## 83. Math Puzzle $4 / 72$

Peter is to be photographed while crossing the image plane at a distance of $50 \mathrm{~m}$ with a speed of $18 \mathrm{~km} / \mathrm{h}$.

How long can the exposure time be at most if the optics have a focal length of 50 $\mathrm{mm}$ and a permissible image contour blur of $0.05 \mathrm{~mm}$ is allowed?

How far would Peter have to be to achieve the same contour blur with a permissible exposure time of $1 / 20 \mathrm{~s}$ and otherwise the same conditions?

(It is assumed for simplification that all projecting light rays pass through a point at the center of the optics.)
In one second, Peter covers five meters. This distance corresponds to the length $x$ on the film. According to the ray theorem, we have

$$
\frac{x}{50 \text{~mm}}=\frac{5 \text{~m}}{50 \text{~m}} ; \quad x=5 \text{~mm} (\text{ distance per second })
$$

With a contour blur of $0.05 \text{~mm}$ caused by this movement, the exposure time must not exceed $\frac{0.05}{5}=\frac{1}{100}$ of a second.

For an exposure time of $\frac{1}{20} \text{~s}=\frac{5}{100} \text{~s}$, the distance of the object (Peter) must be five times greater, i.e., $250 \text{~m}$, to maintain the same contour sharpness.