VerifiedHint: We are given that the point object is moving towards the concave mirror and its velocity when it is $60\,cm$ away from the mirror is $9\,cm/\sec $. We need to find the velocity of the image at that instant. We can first find the position of an image simply by using a mirror formula. Mirror formula relates v(position of the image), u(position of the object) and f(focal length of the mirror). We are already given focal length $24cm$,
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$
From this we will get the position of the image . We know that velocity is the differential of displacement with respect to time.
So if we differentiate the above relation we get
$0 = - \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} - \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}}$
$\dfrac{{df}}{{dt}} = 0$ Because focal length is a constant. We know that (du) velocity of object is $9\,cm/\sec $, $u = 60\,cm$ and value of v we will get from the mirror formula
Substituting this value in the above equation we will easily get $\dfrac{{dv}}{{dt}}$ i.e. velocity of image, keeping in mind the sign convention.
Complete step by step solution:
A point object is moving towards a concave mirror of focal length $24\,cm$
The sign convention that we will be following is positive for right and upward direction and negative for left and downward direction.
We are given that when object is at a distance of $60\,cm$ from the mirror its velocity is $9\,cm/\sec $ towards the mirror
So object distance, $u = - 60\,cm$
Velocity is given as the derivative of displacement with respect to time.
So velocity of object, $\dfrac{{du}}{{dt}} = + 9\,cm/\sec $
Focal length of concave mirror given is $f = - 24cm$
For determining the position of image we will use mirror formula
Mirror formula relates v(position of the image), u(position of the object) and f(focal length of the mirror) through
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$
Substituting the values in above relation we have
$\dfrac{1}{{ - 24}} = \dfrac{1}{{ - 60}} + \dfrac{1}{v}$
$\dfrac{1}{{60}}+\dfrac{1}{{ - 24}} = \dfrac{1}{v}$
$\dfrac{1}{v} = \dfrac{1}{{ - 40}}$
$v = - 40\,cm$
Now we will differentiate the relation
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$
With respect to time
$0 = - \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} - \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}}$
Differential of focal length with respect to time is zero because it is constant value
We have
$v = - 40cm$
$u = - 60cm$
$\dfrac{{du}}{{dt}} = + 9cm/\sec$
Substituting all these values in the relation obtained we get
Velocity of image, $\dfrac{{dv}}{{dt}} = $
$ - \dfrac{{{{\left( { - 40} \right)}^2}}}{{{{\left( { - 60} \right)}^2}}}\left( 9 \right) = - 4\,cm/\sec $
So velocity of image is $4\,cm/\sec $ , away from the mirror
So, option C is correct.
Note: Be very careful with the sign convention, use a single sign convention throughout the question to avoid error.
