Discussion
Consider a convex lens of focal length f. A point object moves towards the lens along its axis between 2f and f. If the speed of the object is \(V_{O}\), then its image would move with speed \(V_{I}\). Which of the following is correct?
Answer: C
The answer is \(V_{I} \gt V_{O}\); the image moves away from the lens.
From the lens equation \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\), differentiating with respect to time gives
\[ \begin{aligned} \frac{dv}{dt} &= \frac{v^{2}}{u^{2}}\frac{du}{dt} \\ &= m^{2}\frac{du}{dt} \end{aligned} \]
For an object between \(2f\) and \(f\) the image is real, beyond \(2f\) on the other side, and magnified (\(|m| \gt 1\)). So the image speed is \(m^{2}\) times the object speed, greater than \(V_{O}\). As the object approaches \(f\) the image runs off towards infinity, so it moves away from the lens.
Equal speeds occur only at \(u = 2f\), where \(|m| = 1\), and a slower image would need \(|m| \lt 1\), which happens only for objects beyond \(2f\).
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