Question

A small object is moving along the principal axis of a concave mirror of focal length $24cm$ towards it. When it is at a distance of $60cm$ from the mirror, its velocity is $9cm/s$. The velocity of its image at that instant is :

• A. $4cm/s$ towards the mirror
• B. $4cm/s$ away from the mirror
• C. $6cm/s$ towards the mirror
• D. $6m/s$ away from the mirror

Answer 1

The correct option is B $4cm/s$ away from the mirror

Given: A small object is moving along the principal axis of a concave mirror of focal length 24 cm towards it. When it is at a distance of 60cm from the mirror, its velocity is $9cm/s$.

To find the velocity of its image at that instant

Solution:

According to the given criteria,

Object distance, $u=60cm$

Focal length of the mirror, $f=24cm$

and object velocity, $\frac{du}{dt}=9cm/s$

Applying mirror formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}⟹\frac{1}{24}=\frac{1}{v}+\frac{1}{60}⟹\frac{1}{v}=\frac{1}{24}-\frac{1}{60}⟹\frac{1}{v}=\frac{5-2}{120}⟹v=40cm$

Also,

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

Differentiating with respect to t, we get

$0=-\frac{1}{v^2}\frac{dv}{dt}-\frac{1}{u^2}\frac{du}{dt}⟹\frac{dv}{dt}=-\frac{v^2}{u^2}.\frac{du}{dt}⟹\frac{dv}{dt}=-\frac{(40{)}^2}{(60{)}^2}\times 9⟹\frac{dv}{dt}=-\frac{16}{36}\times 9⟹\frac{dv}{dt}=-4cm/s$

is the required velocity of the image, here negative sign indicates that the velocity of the image is away from the mirror.