Question

A point object is moving on a principal axis of a 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/sec. What is the velocity of the image at that instant?
A. 5 cm/sec
B. 12 cm/sec
C. 4 cm/sec
D. 9 cm/sec

Answer 1

First, we need to find the distance of image $v$, using mirror’ formula, then we can find the velocity of image of the object $v'$ using the formula of velocities of image and the object for a mirror.

Formula used:
Mirror formula:
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$, where $v$ is distance of image from the mirror, $u$ is distance of object from the mirror and $f$ is focal length of the mirror.
The second formula we will be using will be:
$\dfrac{v'}{{{v}^{2}}}+\dfrac{u'}{{{u}^{2}}}=0$, where, $v'$is velocity of image and $u'$is velocity of the object.

Complete step by step answer:
We can see the following figure of a concave mirror.

We can see that the sign of $f$and $u$is negative as the right side is taken positive.
Now, using the mirror formula:
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
We can put $f=-24$cm and $u=-60$cm and we will get
$\dfrac{1}{v}-\dfrac{1}{60}=-\dfrac{1}{24}$
Or
$\dfrac{1}{v}=-\left( \dfrac{1}{24}-\dfrac{1}{60} \right)$
Which will give us,
$v=-40$cm.
Now, we can put the values of $u$, $v$and object velocity $u'=9$cm/sec in the velocity formula to get the image velocity $v'$.
$\dfrac{v'}{{{v}^{2}}}+\dfrac{u'}{{{u}^{2}}}=0$
By rearranging the expression, we will get
$v'=-\dfrac{{{v}^{2}}}{{{u}^{2}}}u'$
Or
$v'=-\dfrac{{{40}^{2}}}{{{60}^{2}}}\times 9$
Or
$v'=-\dfrac{1600}{3600}\times 9$
$v'=-4$cm/sec or simply 4 cm/sec
Hence, the correct answer is option C.

Additional Information
The formula used for finding the velocities can be derived by differentiating the mirror formula with respect to time. We can derive it as follows,

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
Now, differentiating with respect to time:
$\dfrac{d}{dt}\left( \dfrac{1}{v} \right)+\dfrac{d}{dt}\left( \dfrac{1}{u} \right)=\dfrac{d}{dt}\left( \dfrac{1}{f} \right)$
We know that $f$is constant for a mirror hence, so will be $\dfrac{1}{f}$. Hence, its derivative will be 0. Furthermore, for a variable function we know that, $\dfrac{d}{dt}\left( \dfrac{1}{g(t)} \right)=-\dfrac{1}{g{{(t)}^{2}}}\left( \dfrac{dg(t)}{dt} \right)$.
So, we get
$-\dfrac{1}{{{v}^{2}}}v'-\dfrac{1}{{{u}^{2}}}u'=0$.

Note:
Students can confuse the lens formula and the mirror formula. The sign convention must be kept in mind while finding the position of the image, otherwise the answer will give an incorrect value.