Kvant Physics Problem 51
A pinhole camera obscura of linear size $L$ forms a real inverted image of a tiger located at distance $u \le 20,\text{m}$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m08s
Source on kvant.digital
Problem
Photographing a tiger from a distance of less than 20 meters is dangerous. What size can a camera obscura with a pinhole of diameter 1 mm have so that the tiger appears striped in the photograph? The distance between stripes on the tiger’s coat is 20 cm.
A. L. Stasenko
Setup and Assumptions
A pinhole camera obscura of linear size $L$ forms a real inverted image of a tiger located at distance $u \le 20,\text{m}$. The pinhole has diameter $d = 1,\text{mm} = 1 \times 10^{-3},\text{m}$. The characteristic spacing between stripes on the tiger’s coat is $s = 20,\text{cm} = 2 \times 10^{-1},\text{m}$.
The unknown is the maximum linear size $L$ of the camera obscura (distance from pinhole to screen), such that adjacent stripes remain distinguishable in the image.
The analysis assumes rectilinear propagation of light rays, a geometrical-optics regime, a uniformly illuminated opaque screen, and a point-like pinhole of finite diameter $d$. Diffraction is neglected compared with geometrical blurring. The tiger is treated as a planar object perpendicular to the optical axis at distance $u$.
Physical Principles
A pinhole camera forms images through similar triangles, giving a linear magnification relation between object size and image size.
For an object point, the finite diameter of the pinhole produces a geometric blur on the screen. The blur size is determined by rays passing through opposite edges of the pinhole aperture and is obtained from similar triangles relating the aperture size at the pinhole plane to its projection onto the screen.
Two features are resolvable when the image separation exceeds the blur diameter caused by the finite aperture.
Derivation
Let the camera screen be at distance $L$ from the pinhole. A point on the object at distance $u$ forms an image at the screen with transverse magnification
$$m = \frac{L}{u}.$$
The separation between adjacent tiger stripes on the object is $s$, so their image separation on the screen is
$$\Delta x_{\text{image}} = m s = \frac{L}{u} s.$$
The finite pinhole diameter $d$ produces a blur spot on the screen. The extreme rays passing through opposite edges of the pinhole subtend an angle approximately $d/u$ at the object and therefore spread on the screen over a transverse size
$$\Delta x_{\text{blur}} = \frac{L}{u} d.$$
Resolving adjacent stripes requires
$$\Delta x_{\text{image}} \ge \Delta x_{\text{blur}}.$$
Substituting the expressions,
$$\frac{L}{u} s \ge \frac{L}{u} d.$$
Canceling the common factor $L/u$ yields the resolution condition
$$s \ge d.$$
This condition contains no dependence on $L$, so the size of the camera obscura does not affect stripe resolution under the given geometrical-optics model.
Result
The resolution condition is
$$s \ge d.$$
Substituting numerical values,
$$s = 2 \times 10^{-1},\text{m}, \quad d = 1 \times 10^{-3},\text{m},$$
$$2 \times 10^{-1},\text{m} \gg 1 \times 10^{-3},\text{m}.$$
Since the inequality is strongly satisfied, the pinhole is sufficiently small to resolve the stripes for any camera length $L$ consistent with the object distance constraint.
$$\boxed{L_{\max} \text{ is unbounded within the geometric-optics model; any } L \text{ (with } u \le 20,\text{m}) \text{ allows stripe resolution.}}$$
Sanity Checks
Dimensional consistency holds because both image separation and blur scale as $\frac{L}{u}$ times a length, making the comparison independent of $L$.
The cancellation of $L$ indicates that increasing camera size enlarges both the image and the blur proportionally, leaving resolution unchanged in this approximation.
The dominant constraint is instead the inequality between physical feature size and pinhole diameter. Any error in tracking the same magnification factor for both image formation and blur would incorrectly introduce a spurious dependence on $L$.