Question

An object is at a distacen 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 $5\, m/s$, the speed and direction of the image will be :

• A. $0.92 \times 10^{-3} \; m/s$ away from the lens
• B. $2.26 \times 10^{-3} \; m/s$ away from the lens
• C. $1.16 \times 10^{-3} \; m/s$ towards the lens
• D. $3.22 \times 10^{-3} \; m/s$ towards the lens

Answer 1

Answer: (C)

From lens equation
$\frac{1}{v} - \frac{1}{u} =\frac{1}{f} $
$ \frac{1}{v} - \frac{1}{\left(-20\right)} = \frac{1}{\left(.3\right)} = \frac{10}{3} $
$ \frac{1}{v} = \frac{10}{3} - \frac{1}{20} $
$\frac{1}{v} = \frac{197}{60} ; v = \frac{60}{197} $
$ m= \left(\frac{v}{u}\right) =\frac{\left(\frac{60}{197}\right)}{20} $
velocity of image wrt. to lens is given by vI/L = m2vO/L
direction of velocity of image is same as that of object
vO/L = 5 m/s
$ v_{I/L} = \left(\frac{60\times1}{197\times20}\right)^{ 2} \left(5\right) $
= $1.16 \times 10^{-3} \; m/s$ towards the lens