Number of images formed by two plane mirrors inclined at an angle of $\theta$ is given by $$\frac{360}{\theta} -1 $$

What I think: Inclined mirror forms images in the circle and one image lies in one sector.

No of images = Number of sectors=$\frac{360}{\theta}$

And $1$ is subtracted from $\frac{360}{\theta}$ because a sector is occupied by the object.

I think this is not a proper derivation. How to prove that Inclined mirror forms images in the circle?

I saw an answer but I didn't understand it.

How to derive it formally?

What's correct:
Let $$n=\dfrac{360}{\theta}$$ where $\theta$ is the angle between the two mirrors

If $n$ is even: $$\mathrm{Number\ of\ images}=n-1$$ If $n$ is odd and the object is placed symmetrically: $$\mathrm{Number\ of\ images}=n-1$$ If $n$ is odd and the object is not placed symmetrically: $$\mathrm{Number\ of\ images}=n$$ If $n$ is in decimal then only integral part is taken and above rules are followed.

It should be noted that above the 'number of images' means the number of images formed.

Experiment work:





Plus corner:
I don't think there exists a derivation to the above formulae. Maybe it was found by experiments.
Note: A very tiny change in the angle can spilt the farthest image.

  • $\begingroup$ This is not true if $\frac{360}{\theta}$ is odd and object is not on bisector. $\endgroup$ Sep 24, 2020 at 6:01
  • $\begingroup$ to prove my point, try with an angle of $24^\circ$ or $72^\circ$ and place the object somewhere other than bisector. $\endgroup$ Sep 24, 2020 at 6:04
  • $\begingroup$ I upvoted to reverse a downvote. The OP clearly showed good work that may or may not have been accurate. In fact, regardless of whether the OP's analysis is accurate, even if there had not been a downvote, I would have upvoted: the problem is complicated, the OP showed good work, and his presentation was nicely formatted. $\endgroup$ Sep 24, 2020 at 9:43
  • $\begingroup$ Does this answer your question? How to prove the number of images of two mirrors inclined at $A$ is $360/A -1$ $\endgroup$
    – Jean Marie
    Sep 24, 2020 at 15:50
  • $\begingroup$ For more information read this and this. $\endgroup$
    – Wolgwang
    Mar 2, 2021 at 4:48

1 Answer 1


Reflection of light in mirror is the same as reflection of the world in the mirror. See this

So, just do that... Let us consider your $\theta=60^o$. The original setup looks like this...

enter image description here


enter image description here

There you go! Cheers :)

  • 1
    $\begingroup$ Great response and visuals. Some things to watch out for is in the cases when $\frac{180}{\theta}$ are not integers, which the answer OP linked for briefly discusses. I'm not sure if this is true, but I tried with $\theta=40$ and got 18 images by continuously reflecting (geogebra.org/calculator/smy6jdqf). I believe it makes sense that if $\frac{360}{n}\in Z$ and $v_2(n)>2$, then the number of images is $\frac{720}{\theta}-1$. $\endgroup$ Sep 24, 2020 at 6:11
  • $\begingroup$ I think this was a correct proof. Though this shouldnt be used as an empirical formula. We have a method for it too... And also this doesnt work if the object is on the bisector (bcz you cannot join the reflected object by straight line in that case) or theta is not a divisor of 360. So, go with the method and not formulas in this case. If you wish to see that, you need to post another question regarding it. $\endgroup$ Sep 24, 2020 at 6:15
  • $\begingroup$ @Mayank What do you mean by observing it. I think the explanation Soumyadwip provided is sufficient. It's clear that as you continuously reflect the entire environment, you can only get 5 images because the sectors are a divisor of 180. $\endgroup$ Sep 24, 2020 at 6:18
  • $\begingroup$ @SoumyadwipChanda Could you elaborate why it doesn't work when the object is on the angle bisector? $\endgroup$ Sep 24, 2020 at 6:20
  • $\begingroup$ You can use this simulator $\endgroup$
    – Wolgwang
    Sep 24, 2020 at 6:38

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