. In the figure, the upper mirror has a higher curvature than the lower one, i.e. $b > a$ (in this particular case, I chose $b = 2a$). If a ray impinges vertically on the lower mirror, it is reflected through the focal point. Since this is also the focal point of the small mirror, it will reflect the ray back into vertical direction. This vertical ray is closer to the symmetry axis of the mirrors. Each further round-trip will bring the ray closer to the symmetry axis until it converges to $x_\infty = 0$.

Proof of convergence:

We need to show generally that a vertical ray in the $i$th round-trip impinging on the lower mirror at $x_i$ impinges on the lower mirror after another round-trip at $x_{i+1}$, such that $\left| x_{i+1} \right| < \left| x_i \right|$.
After reflection from the lower mirror, the ray is defined by passing through the points $(x_i, p_a (x_i))$ and $(0, 0)$. That means, it can be described by the straight line $y_i (x) = m_i x$ with slope $m_i = \frac{p_a (x_i)}{x_i}$. It hits $p_b (x)$ at $$x_{i+1} = \frac{m_i \pm \sqrt{{m_i}^2 - 4(-b)\frac{1}{4b}}}{-2b} = \frac{m_i \pm \sqrt{{m_i}^2 + 1}}{-2b} \tag{1}$$ Out of the two possible solutions, only one lies within the parabolic mirror $x \in [-\frac{1}{2b}, \frac{1}{2b}]$. If the slope of the ray is positive $m_i > 0$, it hits the parabola $p_b$ at a positive $x_{i+1}$, that is the "$-$"-solution of the above equation. Similarly, if the slope is negative $m_i < 0$, $x_{i+1}$ is negative, which is the "$+$"-solution. Since the negative-slope cases can be mirrored to positive-slope cases, we focus on these here. Hence, we have $x_i < 0$, $m_i > 0$, $x_{i+1} > 0$ and must show that $-x_i > x_{i+1}$. Using the definition of $m_i$ in Eq. $(1)$, we get $$\begin{align}x_{i+1} &= \frac{\frac{p_a (x_i)}{x_i} - \sqrt{\left( \frac{p_a (x_i)}{x_i} \right)^2 + 1}}{-2b} = \frac{\frac{a {x_i}^2 - \frac{1}{4a}}{x_i} - \sqrt{\left( \frac{a {x_i}^2 - \frac{1}{4a}}{x_i} \right)^2 + 1}}{-2b} \\ &= \frac{\frac{a {x_i}^2 - \frac{1}{4a}}{x_i} - \sqrt{\left( \frac{a {x_i}^2 + \frac{1}{4a}}{x_i} \right)^2}}{-2b} = \frac{\frac{a {x_i}^2 - \frac{1}{4a}}{x_i} - \left| \frac{a {x_i}^2 + \frac{1}{4a}}{x_i} \right|}{-2b} \\ &= \frac{\frac{a {x_i}^2 - \frac{1}{4a}}{x_i} - \frac{a {x_i}^2 + \frac{1}{4a}}{\left| x_i \right|}}{-2b} = \frac{\frac{a {x_i}^2 - \frac{1}{4a}}{x_i} + \frac{a {x_i}^2 + \frac{1}{4a}}{x_i}}{-2b} = \frac{\frac{2a {x_i}^2}{x_i}}{-2b} = -\frac{a}{b} x_i.\end{align}$$ Given the condition $b > a$, we indeed have $-x_i > x_{i+1}$.

Answer 4

If we take a more mathematical theoretical approach, instead of a realistic one..

we could create an infinity box, based on the following limit:
${\displaystyle {\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+\cdots =\sum _{n=1}^{\infty }\left({\frac {1}{2}}\right)^{n}=1.}$
reference: wikipedia

This assumes the laser beam and the thickness of the walls could get infinitely small. The box could then look something like this:

Answer 5

Partial Mathematical Response:

This problem seems to share a lot of properties with the Assassin Puzzle, where a finite number of guards are used to protect a target from bullets that can bounce an infinite number of times. This problem seems similar except that the assassin and target are at the same point!

I shared some links on this answer on the math stack exchange, in particular :

• https://www.math3ma.com/blog/is-the-square-a-secure-polygon - this page outlines the solution to the original problem.

One simplification made is that they assumed that the room in question is a square. This allows the room to be tessellated across the plane and for the bullet to be modeled as a straight line.

Unfortunately, our solution is likely not a square, but perhaps the "guards" in the assassin puzzle could be special points in our problem where a corner could be placed and re-shape our box in such a way that our light does not escape.

This means that perhaps a box with a finite number of curves and no corners could solve the problem above. (The wonderful solutions by @Lezzup and @Kris van Bael assume an infinite number of straight line segments or infinitely smooth curves.) Unfortunately, I am not familiar enough with the math to dive deeper into this question but I leave this information here in hopes someone else with more expertise can!