# How can one prove that the number of images formed by the reflecting surfaces of two plane mirrors at right angles to each other is 3?

How can one prove that the number of images formed by two plane mirrors at right angles to each other is 3?

Is there a mathematical proof for the same ?

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This link has a pretty good answer. –  Curious Aug 31 '12 at 7:50
Hi @Curious - I've converted your answer to a comment because if you're just going to post a link, it doesn't constitute an answer by itself. If you describe the content at the linked page in your answer, so that one doesn't have to click on the link to get the point of your answer, it would be perfectly fine. Feel free to post your answer again if you'd like to do that. –  David Z Aug 31 '12 at 17:33

Sticking to 2D for simplicity, the transformation matrices for reflections in the x = 0 and y = 0 lines are:

$$R_x = \left( \begin{matrix} -1 & 0 \\ 0 & 1 \end{matrix} \right)$$ $$R_y = \left( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right)$$

Any combination of these transformations can be given by $R_x^m R_y^n$ where $m$ and $n$ are integers giving the number of each reflection.

But both the reflections are their own inverses i.e. $R_xR_x = I$ and $R_yR_y = I$. If this isn't intuitively obvious you can prove it by multiplying out the matrices above. So for any integer $m$, $R^m$ is equal to $R$ if $m$ is odd or $I$ if $m$ is even. That means there are only three distinct combinations that are not the identity:

1. $R_x$
2. $R_y$
3. $R_xR_y$

That's why there are three and only three reflections.

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This is cute. :-) –  Ryan Thorngren Aug 31 '12 at 8:13
Actually I should have mentioned that my proof only works because $R_x$ and $R_y$ are commutative i.e. $R_xR_y$ = $R_yR_x$. This is true, and you can prove it by manually multiplying the matrices, but note that in general two reflections are not commutative - two mirrors at right angles is a special case. –  John Rennie Sep 1 '12 at 6:23

You can just draw what the rays look like for a particular observer. That's a fine proof.

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