Three plane mirrors are arranged mutually perpendicular to each other as shown in the figure. If a point object is placed symmetrically between the mirrors, then the number of distinct images formed will be

The correct option is B $7$
Given, all the three mirrors are mutually perpendicular.

We know that when two mirrors are placed perpendicular to each other, total 3 images are formed. Hence considering mirrors $M_2$ and $M_3$, they will form 3 images.
Now, for these $3$ images, $3$ more images will be formed by the base mirror $M_1$.
Again, for the point object, the base mirror will form $1$ image.
$\therefore$ Total number of images $=3+3+1=7$

Alternatively:
For the plane mirrors $M_2$ and $M_3$, the angle between them is $\theta ={90}^{\circ }$.
So,
$n=\frac{{360}^{\circ }}{\theta }=\frac{{360}^{\circ }}{{90}^{\circ }}=4$
As $n$ is even, hence, the number of images formed by $M_2$ and $M_3$ will be, $n-1=4-1=3$.
Simillarly, for these $3$ images, $3$ more images will be formed by the base mirror $M_1$.
Again, for the point object, the base mirror will form $1$ image.
$\therefore$ Total number of images $=3+3+1=7$.