An object is at a distance 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
Answer: (D) $1.16\times10^{-3}$ m/s towards the lens
Image position ($u = -20$ m, $f = 0.3$ m):
$$\frac1v = \frac1f + \frac1u = \frac{1}{0.3} - \frac1{20} \Rightarrow v = \frac{60}{197} \approx 0.3046\ \text{m}$$
Differentiating $\dfrac1v - \dfrac1u = \dfrac1f$ gives $\dfrac{dv}{dt} = \dfrac{v^2}{u^2}\dfrac{du}{dt}$:
$$|v_{\text{image}}| = \frac{(0.3046)^2}{400}\times5 \approx 1.16\times10^{-3}\ \text{m/s}$$
As the object moves away, $v$ decreases towards $f$, so the image moves towards the lens.
Solution by Sreeraj P, M.Sc Physics