Circular Aperture Diffraction and the Airy Pattern

Send a plane wave through a round hole — the pupil of a telescope, the aperture stop of a camera, the back focal plane of a microscope objective — and no matter how perfect the optics, the far-field image of a point source is never a point. It is a small bright disk wrapped in a set of faint, rapidly fading rings: the Airy pattern. The bright core is the Airy disk, and its angular radius is set by a single ratio, $1.22\,\lambda/D$ , that is the most-cited number in all of optics — the line between "resolved" and "blurred" for every rotationally symmetric instrument ever built. The pattern is not an aberration or a flaw to be polished away; it is diffraction itself, the unavoidable price of collecting light through a finite opening. The far-field amplitude of any aperture is the Fourier transform of its shape, and the transform of a uniform disk is not the sine that governs a slit but a Bessel function, $2J_1(x)/x$ — the smooth "sombrero" whose square is the observed irradiance. Its first zero, at $x = 3.8317$ , sits a little farther out than the slit's first null, and dividing that root by $\pi$ produces the famous factor $1.22$ . Because the core holds the overwhelming majority of the light — about $83.8\%$ — a well-made lens delivers most of its energy into a tight spot, and the ring structure fades so fast (the first ring peaks at only $1.75\%$ of the centre) that the disk alone fixes the practical resolution. Every diffraction-limited telescope, every microscope point-spread function, every focused laser spot is an Airy pattern in disguise.

Key takeaways

The Fraunhofer diffraction pattern of a uniformly illuminated circular aperture is the Airy pattern: a bright central disk (the Airy disk) ringed by faint, fast-fading concentric rings.
The far-field amplitude is the Fourier transform of a uniform disk, which evaluates to $2J_1(x)/x$ with $x = (\pi D/\lambda)\sin\theta$ ; the irradiance is $I(\theta) = I_0[2J_1(x)/x]^2$ .
The first dark ring falls at the first zero of $J_1$ , $x_1 = 3.8317$ , giving the Airy-disk angular radius $\sin\theta_1 = 1.22\,\lambda/D$ with $1.22 = 3.8317/\pi$ .
The central disk holds $\approx 83.8\%$ of the transmitted energy; the first bright ring peaks at only $1.75\%$ of the centre and carries $\approx 7.2\%$ — the core dominates.
Key idea: the far-field of a circular pupil is the Fourier transform of a uniform disk, $2J_1(x)/x$ ; its square is the Airy pattern, whose first dark ring at $\sin\theta_1 = 1.22\,\lambda/D$ (with $1.22 = 3.8317/\pi$ , the first $J_1$ zero over $\pi$ ) sets the diffraction limit of every round optical system.

The core insight

The far-field amplitude of a circular aperture is the two-dimensional Fourier transform of a uniform disk, namely $2J_1(x)/x$ with $x = (\pi D/\lambda)\sin\theta$ . Its square is the Airy irradiance, a bright central disk surrounded by faint rings. The first dark ring, fixed by the first zero of the Bessel function $J_1$ at $x_1 = 3.8317$ , lands at $\sin\theta_1 = 1.22\,\lambda/D$ — the single number that sets the resolution of every circular telescope, microscope, and camera.

Where this lives today

The Airy pattern is the working point-spread function of essentially every imaging instrument. Space telescopes such as Hubble and JWST are built to reach their $1.22\,\lambda/D$ limit, and ground observatories deploy adaptive optics with deformable mirrors precisely to claw the star image back down from atmospheric blur to its Airy disk. In microscopy the same pattern, expressed through the numerical aperture, sets the classical resolution that super-resolution methods (STED, PALM/STORM) were invented to beat; the Airy disk is the barrier those Nobel-winning techniques circumvent rather than break. It rules the making of things as much as the seeing of them. Photolithography chases ever-shorter wavelengths — deep ultraviolet, then $13.5$ nm extreme ultraviolet — because the printable feature size scales with the Airy spot, and laser machining, optical tweezers, and confocal scanning all deliver their light into an Airy core whose radius is $1.22\,\lambda N$ . Even consumer cameras meet it daily: the "diffraction-limited aperture" printed in lens reviews is the f-number at which the growing Airy disk first exceeds the pixel pitch.

The physical picture

Treat the illuminated disk as a dense sheet of Huygens wavelets, all in phase across the aperture. Look off-axis at an angle $\theta$ and the wavelets arrive at the observation point with a linear ramp of phase across the pupil; the observed amplitude is their coherent sum, which is exactly the Fourier transform of the aperture evaluated at the spatial frequency $\sin\theta/\lambda$ . For a round pupil the sum has rotational symmetry, so it can depend only on the single radial variable $x = (\pi D/\lambda)\sin\theta$ . Integrating a linear phase around a circle produces a Bessel integral rather than a plain sine — that is the whole reason the answer is $J_1$ and not a sinc. The rings you see are the successive overshoots of $J_1$ as it crosses zero, unevenly spaced because the Bessel zeros are not harmonics of $\pi$ .

Figure

Top-down Airy pattern: a bright central disk ringed by faint rings, with dark rings (J₁ = 0) at radii 1.22, 2.23, 3.24 × λ/D

The rings are not merely faint; they collapse. The disk holds about $84\%$ of the light, the first ring only about $7\%$ , and each successive ring less still, because the ring peaks decay as $x^{-3}$ — steeper than the $x^{-2}$ envelope of a single slit's side lobes. The next figure is the radial cross-section that makes this concrete, with the diffraction-limit number read straight off the first zero.

:::svg[Radial Airy profile [2J₁(x)/x]² versus x: first zero x₁ = 3.8317 ⇒ sinθ₁ = 1.22 λ/D, central disk 83.8% of energy, first ring peak 1.75%]

The core equations

The pattern begins as the general Fraunhofer result — the far-field amplitude is the Fourier transform of the aperture — specialised to a hard circular pupil of diameter $D = 2a$ . Carrying out the transform of a uniform disk (Derivation A) collapses the two-dimensional integral onto a single Bessel function through the aperture's rotational symmetry. The natural coordinate is the dimensionless radial variable $x$ , so that one curve describes every wavelength and aperture at once.

I(\theta) = I_0\left[\frac{2J_1(x)}{x}\right]^2, \qquad x = \frac{\pi D}{\lambda}\sin\theta = ka\sin\theta \qquad (1)

where: $I(\theta)$ — irradiance at diffraction angle $\theta$ ; $I_0$ — on-axis peak irradiance; $J_1$ — first-order Bessel function of the first kind; $D = 2a$ — aperture diameter; $\lambda$ — wavelength; $k = 2\pi/\lambda$ — wavenumber; $x$ — reduced radial coordinate. On axis $x \to 0$ and $2J_1(x)/x \to 1$ , so $I(0) = I_0$ ; the function then falls to its first zero and oscillates. The dark rings sit exactly where $J_1(x) = 0$ . The first such root is not at $\pi$ (as it would be for a slit's sinc) but a little beyond, at $x_1 = 3.8317$ ; converting that root to an angle through $x = (\pi D/\lambda)\sin\theta$ is what produces the celebrated resolution factor. The constant $1.22$ is nothing more fundamental than the first Bessel zero divided by $\pi$ .

J_1(x_1) = 0,\quad x_1 = 3.8317 \;\;\Rightarrow\;\; \sin\theta_1 = \frac{x_1}{\pi}\frac{\lambda}{D} = 1.22\,\frac{\lambda}{D},\qquad \frac{3.8317}{\pi} = 1.2197 \qquad (2)

where: $x_1$ — first nonzero root of $J_1$ ; $\theta_1$ — angular radius of the Airy disk (first minimum), not its diameter. Because $\sin\theta_1 \propto \lambda/D$ , doubling the aperture halves the disk and shorter wavelengths tighten it — the reason telescopes are large and lithography reaches for the ultraviolet. In an imaging system the pattern forms in the focal plane, where the angular radius becomes a physical radius by multiplying through the focal length $f$ . Written with the f-number $N = f/D$ , the Airy-disk radius depends only on wavelength and the relative aperture — a compact rule photographers and lens designers use directly.

r_{\text{Airy}} = f\,\theta_1 = 1.22\,\frac{\lambda f}{D} = 1.22\,\lambda N,\qquad N \equiv \frac{f}{D} \qquad (3)

where: $r_{\text{Airy}}$ — Airy-disk radius in the focal plane; $f$ — focal length; $N = f/D$ — f-number (focal ratio). A slower lens (larger $N$ ) spreads the disk linearly, which is why stopping a camera down eventually softens fine detail. Finally, the fraction of the total transmitted power that lands inside a given ring follows from integrating the irradiance out to a chosen radius (Derivation B). The encircled-energy function has a remarkably clean closed form in terms of the two lowest Bessel functions, and evaluating it at the first dark ring gives the share of light captured by the Airy disk itself.

\eta(x) = 1 - J_0^2(x) - J_1^2(x),\qquad \eta(x_1) = 1 - J_0^2(3.8317) = 0.838 \qquad (4)

where: $\eta(x)$ — fraction of total power within reduced radius $x$ ; $J_0$ — zeroth-order Bessel function, with $J_0(3.8317) = -0.4028$ . Since $J_1(x_1) = 0$ , only the $J_0$ term survives at the first ring: the Airy disk holds $1 - (0.4028)^2 = 0.838$ , about $83.8\%$ of the energy, and the first bright ring adds a further $\approx 7.2\%$ .

Derivation AAiry amplitude from the Fraunhofer integral of a disk

Our aim is to show that the far-field amplitude of a uniformly illuminated circular hole reduces, with no approximation beyond the Fraunhofer condition, to the compact function $2J_1(x)/x$ , and that its first zero delivers the constant $1.22$ . The route is pure integration: start from the two-dimensional Fourier transform of a disk of radius $a$ , exploit the rotational symmetry to do the azimuthal integral in closed form, then do the radial integral with a standard Bessel identity. Nothing about the answer is guessed — the $1.22$ falls out as the first root of $J_1$ divided by $\pi$ .

Step 1. Write the Fraunhofer amplitude as the transform of the pupil, $U(\theta) \propto \iint_{\text{disk}} e^{-ik(x'\sin\theta\cos\varphi + y'\sin\theta\sin\varphi)}\,dx'\,dy'$ , and switch to polar aperture coordinates $x' = \rho\cos\phi$ , $y' = \rho\sin\phi$ . Because the pattern must be independent of the observation azimuth $\varphi$ , set $\varphi = 0$ . Why: a round aperture has no preferred direction, so the whole answer can depend only on the radial spatial frequency — fixing $\varphi$ costs nothing and simplifies the phase to $k\rho\sin\theta\cos\phi$ .

Step 2. Do the azimuthal integral using the integral representation of the Bessel function, $\int_0^{2\pi} e^{-iz\cos\phi}\,d\phi = 2\pi J_0(z)$ with $z = k\rho\sin\theta$ . The imaginary parts cancel by symmetry, leaving a real result: $U(\theta) \propto 2\pi\int_0^a J_0(k\rho\sin\theta)\,\rho\,d\rho$ . Why: integrating a pure phase ramp once around a circle is precisely the definition of $J_0$ — this is the step where the circular geometry, not the slit's geometry, imprints Bessel functions on the answer.

Step 3. Do the radial integral with the identity $\int_0^u t\,J_0(t)\,dt = u\,J_1(u)$ . Substituting $t = k\rho\sin\theta$ and $u = ka\sin\theta = x$ gives $U(\theta) \propto 2\pi\,(a/k\sin\theta)\,J_1(x)$ . Normalising by the on-axis value (proportional to the aperture area $\pi a^2$ ) yields the amplitude $E(\theta) = E_0\,[2J_1(x)/x]$ , whose square is equation (1). Why: one more integration of $J_0$ raises the order to $J_1$ ; the $2/x$ prefactor and the on-axis limit $2J_1(x)/x \to 1$ are exactly what normalisation to the aperture area demands.

Step 4. Locate the first dark ring: the smallest positive root of $J_1(x) = 0$ is $x_1 = 3.8317$ (found numerically, e.g. by bracketing $J_1$ between $3.0$ and $4.5$ ). Setting $x = (\pi D/\lambda)\sin\theta = x_1$ gives $\sin\theta_1 = (x_1/\pi)\lambda/D$ , and $x_1/\pi = 3.8317/3.14159 = 1.2197 \approx 1.22$ . Why: the entire diffraction limit is one Bessel root; the factor $1.22$ is not a physical constant but the geometric fingerprint of a circle, exceeding the slit's $1$ because a round rim redistributes light slightly outward. *The first $J_1$ zero is $3.831706$ , giving $x_1/\pi = 1.21967$ ; the two Bessel identities used above (azimuthal $\int_0^{2\pi}e^{-iz\cos\phi}d\phi = 2\pi J_0(z)$ and radial $\int_0^u t\,J_0(t)\,dt = u\,J_1(u)$ ) can be confirmed symbolically.*

Go deeper · Derivation B · 1 minEncircled energy and why the disk holds 83.8%

Derivation BEncircled energy and why the disk holds 83.8%

The reason a well-corrected lens is useful is that most of the diffracted light is concentrated in the central disk rather than smeared across the rings. Our aim is to prove that share is $83.8\%$ , using the closed-form encircled-energy function $\eta(x) = 1 - J_0^2(x) - J_1^2(x)$ and evaluating it at the first dark ring. This turns a statement about an infinite sum of rings into a single arithmetic step involving one Bessel value, $J_0(3.8317) = -0.4028$ .

Step 1. Integrate the irradiance over a disk of reduced radius $x$ . The cumulative-power integral of $[2J_1/x]^2$ has the exact result $\eta(x) = 1 - J_0^2(x) - J_1^2(x)$ , a standard consequence of the Bessel recurrence relations applied to $\int_0^x [J_1(t)]^2/t\,dt$ . Why: the recurrence relations let the running integral of $J_1^2/t$ telescope into just the two lowest Bessel functions evaluated at the edge — no infinite sum survives.

Step 2. Evaluate at the first dark ring, $x = x_1 = 3.8317$ . There $J_1(x_1) = 0$ by definition, so the $J_1^2$ term vanishes and only $J_0^2$ remains: $\eta(x_1) = 1 - J_0^2(3.8317)$ . Why: the Airy disk is bounded by a zero of $J_1$ , which is exactly what kills one of the two terms and leaves a one-line evaluation.

Step 3. Insert $J_0(3.8317) = -0.4028$ : $\eta(x_1) = 1 - (-0.4028)^2 = 1 - 0.1622 = 0.8378$ . Repeating at the second dark ring ( $x_2 = 7.0156$ , $J_0(x_2) = 0.300$ ) gives $\eta(x_2) = 0.910$ , so the first bright ring carries $0.910 - 0.838 = 0.072$ , about $7.2\%$ . Why: subtracting successive encircled fractions isolates each ring's share, confirming the fast $\propto x^{-3}$ fade — the disk dominates and the rings are a thin halo.

Step 4. Read the practical upshot: roughly six parts in seven of the collected light sit in a spot of angular radius $1.22\,\lambda/D$ , so the Airy disk, not the rings, sets both the delivered spot size and the Strehl-style measures of image quality. Why: because the core is so bright relative to the rings, resolution and focused-spot arguments can treat the disk alone and safely ignore the halo. *Numerically: $\eta(3.8317) = 1 - J_0^2 - J_1^2 = 0.83778$ , with $J_0(3.8317) = -0.40276$ , and the first-ring fraction $0.0722$ .*

Limiting cases

Case	Result	Meaning
on-axis, $x \to 0$	$2J_1(x)/x \to 1$ , $I(0) = I_0$	peak of the pattern; $J_1(x) \approx x/2$ recovers unity
first dark ring	$x_1 = 3.8317 \Rightarrow \sin\theta_1 = 1.22\,\lambda/D$	edge of the Airy disk; sets diffraction-limited resolution
first bright ring peak	$x = 5.1356 \Rightarrow I/I_0 = 0.0175$	first ring only $1.75\%$ of the centre — the halo is dim
large $x$ (ring envelope)	$I \sim (8/\pi)\,x^{-3}\cos^2(x - 3\pi/4)$	ring peaks fade as $x^{-3}$ , steeper than a slit's $x^{-2}$
slit comparison	$I_{\text{slit}} = I_0\,\mathrm{sinc}^2\beta$ , first null $\sin\theta = \lambda/b$	replacing the disk by a slit gives factor $1$ , not $1.22$ — circular symmetry is the difference

Common trap“The Airy radius 1.22 λ/D is the size of the whole bright disk.”

It is the radius to the first dark ring, not the diameter — the full bright disk spans $2.44\,\lambda/D$ , and forgetting the factor of two is one of the commonest resolution blunders. Two more traps live in the same formula. First, the $D$ is the aperture diameter, $D = 2a$ ; plugging in the radius $a$ doubles every answer. Second, the $1.22$ belongs to a circular aperture only: a single slit of width $b$ has its first null at $\sin\theta = \lambda/b$ — factor $1$ , no $1.22$ — because a straight edge gives a sinc, while a round rim gives a Bessel function whose zero sits farther out. Quote $1.22\,\lambda/D$ for round pupils, $\lambda/b$ for slits, and always say whether you mean the disk radius or its diameter.

Worked examples

1 · The diffraction limit of the Hubble Space Telescope (confidence builder)

The Hubble Space Telescope has a primary mirror of diameter $D = 2.4$ m. Working in visible light at $\lambda = 550$ nm, what is the angular radius of the Airy disk it forms for a point star — the finest detail it can in principle resolve?

Step 1. Apply equation (2) directly: $\theta_1 = 1.22\,\lambda/D = 1.22 \times (550\times10^{-9}\,\text{m})/(2.4\,\text{m})$ . Compute the numerator, $1.22 \times 550\times10^{-9} = 6.71\times10^{-7}$ m, then divide by $2.4$ m to get $\theta_1 = \mathbf{2.80\times10^{-7}}$ rad.

Step 2. Convert to astronomer's units. One radian is $206{,}265$ arcseconds, so $\theta_1 = 2.80\times10^{-7} \times 206{,}265 = \mathbf{0.0577''}$ — about six hundredths of an arcsecond.

Step 3. Put it in context: from orbit Hubble actually approaches this limit, because there is no atmosphere to blur the image. A ground telescope of the same size is instead capped by atmospheric "seeing" near $1''$ — more than fifteen times coarser — unless adaptive optics restores the diffraction limit in real time.

Step 4. Note the scaling lever. To halve $\theta_1$ you must either double the mirror or halve the wavelength; this is precisely why the James Webb Space Telescope's $6.5$ m aperture resolves finer than Hubble at the same wavelength, and why X-ray and electron instruments, with their tiny wavelengths, reach far beyond any visible-light limit. θ₁ = 2.80 × 10⁻⁷ rad = 0.0577″.

2 · Airy-disk size in a camera and diffraction softening (exam level)

A camera lens is set to an aperture of $f/2.8$ and used with green light, $\lambda = 550$ nm. Find the radius of the Airy disk in the focal plane. Then find it at $f/16$ , and say what that means for sharpness on a sensor with $\sim 2$ μm pixels.

Step 1. Use the focal-plane form, equation (3), which needs only the wavelength and the f-number: $r_{\text{Airy}} = 1.22\,\lambda N = 1.22 \times (550\times10^{-9}\,\text{m}) \times 2.8$ . Evaluate: $1.22 \times 550\times10^{-9} = 6.71\times10^{-7}$ m, times $2.8$ gives $r_{\text{Airy}} = \mathbf{1.88\ \mu m}$ .

Step 2. At $f/2.8$ the disk (radius $1.88$ μm) is comparable to a $2$ μm pixel, so diffraction is not yet the limiting factor — the lens's aberrations usually dominate wide open. Most of the light, $\mathbf{83.8\%}$ , lands inside this disk by Derivation B.

Step 3. Stop down to $f/16$ : $r_{\text{Airy}} = 1.22 \times 550\times10^{-9} \times 16 = \mathbf{10.7\ \mu m}$ , now five pixels wide. Fine detail smaller than the disk simply cannot be recorded, no matter how many megapixels the sensor has — the image goes soft. This is "diffraction softening", and it is why lenses are rarely sharpest at their smallest aperture.

Step 4. The trade-off is unavoidable: closing the aperture deepens depth of field but enlarges the Airy disk in exact proportion to $N$ . The sweet spot is the aperture where the growing Airy disk just reaches the pixel pitch — beyond it, resolution is being spent on depth of field. r = 1.88 μm at f/2.8; 10.7 μm at f/16; central disk 83.8% of the light.

Physics of the portrait

The bright core has carried one name since the mid-nineteenth century — the Airy disk — after the astronomer who first did the integral and found the rings.

Origins · 1 minairy disk and how this was discovered

airy disk

the diffraction-limited core of a circular pupil · Fourier optics & imaging

The Airy disk is the central bright spot of the circular-aperture diffraction pattern, bounded by the first dark ring at angular radius $1.22\,\lambda/D$ and holding about $83.8\%$ of the light. It is the incoherent point-spread function of every rotationally symmetric optical system, so its size is literally the resolution of the instrument. The name honours George Biddell Airy (1801–1892), who in 1835, while Plumian Professor at Cambridge and shortly before becoming Astronomer Royal, published "On the Diffraction of an Object-Glass with Circular Aperture" and carried the Fraunhofer integral of a uniform disk through to the Bessel-function result — the first correct calculation of a telescope's diffracted image. Airy ran the Royal Observatory at Greenwich for forty-six years, standardised the Greenwich meridian that later fixed the world's prime meridian, and worked on the astigmatism of his own eye (designing the first cylindrical spectacle lens). His disk is the reason "diffraction-limited" is the highest compliment an optical system can earn.

History

1818

— Fresnel's prize memoir establishes the wave theory of diffraction

1824

— Bessel systematises the functions later used to describe the circular-aperture pattern

1835

— George Biddell Airy computes the diffraction pattern of a circular aperture; its bright core becomes the "Airy disk"

1873

— Abbe formulates the diffraction theory of image formation and microscope resolution

1879

— Rayleigh adopts the Airy first dark ring as the resolution criterion

Going further

Texts. Hecht, Optics, 5th ed., §10.2.5 (Circular Apertures) · Born & Wolf, Principles of Optics, 7th ed., §8.5.2 (The Airy pattern) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., Ch. 4 (Fourier optics) · Foot, Atomic Physics, §A.3 (diffraction and resolution) · Abramowitz & Stegun, Handbook of Mathematical Functions, §9 (Bessel functions and their zeros). The Airy amplitude and intensity follow from the Fraunhofer integral of a uniform disk via $\int_0^{2\pi} e^{iz\cos\phi}\,d\phi = 2\pi J_0(z)$ and $\int_0^u t\,J_0(t)\,dt = u\,J_1(u)$ . The first $J_1$ zero $x_1 = 3.8317$ gives the constant $3.8317/\pi = 1.2197$ . The encircled energy $1 - J_0^2 - J_1^2$ with $J_0(3.8317) = -0.4028$ yields $0.838$ . The focal-plane radius is $1.22\,\lambda f/D$ , and the Hubble example $1.22 \times 550\,\text{nm}/2.4\,\text{m}$ is directly recomputable. Note on the ring envelope: it decays as $(8/\pi)\,x^{-3}\cos^2(x-3\pi/4)$ , not $(4/\pi)\,x^{-3}$ — the correct coefficient follows from the large- $x$ asymptotic form $J_1(x) \sim \sqrt{2/\pi x}\,\cos(x-3\pi/4)$ .

Builds on. Fraunhofer Diffraction · Fresnel–Kirchhoff Diffraction Integral · Single-Slit Diffraction · Fresnel Number. Leads to. Rayleigh Resolution Criterion · Alternative Resolution Criteria (Sparrow, Abbe) · Rectangular Aperture Diffraction.

Why is the pattern made of Bessel functions instead of the sines that govern a slit?

Because the shape of the aperture is different, and the far field is its Fourier transform. A slit is a one-dimensional rectangle, whose transform is a sinc (a ratio of sines). A round pupil forces you to integrate a linear phase ramp around a circle, and that azimuthal integral is, by definition, a Bessel function $J_0$ ; one more radial integration raises it to $J_1$ . The circular symmetry is the whole story — it is what replaces $\sin\beta/\beta$ with $2J_1(x)/x$ and pushes the first zero from $\pi$ out to $3.8317$ , giving the $1.22$ .

Is the Airy pattern the same thing as the point-spread function?

For an ideal, aberration-free circular system, yes: the incoherent point-spread function — the image of a single point source — is the Airy pattern. Real optics add aberrations that redistribute light from the core into the rings, lowering the Strehl ratio (the fraction of ideal peak intensity retained). So the Airy pattern is the best case; any actual PSF is a smeared version of it. When people say a lens is "diffraction-limited", they mean its aberrations are small enough that the PSF is essentially the Airy pattern rather than something broader.

Does the Airy formula apply right up against the aperture, or only far away?

Only in the far field (or in the focal plane of a lens, which optically brings the far field to a finite distance). The Airy result is the Fraunhofer limit, valid when the Fresnel number $N_F = a^2/\lambda z \ll 1$ . Close behind the aperture, where $N_F \gtrsim 1$ , the wavefront's curvature across the pupil cannot be dropped and the pattern is a Fresnel diffraction pattern — structured, oscillating on-axis, and needing the full Fresnel integral. A lens is the standard trick for observing the clean Airy pattern in a compact setup: its focal plane is the far field.

Why does stopping a camera lens down make images softer if it also increases depth of field?

Two effects fight as you close the aperture. Geometrically, a smaller opening deepens the depth of field — more of the scene is acceptably focused. But diffraction enlarges the Airy disk in exact proportion to the f-number, $r_{\text{Airy}} = 1.22\,\lambda N$ , so once that disk grows past the pixel pitch (or the film grain), fine detail blurs regardless of focus. The result is a sweet spot: sharpness improves as you stop down from wide open (aberrations shrink), then declines once diffraction takes over. On most cameras that turnover is around $f/8$ to $f/11$ .

How can super-resolution microscopy beat the Airy limit if it is set by diffraction?

It never violates diffraction; it sidesteps the criterion. The Airy disk still forms, but techniques like STED shrink the effective emitting region with a second, doughnut-shaped beam, while PALM/STORM switch individual molecules on one at a time and fit each isolated Airy spot to find its centre to a few nanometres — far finer than the disk's width. Both exploit extra information (nonlinearity, or sparse single-emitter blinking) that the plain resolution criterion assumes you do not have. Diffraction sets the size of the blob; these methods locate what is inside it.

Why is the first ring so much fainter than the central disk?

Because the encircled energy piles up almost entirely inside the first dark ring. The disk holds about $83.8\%$ of the light and peaks at $I_0$ ; the first bright ring peaks at only $1.75\%$ of that, and even integrated over its full annulus carries just $\approx 7.2\%$ . The ring peaks then fall as $x^{-3}$ , so each successive ring is dramatically dimmer than the last. Physically, the aperture edge — which is what produces the rings — is a small perturbation on the bright, in-phase core, so the halo it creates is weak. This is exactly why the Airy disk alone sets the practical spot size.

Does the 1.22 change if the aperture is partly blocked, like a telescope's secondary mirror?

Yes. The clean $1.22$ assumes a uniformly illuminated, unobstructed circular pupil. A central obstruction (a reflecting telescope's secondary mirror and its vanes) changes the pupil function, slightly narrowing the central disk but pumping more energy into the rings. Apodisation — deliberately tapering the pupil transmission — does the opposite, suppressing the rings at the cost of a wider core. The Airy pattern is the reference case; real pupils are engineered departures from it.