A $f=10 \mathrm{~cm}$ focal length converging lens is approached by an object moving along the line determined by the two foci at a speed of $2 \mathrm{~m} / \mathrm{s}$. What is the relative speed of the object and its image when the object distance is $t=30 \mathrm{~cm}$?
A lens with a focal length of $f$ maps a point object located at a distance $t$ from it, according to the lens equation, to a distance

$$
k=\frac{f \cdot t}{t-f}
$$

away. The speed of the image point, by definition, is given by the first derivative of (1) with respect to time $\tau$ (excluding the case $t=f$). When differentiating, we consider that in (1) the object distance $t$ is a function of time, while $f$ is independent of time. According to the chain rule for differentiating composite functions:

$$
v_{k}=\frac{\mathrm{d} k}{\mathrm{~d} \tau}=\frac{\mathrm{d} k}{\mathrm{~d} t} \cdot \frac{\mathrm{d} t}{\mathrm{~d} \tau}=\frac{f(t-f)-f t}{(t-f)^{2}} \cdot \frac{\mathrm{d} t}{\mathrm{~d} \tau}=-\frac{f^{2}}{(t-f)^{2}} \cdot \frac{\mathrm{d} t}{\mathrm{~d} \tau}
$$

If the object moves towards the lens (the value of $t$ decreases), then the speed of the object point is:

$$
v_{t}=-\frac{\mathrm{d} t}{\mathrm{~d} \tau}
$$

Substituting (3) into (2) gives the speed of the image point:

$$
v_{k}=\frac{f^{2}}{(t-f)^{2}} \cdot v_{t}
$$

The relative speed of the object and the image is:

$$
\Delta v=v_{t}-v_{k}
$$

Using the numerical values:

$$
v_{k}=\frac{10^{2}}{(30-10)^{2}} \cdot 200 \mathrm{~cm} / \mathrm{s}=50 \mathrm{~cm} / \mathrm{s}, \Delta v=200 \mathrm{~cm} / \mathrm{s}-50 \mathrm{~cm} / \mathrm{s}=1.5 \mathrm{~m} / \mathrm{s}
$$

Filep János (Bp., I. István Gymn., III. o. t.)

Remarks. 1. In several solutions, it was incorrectly assumed without any reason that the image point also moves with a constant speed, which is not true according to (2).

2. A common mistake was the neglect of signs, leading to a relative speed of $2.5 \mathrm{~m} / \mathrm{s}$.