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Say we pointed a laser that is projected into a $90^{\circ}$corner of a mirror made by perpedicular walls. Ray at 90 degree corner

I know that you cant use tangent since a derivative of boundary of mirror would not exist at a corner. But this didn't convince me that a reflection could not exist.

So I imagined projecting a incident ray of the same slope as close as possible to the corner.

enter image description here

I thought that the ray of incidence would be in the same as the ray of incidence except the reflected ray moves in the opposite direction.

RefletedRay

However, I am unsure if this is correct as I do not have a corner mirror. I'm also not sure if a reflection can exist in a corner.

So am I correct or do laws of physics state otherwise? And if I am not correct where would the relfected ray project?

Arbuja
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    This is a typical situation where the ray approximation fails and you would have to use the wave equations to calculate what happens. – CuriousOne Jun 28 '16 at 00:58
  • I have little understanding in physics as I am not sure where to start. I have sufficient knowledge of pure mathematics so I took this approach. I will what I can learn about wave equations. – Arbuja Jun 28 '16 at 01:12
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    Unfortunately there is no easy theory to this. You can look at papers like https://www.math.wisc.edu/~jin/PS/corner.pdf and this thesis to get a few ideas and pointers where to look further. – CuriousOne Jun 28 '16 at 01:36
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    @CuriousOne Thanks. I am amazed that problem like this is very complicated. – Arbuja Jun 28 '16 at 01:39
  • @CuriousOne In the paper, the reflection of light is estimated on an outward corner rather than an inward. The inward corner may not have the same principles. – Arbuja Jun 28 '16 at 13:16
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    The principles are the same, the result may differ somewhat (but there may be a geometric transformation that gets you from one case to the other, I am not sure about that). You basically have to follow the mathematical methods developed in these papers (also follow the citations) to derive the effective scattering caused by a sharp edge. – CuriousOne Jun 28 '16 at 13:21
  • @CuriousOne I don't think the ray approximation necessarily breaks down here. It fails for your example because the predicted scattering changes discontinuously with the position of the incident light beam. But that's not true in this example. – knzhou Jun 28 '16 at 21:16
  • @CuriousOne I agree that generically you have to go to the wave equations, though. – knzhou Jun 28 '16 at 21:17
  • @knzhou: One does have to describe the scattering on the edge, though. It's an interesting question if one can use the solution from the ray approximation and then superimpose the scattering of the edge. It sounds reasonable to believe that this may be a good approximation, maybe even a correct solution, but I don't know enough about these problems to make that claim. – CuriousOne Jun 28 '16 at 22:01
  • @CuriousOne : Your comments really belong in an Answer (http://meta.physics.stackexchange.com/questions/8821/the-practice-of-answering-a-question-in-its-comments-area). Then the OP can indicate if he is satisfied with your explanation. Otherwise it looks as though the Question has no Accepted Answer. – sammy gerbil Jun 29 '16 at 06:35

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I think you are perhaps confusing reflection at a corner with a "corner reflector" (https://en.wikipedia.org/wiki/Corner_reflector). The latter is a type of mirror which reflects light back to where it came from, as shown in your diagrams. The reflection comes from the 3 perpendicular mirrors, rather than at the join between them.

There is no "specular" (mirror-like) reflection exactly in the corner formed by 3 mirrors, or along an edge formed by 2 mirrors. You will not see an image in such a corner or edge no matter what direction you look at it. Light which strikes at or very close to the corner or edge will be scattered or diffracted in a number of directions, because of the wave nature of light. Scattering depends on the details of how the corner is made, what imperfections it has, and the wavelength of the light. This is "diffuse" reflection, which occurs from microscopically irregular surfaces. If the join is very well made you may not notice by eye the actual corner or edge in the reflected image (unless you magnify it). Otherwise, if you do notice anything it will just be a thin dark line or spot, or (in magnification) a blurring of the image.

I think that your guess is wrong : a ray pointed into a corner will not be reflected directly back upon itself, even if the join is perfect on sub-wavelength scales. If that were the case you would not be able to see the edge or corner at all.

User123
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sammy gerbil
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    I plan on buying a corner reflector with a well-made edge and pointing a laser to see where it reflects. – Arbuja Jun 29 '16 at 00:13
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    @Arbuja : Yes, that would make an interesting experiment. The effect of the edge will be easier to identify with monochromatic coherent light. – sammy gerbil Jun 29 '16 at 06:55