Question

A point object is moving on the principal axis of
concave mirror of focal length 24 cm towards the
mirror. When it is at a distance of 60 cm from
the mirror, its velocity is 9 cm/s. What is the
velocity of the image at that instant :
(a) 5 cm/s towards mirror
(b) 4 cm/s towards mirror
(c) 4 cm/s away from mirror
(d) 9 cm/s away from mirror​

Answer 1

Given :

A point object is moving towards a concave  mirror  on its  principal axis  .

Focal length of the concave mirror :

f = -24 cm , since focal length is negative for concave mirror

Velocity of object when it is at a distance of 60 cm from the mirror :

$v_{obj}$= 9 $\frac{cm}{s}$

To Find :

Velocity of the image at the given instant = ?

Solution :

Here 1st we have to find the distance of image from the mirror .

From the mirror formula we have :

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$    -(1)

So using this formula we can have the v i.e. distance of image as :

$\frac{1}{v} = \frac{-1}{24}+\frac{-1}{60}$

On solving this we get v = -40 cm

On differentiating eq 1 we get :

$\frac{-1}{v^{2}} .\frac{dv}{dt} - \frac{1}{u^{2} }. \frac{du}{dt} = o$

Or, $\frac{1}{v^{2}} .\frac{dv}{dt} + \frac{1}{u^{2} }. \frac{du}{dt} = o$

Or, $\frac{v_{i}}{40^{2} } +\frac{9}{60^{2} } = 0$

Or, $v_{i} = \frac{-9}{60^{2} } \times 40^{2}$

⇒$v_{i$ =$-4 \frac{m}{s}$

So the velocity of image at the given instant is 4 cm per s towards the mirror .

Hence correct option is (b). 4 cm per sec towards the mirror .