
Physics III course sequence · Interactive lesson
Circular apertures & the diffraction limit
Every lens, pupil and telescope is round, so this is the lesson where a Bessel zero becomes a design number. We integrate over the disc to get 2J₁(x)/x, turn its first zero into 1.22 λ/D and the Rayleigh criterion, and then ask what really limits an eye, a camera, a telescope and a microscope, which is rarely diffraction.
UPIII-05.08 · Lesson 8 of 9 in this unit. Position is not a completion score.Build the model
Every round lens draws a point as an Airy disc; the 1.22 is a Bessel zero, not a fudge.
The key idea
The diffraction limit of a circular aperture is the angular radius of the Airy disc it turns every point source into, θ₁ = 1.22 λ/D, and the Rayleigh criterion adopts that angle as the smallest separation at which two equal point sources are conventionally counted as resolved.
Example. A 20 cm telescope at 550 nm has θ₁ = 1.22 × 550 × 10⁻⁹ m ÷ 0.20 m = 3.4 × 10⁻⁶ rad = 0.69 arcsec: a double star 0.69″ apart shows two peaks with a 27% dip between them, while at 0.50″ the dip is gone and the pair looks like one elongated star.
The complete physical model and assumptions
A circular aperture is the shape every lens, pupil and telescope actually has, so its far-field pattern is the one that sets real instruments' limits. Integrating the Fraunhofer phasor sum over a disc instead of a slit replaces sin x/x by 2J₁(x)/x, with x = π D sin θ/λ: the same physics, a different geometry, and a Bessel function instead of a sine because a disc has less area near its edges than a slit does. The intensity, the Airy pattern, has its first zero where J₁ does, at x = 3.8317, which is the whole origin of the number 1.22 in θ₁ = 1.22 λ/D. Rayleigh's criterion then declares two equal incoherent sources resolved when one peak sits on the other's first dark ring; the summed profile dips to 73.5% of its peaks there, a visible but arbitrary threshold. Everything after that is trade-off: shorter λ or larger D shrinks the disc, a detector must sample it, a microscope is limited by the cone it collects (NA = n sin α) rather than by its diameter, and in practice atmospheric seeing, aberration or photon noise blurs the image before diffraction gets the chance. The model stops at the assumptions it was built on: scalar, monochromatic, far-field and incoherent. Coherent sources add in amplitude, not intensity, and a single isolated emitter can be located far more finely than θR, because Rayleigh is a statement about separating two overlapping sources, not about locating one.
1Integrate over a disc and a Bessel function appears
In the Fraunhofer limit the far-field amplitude is the phasor sum over the aperture, E(θ) ∝ ∫∫ exp(i k ξ sin θ) dA, with k = 2π/λ and ξ the coordinate across the aperture in the plane of θ. For a disc of radius a use polar coordinates, ξ = ρ cos φ. The angular integral ∫₀(2π) exp(i k ρ sin θ cos φ) dφ equals 2π J₀(k ρ sin θ), a standard result, and the radial integral ∫₀a J₀(k ρ sin θ) ρ dρ equals a J₁(k a sin θ)/(k sin θ) by the identity d[u J₁(u)]/du = u J₀(u). Dividing by the on-axis value π a² gives E(θ)/E(0) = 2 J₁(x)/x with x = k a sin θ = π D sin θ/λ, and the intensity is its square. Two checks: x is dimensionless, a length times a wavenumber, and since J₁(x) ≈ x/2 for small x, 2J₁(x)/x → 1 on axis as it must. Run the same procedure on a slit and sin(x)/x comes out instead. The slit's pattern is a sine because every strip of the slit has the same length, whereas a disc's chords shrink towards its rim, weighting the edges less and pushing the zeros outward into a Bessel function.
2Read the numbers off the Airy pattern
J₁ vanishes at x = 3.8317, 7.0156 and 10.1735, so the dark rings sit at sin θ = 1.220, 2.233 and 3.238 λ/D; the first is the famous 1.22. Between them the secondary maxima at x = 5.136 and 8.417 carry only 1.75% and 0.42% of the central intensity, and integrating the pattern puts 83.8% of the power inside the first dark ring and 91.0% inside the second, which is why the central disc is what an eye or detector registers. Put in the Hubble Space Telescope, D = 2.4 m at 550 nm: θ₁ = 1.22 × 5.50 × 10⁻⁷ ÷ 2.4 = 2.8 × 10⁻⁷ rad, and multiplying by 206 265 arcsec rad⁻¹ gives 0.058″. Halve λ or double D and the whole pattern shrinks by two; its shape never changes. That fixed shape is the point: a lens of any quality draws a point as an Airy disc of this size, and no amount of polishing removes it.
3Rayleigh's criterion is a convention with a number attached
Two incoherent sources, different stars or different fluorescent molecules, have no fixed phase relation, so their intensities add: I(θ) = A(θ − Δθ/2) + A(θ + Δθ/2), with A the Airy profile. At Rayleigh's separation Δθ = θ₁ the midpoint lies x = 3.8317/2 = 1.916 from each peak, where 2J₁(x)/x = 0.6062, so the midpoint intensity is 2 × 0.6062² = 0.735 of a single peak: a 26.5% dip. Move the sources closer and the dip shallows; at Δθ = 0.947 λ/D the second derivative at the midpoint vanishes and the two humps merge into one flat-topped blob, the Sparrow limit. Dawes's empirical visual limit for double stars, 116″/D(mm), works out to 1.00 λ/D at 562 nm, in between. None of these is a wall. With enough photons a two-source model fitted to the profile separates a pair well below θ₁, and a single isolated emitter can be located to a small fraction of its disc, which is what localisation microscopy does. The criterion is also only for incoherent sources: two coherent in-phase sources add in amplitude and show no dip at all at θ₁, while an anti-phase pair shows a true zero between them.
4Trade wavelength against aperture, then check the detector
For an eye with a 3 mm daylight pupil at 550 nm, θ₁ = 2.2 × 10⁻⁴ rad = 0.77 arcmin; measured acuity is about 1 arcmin, so the retina's cone spacing and the lens's aberrations, not diffraction, set the last factor. In a camera the angular limit becomes a length on the sensor, r₁ = f θ₁ = 1.22 λ N: at f/8 and 550 nm the Airy disc is 2 × 5.37 = 10.7 μm across, at f/5.6 it is 7.5 μm. The finest period the lens passes is λN, so a sensor with 3.8 μm pixels is pixel-limited at f/5.6 and diffraction-limited beyond about f/14, where λN/2 reaches the pixel pitch; that is why stopping down for depth of field softens fine detail. At radio wavelengths the same formula is punishing: a 100 m dish at λ = 21 cm has θ₁ = 2.6 mrad = 8.8 arcmin, worse than the unaided eye, so radio astronomy replaces D by the baseline between dishes and synthesises the aperture by interferometry. In every case check the limit: θ₁ → 0 as λ → 0 recovers geometrical optics, where a point images to a point.
5A microscope is limited by its cone, not its diameter
Apply r₁ = 1.22 λ L/D in the image plane of an objective, L being the distance from the lens to that image, and divide by the magnification M to find the equivalent distance in the specimen. Abbe's sine condition, n sin α = M n′ sin α′, relates the object-side half-angle α to the small image-side angle α′ ≈ D/(2L), so 1.22 λ L/(D M) becomes dₘᵢₙ = 0.61 λ/(n sin α) = 0.61 λ/NA. The numerical aperture is a cone, not a diameter: a wide barrel with a long working distance collects a narrow cone and resolves poorly. Since sin α ≤ 1, a dry objective has NA < 1 (0.95 in practice, giving dₘᵢₙ = 0.61 × 550/0.95 = 353 nm), and only an immersion liquid matching the coverslip lets the steep rays through: oil at n = 1.515 allows NA = 1.40 and dₘᵢₙ = 240 nm, and violet 405 nm light in the same objective reaches 176 nm. Image brightness scales as NA², but that is a different statement from resolution, and a higher-power objective with the same NA shows nothing new.
6Something else usually gets there first
Real images are blurred by several effects at once, and for roughly Gaussian blurs the widths add in quadrature. Through the atmosphere a wavefront stays flat only over a patch of size r₀ ≈ 10 cm at 500 nm, so a long-exposure star image has a seeing width near 0.98 λ/r₀ ≈ 1″ whatever the aperture; an 8 m telescope without adaptive optics resolves no better than a 12 cm one, it only collects some 4000 times more light. Aberration is bounded by the Maréchal criterion: an RMS wavefront error of λ/14 leaves the Strehl ratio, the peak intensity relative to the perfect Airy peak, at exp(−(2π/14)²) = 0.82, the usual threshold for calling a system diffraction limited. Noise is the last gate. With N photons in each peak the shot noise is √N, so the 26.5% Rayleigh dip is a 3σ detection only when 0.265 N ≥ 3√N, that is N ≥ 130 photons per source; below that the two sources are unresolved by any criterion. Diffraction is the floor beneath all three, but it is the floor you reach last.
Key equations
Read the units, assumptions and limits before substituting.
Airy pattern of a circular aperture
I(θ) = I₀ [2 J₁(x)/x]², x = (π D/λ) sin θ
The disc's counterpart of the slit's sinc²: integrating phasors over a circle turns the sine into a Bessel function, and 2J₁(x)/x → 1 as x → 0 fixes the normalisation.
I₀ on-axis intensity (W m⁻²); θ angle from the axis (rad); D aperture diameter and λ wavelength in the same unit (m); J₁ the first-order Bessel functionFirst dark ring
sin θ₁ = 3.8317 λ/(π D) = 1.220 λ/D
The 1.22 is a Bessel zero, not a fudge: a disc spreads light wider than a slit of width D (first zero at λ/D) because less of its area lies near the edges.
3.8317 is the first zero of J₁; θ₁ in rad; 83.8% of the transmitted power falls inside this ringRayleigh criterion
θR = 1.22 λ/D
The peak of one source sits on the first dark ring of the other, and the summed intensity dips to 73.5% of its peaks. A convention, not a law.
θR smallest resolvable angular separation (rad) of two equal, incoherent point sources; the small-angle form of sin θ₁Airy radius on a detector
r₁ = 1.22 λ f/D = 1.22 λ N
Sets the useful pixel: at f/8 and 550 nm the disc is 10.7 μm across and the lens passes no period finer than λN = 4.4 μm, so pixels below λN/2 = 2.2 μm buy nothing.
f focal length (m); N = f/D the f-number (dimensionless); r₁ comes out in the units of λMicroscope resolution
dₘᵢₙ = 0.61 λ/NA, NA = n sin α
0.61 = 1.22/2 with D/(2f) → sin α: the collecting cone sets the limit, not the lens diameter, and NA > 1 needs an immersion liquid.
dₘᵢₙ smallest resolvable separation in the specimen (m); n index of the immersion medium; α half-angle of the accepted coneSparrow limit
θS = 0.947 λ/D
Between θS and θR a shallow dip exists; below θS the pair is one blob. Fitting a two-source model to a high-signal profile can still separate them.
the separation at which the central dip just vanishes, ∂²I/∂θ² = 0 at the midpoint; same units as θRInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start near Δθ/θR = 1: the sum dips to 0.735 I₀ between two peaks. Slide the separation down to the Sparrow readout, about 0.78 θR, and the dip flattens out, then the pair melts into one blob twice as bright. Hold the separation and double D or halve λ: the arrow shrinks and the peaks split again.
Live results
- RAYLEIGH θR = 1.22 λ/D
- 0.69 arcsec
- SEPARATION Δθ / θR
- 1.01
- SPARROW θS = 0.947 λ/D
- 0.54 arcsec
- AIRY RADIUS AT f = 1.0 m
- 3.35 μm
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Aperture diameter D: 20 cm; Wavelength λ: 550 nm; Angular separation Δθ: 0.70 arcsec
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
First-year students treat magnification as resolution: if the eyepiece shows two stars as one blob, use a stronger eyepiece. Magnification scales the Airy discs and their separation by the same factor, so the ratio Δθ/θ₁ that decides whether a dip exists never changes; past a point the image is only a bigger, dimmer blob, which astronomers call empty magnification. The useful ceiling is set by the eye: it resolves about 1 arcmin = 2.9 × 10⁻⁴ rad, and a telescope's θ₁ = 1.22 × 550 × 10⁻⁹/D reaches that angle at a magnification of roughly 0.43 per millimetre of aperture, so a 100 mm telescope has shown everything it can at 43× and gains only viewing comfort up to about 200×. The same confusion appears in microscopy as the belief that a 100× objective beats a 60× one; only its NA can.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A phone camera has focal length 6.9 mm, a fixed f/1.8 aperture and a sensor with 1.0 μm pixels. For 550 nm light, find the aperture diameter, the angular resolution, the radius of the Airy disc on the sensor, and the smallest separation it can resolve on a wall 50 m away.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- D = f/N = 6.9 mm ÷ 1.8 = 3.83 mm.
- θ₁ = 1.22 λ/D = 1.22 × 550 × 10⁻⁹ m ÷ 3.83 × 10⁻³ m = 1.75 × 10⁻⁴ rad, about 36 arcsec. Dimensions: a length over a length, so the radian is honest.
- On the sensor r₁ = f θ₁ = 6.9 × 10⁻³ m × 1.75 × 10⁻⁴ = 1.21 × 10⁻⁶ m = 1.21 μm, the same as 1.22 λ N = 1.22 × 0.55 μm × 1.8. The disc is 2.4 μm across, about 2.4 pixels; the finest period the lens passes is λN = 0.99 μm, but 1.0 μm pixels record no period finer than 2 × 1.0 = 2.0 μm, so they still undersample the lens by a factor of 2; only pixels down to λN/2 = 0.5 μm capture everything it passes, and anything smaller would add nothing.
- At L = 50 m the resolvable separation is Δx = L θ₁ = 50 m × 1.75 × 10⁻⁴ = 8.8 × 10⁻³ m.
Answer. D = 3.8 mm; θ₁ = 1.75 × 10⁻⁴ rad; r₁ = 1.21 μm (a 2.4 μm disc on 1.0 μm pixels); 8.8 mm at 50 m.
MediumExample 2
A 0.35 m amateur telescope observes at 550 nm on a night when the seeing blurs stars to a full width of 1.5″. Find its Rayleigh limit in arcseconds, the aperture that would just match the seeing, the combined blur width (add the two in quadrature), and whether a binary star 0.60″ apart can be split in a long exposure.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- θR = 1.22 λ/D = 1.22 × 550 × 10⁻⁹ ÷ 0.35 = 1.92 × 10⁻⁶ rad; × 206 265 arcsec rad⁻¹ = 0.395″.
- Seeing-matched aperture: set 1.22 λ/D = 1.5″ = 7.27 × 10⁻⁶ rad, so D = 6.71 × 10⁻⁷ ÷ 7.27 × 10⁻⁶ = 0.092 m. Every centimetre beyond 9 cm buys light (the 0.35 m collects (0.35/0.092)² = 14 times more) but no long-exposure resolution.
- Combined width ≈ √(0.395² + 1.5²) = √(0.156 + 2.25) = 1.55″; diffraction adds only 3% to the blur.
- The 0.60″ pair needs D ≥ 1.22 × 550 × 10⁻⁹ ÷ (0.60 × 4.85 × 10⁻⁶) = 0.23 m by Rayleigh, which the 0.35 m satisfies, but 0.60″ sits far inside a 1.55″ blur: unresolved in a long exposure. Only exposures short enough to freeze the seeing, about 10 ms, or adaptive optics, recover the 0.4″ the aperture can deliver.
Answer. θR = 0.40″; seeing-matched D ≈ 9 cm; combined blur 1.55″; the 0.60″ binary passes Rayleigh's test yet is unresolved in a long exposure.
HardExample 3
Two fluorescent beads 250 nm apart emit at 520 nm and are imaged by an oil-immersion objective (n = 1.515, NA = 1.40). Find dₘᵢₙ, the half-angle of the collected cone, and the depth of the dip between the two images, taking J₁(2.11) = 0.567 and J₁(4.23) = −0.149. Then decide whether a dry NA 0.95 objective would show any dip at all.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- dₘᵢₙ = 0.61 λ/NA = 0.61 × 520 nm ÷ 1.40 = 227 nm, so 250 nm is 1.10 dₘᵢₙ: resolved by Rayleigh, barely. The cone: sin α = NA/n = 1.40/1.515 = 0.924, so α = 67.5°.
- In the specimen plane the Airy argument is x = 2π NA r/λ (its first zero, x = 3.8317, gives back r = 0.61 λ/NA). The bead separation is s = 2π × 1.40 × 250 ÷ 520 = 4.23, so the midpoint lies x = 2.11 from each bead.
- A(2.11) = [2 × 0.567 ÷ 2.11]² = 0.537² = 0.288, so the midpoint intensity is 2 × 0.288 = 0.576 I₀. A(4.23) = [2 × 0.149 ÷ 4.23]² = 0.0705² = 0.0050, so each peak is 1.005 I₀.
- Dip ratio = 0.576 ÷ 1.005 = 0.57: the intensity between the beads falls to 57% of the peaks, deeper than Rayleigh's 73.5% because 1.10 dₘᵢₙ lies beyond the criterion.
- Dry objective: dₘᵢₙ = 0.61 × 520 ÷ 0.95 = 334 nm > 250 nm, and s = 2π × 0.95 × 250 ÷ 520 = 2.87, below the Sparrow value 2.98. There is no dip at all: the pair is one elongated blob, 1.17 I₀ bright at its centre.
Answer. dₘᵢₙ = 227 nm, α = 67.5°, and the dip falls to 57% of the peaks with the oil objective; the dry objective (dₘᵢₙ = 334 nm, s = 2.87 < 2.98) shows no dip.
Quick check
Select an answer and explain it before checking.
Review & continue
The diffraction limit of a circular aperture is the angular radius of the Airy disc it turns every point source into, θ₁ = 1.22 λ/D, and the Rayleigh criterion adopts that angle as the smallest separation at which two equal point sources are conventionally counted as resolved.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics III completion tasks.