Diffraction and the Optimum Pinhole

The previous lesson said that a distant point blurs to a disc exactly the size of the hole, so a smaller hole always makes a sharper picture. Everyone who has ever pierced three pieces of foil knows that this stops being true. Past some diameter the image gets softer as the hole gets smaller, and dimmer at the same time, which is the worst of both worlds.

That failure is not a flaw in the geometry. The geometry is correct as far as it goes; it simply leaves out that light is a wave, and a wave squeezed through a small opening spreads. This page works out how much it spreads, sets that spreading against the geometric blur, finds the diameter where their sum is least — and then shows that three of the people who did this got three different answers, for reasons worth understanding.

Light at the aperture

Huygens’ construction treats every point of a wavefront as the source of a secondary wavelet, and the wave a moment later as the envelope of all of them. Put an opaque screen with a hole in it across a plane wavefront and the construction says something immediately: the wavelets that would have cancelled the sideways spreading — the ones from the parts of the wavefront you have just blocked — are no longer there to do it. What emerges from the hole is not a beam with the shape of the hole. It is the sum of wavelets from every point of the hole, and that sum depends on direction.

Huygens’ principle also gives the size of the effect without any calculation. Look at a direction θ away from the axis. A wavelet from one edge of the hole and a wavelet from the other edge travel path lengths that differ by d sin θ. When that difference is comparable with a wavelength, the two arrive in opposite phase and cancel; when it is small compared with a wavelength they arrive together and add. So the light stays concentrated within an angle of roughly λ/d, and spreads outside it. The narrower the hole, the wider the spread — the opposite of what geometry alone predicts, and the whole of the problem.

For a circular hole the exact answer needs a Bessel function rather than a sine, and the result is the Airy pattern: a bright central disc surrounded by faint rings. OpenStax’s treatment gives the first dark ring at

sin θ₁ = 1.22 λ / d

Angular radius of the Airy disc

θ₁ is the angle from the axis to the first minimum, λ the wavelength and d the hole diameter. The 1.22 is the first zero of the Bessel function J₁ divided by π, and it is the only place a number appears in this page that cannot be derived in a line.

The Airy pattern: relative intensity across the diffraction image of a point

• Airy pattern, (2J₁(x)/x)²

Show the numbers behind this plot

A curve of relative intensity against the quantity d sin theta over lambda, starting at 1.0 on the axis and falling smoothly: 0.976 at 0.1, 0.905 at 0.2, 0.797 at 0.3, 0.665 at 0.4, 0.521 at 0.5, 0.381 at 0.6, 0.256 at 0.7, 0.154 at 0.8, 0.080 at 0.9, 0.033 at 1.0, 0.008 at 1.1, and reaching zero at 1.22, which is the first dark ring. Beyond it the curve rises again to a first bright ring peaking at only about 0.017, that is 1.7 per cent of the central intensity, near 1.65, falls to a second zero at 2.23, and rises to a second ring of about 0.004. The pattern is therefore overwhelmingly a central disc: essentially all of the light is inside the first dark ring, and the rings outside it are faint enough to matter only where a very bright highlight sits against a dark surround.

Series	d sin θ / λ	Relative intensity
Airy pattern, (2J₁(x)/x)²	0.00	1.00
Airy pattern, (2J₁(x)/x)²	0.10	0.98
Airy pattern, (2J₁(x)/x)²	0.20	0.91
Airy pattern, (2J₁(x)/x)²	0.30	0.80
Airy pattern, (2J₁(x)/x)²	0.40	0.67
Airy pattern, (2J₁(x)/x)²	0.50	0.52
Airy pattern, (2J₁(x)/x)²	0.60	0.38
Airy pattern, (2J₁(x)/x)²	0.70	0.26
Airy pattern, (2J₁(x)/x)²	0.80	0.15
Airy pattern, (2J₁(x)/x)²	0.90	0.08
Airy pattern, (2J₁(x)/x)²	1.00	0.03
Airy pattern, (2J₁(x)/x)²	1.10	0.01
Airy pattern, (2J₁(x)/x)²	1.20	0.00
Airy pattern, (2J₁(x)/x)²	1.30	0.00
Airy pattern, (2J₁(x)/x)²	1.40	0.01
Airy pattern, (2J₁(x)/x)²	1.50	0.01
Airy pattern, (2J₁(x)/x)²	1.60	0.02
Airy pattern, (2J₁(x)/x)²	1.70	0.02
Airy pattern, (2J₁(x)/x)²	1.80	0.01
Airy pattern, (2J₁(x)/x)²	1.90	0.01
Airy pattern, (2J₁(x)/x)²	2.00	0.01
Airy pattern, (2J₁(x)/x)²	2.10	0.00
Airy pattern, (2J₁(x)/x)²	2.20	0.00
Airy pattern, (2J₁(x)/x)²	2.30	0.00
Airy pattern, (2J₁(x)/x)²	2.40	0.00
Airy pattern, (2J₁(x)/x)²	2.50	0.00

Computed from the standard circular-aperture expression rather than measured. The horizontal scale is universal: it applies to any hole diameter and any wavelength, which is why the first dark ring is always at 1.22 in these units. The curve is drawn to show the shape, not measured from a real material. Your own materials will differ, and measuring them is what the sensitometry part of the course is for.

The Airy disc on the film

The film sits f behind the hole, and for the small angles involved sin θ ≈ tan θ, so the radius of the disc at the film is f sin θ₁ and the diameter is twice that:

b_diff = 2.44 λ f / d = 2.44 λ N

Diffraction blur at the film

where N = f/d is the effective f-number from the previous lesson. Two things about this expression deserve a moment.

First, written as 2.44 λ N it says the diffraction blur depends only on the f-number and the wavelength — a fact that governs every camera, not just this one. At f/8 with green light it is 0.011 mm, which is why a good lens is sharp; at f/200 it is 0.27 mm, which is why a pinhole is not.

Second, written as 2.44 λ f/d it says the blur is inversely proportional to the hole. The geometric blur was directly proportional to it. That opposition is the whole argument.

Two blurs, and where their sum is least

Set the two side by side, for a distant subject:

b_geom = d and b_diff = 2.44 λ f / d

The two blurs

One rises with d, the other falls as 1/d, so their sum has a minimum. Differentiate b = d + 2.44 λ f/d and set the result to zero: 1 − 2.44 λ f/d² = 0, so

d_opt = √(2.44 λ f) ≈ 1.56 √(λ f)

The optimum diameter

At the minimum the two blurs are equal — put d² = 2.44λf back into either expression and you get the same number — so the total blur there is exactly 2d_opt. That is a useful thing to remember: at the best hole size, half your unsharpness is the hole and half is the wave.

Combining the blurs in quadrature instead, √(b_geom² + b_diff²), which is the better model if you think of them as independent spreading processes, gives exactly the same optimum diameter, though a smaller total. The optimum is robust to how you add them; only the predicted sharpness changes.

How the two blurs really combine

Neither addition is right, and it is worth knowing why, because it tells you how much to trust the sharpness figures later on this page.

What actually reaches the film from a single object point is the geometric disc convolved with the Airy pattern: each point of the disc is smeared by diffraction, and the results overlap. The width of a convolution is always less than the sum of the two widths and more than the larger of them, so the plain sum is an upper bound and quadrature — which would be exact only if both spreads were Gaussian, and neither is — is a good lower bound. At the optimum the sum gives 2d and quadrature gives √2 d, so for the 50 mm design the true blur lies somewhere between 0.33 and 0.47 mm. This course quotes the upper bound throughout, because a camera that turns out sharper than predicted is a better surprise than the other kind.

The shape matters as much as the width, and it changes as you move along the curve. The geometric contribution is a flat-topped disc: a uniformly lit hole projects uniform illumination right out to a hard edge. The diffraction contribution is peaked, with rings. So a geometry-dominated pinhole image is soft in a plain, even way with definite edges to its unsharpness, while a diffraction-dominated one has a bright core and a faint skirt that reaches much further, which the eye reads as glow rather than as blur. Two negatives with the same measured blur diameter can therefore look quite different, and that is a large part of why the optimum is an argument rather than a calculation.

Geometric blur, diffraction blur and their sum, against hole diameter

• Geometric blur, = d
• Diffraction blur, = 2.44 λ f / d
• Sum of the two

Show the numbers behind this plot

Three curves against pinhole diameter from 0.10 to 0.60 mm, computed for a focal distance of 50 mm and a wavelength of 450 nanometres. The geometric blur is a straight line through the origin: it equals the hole diameter, so 0.10 mm at 0.10 mm and 0.60 mm at 0.60 mm. The diffraction blur is a falling hyperbola: 0.549 mm at a hole of 0.10 mm, 0.366 at 0.15, 0.275 at 0.20, 0.234 at 0.234, 0.183 at 0.30, 0.137 at 0.40 and 0.092 at 0.60. The two cross at 0.234 mm, where both equal 0.234 mm. Their sum is a broad U-shaped curve: 0.649 mm at a hole of 0.10, 0.516 at 0.15, 0.475 at 0.20, a minimum of 0.469 at 0.234, 0.470 at 0.25, 0.483 at 0.30, 0.537 at 0.40 and 0.692 at 0.60. The important feature is how flat the bottom is: anywhere between 0.20 and 0.30 mm the total is within three per cent of the best value, so a ten or twenty per cent error in the hole is invisible, while going down to 0.15 mm or up to 0.50 mm costs ten and thirty per cent respectively.

Series	Pinhole diameter d, mm	Blur at the film, mm
Geometric blur, = d	0.10	0.10
Geometric blur, = d	0.13	0.13
Geometric blur, = d	0.15	0.15
Geometric blur, = d	0.17	0.17
Geometric blur, = d	0.20	0.20
Geometric blur, = d	0.23	0.23
Geometric blur, = d	0.25	0.25
Geometric blur, = d	0.30	0.30
Geometric blur, = d	0.35	0.35
Geometric blur, = d	0.40	0.40
Geometric blur, = d	0.45	0.45
Geometric blur, = d	0.50	0.50
Geometric blur, = d	0.60	0.60
Diffraction blur, = 2.44 λ f / d	0.10	0.55
Diffraction blur, = 2.44 λ f / d	0.13	0.44
Diffraction blur, = 2.44 λ f / d	0.15	0.37
Diffraction blur, = 2.44 λ f / d	0.17	0.31
Diffraction blur, = 2.44 λ f / d	0.20	0.28
Diffraction blur, = 2.44 λ f / d	0.23	0.23
Diffraction blur, = 2.44 λ f / d	0.25	0.22
Diffraction blur, = 2.44 λ f / d	0.30	0.18
Diffraction blur, = 2.44 λ f / d	0.35	0.16
Diffraction blur, = 2.44 λ f / d	0.40	0.14
Diffraction blur, = 2.44 λ f / d	0.45	0.12
Diffraction blur, = 2.44 λ f / d	0.50	0.11
Diffraction blur, = 2.44 λ f / d	0.60	0.09
Sum of the two	0.10	0.65
Sum of the two	0.13	0.56
Sum of the two	0.15	0.52
Sum of the two	0.17	0.49
Sum of the two	0.20	0.47
Sum of the two	0.23	0.47
Sum of the two	0.25	0.47
Sum of the two	0.30	0.48
Sum of the two	0.35	0.51
Sum of the two	0.40	0.54
Sum of the two	0.45	0.57
Sum of the two	0.50	0.61
Sum of the two	0.60	0.69

Computed for f = 50 mm and λ = 450 nm, the region blue-sensitive paper works in. Nothing here is measured. Notice the flatness of the bottom: it is the reason a century of authors could disagree about the constant without anyone being able to see the difference. The curve is drawn to show the shape, not measured from a real material. Your own materials will differ, and measuring them is what the sensitometry part of the course is for.

The Fresnel-zone view: the same answer from the other side

There is a second way of seeing why an optimum exists, and it explains why the various answers cannot differ by much.

Stand at the point on the film where the image of a distant axial point falls, and look back at the hole. Light from a ring of radius r in the aperture travels a distance √(r² + f²), which for small r is about f + r²/2f. So the ring’s contribution arrives late by r²/2f compared with the centre. Contributions that are late by half a wavelength arrive in antiphase and cancel the central ones. Set r²/2f = nλ/2 and the boundaries between successive half-period zones are at

r_n = √( n λ f )

Fresnel zone radii

A hole of radius r therefore admits r²/λf half-period zones — the Fresnel number — or, in terms of diameter, d²/(4λf). Now the argument writes itself:

• Fewer than about one zone: everything in the aperture arrives within half a wavelength, so it all adds. The image point is as compact as it can be, but the aperture is small, so the diffraction spread λ/d is wide and there is little light.
• More than one zone: the second zone arrives in antiphase with the first and cancels part of it, and the light that was cancelled has to go somewhere: it goes into the surrounding rings. Opening the hole further stops improving the concentration and starts scattering.

So the sharpest hole is the one admitting about one half-period zone, and every published optimum is a different opinion about how much of a second zone is tolerable. Convert the constants and the whole family collapses onto a small range:

Criterion	d = k√(λf)	Half-period zones admitted
Quarter-wave path error (Rayleigh 1889; Petzval’s minimum)	k = 1.41	0.50
Equal geometric and Airy blur	k = 1.56	0.61
Rayleigh’s own photographic result (1891)	k = 1.90	0.90
Exactly one zone	k = 2.00	1.00

Half-period zones seen from a point on the film

1. 1Zone 1, radius √(λf) — all of it adds; a pinhole is roughly this disc
2. 2Zone 2, out to √(2λf) — arrives half a wave late, so it subtracts
3. 3Zone 3, out to √(3λf) — adds again — hence the alternation
4. 4The image point P — all the path differences are measured to here

The zone boundaries crowd together as you go out, because r grows as √n while the area of each ring stays nearly constant. Block the even zones and you have a zone plate.

Deeper: what “sharp” means, and why the criterion decides the constant

Before reading the disagreement, it is worth being clear that it is not an arithmetic disagreement. Every author below can differentiate. They differ because “the sharpest hole” is not a well-formed question until you say what quantity you are extremising, and there are at least four reasonable candidates.

Minimise the width of the point spread. Take the blur diameter — the disc plus the Airy disc — and make it as small as possible. This is Petzval’s criterion and the equal-blur criterion, and it weights the far tails of the pattern heavily, because the first dark ring is defined by where the light finally runs out rather than by where most of it is.

Maximise the concentration of light at the image point. Ask instead for the aperture that puts the most light into the smallest core, which is what a lens is for. Rayleigh’s quarter-wave argument answers this one: open the hole until the extreme path error across it reaches λ/4, at which point, by his own earlier result, a lens would no longer measurably improve the definition. Nothing in that argument mentions blur diameter at all.

Maximise contrast at the spatial frequencies that carry the subject. A modern treatment would compute the modulation transfer function of the aperture and choose the diameter that maximises modulation at, say, one or two line pairs per millimetre. That is a different optimisation again, and because a bigger hole moves more energy into the core at low frequencies, it tends to prefer larger apertures. This course has not computed it and quotes no constant for it; it is named here so that you know a fourth answer exists.

Ask which print somebody preferred. This is what Rayleigh actually did in 1891, and it is the only one of the four that involves a photograph. It bundles everything — the point-spread shape, the material’s own resolving power, the contrast of the paper, the viewing distance and the taste of the judge — into a single verdict, and it is both the most relevant criterion to a photographer and the least reproducible.

The constants these criteria actually produce run from 1.41 for the first two, through 1.56 when the width criterion is applied with the true Airy diameter rather than Petzval’s cruder estimate, to 1.90 for the one Rayleigh reached by looking at photographs. The course does not claim to know a mechanism that orders them; what it claims is the thing that matters for your notebook. The constant is a statement about what you are trying to do, not a fact about optics, and a source that gives you a number without telling you its criterion has withheld the only part that could have helped you choose.

The formula family, and why it is a genuine dispute

⁇The optimum-pinhole constant: three criteria, three answers

Everything in this box is quoted from the original documents, which this course has read in facsimile. The disagreement is real, and it is a disagreement about what “best” means, not about arithmetic.

Petzval, 1857 — minimise a summed blur. In his report on dioptric investigations to the Vienna Academy, Josef Petzval gives the diffraction patch of a very small hole as D = Aλ/ρ, with A the screen distance and ρ the aperture radius. For a larger hole, he says, the hole’s own diameter must be added, giving D = 2ρ + Aλ/ρ. He differentiates, sets the derivative to zero, and obtains ρ = √(Aλ/2), so that the diameter is d = √(2fλ) ≈ 1.41√(fλ), with a minimum blur of 2√(2Aλ). His worked case is A = 11 Vienna inches, for which he gets a radius of 0.01 inch and a blur of 0.042 inch. Note what his criterion assumes: that the blur in the intermediate case is the plain sum of the two extreme cases.

Rayleigh, 1889 — the largest hole a lens would not improve. In a British Association note Rayleigh gives 2ρ² = fλ, which is the same 1.41√(fλ), from a completely different argument: a lens exists to correct the phase differences across the aperture, and he had shown earlier that an extreme phase error of a quarter of a wavelength “produces no appreciable deterioration in the definition”. Open the hole until the error reaches a quarter wave and the lens is no longer missed. He works his own example — a hole of a fifth of an inch, so the sun’s image on a screen beyond about 66 feet is as well defined as the direct view.

Rayleigh, 1891 — and then he measured it. In On Pin-hole Photography he quotes Petzval, says his theory “can hardly be regarded as sound” because adding the two extreme cases is inadmissible, remarks that it is “very remarkable” that it nevertheless produces the right relation, and then does the diffraction properly using Lommel’s 1884 results. He also pierces six apertures in sheet zinc, from 0.0210 to 0.0366 inch, and photographs a test object with each. The best photographic result gives (2r)²/f = 1.52 × 10⁻⁴ cm, from which he derives, if you divide through by the effective wavelength he back-calculates for his own plates, 4.2 × 10⁻⁵ cm, a constant of 1.90√(fλ). He gives the practical form directly: at f = 12 inches, 2r = 0.027 inch; at f = 84 inches, 2r = 0.071 inch.

This is where “Rayleigh’s 1.9” comes from, and it is genuinely his — but it is his measurement of what his plates and his eye preferred, not his theory, and it is tied to a photographic effective wavelength of 420 nm rather than to any wavelength you might be working at. Anyone who quotes 1.9 as “the Rayleigh criterion” has conflated a measurement with the phase criterion that gives 1.41.

Young, 1971 and 1989 — cited everywhere; not read here. Matt Young’s Pinhole optics, Applied Optics 10(12), pp. 2763–2767 (1971), and his later treatments in the American Journal of Physics and The Physics Teacher, are the modern references, and a constant near 1.5 to 1.6 is widely attributed to him. This course has read only the 1971 abstract, which confirms the paper’s subject and states no constant, and has not obtained the full text of any of the three. It therefore does not put a number in Young’s mouth. What it does instead is derive the constant that follows from the modern criterion — set the geometric blur equal to the Airy diameter 2.44λf/d rather than to Petzval’s cruder 2λf/d — which gives 1.56√(fλ), and label it by its criterion rather than by a person. If you can read Young, check him against this.

What this course uses. The range 1.4 to 1.9, not a single number, with 1.56 as the working value when one number is needed. The plot above shows why that is not a cop-out: across the whole of that range the total blur varies by about three per cent, which nobody can see. The dispute has survived a hundred and seventy years precisely because it does not matter very much.

Optimum hole diameter against focal distance, for the constants and the two wavelengths

• 1.41 √(λf) — quarter-wave, at 450 nm
• 1.56 √(λf) — equal blurs, at 450 nm
• 1.90 √(λf) — Rayleigh measured, at 450 nm
• 1.56 √(λf) at 550 nm, for comparison

Show the numbers behind this plot

Four square-root curves of optimum hole diameter against focal distance from 10 to 300 mm. Three are at 450 nanometres, the wavelength blue-sensitive paper works at, and together they form a band. The lowest is the quarter-wave criterion with constant 1.41: 0.095 mm at 10 mm focal distance, 0.212 at 50 mm, 0.300 at 100 mm, 0.424 at 200 mm and 0.520 at 300 mm. The middle is the equal-blur criterion with constant 1.56: 0.105 mm at 10 mm, 0.234 at 50, 0.331 at 100, 0.469 at 200 and 0.574 at 300. The highest at 450 nanometres is Rayleigh's photographic result with constant 1.90: 0.127 mm at 10 mm, 0.285 at 50, 0.403 at 100, 0.570 at 200 and 0.698 at 300. The fourth curve is the equal-blur criterion again but at 550 nanometres, the conventional design wavelength for panchromatic film: 0.116 mm at 10 mm, 0.259 at 50, 0.366 at 100, 0.518 at 200 and 0.635 at 300. It lies inside the 450-nanometre band, between the 1.56 and 1.90 curves, which is the point: changing the wavelength from blue to green moves the answer by less than the disagreement between authors does. Every curve grows as the square root of the focal distance, so doubling the focal distance multiplies the hole by only 1.41.

Series	Focal distance f, mm	Optimum hole diameter, mm
1.41 √(λf) — quarter-wave, at 450 nm	10.00	0.10
1.41 √(λf) — quarter-wave, at 450 nm	20.00	0.13
1.41 √(λf) — quarter-wave, at 450 nm	30.00	0.16
1.41 √(λf) — quarter-wave, at 450 nm	50.00	0.21
1.41 √(λf) — quarter-wave, at 450 nm	75.00	0.26
1.41 √(λf) — quarter-wave, at 450 nm	100.00	0.30
1.41 √(λf) — quarter-wave, at 450 nm	150.00	0.37
1.41 √(λf) — quarter-wave, at 450 nm	200.00	0.42
1.41 √(λf) — quarter-wave, at 450 nm	250.00	0.47
1.41 √(λf) — quarter-wave, at 450 nm	300.00	0.52
1.56 √(λf) — equal blurs, at 450 nm	10.00	0.10
1.56 √(λf) — equal blurs, at 450 nm	20.00	0.15
1.56 √(λf) — equal blurs, at 450 nm	30.00	0.18
1.56 √(λf) — equal blurs, at 450 nm	50.00	0.23
1.56 √(λf) — equal blurs, at 450 nm	75.00	0.29
1.56 √(λf) — equal blurs, at 450 nm	100.00	0.33
1.56 √(λf) — equal blurs, at 450 nm	150.00	0.41
1.56 √(λf) — equal blurs, at 450 nm	200.00	0.47
1.56 √(λf) — equal blurs, at 450 nm	250.00	0.52
1.56 √(λf) — equal blurs, at 450 nm	300.00	0.57
1.90 √(λf) — Rayleigh measured, at 450 nm	10.00	0.13
1.90 √(λf) — Rayleigh measured, at 450 nm	20.00	0.18
1.90 √(λf) — Rayleigh measured, at 450 nm	30.00	0.22
1.90 √(λf) — Rayleigh measured, at 450 nm	50.00	0.28
1.90 √(λf) — Rayleigh measured, at 450 nm	75.00	0.35
1.90 √(λf) — Rayleigh measured, at 450 nm	100.00	0.40
1.90 √(λf) — Rayleigh measured, at 450 nm	150.00	0.49
1.90 √(λf) — Rayleigh measured, at 450 nm	200.00	0.57
1.90 √(λf) — Rayleigh measured, at 450 nm	250.00	0.64
1.90 √(λf) — Rayleigh measured, at 450 nm	300.00	0.70
1.56 √(λf) at 550 nm, for comparison	10.00	0.12
1.56 √(λf) at 550 nm, for comparison	20.00	0.16
1.56 √(λf) at 550 nm, for comparison	30.00	0.20
1.56 √(λf) at 550 nm, for comparison	50.00	0.26
1.56 √(λf) at 550 nm, for comparison	75.00	0.32
1.56 √(λf) at 550 nm, for comparison	100.00	0.37
1.56 √(λf) at 550 nm, for comparison	150.00	0.45
1.56 √(λf) at 550 nm, for comparison	200.00	0.52
1.56 √(λf) at 550 nm, for comparison	250.00	0.58
1.56 √(λf) at 550 nm, for comparison	300.00	0.64

Computed from d = k√(λf); nothing plotted here was measured by this course. The vertical spread of the 450 nm band at any focal distance is the whole of the historical disagreement. The curve is drawn to show the shape, not measured from a real material. Your own materials will differ, and measuring them is what the sensitometry part of the course is for.

Wavelength: why paper wants a smaller hole than film

λ appears under a square root, so the optimum diameter goes as √λ, and the choice of λ is a choice about the material, not about the light.

Part IV established what each class of material actually responds to: an undyed silver halide emulsion, which is what ordinary photographic paper and a student-coated plate are, works in the blue and the ultraviolet and stops around 500 nm; ILFORD describe their chloro-bromide papers as blue sensitive with a slight sensitivity to green. Panchromatic film responds across the visible, and the conventional design wavelength for it is 550 nm, near the middle of the visible band and near the peak of daylight vision. So:

• Blue-sensitive paper and home-coated emulsions: work at about 450 nm.
• Panchromatic film: work at about 550 nm by convention.

At f = 50 mm that is 0.234 mm against 0.259 mm — the paper’s hole is 10 per cent smaller, and the f-number 10 per cent larger, worth 0.29 stop of exposure. Small. But it points the right way, and it has a consequence worth stating plainly: one hole cannot be optimal for both. A hole sized for paper is slightly undersized for film, which puts film into the diffraction-dominated half of the curve, so the same camera renders film a little softer than it renders paper. That is a real, predictable, testable difference, and it is one of the things the pinhole-diameter series in this part can show.

🕰Rayleigh measured the effective wavelength, and it was blue

Rayleigh did not assume a photographic wavelength; he worked it out. Comparing his eye observations, made through orange-red glass at a mean wavelength he measured against a soda flame as 6.23 × 10⁻⁵ cm, with his photographic results on Ilford plates, he found the photographically effective wavelength to be 4.2 × 10⁻⁵ cm — 420 nm — “a little less than that belonging to Fraunhofer’s line G”, and remarked that the difference between the eye and the plate “is readily explained by the smaller effective wavelength in the former case”. A blue-sensitive plate in 1891 and a blue-sensitive paper today are the same kind of material, and they give the same answer.

Petzval’s wavelengths, by contrast, do not survive the conversion. He gives red as one fifty-thousandth of an inch and violet as one hundred-thousandth. Taking the English inch, that is 508 nm and 254 nm; taking the Viennese inch of the period, 527 nm and 263 nm. Either way his “red” is what we would call green and his “violet” is deep ultraviolet. His method is sound and his arithmetic is right; his wavelength scale is not one you should carry into a calculation.

What a pinhole actually resolves

At the optimum, total blur is 2d_opt. Take a line pair to need two blur widths — one for the dark line, one for the light one — so the limiting resolving power is 1/(2b):

Focal distance	dopt at 450 nm	Total blur	Resolution	Angular blur
25 mm	0.166 mm	0.33 mm	1.5 lp/mm	13.3 mrad
50 mm	0.234 mm	0.47 mm	1.1 lp/mm	9.4 mrad
100 mm	0.331 mm	0.66 mm	0.75 lp/mm	6.6 mrad
200 mm	0.469 mm	0.94 mm	0.53 lp/mm	4.7 mrad

Those are the pessimistic figures, since the total blur here is the summed bound; using the quadrature bound instead would multiply every resolution by √2, giving 2.1, 1.5, 1.1 and 0.75 lp/mm. Either way the answer is the same order.

One to two line pairs per millimetre. A diffraction-limited lens at f/8, computed the same way from 2.44λN, gives about 47 lp/mm — some forty times better. Petzval reached the same conclusion in 1857 by a different route, reckoning a good 3-inch portrait objective of 11-inch focus to be “about 180 times superior in sharpness to the camera obscura without glass”, and noting that the corresponding light intensities stand as 1 to 32,400.

Notice the last column. Resolution in lines per millimetre gets worse as the focal distance grows, but the image gets bigger in proportion, and the angular blur — which is what determines how much of the subject you have recorded — improves as 1/√f. This is exactly Rayleigh’s argument for long pinholes, and he demonstrated it with an aperture of 0.07 inch at seven feet of focus, photographing a group of cedars on 12 × 10-inch plates in about an hour and a half.

🧮How big can you print it?

Petzval supplies the criterion: a blur spot stops being visible as a spot when it subtends about one minute of arc at the eye. One minute of arc is 2.91 × 10⁻⁴ radians, so a blur of b millimetres is invisible from b/2.91 × 10⁻⁴ millimetres away.

For the 50 mm design, b = 0.47 mm, so a contact print looks sharp from about 1.6 m and soft at arm’s length. Enlarge it twice and the blur doubles: sharp only from 3.2 m. Petzval, working the identical sum on his own 1.1 mm blur, concluded that his image could be viewed properly only from about two Klafter — very nearly four metres — and that it “will of course tolerate no enlargement at all”.

The practical consequence is the one this part builds around: pinhole negatives are made to be contact printed, and if you want a big print you make a big negative, which is why 4 × 5 inches is the smallest format worth bothering with and why the paper negative is the right first material.

Going off the optimum deliberately

The optimum is where the sum of the blurs is least. It is not where every picture wants to be.

An oversized hole — say twice the optimum — roughly doubles the total blur but is four times the area, so it is two stops faster, and it puts you firmly in the geometric half of the curve. That matters aesthetically as well as practically: a geometry-dominated point spread is a flat-topped disc, so the image is soft in an even, plain way, with hard-edged highlights. Two stops is the difference between a moving cloud recording as a cloud and recording as a smear, and it is often worth the softness.

An undersized hole is a trap. Halving the diameter costs two stops and increases the blur, because you have moved into the diffraction-dominated half. What you buy for that price is a characteristic glow: the point spread is now an Airy pattern with rings, so bright highlights bleed haloes into their surroundings, and since the ring positions scale with λ the haloes are faintly coloured on colour material. Photographers who want that effect should choose it knowing that “smaller is sharper” stopped being true at the bottom of the curve.

The thing never to do is to undersize a hole in the belief that it is sharper. That belief is the single most common pinhole error, and the plot above is its refutation.

🧪See the turning point for yourself, in one afternoon

Part I told you to pierce three apertures of roughly 0.3, 0.6 and 1.2 mm in foil for the box camera obscura, and to rank them by eye. That ranking was qualitative and the box had no film in it. Now it can be made quantitative, and the interesting version needs holes on both sides of the optimum.

Make or select four plates spanning about 0.15, 0.25, 0.35 and 0.50 mm, and use them at a single focal distance on a single subject with a good edge in it — a roof line against sky, or a printed test chart at a stated distance. Correct the exposure for each hole by the square law, so that all four negatives print alike and only the sharpness differs, and record the times.

Predict first. At f = 50 mm on paper the sum-of-blurs model says 0.52, 0.47, 0.51 and 0.61 mm respectively, so 0.25 and 0.35 should be indistinguishable, 0.15 should be slightly worse than either, and 0.50 should be clearly the softest. Then look. The prediction that most often fails is the first one: many people find the 0.15 mm negative noticeably worse than the model says, which is usually the hole and not the physics — a very small hole is the hardest to pierce round.

The commissioning experiment later in this part runs this properly, with a control and a fixed subject. Doing it roughly now, before you have committed to a hole for the build, is worth an afternoon.

Zone plates and pinhole sieves

The zone diagram above contains a better idea than a hole. If the even-numbered zones are the ones that cancel, block them. What is left is a zone plate: a set of transparent rings with boundaries at r_n = √(nλf), alternately open and opaque, so that everything reaching the image point arrives in phase.

For f = 50 mm and λ = 450 nm the zone radii are 0.150 mm, 0.212 mm, 0.260 mm, 0.300 mm and so on — tiny, and getting closer together as they go out, which is why a zone plate is a photographic reproduction job rather than a piercing job.

A zone plate for f = 50 mm at 450 nm, and how its radii are set

1. 1Zone 1, clear, r = √(λf) = 0.150 mm — the same disc a pinhole would be
2. 2Zone 2, opaque, to √(2λf) = 0.212 mm — this is the half-wave-late light; blocking it is the whole trick
3. 3Zone 3, clear, to √(3λf) = 0.260 mm — in phase with zone 1 again
4. 4Zone 4, opaque, to √(4λf) = 0.300 mm — radii crowd as √n

Radii computed for f = 50 mm and λ = 450 nm. The drawing is enlarged: the whole mask here is well under a millimetre across.

The speed gain is real and easy to see in outline: taken out to ten zones, the open area of the plate is several times that of the optimum pinhole for the same focal distance, and the geometry above puts that at roughly three stops. The contrast loss is equally real and comes from the same structure. A zone plate is a diffraction grating in the round, and a grating sends light into several orders at once. Only one of those orders forms the image at f; the rest form other foci, and light that is not diffracted at all passes straight through. All of it lands on the film as non-image light, which is the definition of flare. Add the wavelength dependence — the radii are right for one λ only, so a zone plate is genuinely in focus for one colour — and the result is the soft, glowing, low-contrast rendering zone plates are used for. This course has not built or measured one and quotes no measured speed or contrast figure; Part VII offers the making as an optional exercise.

A pinhole sieve applies the same idea with an array of small holes positioned so that their contributions arrive in phase at the film, rather than with continuous rings. The course has not made or tested one, and says no more about it than that it exists and rests on the same zone arithmetic.

📓What the course could not verify

Whether commercially sold laser-drilled pinholes are specified against a stated design wavelength and a stated criterion could not be established from manufacturer data. If you buy one, ask, and record the answer in your pinhole register beside the measured diameter — because a plate sold as “optimum for 50 mm” is meaningless until you know which constant and which wavelength were used, and the answer could be anything from 0.21 mm to 0.29 mm.

🧮DO THE MATHS: the optimum for three focal distances at two wavelengths

Use d = k√(λf) with k = 1.56, the equal-blur criterion, and then N = f/d. Wavelengths in millimetres: 450 nm = 4.5 × 10⁻⁴ mm, 550 nm = 5.5 × 10⁻⁴ mm.

At 25 mm. λf = 4.5 × 10⁻⁴ × 25 = 0.01125, √ = 0.1061, × 1.56 = 0.166 mm, so N = 25/0.166 = f/151. At 550 nm: √(5.5 × 10⁻⁴ × 25) = 0.1173, × 1.56 = 0.183 mm, N = f/136.

At 50 mm. 450 nm: 0.234 mm, N = f/213. 550 nm: 0.259 mm, N = f/193.

At 100 mm. 450 nm: 0.331 mm, N = f/302. 550 nm: 0.366 mm, N = f/273.

f	d at 450 nm	N	d at 550 nm	N
25 mm	0.166 mm	f/151	0.183 mm	f/136
50 mm	0.234 mm	f/213	0.259 mm	f/193
100 mm	0.331 mm	f/302	0.366 mm	f/273

Read the table three ways. Down a column: quadrupling the focal distance only doubles the hole (√4 = 2) but doubles the f-number as well, so exposure rises fourfold — the f ∝ exposure law from the geometry lesson. Across a row: green light wants a 10 per cent bigger hole than blue, worth 0.29 stop. And against the range in the contested block: the whole 1.41-to-1.90 disagreement at 50 mm spans 0.21 to 0.29 mm, which is wider than the difference between blue and green — so if you are going to argue about a tenth of a millimetre, argue about the criterion first.

Now do it for your own camera. Take the focal distance you intend, use 450 nm if you will expose paper and 550 nm if you will expose panchromatic film, work the three constants, and write the range in your notebook. That range, not a single number, is what the pinhole lab is aiming at.

Recap

• Light spreads at a small aperture because the wavelets that would have cancelled the spreading have been blocked. For a circular hole the pattern is an Airy disc with faint rings, with the first dark ring at sin θ = 1.22 λ/d.
• On the film that disc has diameter 2.44 λ f/d = 2.44 λ N: it grows as the hole shrinks, where the geometric blur shrinks with it.
• Their sum is least at d = √(2.44 λ f) ≈ 1.56 √(λf), where the two blurs are equal and the total is twice the hole. The minimum is very flat.
• The Fresnel-zone picture says the same thing: the best hole admits about one half-period zone, and every published constant corresponds to between half a zone and one.
• Petzval (1.41), Rayleigh’s quarter-wave criterion (1.41) and Rayleigh’s own photographic result (1.90) are three different criteria, not three attempts at one number. The course uses the range and 1.56 as its working value.
• λ is a property of the material: 450 nm for blue-sensitive paper, 550 nm by convention for panchromatic film. Paper wants a hole 10 per cent smaller.
• A pinhole resolves one to two line pairs per millimetre, which is why its negatives are contact printed.
• Going over the optimum buys speed for a predictable, plain softness. Going under it costs speed and sharpness, and buys a diffraction glow. A zone plate blocks the cancelling zones, gaining speed and losing contrast.

Check your understanding

Question 1. Compute the optimum pinhole diameter for a focal distance of 60 mm at 450 nm and at 550 nm using the equal-blur constant 1.56, and give the resulting f-numbers.

• 0.256 mm (f/234) at 450 nm and 0.284 mm (f/211) at 550 nm
• 0.161 mm (f/373) at 450 nm and 0.178 mm (f/337) at 550 nm
• 0.312 mm (f/192) at 450 nm and 0.343 mm (f/175) at 550 nm
• 0.256 mm (f/234) at both, because the wavelength cancels

Show the answer and why

Answer: 0.256 mm (f/234) at 450 nm and 0.284 mm (f/211) at 550 nm

Work in millimetres: 450 nm is 4.5 × 10⁻⁴ mm. λf = 4.5 × 10⁻⁴ × 60 = 0.027, whose square root is 0.1643; times 1.56 gives 0.256 mm, and N = 60/0.256 = 234. At 550 nm, λf = 0.033, root 0.1817, times 1.56 gives 0.284 mm, and N = 60/0.284 = 211. Option 2 forgets the constant and uses √(λf) alone; option 3 uses 1.9 rather than 1.56. Option 4 is the common misreading: λ does not cancel, it sits under the square root, so a 22 per cent change in wavelength moves the hole by 10 per cent.

Question 2. Pinholes of 0.25 mm and 0.45 mm are used at a focal distance of 60 mm on blue-sensitive paper. Which gives the sharper negative, and roughly by how much?

• The 0.45 mm hole, because a bigger hole always gathers more light and therefore records more detail
• The 0.25 mm hole, by about 15 per cent in total blur: 0.513 mm against 0.596 mm
• The 0.25 mm hole, by a factor of nearly two
• They are identical, because both are within the historical range of the optimum

Show the answer and why

Answer: The 0.25 mm hole, by about 15 per cent in total blur: 0.513 mm against 0.596 mm

Compute both blurs for each. At d = 0.25 mm: geometric 0.250 mm, diffraction 2.44 × 4.5 × 10⁻⁴ × 60 / 0.25 = 0.0659/0.25 = 0.263 mm, total 0.513 mm. At d = 0.45 mm: geometric 0.450, diffraction 0.0659/0.45 = 0.146, total 0.596 mm. So the smaller hole wins by about 15 per cent — but notice how little that is for nearly a doubling of diameter, and notice that the 0.45 mm hole is 3.2 times the area and therefore about 1.7 stops faster. The optimum for 60 mm at 450 nm is 0.256 mm, so the 0.25 mm hole is essentially on it, which is why the margin is small: you are comparing the bottom of the curve with a point some way up its right-hand side.

Question 3. A website states: "the optimum pinhole for 50 mm is exactly 0.30 mm". Which qualifications does that sentence omit?

• Which criterion is being used — the quarter-wave, equal-blur and Rayleigh-measured constants give 0.21, 0.23 and 0.29 mm
• Which wavelength, and therefore which material: blue-sensitive paper and panchromatic film do not want the same hole
• That the minimum is flat, so "exactly" is meaningless — anything from 0.20 to 0.30 mm is within three per cent
• That the optimum depends on the subject distance

Show the answer and why

Answer: Which criterion is being used — the quarter-wave, equal-blur and Rayleigh-measured constants give 0.21, 0.23 and 0.29 mm, Which wavelength, and therefore which material: blue-sensitive paper and panchromatic film do not want the same hole, That the minimum is flat, so "exactly" is meaningless — anything from 0.20 to 0.30 mm is within three per cent

The first three are all real omissions, and each of them is larger than the precision the word "exactly" claims. The fourth is not: the optimum comes from setting d against 2.44λf/d, and neither term contains the subject distance. Subject distance does enter the geometric blur through the bracket (1 + f/u), so a very close subject blurs more, but it does not move the diameter at which the two effects balance. The best short answer to a sentence like that one is to ask which constant, at which wavelength, and to point out that the answer is a range.

Question 4. Why is a zone plate faster than a pinhole of the same focal distance, yet gives a lower-contrast image?

• It uses a thinner material, so it vignettes less
• It has far more open area, because it passes every zone that adds in phase instead of only the first; but it is a circular grating, so it also sends light into other orders and passes undiffracted light, all of which lands on the film as non-image light
• It focuses the light like a lens, and lenses always have lower contrast than pinholes
• It admits more zones, so the geometric blur is larger and the picture is both brighter and softer

Show the answer and why

Answer: It has far more open area, because it passes every zone that adds in phase instead of only the first; but it is a circular grating, so it also sends light into other orders and passes undiffracted light, all of which lands on the film as non-image light

The speed comes from area: a pinhole is one half-period zone, and a zone plate taken out to ten zones opens five of them, several times the area, worth roughly three stops on the geometry. The contrast loss comes from the same structure seen as a grating: a grating diffracts into several orders, only one of which forms the image at f, and light that passes without being diffracted forms a uniform veil. Non-image light on the film is flare, and flare lifts the shadows and compresses the scale. The wavelength dependence of the zone radii adds to it, since the plate is exactly in tune for one colour only.

Question 5. A student halves the diameter of a working pinhole "to make it sharper". What actually happens?

• The picture is sharper and two stops darker
• The picture is darker by two stops and, if the original hole was near the optimum, also softer, because the diffraction blur has doubled
• Nothing changes except the exposure, because diffraction affects only the corners
• The picture is sharper in the centre and softer at the edges

Show the answer and why

Answer: The picture is darker by two stops and, if the original hole was near the optimum, also softer, because the diffraction blur has doubled

Area falls by four, so it is two stops darker: that part of the intuition is right. But the diffraction blur is 2.44λf/d, which doubles when d halves, while the geometric blur only halves. Starting at the optimum, where the two are equal at value b, the new total is b/2 + 2b = 2.5b against 2b — a 25 per cent increase in blur for a two-stop loss of speed. This is the single most common pinhole mistake, and the two-blur plot is its refutation. Note also that the falloff at the edges is a separate matter governed by cos⁴ and the plate thickness, and it is not what diffraction does.

Question 6. Rayleigh is often credited with the optimum-pinhole constant 1.9. What is the correct account of where that number comes from?

• It is his theoretical criterion, derived from the quarter-wave tolerance
• It is Young's number, misattributed to Rayleigh
• It comes from his 1891 photographic experiments: the best of six zinc apertures gave (2r)²/f = 1.52 × 10⁻⁴ cm, and dividing by the 420 nm effective wavelength he back-calculated for his own plates yields 1.9. His theoretical quarter-wave criterion gives 1.41
• It is Petzval's constant, which Rayleigh quoted and was later credited with

Show the answer and why

Answer: It comes from his 1891 photographic experiments: the best of six zinc apertures gave (2r)²/f = 1.52 × 10⁻⁴ cm, and dividing by the 420 nm effective wavelength he back-calculated for his own plates yields 1.9. His theoretical quarter-wave criterion gives 1.41

Rayleigh gives two different numbers for two different questions, and conflating them is what causes the confusion. His phase criterion — open the hole until the extreme path error reaches a quarter of a wavelength, at which point a lens would not improve the image — gives 2ρ² = fλ, that is 1.41√(fλ), and is the same relation Petzval reached by minimising a summed blur. Then he pierced six apertures in sheet zinc, photographed a test object through each, and found the best gave (2r)²/f = 1.52 × 10⁻⁴ cm; his own back-calculation of the photographically effective wavelength gave 4.2 × 10⁻⁵ cm, and the ratio of those two is 3.6, whose square root is 1.90. So 1.9 is Rayleigh's, but it is a measurement on his plates at their effective wavelength, not a theory.

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