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Consider 2 mirrors which intersect at an angle. If we draw a light ray intersecting the mirrors at their point of intersection, how will it reflect?

What will be the normal at that point?

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Independant of votes you may accept an answer which helped you most. You get 2 reputation and the question no longer is in the unanswered questions list. –  Stefan Bischof Jun 13 at 7:21

3 Answers 3

It is overdefined geometry. By most optics ray tracing simulation software it will be treated as absorption.

The intersection point belongs to two interfaces at once. Praxis differs

  1. Two separate mirrors are likely to produce scattered light, as mentioned by @Carl Witthoft E.g. Glas substrate, which is cemented at the interface to a second mirror. A light ray at the cement will not hit a mirror and produce stray light.
  2. In manufactured mirrors, like optics injection molding, there will be a curve connecting the two mirror plances. This tiny curve forms a concave mirror.

Depending on your application you should stick either to advance your model with a curved edge or assume this ray is absorbed. In a ray tracing simulation this one ray will not affect simulated irradiance. However you may take a closer look if performing a stray light analysis.

Thanks for the comments

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Not quite: any manufactured mirror made of two bonded mirrors, or made from a prism with silvered surfaces (see "pentaprism") will have an edge with defined roughness, rather than curved. What you get, naturally, is both internal and external scatter. –  Carl Witthoft Jun 10 at 11:58
    
I think it would be better to include in your post that you do not mean Ray Tracing from Computer Graphics but from Optics. At least this is what I guess you mean, because in Computer Graphics' Ray Tracing there is no rule that dictates this behaviour. –  phresnel Jun 10 at 12:23
    
@CarlWitthoft I tried to cover your reasonable arguments. Feel free to edit. –  Stefan Bischof Jun 10 at 12:35
    
Stefan, looks great. Thanks for adding that info to the answer. –  Carl Witthoft Jun 10 at 13:32

There is no physical beam that is infinitely thin.

In fact in the quantum physical world each free space light beam extends infinitely in space. It falls off rapidly, but there is a probability finding the photon at arbitrarily far distance.

So, since each beam has a certain thickness, even in case we assume the angle is infinitely sharp the behavior is well defined. In the geometrical approximation, by just separate reflection of the two beam parts hitting the two mirrors.

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+1 But the specifically asked within geometric optics. I agree however, that the application of a deeper concept than geometric optics may be applied. Depends on application. –  Stefan Bischof Jun 10 at 13:27

It is not well defined, but that is not a problem because no rays hit the point precisely. If you are a little on one side or the other you know what happens. It is like choosing a uniform random number on $[0,1]$. The chance of each number is zero, which is what it takes to hit the corner exactly.

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Theoretically, assume that that the light ray hits the point precisely. Then what will happen (hypothetically)? –  Akshit Jun 10 at 3:50
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It is not well defined. You can take the limit as it approaches from either side and you will find a discontinuity. That tells you that you need to define the situation at another level of depth before you get a reasonable answer. –  Ross Millikan Jun 10 at 4:05

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