# Question 382fc

Sep 16, 2017

We know the following relation between number of images $N$ formed by two inclined mirrors and their angle of inclination $\theta$

$N = {360}^{\circ} / \theta - 1$

In our problem $N = 2$

So we have

$2 = {360}^{\circ} / \theta - 1$

$\implies \theta = {360}^{\circ} / 3 = {120}^{\circ}$

The ray diagram of image formation is shown in above figure where two images are ${O}_{1} \mathmr{and} {O}_{2}$ and the is object $O$. These three form the equilateral triangle ${O}_{1} {O}_{2} O$

Given that the minimum distance between mirror and point source be O_1O=O_2O= √3cm#

The length of each side of the equilateral triangle $= 2 \sqrt{3} c m$

then the square of the side of the equilateral triangle is $= {\left(2 \sqrt{3}\right)}^{2} c {m}^{2} = 12 c {m}^{2}$