A point object P  (at distance of 15 cm from lens) is approaching towards a thin and silvered equiconvex lens of focal length 20 cm and refractive index 1.5 with a speed of  $4 cm{s}^{-1}$ as shown in figure. Find velocity of image.

$\begin{array}{l}\frac{1}{f}=(\mu -1)(\frac{1}{R}+\frac{1}{R})\\ \Rightarrow \frac{1}{20}=(1.5-1)\frac{2}{R}\\ \therefore R=20cm\end{array}$

Power of equivalent mirror is $P=2P_1+P_m=\frac{2}{20}-\frac{1}{-10}=\frac{1}{10}+\frac{1}{10}=\frac{1}{5}$

$\begin{array}{l}\therefore \frac{-1}{F}=\frac{1}{5}\\ \therefore F=-5cm\\ Combination behaves like Concave Mirror\\ \because \frac{1}{v}+\frac{1}{u}=\frac{1}{F}\\ or\frac{1}{v}-\frac{1}{15}=-\frac{1}{5}\\ Or \frac{1}{v}=\frac{1}{15}-\frac{1}{5}=\frac{1-3}{15}=\frac{-2}{15}\\ \therefore v=\frac{-15}{2}cm\\ \therefore v_{image}=\frac{-v^2}{u^2}\frac{du}{dt}=-\frac{{(-\frac{15}{2})}^2}{{(-15)}^2}\times 4=-1cm{s}^{-1}\end{array}$