A diverging lens of focal length –10 cm is moving towards right with a velocity 5 m/s. An object, placed on Principal axis is moving towards left with a velocity 3 m/s. Find the velocity of image at the instant when the lateral magnification produced is 1/2. All velocities are with respect to ground.

3 m/s towards right

The correct answer is Option 1: 3 m/s towards right

For a lens,

$m = \frac{v}{u}$

Given magnification,

$m = \frac{1}{2}$

For image velocity,

$V_{IL} = m^2 V_{OL}$

Velocity of object with respect to lens:

$V_{OL} = (-3) - (+5) = -8 \text{ m/s}$

So,

$V_{IL} = \left(\frac{1}{2}\right)^2 (-8)$

$V_{IL} = -2 \text{ m/s}$

Negative sign shows that the image moves towards left with respect to the lens.

Now lens itself moves towards right with 5 m/s.

Therefore, velocity of image with respect to ground:

$V_I = 5 - 2 = 3 \text{ m/s}$

Hence, the image moves towards right with speed 3 m/s.