An observer is standing at a point O, at a distance of 100 cm from a convex lens of focal length 50 cm. A plane mirror is placed behind the lens at a distance of 150 cm from lens. The magnitude of velocity of her own image as seen by the observer, at the moment when the mirror just starts moving with a velocity of 10 cm/s.

${\overrightarrow{V}}_I=2{\overrightarrow{V}}_m-{\overrightarrow{V}}_0\Rightarrow {\overrightarrow{V}}_I=2{\overrightarrow{V}}_m$
$\therefore$  velocity of image in the plane mirror will be 20 cm/s in the direction of motion of mirror.

First refraction at lens, Image is formed at I

$\begin{array}{l}\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\text{}\\ \Rightarrow \frac{1}{50}=\frac{1}{v}-\frac{1}{\left(-100\right)}\text{}\\ \Rightarrow v=100\end{array}$

Plane Mirror will form image at I'

This I' will behave as object for, 2nd refraction in lens.

$\begin{array}{l}u=-200cm\\ \frac{1}{\text{f}}=\frac{1}{v}-\frac{1}{u}\\ \Rightarrow \frac{1}{50}=\frac{1}{v}-\frac{1}{-200}\\ \Rightarrow v=\frac{200}{3}\end{array}$

$\therefore \left|\frac{dV}{dt}\right|=\frac{{\text{v}}^2}{{\text{u}}^2}\left|\frac{du}{dt}\right|=\frac{{\left(\frac{200}{3}\right)}^2}{{200}^2}\left|\frac{du}{dt}\right|=\frac{20}{9}cm/\text{s}$