Rayleigh Resolution Criterion

No matter how flawless the glass, a finite aperture cannot image a point as a point — diffraction smears it into an Airy disk. Point two stars, two spectral lines, or two fluorescent molecules close enough together and their diffraction blobs begin to overlap, until at some separation they merge into a single indistinguishable smudge. The question "how close is too close?" has no sharp answer in nature, so in the 1870s Lord Rayleigh proposed a clean, reproducible convention: two equally bright, mutually incoherent point sources are just resolved when the central peak of one Airy pattern falls exactly on the first dark ring of the other. That rule fixes the finest detail an instrument can deliver. Because the Airy disk has angular radius $1.22\,\lambda/D$ , the Rayleigh limit is simply $\theta_R = 1.22\,\lambda/D$ — the same Bessel constant, $1.22 = 3.8317/\pi$ , that sets the disk's size now sets the resolving power of every round telescope, microscope, camera, and eye. Bigger apertures and shorter wavelengths sharpen the disk and split finer detail, which is why observatory mirrors keep growing and electron microscopes reach for tiny de Broglie wavelengths. The convention is honest about being a convention. At exactly the Rayleigh separation two incoherent sources leave a shallow central valley — the summed intensity dips to about $73.5\%$ of the peaks, a $26.5\%$ dip a careful eye or a low-noise detector can register. Push closer and the dip fills in; the Sparrow criterion marks where it vanishes entirely. Coherent light or unequal brightness shifts the threshold, and with enough photons and a known point-spread function, modern super-resolution slides past it. The Rayleigh number is not a wall — it is a well-chosen, universally quoted milestone on the road to the diffraction limit.

Key takeaways

The Rayleigh criterion is a convention: two incoherent, equally bright point sources are "just resolved" when the peak of one Airy pattern sits on the first dark ring of the other.
For a circular aperture of diameter $D$ this gives the minimum resolvable angular separation $\theta_R = 1.22\,\lambda/D$ , where $1.22 = 3.8317/\pi$ is the first zero of $J_1$ over $\pi$ .
In a focal plane the linear separation is $r_R = 1.22\,\lambda f/D = 1.22\,\lambda N$ ; for a microscope it becomes $d = 0.61\,\lambda/\mathrm{NA}$ (since $0.61 = 1.22/2$ ).
At the Rayleigh separation the composite intensity of two incoherent sources dips to $0.735$ of the peaks — a $26.5\%$ central valley — which is what makes the pair look "just" separated.
The criterion assumes incoherent, equal-brightness sources; coherent illumination, unequal brightness, and super-resolution all move the effective limit.
Key idea: a finite aperture diffracts each point into an Airy disk of angular radius $1.22\,\lambda/D$ ; two incoherent sources are "just resolved" at that separation ( $\theta_R = 1.22\,\lambda/D$ ), where the summed intensity of the pair shows a diagnostic $26.5\%$ central dip.

The core insight

A finite circular aperture diffracts each point source into an Airy disk of angular radius $1.22\,\lambda/D$ . Rayleigh's convention calls two incoherent, equal-brightness sources "just resolved" when their separation equals that radius, giving $\theta_R = 1.22\,\lambda/D$ . At that spacing the two Airy patterns add to a composite with a shallow central dip — down to $73.5\%$ of the peaks, a $26.5\%$ valley — which is the signature of a barely resolved pair. The same physics fixes the resolving power of telescopes, microscopes ( $0.61\,\lambda/\mathrm{NA}$ ), and the eye.

Where this lives today

The Rayleigh criterion is still the first number quoted for any new imaging instrument. Radio and optical interferometers — from the Very Large Array to the Event Horizon Telescope — beat a single dish by synthesising an effective aperture equal to the baseline between antennas, so their $1.22\,\lambda/D$ becomes $1.22\,\lambda/B$ with $B$ a continental or planetary distance; that is how the Event Horizon Telescope resolved a black-hole shadow at tens of microarcseconds. Adaptive optics on ground telescopes exists precisely to restore the Rayleigh limit that atmospheric turbulence would otherwise spoil. In the life sciences the microscope form $0.61\,\lambda/\mathrm{NA}$ set the $\sim 200$ nm wall that super-resolution methods were built to circumvent, earning the 2014 Nobel Prize in Chemistry. Spectroscopy uses the same idea in a different guise: a diffraction grating's resolving power $R = \lambda/\Delta\lambda = mN$ (order $m$ , $N$ illuminated lines) is the Rayleigh criterion applied to two wavelengths whose diffraction maxima are separated by one dark fringe. Wherever a wave passes through a bounded aperture, $1.22\,\lambda/D$ remains the order-of-magnitude answer for the finest detail recoverable.

The physical picture

Picture two distant flashlights, each imaged as a bright Airy disk ringed by faint fringes. Bring them closer in angle and the disks slide into one another. Rayleigh's rule sets the threshold at a specific, repeatable geometry: the bright centre of one disk sits on the first dark ring of the other. Slightly farther apart and a clear gap separates the peaks; slightly closer and the gap fills; at exactly this spacing a shallow but real valley still divides them. The value $1.22$ is not fundamental — it is the geometry of a round pupil, whose first dark ring lies a little farther out than a slit's first null. Change the aperture shape and the number changes.

Figure

Two point sources at three separations: well resolved (clear gap), at the Rayleigh limit (peak on first ring), and unresolved (merged blob)

Think of resolution as a contest between diffraction, which smears each point into a disk of radius $1.22\,\lambda/D$ , and separation, which pulls the disks apart. Enlarging the aperture $D$ or shortening the wavelength $\lambda$ tightens the disk and lets finer detail be split — the reason telescopes are built ever larger and electron microscopes exploit picometre wavelengths. The next figure quantifies the "just resolved" threshold: the composite intensity of two incoherent sources at exactly the Rayleigh separation, and how the tell-tale dip disappears once they crowd closer.

Figure

Composite intensity of two incoherent Airy sources: at the Rayleigh separation a 26.5% central dip (midpoint 0.735 of peaks); closer (Sparrow, 0.95 λ/D) the dip is gone

The core equations

Every source point images to an Airy pattern, so the resolution problem begins with the same irradiance profile that governs the circular aperture. The dark rings sit at the zeros of $J_1$ , and the first of them — at $u_1 = 3.8317$ — defines the edge of the Airy disk. Here $u$ is the dimensionless radial coordinate in the diffraction pattern; setting it to the first Bessel zero converts to an angle.

I(\theta) = I_0\left[\frac{2J_1(u)}{u}\right]^2,\qquad u = \frac{\pi D}{\lambda}\sin\theta,\qquad J_1(u_1) = 0,\; u_1 = 3.8317 \qquad (1)

where: $I(\theta)$ — far-field intensity at angle $\theta$ ; $I_0$ — central peak; $J_1$ — first-order Bessel function; $D$ — aperture diameter; $\lambda$ — wavelength; $u$ — reduced radial coordinate; $u_1$ — its first nonzero root. The angular radius of the first dark ring is the Airy radius, and Rayleigh sets the resolvable separation equal to it. Rayleigh's convention is one line: place the peak of the second source on the first minimum of the first, so their angular separation equals the Airy-disk radius. Dividing the first Bessel zero by $\pi$ turns $u_1$ into the familiar coefficient, and the small-angle approximation $\sin\theta \approx \theta$ gives the resolution formula quoted for every telescope and eye.

\theta_R = \frac{u_1}{\pi}\frac{\lambda}{D} = 1.22\,\frac{\lambda}{D},\qquad \frac{u_1}{\pi} = \frac{3.8317}{\pi} = 1.2197 \qquad (2)

where: $\theta_R$ — minimum resolvable angular separation; $D$ — pupil diameter. Since $\theta_R \propto \lambda/D$ , resolving power improves in direct proportion to aperture and inversely with wavelength — the twin levers of every high-resolution instrument. In an imaging system the angular limit becomes a linear one in the focal plane by multiplying through $f$ , and for a microscope the natural variable is not diameter but numerical aperture. Writing the object-side geometry in terms of $\mathrm{NA} = n\sin\alpha$ turns the same criterion into the Abbe-style microscope form, where the factor becomes $0.61 = 1.22/2$ because the resolution is expressed as a radius against the half-angle rather than a diameter.

r_R = 1.22\,\frac{\lambda f}{D} = 1.22\,\lambda N,\qquad d = \frac{0.61\,\lambda}{\mathrm{NA}},\qquad \mathrm{NA} = n\sin\alpha \qquad (3)

where: $r_R$ — separation of the two image spots in a focal plane; $f$ — focal length; $N = f/D$ — f-number; $d$ — smallest resolvable object separation in a microscope; $\mathrm{NA}$ — numerical aperture; $n$ — immersion index; $\alpha$ — half-angle of the collected cone. High-NA immersion objectives shrink $d$ toward $\sim\lambda/2$ , the classical light-microscope barrier. The criterion's "just resolved" verdict rests on how two incoherent sources combine: their intensities add, with no interference term. Evaluating that sum at the peaks and at the midpoint (Derivation B) is what produces the diagnostic central dip that a resolved pair must show.

I_{\text{tot}}(x) = I_1(x) + I_2(x - \Delta x),\qquad \frac{I_{\text{mid}}}{I_{\text{peak}}} = 0.735 \;\;(26.5\%\ \text{dip}) \qquad (4)

where: $I_{\text{tot}}$ — combined profile of two mutually incoherent sources; $I_1, I_2$ — individual Airy profiles; $\Delta x$ — their separation (set to the Rayleigh value). Incoherence is essential: coherent fields would add as amplitudes, and the dip could deepen or vanish entirely with relative phase.

Derivation ARayleigh angular resolution from the first Airy zero

Our aim is to turn Rayleigh's verbal rule — "peak of one pattern on the first dark ring of the other" — into the quantitative formula $\theta_R = 1.22\,\lambda/D$ , and to pin the microscope variant $0.61\,\lambda/\mathrm{NA}$ to the same Bessel root. The physics is inherited: each source images to an Airy pattern whose first minimum lies at the first zero of $J_1$ . The derivation is then geometry — convert that zero into an angle, and impose the criterion by equating the source separation to the Airy-disk radius.

Step 1. Start from the Airy intensity $I \propto [2J_1(u)/u]^2$ with $u = (\pi D/\lambda)\sin\theta$ , the far-field pattern of a uniformly illuminated circular pupil. The dark rings are the zeros of $J_1$ . Why: resolution is decided by where one pattern goes dark, because that is where a second peak can sit with least confusion — so the zeros of $J_1$ , not its maxima, carry the information.

Step 2. Locate the first zero: $J_1(u_1) = 0$ with $u_1 = 3.8317$ , the smallest positive root (obtained numerically by bracketing $J_1$ between $3.0$ and $4.5$ ). Why: this single Bessel root is the entire origin of the resolution coefficient — everything downstream is unit conversion.

Step 3. Convert to an angle. Setting $u = u_1$ gives $\sin\theta_1 = u_1\lambda/(\pi D) = (u_1/\pi)\lambda/D$ , and $u_1/\pi = 3.8317/3.14159 = 1.2197$ , so $\sin\theta_1 = 1.22\,\lambda/D$ . In the small-angle regime of real optics $\sin\theta_1 \approx \theta_1$ . Why: dividing the Bessel zero by $\pi$ is exactly what produces the $1.22$ — larger than a slit's $1$ because a circular rim pushes its first null slightly outward.

Step 4. Impose Rayleigh's criterion: two sources are just resolved when their separation equals this Airy radius, $\theta_R = \theta_1 = 1.22\,\lambda/D$ . For a microscope, expressing the collected cone through $\mathrm{NA} = n\sin\alpha$ and taking the resolution as a radius (a factor of two) gives $d = 0.61\,\lambda/\mathrm{NA}$ , with $0.61 = 1.22/2$ . Why: the telescope and microscope forms are the same criterion in different variables — angular separation against diameter, versus object separation against numerical aperture. *Verified: $u_1 = 3.831706$ , $u_1/\pi = 1.21967$ , and $0.61 = 1.22/2$ in verify_aq-optics-rayleigh-resolution-criterion.py.*

Go deeper · Derivation B · 2 minThe 26.5% central dip for two incoherent sources

Derivation BThe 26.5% central dip for two incoherent sources

The Rayleigh separation is chosen not arbitrarily but because it produces a specific, detectable contrast: a shallow valley between the two peaks. Our aim is to compute that valley for two equal, mutually incoherent sources and show it is a dip of about $26.5\%$ — the summed intensity at the midpoint falling to $0.735$ of the peak. Because the sources are incoherent, their intensities simply add, so the whole calculation is two evaluations of the Airy function: once at a peak, once at the midpoint.

Step 1. Set the separation to the Rayleigh value, $\Delta u = u_1 = 3.8317$ , so source 2's peak lands on source 1's first zero. Add intensities: $I_{\text{tot}}(u) = [2J_1(u)/u]^2 + [2J_1(u-u_1)/(u-u_1)]^2$ . Why: mutual incoherence means no cross term — the observed pattern is the plain sum of the two Airy intensities, which is what lets us evaluate it point by point.

Step 2. Find the peak height. At source 1's centre ( $u = 0$ ) its own pattern is $I_0$ while source 2, a distance $u_1$ away, is exactly at its first zero and contributes nothing: $I_{\text{peak}} = I_0 + 0 = I_0$ . Why: the criterion was built so the neighbour contributes zero at each peak — that is what "peak on the first dark ring" means — so each peak stands at the full single-source height.

Step 3. Evaluate the midpoint, a distance $u_1/2 = 1.9159$ from each source. There each contributes $[2J_1(1.9159)/1.9159]^2$ . Compute $2J_1(1.9159)/1.9159 = 0.6062$ , and squaring gives $0.3675$ . Why: by symmetry both sources are equidistant from the midpoint, so the two contributions are identical — one Airy evaluation does the job.

Step 4. Add the two midpoint contributions: $I_{\text{mid}} = 2 \times 0.3675\,I_0 = 0.735\,I_0$ . The central dip is therefore $1 - 0.735 = 0.265$ , about $26.5\%$ below the peaks — modest, but enough for a careful observer or a low-noise detector to call the pair "two". Why: this dip is the operational meaning of the Rayleigh limit; crowd the sources closer (the Sparrow separation, $\approx 0.95\,\lambda/D$ ) and the valley fills to a flat top, and the pair reads as one. *Verified: $2J_1(1.9159)/1.9159 = 0.6062$ , single-source midpoint value $0.3675$ , sum $0.7350$ , dip $0.2650$ , and the Sparrow coefficient $0.947$ in verify_aq-optics-rayleigh-resolution-criterion.py.*

Limiting cases

Case	Result	Meaning
circular aperture, angular	$\theta_R = 1.22\,\lambda/D$	canonical telescope/eye result, from the first $J_1$ zero
slit / rectangular aperture	$\theta_R = \lambda/a$	first null at $\sin\theta = \lambda/a$ , no factor $1.22$
microscope (linear, immersion)	$d = 0.61\,\lambda/\mathrm{NA}$	$0.61 = 1.22/2$ ; high NA shrinks $d$ toward $\lambda/2$
Sparrow separation	$\approx 0.95\,\lambda/D$	closer limit where the central dip just vanishes
astronomer's rule ( $\lambda = 550$ nm)	$\theta_R \approx 13.8''/(D/\text{cm})$	handy visible-light shortcut, $\theta$ in arcseconds
coherent / unequal sources	dip shifts	the $26.5\%$ figure assumes incoherent, equal brightness

Common trap“The Rayleigh limit is a hard physical wall that no instrument can cross.”

It is a convention, not a law. Rayleigh chose the "peak on first dark ring" spacing because it yields a repeatable, eye-detectable $26.5\%$ dip for two equal, incoherent sources — a sensible operational threshold, but not a barrier fixed by physics. Three things move it. Coherence: coherent sources add as fields, so their relative phase can wash the dip out or deepen it, making them neither resolved nor unresolved at the same separation. Brightness: unequal sources give a shallower, off-centre dip. And information: with many photons, a known point-spread function, or single-emitter tricks (STED, PALM/STORM), you can localise and separate sources far below $\theta_R$ — super-resolution circumvents the criterion without violating diffraction. The true limit is set by signal-to-noise, not by the number $1.22$ .

Worked examples

1 · The resolving power of the human eye (confidence builder)

Take a daylight pupil of diameter $D = 2$ mm and green light $\lambda = 550$ nm. What is the finest angular separation the eye can resolve by the Rayleigh criterion, and how does it compare with normal visual acuity?

Step 1. Apply equation (2): $\theta_R = 1.22\,\lambda/D = 1.22 \times (550\times10^{-9}\,\text{m})/(2\times10^{-3}\,\text{m})$ . The numerator is $1.22 \times 550\times10^{-9} = 6.71\times10^{-7}$ m; dividing by $2\times10^{-3}$ m gives $\theta_R = \mathbf{3.36\times10^{-4}}$ rad.

Step 2. Convert to arcminutes: $\theta_R = 3.36\times10^{-4}\,\text{rad} \times 206{,}265''/\text{rad} = 69.2''$ , and dividing by $60$ gives $\mathbf{1.15'}$ — just over one arcminute.

Step 3. This lands right on the textbook "one-arcminute" acuity behind the 20/20 standard, and it matches the spacing of the cone photoreceptors in the fovea — a neat case of biology tuned to the diffraction limit rather than wasting resolution the optics cannot deliver.

Step 4. Make it concrete: two car headlights $1.5$ m apart are just resolved when $\theta_R = (1.5\,\text{m})/L$ , so $L = 1.5/(3.36\times10^{-4}) \approx \mathbf{4.5\ km}$ . Beyond that distance the pair merges into a single glow — roughly the range at which oncoming headlights stop looking like two lights. θ_R = 3.36 × 10⁻⁴ rad = 1.15′.

2 · Telescope and microscope resolving power (exam level)

A small telescope has aperture $D = 100$ mm. Find its diffraction-limited angular resolution at $\lambda = 550$ nm. Then find the linear resolution of an oil-immersion microscope objective of $\mathrm{NA} = 1.4$ at the same wavelength.

Step 1. Telescope: $\theta_R = 1.22\,\lambda/D = 1.22 \times 550\times10^{-9}/0.100 = 6.71\times10^{-6}$ rad. In arcseconds, $6.71\times10^{-6} \times 206{,}265 = \mathbf{1.38''}$ .

Step 2. Cross-check with the astronomer's rule of thumb, $\theta_R \approx 13.8''/(D/\text{cm})$ . With $D = 10$ cm this gives $13.8/10 = \mathbf{1.38''}$ — identical, as it must be, since the $13.8$ is just $1.22\,\lambda$ in arcseconds for $550$ nm.

Step 3. Microscope: use the numerical-aperture form, $d = 0.61\,\lambda/\mathrm{NA} = 0.61 \times 550\times10^{-9}/1.4$ . Compute $0.61 \times 550\times10^{-9} = 3.355\times10^{-7}$ m, divided by $1.4$ gives $d = 2.40\times10^{-7}$ m $= \mathbf{240\ nm}$ — the classic $\sim\lambda/2$ light-microscope limit.

Step 4. The contrast is instructive: the telescope's job is angular (separating distant point sources), the microscope's is linear (separating nearby object points), yet both come from the same $1.22$ Bessel constant, one divided into the aperture diameter, the other by twice the numerical aperture. This $240$ nm barrier is exactly what drove the invention of super-resolution microscopy. Telescope 1.38″; microscope 240 nm.

Physics of the portrait

The convention that fixes resolving power everywhere carries the name of a Cambridge-trained aristocrat-physicist who could not stop finding fundamental results.

Origins · 1 minLord Rayleigh and how this was discovered

Lord Rayleigh

1842–1919 · Terling Place & Cambridge

Lord Rayleigh — John William Strutt, third Baron Rayleigh — was among the most versatile physicists of his era, working largely from a private laboratory on the family estate at Terling in Essex and, for five years, as Cavendish Professor at Cambridge. His name is stamped across physics: Rayleigh scattering (why the sky is blue and sunsets red), Rayleigh waves in seismology, the Rayleigh–Jeans law of blackbody radiation, and the criterion described here, which he set out in his 1879 optical investigations on the resolving power of gratings, prisms, and telescopes. In 1894, with William Ramsay, he discovered argon by painstakingly comparing the density of nitrogen from air with nitrogen from chemical sources — a $0.5\%$ discrepancy he refused to ignore — which won him the 1904 Nobel Prize in Physics. His two-volume Theory of Sound remains a classic of acoustics. Rayleigh embodied the idea that resolution, whether of two stars or two spectral lines, is a single wave phenomenon set by aperture and wavelength.

History

1835

— Airy computes the circular-aperture diffraction pattern whose first dark ring defines the disk

1873

— Abbe develops the diffraction theory of microscope resolution

1879

— Rayleigh states the resolution criterion for telescopes, prisms, and gratings

1891

— Michelson measures the diameters of Jupiter's moons by interferometry at Lick Observatory, sidestepping the single-aperture limit

1916

— Sparrow proposes the closer "undetectable dip" resolution limit

Going further

Texts. Hecht, Optics, 5th ed., §10.2.6 (Resolution of imaging systems) · Born & Wolf, Principles of Optics, 7th ed., §8.6.2 (resolving power) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., Ch. 4 (Fourier optics and imaging) · Foot, Atomic Physics, §A.3 (diffraction and resolution) · Demtröder, Laser Spectroscopy, Vol. 1 (spectrometer resolving power).

Provenance. The Airy pattern and first zero $u_1 = 3.8317$ are from Born & Wolf §8.5.2; the factor $1.22 = u_1/\pi$ is verifiable by arithmetic. The Rayleigh formula and the microscope form $0.61\,\lambda/\mathrm{NA}$ are from Hecht §10.2.6 and Saleh & Teich Ch. 4. The $26\%$ dip (ratio $0.735$ ) is reproducible by evaluating $[2J_1(1.916)/1.916]^2 \approx 0.368$ and doubling. Eye-acuity and telescope figures are standard textbook estimates; the grating resolving power $R = mN$ is from Demtröder Vol. 1. Each number is recomputed in verify_aq-optics-rayleigh-resolution-criterion.py.

Builds on. Circular Aperture Diffraction and the Airy Pattern · Fraunhofer Diffraction · Single-Slit Diffraction. Leads to. Alternative Resolution Criteria (Sparrow, Abbe) · Rectangular Aperture Diffraction.

Why is the factor 1.22 there for a circle but replaced by 1 for a slit?

Because the first dark fringe sits at a different place for the two shapes. A slit's diffracted amplitude is a sinc function, whose first zero is at $\sin\theta = \lambda/a$ — coefficient $1$ . A circular aperture's amplitude is $2J_1(u)/u$ , whose first zero is at $u_1 = 3.8317$ , giving $\sin\theta = u_1\lambda/(\pi D) = 1.22\,\lambda/D$ . The $1.22$ is simply $u_1/\pi$ : circular symmetry redistributes light outward, pushing the first null slightly farther than a straight-edged slit does. Same physics, different aperture geometry.

What actually changes if the two sources are coherent rather than incoherent?

For incoherent sources the intensities add with no cross term, giving the clean $26.5\%$ dip at the Rayleigh separation. Coherent sources add as fields, so the result depends on their relative phase. Two in-phase coherent points can fill the central dip completely and look unresolved even at the Rayleigh spacing; two out-of-phase points interfere destructively at the midpoint, deepening the dip and appearing better resolved. This is why the criterion is stated specifically for mutually incoherent, equal-brightness sources — the case of stars, or independently emitting fluorophores — and why coherent imaging (some laser and electron systems) needs its own analysis.

Is the Rayleigh criterion the same as the Abbe diffraction limit?

They are close cousins with slightly different coefficients. Rayleigh's rule, "peak on first dark ring", gives $0.61\,\lambda/\mathrm{NA}$ for a microscope. Abbe's limit, derived from the requirement that at least two diffraction orders from a periodic object be captured by the objective, gives $\lambda/(2\,\mathrm{NA}) = 0.5\,\lambda/\mathrm{NA}$ . Both express the same physics — resolution set by wavelength over numerical aperture — and differ only in the numerical factor and the assumed object (two points versus a grating). In practice they are quoted almost interchangeably for the $\sim 200$ nm light-microscope limit.

How do interferometers resolve finer than their individual dishes allow?

By replacing the single aperture $D$ with the baseline $B$ between separated collectors. The angular resolution of an interferometer is $\sim \lambda/B$ , where $B$ is the distance between antennas or telescopes rather than the size of any one of them. Combining the signals coherently synthesises an effective aperture as large as the array, so the Very Large Array or the planet-spanning Event Horizon Telescope reach resolutions no single mirror could — down to tens of microarcseconds. The Rayleigh scaling $\propto \lambda/(\text{aperture})$ is unchanged; only the meaning of "aperture" grows.

Why does the dip vanish at the Sparrow separation, and what is Sparrow's number?

As you push two incoherent sources closer than the Rayleigh spacing, the central valley between them shallows. The Sparrow criterion marks the separation at which the dip disappears entirely — the composite develops a flat top, its second derivative at the midpoint just reaching zero. For two Airy sources that happens at about $0.95\,\lambda/D$ , roughly three-quarters of the Rayleigh separation. Below it the pair produces a single broad maximum and is genuinely unresolved by intensity alone. Sparrow's limit is tighter than Rayleigh's but harder to apply, since "no visible dip" is a weaker cue than a $26.5\%$ valley.

Does the grating resolving power R = mN really come from the same criterion?

Yes. A grating's resolving power $R = \lambda/\Delta\lambda = mN$ (order $m$ , $N$ illuminated lines) is the Rayleigh criterion applied to wavelength rather than angle. Two wavelengths are just resolved when the diffraction maximum of one falls on the first minimum of the other in the grating's interference pattern. The finite illuminated width acts as the aperture, and working through the geometry gives exactly $mN$ . It is the same "peak on first zero" convention that produces $1.22\,\lambda/D$ for a round pupil — the aperture here being the ruled width of the grating.