Question

An object is at a distance of $20\text{m}$ from a convex lens of focal length $0.3\text{m}$. The lens forms an image of the object. If the object moves away from the lens at a speed of $5{\text{ms}}^{-1}$, the speed and direction of the image will be:

• A. $2.26\times {10}^{-3}{\text{ms}}^{-1}$ away from the lens
• B. $0.92\times {10}^{-3}{\text{ms}}^{-1}$ away from the lens
• C. $3.22\times {10}^{-3}{\text{ms}}^{-1}$ towards the lens
• D. $1.16\times {10}^{-3}{\text{ms}}^{-1}$ towards the lens

Answer 1

The correct option is D $1.16\times {10}^{-3}{\text{ms}}^{-1}$ towards the lens

By the lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}-\frac{1}{(-20)}=\frac{10}{3}$

$\frac{1}{v}=\frac{10}{3}-\frac{1}{20}$

$\frac{1}{v}=\frac{197}{60}\Rightarrow v=\frac{60}{197}\text{m}$

Magnification of lens $\left(m\right)$ is given by,

$m=\left(\frac{v}{u}\right)=\frac{\left(\frac{60}{197}\right)}{20}$

The velocity of image with respect to lens is given by

$V_I=-m^2{V}_O$

Negative sign indicates that the direction of velocity of the image is the opposite to that of the object.

$\text{Given that.}{V}_O=5{\text{ms}}^{-1}$

$\Rightarrow V_I={\left(\frac{60\times 1}{197\times 20}\right)}^2\left(5\right)$

$V_I=1.16\times {10}^{-3}{\text{ms}}^{-1}$ towards the lens

Hence, option $\left(D\right)$ is correct.