Question

An object is at a distance of $20\,m$ from a convex lens of focal length $0.3\,m$. The lens forms an image of the object. If the object moves away from the lens at a speed of $5\,m{s^{ - 1}}$, the speed and direction of the image will be:
(A) ${0.9210^3}\,m{s^{ - 1}}$ away from the lens
(B) $2.26 \times {10^{ - 3}}\,m{s^{ - 1}}$ away from the lens
(C) $1.16 \times {10^3}\,m{s^{ - 1}}$ towards from the lens
(D) $3.22 \times {10^3}\,m{s^{ - 1}}$ towards from the lens

Answer 1

Hint: The velocity of the image is determined by using the velocity of the image with respect to the lens formula. In this formula, the magnification is required, the magnification is determined by using the magnification formula. Then the velocity of the image is determined.

Additional Formula:
The focal length is given by,
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Where, $f$ is the focal length of the lens, $v$ is the distance of the image from the lens and $u$ is the distance of the object from the lens.
The magnification of the lens is given by,
$m = \dfrac{v}{u}$
Where, $m$ is the magnification of the lens, $v$ is the distance of the image from the lens and $u$ is the distance of the object from the lens.
The velocity of the image with respect to the lens is given by,
${V_I} = {m^2}{V_O}$
Where, ${V_I}$ is the velocity of the image, $m$ is the magnification of the lens and ${V_O}$ is the velocity of the object.

Complete step by step answer:
Given that,
The distance of the object from the lens is, $u = 20\,m$,
The focal length of the lens is, $f = 0.3\,m$,
The velocity of the object is, ${V_O} = 5\,m{s^{ - 1}}$
Now,
The focal length is given by,
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\,...................\left( 1 \right)$
By substituting the distance of the object from the lens and the focal length of the lens in the above equation (1), then
$\dfrac{1}{{0.3}} = \dfrac{1}{v} - \dfrac{1}{{ - 20}}$
By rearranging the terms in the above equation, then
$\dfrac{1}{{0.3}} = \dfrac{1}{v} + \dfrac{1}{{20}}$
By keeping the term $v$ in one side, then
$\dfrac{1}{v} = \dfrac{1}{{0.3}} - \dfrac{1}{{20}}$
The above equation is also written as,
$\dfrac{1}{v} = \dfrac{{10}}{3} - \dfrac{1}{{20}}$
By cross multiplying the terms in the above equation, then
$\dfrac{1}{v} = \dfrac{{200 - 3}}{{60}}$
By subtracting the terms in the above equation, then
$\dfrac{1}{v} = \dfrac{{197}}{{60}}$
By taking reciprocal on both sides, then
$v = \dfrac{{60}}{{197}}\,m$
Now,
The velocity of the image with respect to the lens is given by,
${V_I} = {m^2}{V_O}\,...............\left( 2 \right)$
By using the magnification formula in the above equation, then
${V_I} = {\left( {\dfrac{v}{u}} \right)^2}{V_O}$
By substituting the distance of the image from the lens, the distance of the object from the lens and velocity of the object in the above equation, then
${V_I} = {\left( {\dfrac{{\left( {\dfrac{{60}}{{197}}} \right)}}{{20}}} \right)^2} \times 5$
By rearranging the terms in the above equation, then
${V_I} = {\left( {\dfrac{{60}}{{197 \times 20}}} \right)^2} \times 5$
By multiplying the terms in the above equation, then
${V_I} = {\left( {\dfrac{{60}}{{3940}}} \right)^2} \times 5$
By dividing the terms in the above equation, then
${V_I} = {\left( {0.015} \right)^2} \times 5$
By squaring the terms in the above equation, then
${V_I} = 2.3 \times {10^{ - 4}} \times 5$
By multiplying the terms in the above equation, then
${V_I} = 1.16 \times {10^{ - 3}}\,m{s^{ - 1}}$
The final answer is positive, so the direction is towards the centre.
Hence, the option (C) is the correct answer.

Note:The magnification of the lens is directly proportional to the distance of the image from the lens and inversely proportional to the distance of the object from the lens. And the velocity of the image is directly proportional to the magnification and the velocity of the lens.