Question

An object approaches a convergent lens from the left of the lens with a uniform speed $5m/s$ 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

Answer: (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
$\left(d{V}_i\right)/dt=\left(d{V}_i/du\right)\times (du/dt)$
$=2(f/f+u)f\ln (f+u)\times 5$ (because $du/dt=5m/s)$
$\Rightarrow \left(d{V}_i\right)/dt=10f^2\ln (f+u)/(f+u)$
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