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Question: A point object is moving on the principal axis of concave mirror of focal length \(24\,cm\) towards ...

A point object is moving on the principal axis of concave mirror of focal length 24cm24\,cm towards the mirror when it is at the distance of 60cm60\,cm from the mirror, its velocity is 9cm/sec9\,cm/\sec . What is the velocity of the image at that instant?

A) 4cm/sec4\,cm/\sec towards the mirror

B) 9cm/sec9\,cm/\sec towards the mirror

C) 4cm/sec4\,cm/\sec away from the mirror

D) 9cm/sec9\,cm/\sec away from the mirror

Explanation

Solution

We are given that the point object is moving towards the concave mirror and its velocity when it is 60cm60\,cm away from the mirror is 9cm/sec9\,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 24cm24cm,

1f=1u+1v\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=1v2dvdt1u2dudt0 = - \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} - \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}}

dfdt=0\dfrac{{df}}{{dt}} = 0 Because focal length is a constant. We know that (du) velocity of object is 9cm/sec9\,cm/\sec , u=60cmu = 60\,cm and value of v we will get from the mirror formula

Substituting this value in the above equation we will easily get dvdt\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 24cm24\,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 60cm60\,cm from the mirror its velocity is 9cm/sec9\,cm/\sec towards the mirror

So object distance, u=60cmu = - 60\,cm

Velocity is given as the derivative of displacement with respect to time.

So velocity of object, dudt=+9cm/sec\dfrac{{du}}{{dt}} = + 9\,cm/\sec

Focal length of concave mirror given is f=24cmf = - 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

1f=1u+1v\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}

Substituting the values in above relation we have

124=160+1v\dfrac{1}{{ - 24}} = \dfrac{1}{{ - 60}} + \dfrac{1}{v}

160+124=1v\dfrac{1}{{60}}+\dfrac{1}{{ - 24}} = \dfrac{1}{v}

1v=140\dfrac{1}{v} = \dfrac{1}{{ - 40}}

v=40cmv = - 40\,cm

Now we will differentiate the relation

1f=1u+1v\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}

With respect to time

0=1v2dvdt1u2dudt0 = - \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=40cmv = - 40cm

u=60cmu = - 60cm

dudt=+9cm/sec\dfrac{{du}}{{dt}} = + 9cm/\sec

Substituting all these values in the relation obtained we get

Velocity of image, dvdt=\dfrac{{dv}}{{dt}} =

(40)2(60)2(9)=4cm/sec- \dfrac{{{{\left( { - 40} \right)}^2}}}{{{{\left( { - 60} \right)}^2}}}\left( 9 \right) = - 4\,cm/\sec

So velocity of image is 4cm/sec4\,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.