Question

An object A is moving towards a converging lens as shown in the figure. If object is moving with speed $5\text{m/s}$, find the average velocity of image in next 4 seconds.

• A. $5\text{cm/sec}$
• B. $\frac{5}{3}\text{cm/sec}$
• C. $15\text{cm/sec}$
• D. $20\text{cm/sec}$

Answer 1

The correct option is C $15\text{cm/sec}$


In next 4 sec, object will get nearer to lens by 20 cm.
Then, $u=-40cm$
and $f=30cm$
Using lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}-\frac{1}{-40}=\frac{1}{30}$
$\frac{1}{v}=\frac{1}{30}-\frac{1}{40}\Rightarrow \frac{1}{120}$
$v=120cm$


So the image moves $60cm$ in $4s$.

$\therefore v_{image}=\frac{dx}{dt}=\frac{60cm}{4sec}=15\text{cm/sec}$