# How does this trick with mirrors work?

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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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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@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 at 0:04