Mirror system in geometrical optics

[Hak]

Homework Statement

A mirror system consists of two perfectly reflecting half-planes arranged to form an angle of 45°. Let two points, A and B, be given within the angle formed by the two mirrors. Prove that regardless of their position, there are no optical trajectories from A to B with more than 4 reflections.

Relevant Equations

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I tried solving it in this way, distinguishing 3 cases and passing the first ray through the point located at a greater distance from the vertex of the angle formed by the two mirrors:

1) the angle of incidence of the ray on the mirror inclined 45° to the horizontal is 45° or 0°: in the first case the ray is parallel to the mirror placed on the horizontal, reflects perpendicular to it, then returns back following the same initial path; in the second case, after the second reflection the ray reflects parallel to the inclined mirror

2) the angle of incidence is greater than 45°: after the fourth reflection the ray begins to diverge from the inclined mirror.

3) the angle of incidence is less than 45°: after the third reflection the ray begins to diverge with respect to the mirror placed on the horizontal.

This is a purely intuitive demonstration, however, I should explain why the ray diverges. Perhaps because in case 2) after the fourth reflection the angles that the rays form with the horizontal are less than 45°. In case 3), on the other hand, the rays after the third reflection form angles greater than 45° with respect to the normal to the inclined mirror.

I do not see a quick and clean way to prove this. Do you have any hints?

 

[haruspex]

Replicate the actual octant into eight octants around the origin. Throw away the mirrors. Just consider how a ray moves through the octants.

 

[Hak]

Replicate the actual octant into eight octants around the origin. Throw away the mirrors. Just consider how a ray moves through the octants.

Thank you very much. I followed your advice, but I don't know if my process is right.

First, let us imagine that the two mirrors form an octant of a cube, with the origin at the vertex of the angle. Then, we can replicate this octant into eight octants around the origin. Each octant has a different color and a label from 1 to 8.

Now, we can throw away the mirrors and just consider how a ray moves through the octants. A ray that starts from point A in octant 1 can only reach point B in octant 8 if it crosses four planes that separate the octants. For example, one possible path is 1-2-4-7-8. This path corresponds to four reflections on the mirrors.

We can prove that there is no path with more than four reflections by using a parity argument. Let us assign a parity (even or odd) to each coordinate of a point in the octants. For example, point A has coordinates $(x,y,z)$, where $x$ is even, $y$ is odd, and $z$ is odd. Point B has coordinates $(-x,-y,-z)$, where x is odd, $y$ is even, and $z$ is even.

A ray that crosses a plane changes the parity of one coordinate. For example, a ray that crosses the $xy$-plane changes the parity of $z$. A ray that starts and ends at points with different parities must cross at least three planes. A ray that starts and ends at points with the same parities must cross at least four planes.

Since point A and point B have different parities for all three coordinates, a ray from A to B must cross at least three planes. Therefore, there are no paths with more than four reflections.

How can this explanation be correct? Can you provide more correct and detailed information?

 

[haruspex]

imagine that the two mirrors form an octant of a cube

No, that's not what I meant. I was using octant in a 2D sense, i.e. a 45° wedge.

 

[Hak]

No, that's not what I meant. I was using octant in a 2D sense, i.e. a 45° wedge.

I didn't fully understand your method. I'll try to explain what I had thought, but I don't think it's relevant to what you recommended. I would appreciate any other suggestions on this problem.

Suppose we have a ray of light that travels from point A to point B in the octant formed by the two half-planes. We can label this ray as AB. Now, imagine that we replicate this octant into eight octants around the origin, as suggested by the hint. We can label these octants as I, II, III, …, VIII in a counterclockwise order, starting from the original one. We can also label the corresponding points A and B in each octant as $A_1$, $A_2$, $A_3$, …, $A_8$ and $B_1$, $B_2$, $B_3$, …, $B_8$.

Now, let us see what happens when we introduce the mirrors. Each mirror will reflect the ray of light into another octant. For example, if the ray AB hits the mirror along the $x$-axis, it will be reflected into octant II and become $A_2B_2$. Similarly, if the ray AB hits the mirror along the y-axis, it will be reflected into octant IV and become $A_4B_4$. In general, if the ray AB hits the mirror along the x-axis n times and the mirror along the y-axis m times, it will be reflected into octant $(n+m)$ mod 8 and become $A_n+B_n$.

We can consider two cases:

Case 1: A and B are in different octants. In this case, we can see that there is only one possible way for a ray of light to travel from A to B without hitting any mirror. This is because any reflection will change the octant of the ray and make it miss B. Therefore, there are no optical trajectories from A to B with any reflections in this case.

Case 2: A and B are in the same octant. In this case, we can see that there are multiple possible ways for a ray of light to travel from A to B with different numbers of reflections. However, we can also see that there is a limit on how many reflections are possible. This is because each reflection will change the direction of the ray by 90°. Therefore, after four reflections, the ray will return to its original direction and repeat its previous path. This means that there are no optical trajectories from A to B with more than four reflections in this case.

What do you think? I can't prove it except intuitively. What more can you tell me?

 

[haruspex]

Each mirror will reflect the ray of light into another octant. For example, if the ray AB hits the mirror along the x-axis, it will be reflected into octant II and become A2B2.

Not sure how you are thinking of the ray being reflected into the next octant.
Consider a ray from A striking the X axis in the first octant at point P. Instead of drawing its reflection in the first octant, as a line PQ say, draw the reflection of PQ (PQ') in the X axis.
What is the relationship between the lines AP and PQ'?

 

[Hak]

Not sure how you are thinking of the ray being reflected into the next octant.
Consider a ray from A striking the X axis in the first octant at point P. Instead of drawing its reflection in the first octant, as a line PQ say, draw the reflection of PQ (PQ') in the X axis.
What is the relationship between the lines AP and PQ'?

To answer your question, I will use the hint you provided and try to explain the steps of the proof. First, let us draw a diagram of the mirror system and the points A and B in the first octant, as shown below:

The two mirrors are represented by the x-axis and the y-axis, and the angle between them is 45°. The points A and B are arbitrary points in the first octant, and we want to find the optical trajectories from A to B with reflections on the mirrors.

As suggested by the hint, we can replicate the actual octant into eight octants around the origin by extending the x-axis and the y-axis. Then, we can throw away the mirrors and consider how a ray moves through the octants. For example, a ray from A that strikes the x-axis at point P will reflect into the fourth octant as a line PQ. Instead of drawing this reflection, we can draw the reflection of PQ (PQ’) in the x-axis, which will be in the second octant. This is equivalent to imagining that there is no mirror at the x-axis, and the ray continues its path through the second octant.

The relationship between the lines AP and PQ’ is that they are parallel to each other, since they have the same slope. This is because the angle of incidence equals the angle of reflection at point P.

Is it correct?

 

[haruspex]

The relationship between the lines AP and PQ’ is that they are parallel

More than that, they are one straight line. What about the next 'reflection' of the ray?

 

[Hak]

More than that, they are one straight line.

Why?

 

[haruspex]

Why?

Say AP is angled 60° below the horizontal, i.e. it goes right and down at 30° to the vertical. With a mirror at the X axis, PQ would rise to the right at 30° to the vertical. Reflecting that line in the X axis produces PQ', descending to the right at 30° to the vertical. APQ' lie in a straight line. This is the path taken without the mirror.