Three identical flat mirrors are placed mutually perpendicular to each other. The number of distinct images that will be formed due to a point object placed at a point formed by the intersection of lines joining their centres will be (Nearest Integer value)

Let us consider, mirror-I and mirror-II:

Angle between the mirrors be $\theta ={90}^{\circ }$

To find the number of images formed by the system of two mirrors,

we find , $\frac{360}{\theta }=\frac{360}{90}=4$ (even number)

$\therefore$ Number images formed by the system of two mirrors which are perpendicular to each other ,
$\left(n\right)=\frac{360}{90}-1\Rightarrow n=3$

Since, mirror-III is mutually perpendicular to mirrors I and II , these three images formed, will form $3$ images with plane mirror-III.

[ Images formed by two mirrors (I and II) act as objects to the mirror-III]

Since the object is placed in front of mirror - I ,mirror - II and mirror-III, it will form its image in mirror-III along with the other $3$ images.

Total number of images formed by the system of three mutually perpendicular plane mirrors $=3+3+1=7$

All the images formed are concyclic.

Note:- Image of an object formed by a plane mirror, can be an object for other plane mirror or extension of plane mirror.