An $f=0.2 \mathrm{~m}$ focal length lens is approached by an object moving along the optical axis at a speed of $3 \mathrm{~m} / \mathrm{s}$. How fast is the image moving relative to the object when the object is 0.5 meters from the lens?
Let's denote the object distance by $x$, the image distance by $y$, and the time by $t$! According to the imaging law,

$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{f}
$$

from which

$$
y(x)=\frac{x f}{x-f}
$$

If the object approaches the lens at a speed of $v_{t}$, then in a small $\Delta t$ time,

$$
\Delta x=-v_{t} \cdot \Delta t
$$

it moves. The displacement of the image during this time is

$$
\Delta y=y(x+\Delta x)-y(x) \approx y^{\prime}(x) \cdot \Delta x
$$

and the speed of the image is

$$
v_{k}=\frac{\Delta y}{\Delta t}=-v_{t} \cdot y^{\prime}(x)
$$

The derivative of the function $y(x)$ can be obtained by applying the rules of differentiation, or directly from the first equation and

$$
\frac{1}{x+\Delta x}+\frac{1}{y+\Delta y}=\frac{1}{f}
$$

by taking the difference of the corresponding sides:

$$
y^{\prime}(x)=-\frac{f^{2}}{(x-f)^{2}}
$$

from which

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

and the relative speed of the image and the object is

$$
v_{\mathrm{rel}}=v_{k}-v_{t}=v_{t}\left(\frac{f^{2}}{(x-f)^{2}}-1\right)=-\frac{5}{3} \frac{\mathrm{m}}{\mathrm{s}}
$$

Based on the work of Gábor Borgulya (Pécs, Nagy L. Gymnasium, 4th grade)

Remarks 1. The obtained relationship can be further transformed:

$$
\begin{aligned}
& v_{\text {rel }}=v_{t} \frac{x(2 f-x)}{(x-f)^{2}} \\
& 1993-03-141-1 . \mathrm{eps}
\end{aligned}
$$

from which it can be read that as long as the object is outside the double focal distance, the image approaches the object, when $x=2 f$, the speed of the object and the image are the same, and if $f0$, then the image and the object move away from each other.

János Szép (Szolnok, Verseghy F. Gymnasium, 3rd grade)

2. The relative speed of the object and the image as a function of the object distance $x$ is plotted, and we obtain the relationship shown in the figure.