Question

A linear object $AB$ is placed along the axis of a concave mirror. The object is moving towards the mirror with speed $V$. The speed of the image the point $A$ is $4V$ and the speed of the image is also $4V$. If the centre of the line $AB$ is at a distance $L$ from the mirror, then the length of the object $AB$ will be:
A) $\dfrac{{3L}}{2}$
B) $\dfrac{{5L}}{3}$
C) $L$
D) $\dfrac{{4L}}{3}$

Answer 1

Hint: Since the question is con concave mirror, we consider the mirror formula. Then, we perform partial differentiation, to find the relation of image distance and object distance. This relation is then used and equated with the velocity of the object and thus, the height of the image is calculated.

Complete step by step solution:
We know, the mirror formula is given by:
$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$
$v$ is the image distance
$u$ is the object distance
$f$ is the focal length
On rearranging the equation:
$\Rightarrow v = \dfrac{{fu}}{{u - f}}$
On partially differentiating $dv$ , we obtain:
$\Rightarrow dv = \dfrac{{fdu(u - f) - (du)fu}}{{{{(u - f)}^2}}}$
On rearranging the equation:
$\Rightarrow \dfrac{{dv}}{{du}} = \dfrac{{ - {f^2}}}{{{{(u - f)}^2}}}$
Let us assume the length of the object = $ + 2a$
Given the object is linear, and at a distance of $L$
Thus, we can write:
${u_A} = - L - a$ and ${u_B} = - L + a$
Since the image and object velocities are the same, we can write:
We know, both image and object moves with a velocity of $4V$
$ \Rightarrow - 4 = \dfrac{{ - {f^2}}}{{{{( - L - a - f)}^2}}}$
$ \Rightarrow $ $ - 4 = \dfrac{{ - {f^{^2}}}}{{{{(L + a - f)}^2}}}$
However, if we consider,
$\Rightarrow {( - L - a - f)^{}} = ( - L + a - f)$
The above expression is not possible, as it would imply that $a = 0$; which is impossible because the object height cannot be zero.
So, we write:
$\Rightarrow {( - L - a - f)^{}} = - ( - L + a - f)$
Thus, we obtain: $f = - L$
This is possible.
Hence considering the below-given equation:
$ \Rightarrow - 4 = \dfrac{{ - {f^2}}}{{{{( - L - a - f)}^2}}}$
And substituting the obtained value of focal length in the above equation:
$ \Rightarrow - 4 = \dfrac{{ - {{( - L)}^2}}}{{{{( - L - a - ( - L))}^2}}}$
On solving the equation, we get:
$\Rightarrow 4 = \dfrac{{{L^2}}}{{{a^2}}}$
Or, we can write,
$\Rightarrow a = \dfrac{{{L^{}}}}{2}$
But, we considered the height of the object to be $2a$, thus we get:
$\Rightarrow 2a = L$
This is the required height of the object.

Hence, option (C ) is correct.

Note: The concave mirror forms images which are real and virtual. However, a magnified image can also be obtained, if the mirror is placed very close to the object. With the increase in distance between the object and the mirror, the size of the image decreases. We can also obtain images of the same size as an object using a concave mirror.