Kvant Physics Problem 51
A camera obscura is modeled as a rectangular box of length $L$ with a small circular aperture of diameter $d = 1,\text{mm} = 1\times 10^{-3},\text{m}$.
Verified: yes
Verdicts: PASS + PASS
Solve time: 3m32s
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 1 mm diameter aperture have so that the tiger appears striped in the photograph? The distance between the stripes on the tiger’s coat is 20 cm.
A. L. Stasenko
Setup and Assumptions
A camera obscura is modeled as a rectangular box of length $L$ with a small circular aperture of diameter $d = 1,\text{mm} = 1\times 10^{-3},\text{m}$. A tiger is located at distance $R = 20,\text{m}$ from the aperture. The tiger’s coat contains alternating bright and dark stripes with spatial period $S = 20,\text{cm} = 2\times 10^{-1},\text{m}$.
The image is formed on a screen placed at the back wall of the camera obscura, a distance $L$ from the aperture. Geometrical optics is used, treating light as rays. The finite aperture produces geometric blur but diffraction effects are neglected since no wavelength is provided and the aperture is macroscopic compared to optical wavelengths.
The unknown is the maximal camera obscura length $L_{\max}$ for which the stripe pattern remains resolvable in the image.
Physical Principles
The pinhole projection produces a linear magnification relation between object size and image size,
$$m = \frac{L}{R}.$$
The separation of adjacent stripes in the image is given by the projected spacing
$$\Delta x_{\text{img}} = m S.$$
A finite aperture produces a geometric blur spot on the screen whose diameter is determined by similar triangles between the aperture and the screen,
$$b = \frac{d}{R} L.$$
Resolution requires that the blur diameter be smaller than the image separation of adjacent stripes,
$$b < \Delta x_{\text{img}}.$$
Derivation
The image separation between adjacent stripes follows from linear scaling,
$$\Delta x_{\text{img}} = \frac{L}{R} S.$$
The blur size caused by the finite aperture is obtained by considering rays entering from opposite edges of the aperture. These rays form a cone that expands linearly with screen distance, giving
$$b = \frac{d}{R} L.$$
The condition for resolving stripes is
$$\frac{d}{R} L < \frac{L}{R} S.$$
Both sides contain the same geometric factor $\frac{L}{R}$, which cancels,
$$d < S.$$
The inequality contains no dependence on $L$, so resolution depends only on the ratio of aperture diameter to stripe spacing.
Result
The resolving condition reduces to
$$d < S.$$
Substituting numerical values,
$$1\times 10^{-3},\text{m} < 2\times 10^{-1},\text{m}.$$
This inequality is satisfied, and it does not restrict the camera length $L$. Therefore,
$$L_{\max} = \infty.$$
Boxed result:
$$\boxed{L_{\max} = \infty}$$
Sanity Checks
Dimensional consistency holds since both $d$ and $S$ carry units of length and appear in a direct comparison.
The cancellation of $L$ is physically consistent because both the geometric blur and the geometric magnification scale linearly with the camera length in a pinhole model, preserving relative contrast.
The dominant scale comparison is $d/S = 10^{-3}/(2\times 10^{-1}) = 5\times 10^{-3}$, which lies far below unity, confirming that the aperture is sufficiently small to preserve stripe contrast for any finite projection distance.