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.
Answer & explanation
Correct answer: option 1
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.