Question

An object approaches a convergent lens from the left of the lens with a uniform speed $5 \,m/s$ and stops at the focus, the image

• A. Moves away from the lens with an uniform speed 5 m/s
• B. Moves away from the lens with an uniform acceleration
• C. Moves away from the lens with a non- uniform acceleration
• D. Moves towards the lens with a non-uniform acceleration

Answer 1

The Correct Option is C

Velocity of image in convex lens $V _{ i }=( f /( f + u ))^{2} V_o$
and image will move from focus to infinity (i.e. away from the lens).
From the formula, velocity of image
$\left( dV _{ i }\right) / dt =\left( d V _{ i } / du \right) \times( du / dt )$
$=2( f / f + u ) f \ln ( f + u ) \times 5$ (because $\left.du / dt =5 m / s \right)$
$\Rightarrow\left( dV _{ i }\right) / dt =10 f ^{2} \ln ( f + u ) /( f + u )$
We can see that acceleration of image $\propto \ln (f+u)$,
i.e varies with distance of object
So the image moves away from the lens with a non-uniform acceleration

Answer 2

When an object moves towards a converging lens at a uniform speed, the image formed will change in a way that depends on the position of the object relative to the focus. As the object approaches the focus, the image initially moves towards the lens at a higher speed, but as the object nears the focus, the rate of change of the image's position decreases. The image accelerates away from the lens, but the acceleration is not uniform due to the changing relative speed as the object gets closer to the focus.
Thus, the image moves away from the lens with a non-uniform acceleration.

Therefore, the correct answer is (C).