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Imagine two mirrors, set touching each other at right angles to one another. There is a 90 degree arc in which reflections can be seen, and a person standing in that arc can see himself reflected in one or the other mirror. The reflection(s) will move as the person moves, as with any other mirror, and that much makes sense to me.

But I have observed that there is another reflection. With mirrors positioned this way, no matter where I stand within the 90 degree arc, I can always see myself reflected exactly at the line where the two mirrors meet.

It would be nice to understand the principles behind this. Anyone know why that works that way?

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up vote 4 down vote accepted

The reason is simple geometry. Consider the following image.

Then you can clearly see the image at $O_{3}$ is due to double reflection from mirrors $M'$ and $M$. In general all the images including the object will be on a circle centered around the intersection point between two mirrors. For two mirrors inclined at $n^{0}$ with respect to each other.The number of images are $\frac{360}{n}-1$. I do not know the proof of the last formula. As far as I know it is because reflection of one mirror to another forms a virtual mirror and the number of virtual mirrors are $\frac{360}{n}$ Final image from both the mirror coincides hence we have to subtract $1$.

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Without a diagram I'm not sure exactly what phenomena you're talking about but what you describe sounds like the effect achieved with Retroreflectors

A retroreflector uses three mirrors however it would work just fine with just two as long as you restricted your viewing to just the 90 degree arc. You will always see the line between the mirrors in the middle of the reflection because it is the interaction between the two mirrors that causes incident light to always be returned. The explanation is simple geometry (borrowed from the above linked Wikipedia article):

retroreflector diagram

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Do you have a medicine cabinet in your bathroom, with multiple mirrored doors? Open one of them out to 90 degrees and stand somewhere inside the 90 degree arc. No matter where you move, you will always see half of your face on either side of the line where the two mirrors intersect, in addition to the "straight" reflections you'd see from looking into the mirror directly. That's not a diagram, but it's a very simple way to understand it. – Mason Wheeler Apr 17 '13 at 4:21
@MasonWheeler okay what you describe is exactly the 1 right-angle mirror the image. – Brandon Enright Apr 17 '13 at 4:47
All right, so it's a retroreflector. But, as far as I can see at least, neither this answer nor the linked Wikipedia article actually answers the question: When I look into a 90 degree retroreflector, no matter where I'm looking into it from, why do I always see a reflection of myself at the point where the two mirrors touch? – Mason Wheeler Apr 17 '13 at 11:14
@MasonWheeler: Retroreflectors make you see yourself because $\sigma_x\sigma_y\sigma_z\mathbf{v}=-\mathbf{v}$ for any vector $\mathbf{v}$, where the $\sigma_\alpha$ are reflection operators. – DumpsterDoofus Mar 3 '14 at 0:04

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