In the figure shown consider the first reflection at the plane mirror and second at the convex mirror. AB is object.

1^st image$\left({\text{A}}_{\text{1}}{\text{B}}_{\text{1}}\right)$  is formed due to plane mirror to the left of it with magnification 1. It is virtual. This${\text{A}}_{\text{1}}{\text{B}}_{\text{1}}$  acts as real object to the concave mirror of focal length

$\text{f}=\frac{\text{R}}{\text{2}}=\frac{120}{2}=60\text{cm}$

${\text{A}}_2{\text{B}}_{\text{2}}$ is the image of ${\text{A}}_{\text{1}}{\text{B}}_{\text{1}}$ formed by convex mirror

$\text{AB}={\text{A}}_{\text{1}}{\text{B}}_{\text{1}}=30\text{cm}$

For ${\text{A}}_1:\text{u}=-60\text{cm,}\text{f}=+60\text{cm,}\text{V}?$

$\frac{\text{1}}{\text{V}}+\frac{\text{1}}{\text{u}}=\frac{\text{1}}{\text{f}}\Rightarrow \frac{\text{1}}{\text{V}}-\frac{1}{60}$

$=\frac{1}{60}\Rightarrow \text{V}=+30\text{cm}\left(\text{P}\to {\text{A}}_{\text{2}}\right)$

For${\text{B}}_{\text{1}}:\frac{1}{\text{V}}+\frac{\text{1}}{\text{u}}=\frac{1}{\text{f}}\Rightarrow \frac{1}{\text{V}}-\frac{1}{90}=\frac{1}{60}$

$\Rightarrow \text{V}=+36\text{cm}\left(\text{P}\to {\text{B}}_{\text{2}}\right)$

Length of ${\text{A}}_{\text{2}}{\text{B}}_{\text{2}}=36-30=6\text{cm}$

$\frac{{\text{A}}_{\text{2}}{\text{B}}_{\text{2}}}{\text{AB}}=\frac{6}{30}$

$\Rightarrow {\text{A}}_{\text{2}}{\text{B}}_{\text{2}}=\frac{\text{AB}}{\text{5}}$

In the figure, it can be observed that both $\text{A}-\text{B}$ and${\text{A}}_{\text{2}}-{\text{B}}_{\text{2}}$  are left

to right i.e., ${\text{A}}_{\text{2}}{\text{B}}_{\text{2}}$  is erected w.r.t. AB. ${\text{A}}_{\text{2}}{\text{B}}_{\text{2}}$ is virtual

C) Velocity of image formed by a mirror is always opposite to the

object. AB moves left wards$\Rightarrow {\text{A}}_{\text{1}}{\text{B}}_{\text{1}}$  moves right wards$\Rightarrow {\text{A}}_{\text{2}}{\text{B}}_{\text{2}}$

moves left wards i.e., towards the mirror