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A point object is moving on the principal axis of concave mirror of focal length $24\,cm$ towards the mirror when it is at the distance of $60\,cm$ from the mirror, its velocity is $9\,cm/\sec $ . What is the velocity of the image at that instant?

A) $4\,cm/\sec $ towards the mirror

B) $9\,cm/\sec $ towards the mirror

C) $4\,cm/\sec $ away from the mirror

D) $9\,cm/\sec $ away from the mirror

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Last updated date: 17th Jun 2024
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Answer
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Hint: 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.