Question

Find the velocity of images formed due to a point object $O$ w.r.t. ground if the mirror is at rest. All speeds given are w.r.t a stationary ground observer

• A. $25\text{m/s}\text{towards mirror}$
• B. $45\text{m/s}\text{towards mirror}$
• C. $25\text{m/s}\text{away from mirror}$
• D. $45\text{m/s}\text{away from mirror}$

Answer 1

The correct option is B $45\text{m/s}\text{towards mirror}$

For the velocity component along the principal axis, $\left(V_{IM}\right)=-\frac{v^2}{u^2}\left(V_{OM}\right)$
where $V_{IM}=$Velocity of image w.r.t mirror.
$V_{OM}=$Velocity of object w.r.t mirror.

Apply mirror equation:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}+\frac{1}{-20}=\frac{1}{-30}\Rightarrow v=+60\text{cm}\Rightarrow m=-\frac{v}{u}=\frac{-60}{-20}=3$
Since mirror is at rest (given that):
$\left(V_{IM}\right)=-{\left(3\right)}^2\times \left(V_{OM}\right)=-9\times 5=-45\text{m/s}\Rightarrow {\overrightarrow{V}}_{IG}={\overrightarrow{V}}_{IM}+{\overrightarrow{V}}_{MG}=-45+0=-45\text{m/s}$
That means images approaches the mirror with velocity $45\text{m/s}$.