Optics 025 · Imaging, instruments, and visual systems

Camera Obscura

A 2D optical-bench section connects finite-distance inversion, movable screen magnification, exact aperture-footprint geometry, circular-aperture diffraction scale, throughput in exposure stops, and a live three-curve sharpness envelope to one physical pinhole.

Asymptotic crossover and effective f-number

d_{\times},\ N_{\mathrm{eff}}

$0.50$

Geometric, Airy, and RSS blur diameters

b_g,\ b_A,\ b_{\mathrm{RSS}}

$50\%$

Fresnel number and relative exposure

N_F,\ \Delta\mathrm{EV}

$0.00\pi$

Model regime $\text{valid model regime}$

The geometric footprint and far-field Airy diameter are valid asymptotic scales. Their RSS bridge does not solve full near-field Fresnel diffraction. Flux losses beyond aperture area, vignetting, material scatter, finite source spectrum, screen response, and detector sampling are omitted.

Physics tutorial

Audit the physics of Camera Obscura

BackgroundA camera obscura is first a finite-distance central projection. If the object is a distance from the hole and the screen is another distance behind it, similar triangles give $y'=-\frac{L}{S}y$ . The minus sign is the inversion; moving the screen changes image scale even though no lens is present.

Why it mattersHow do finite-object projection, aperture footprint, diffraction, and exposure compete in a real pinhole camera?

Start with the essentials

Focus question: How do finite-object projection, aperture footprint, diffraction, and exposure compete in a real pinhole camera?
One-sentence intuition: A finite hole adds a geometric footprint while wave propagation adds a competing diffraction scale: $b_g=d\!\left(1+\frac{L}{S}\right),\qquad b_A\approx\frac{2.44\lambda L}{d}$ . Equating these two asymptotes defines the marked crossover, but the near-field transition itself requires Fresnel propagation and does not have one universal numerical prefactor.

Core mathematical model

Finite-distance inversion

m=\frac{y'}{y}=-\frac{L}{S}

Every solid chief ray crosses the same physical hole. The three object samples therefore reverse vertical order on the screen, and screen motion changes their separation by the same ratio.

Finite-aperture footprint

b_g=d\!\left(1+\frac{L}{S}\right)

The extra finite-object factor comes from tracing both aperture edges, not from assigning the hole diameter directly to the screen blur. It tends to the hole diameter only for a very distant object.

Diffraction scale and crossover

b_A\approx\frac{2.44\lambda L}{d},\qquad d_{\times}=\sqrt{\frac{2.44\lambda L}{1+L/S}}

The first expression is the circular-aperture far-field first-zero diameter. The second merely equates the two asymptotic diameters; the interface deliberately calls it a crossover rather than an exact optimum.

Common difficulties

Treating one pinhole constant as universal

Typical misconceptionThe sharpest hole always equals one fixed coefficient times the square root of wavelength and screen distance.

Better mental modelThat form is a useful scaling law, but its coefficient depends on the sharpness criterion, object distance, spectrum, and Fresnel regime. This Lab exposes two defensible asymptotes, their crossover, and the Fresnel number instead of hiding those dependencies.

Run the experiment

01
Scene 1: Geometric footprint
Start with the geometric-footprint scene. Increase the physical hole diameter and compare the brighter-area exposure change with the widening screen spot clouds.
What to observe: Doubling hole diameter adds two exposure stops by area but doubles the distant-object geometric footprint. Brightness and sharpness do not improve together.
02
Scene 2: Asymptotic crossover
Enter the crossover scene. Drag the screen and verify that image magnification, finite-object geometric blur, diffraction scale, and the marked crossover all move together.
What to observe: Moving the screen farther increases image size and both blur scales. The finite-object geometric term grows slightly faster than simply copying the hole diameter.
03
Scene 3: Wavelength-resolved diffraction
Select wavelength-resolved diffraction, shrink the hole, and compare the blue, green, and red first-zero rings with the rising effective f-number.
What to observe: In the smallest-hole regime the red ring is largest because diffraction scale grows with wavelength. This chromatic ordering exists even though chief-ray projection is wavelength independent.