Physics · Diffraction and resolution

Rayleigh criterion for a circular aperture

Rayleigh’s criterion sets two incoherent point-image separation to the first-minimum radius of a circular-pupil Airy pattern.

AdvancedUniversity/collegeBachelor foundations

Create flashcards or a quiz I already have an account

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Formula

Calculate

θ_{m i n} \approx 1.22 \frac{λ}{D}

λ,D in the same length unit, preferably meters; θmin in radians.1 rad≈206264.806 arcseconds.

Variables & units

Variables & units
Symbol	Meaning	Unit
$θmin$	Rayleigh separation of the specified incoherent point sources, in radians	rad
$λ$	Wavelength in propagation medium; approximately vacuum wavelength in air	m
$D$	Clear diameter of the uniformly illuminated unobstructed circular pupil	m

LaTeX

\theta_{\min}\approx1.22\frac{\lambda}{D}

QUANTA STUDYquanta-study.de

Two true Bessel-Airy sections centered at q = 0 and 1. Their linear intensity sum is about 0.735 at the midpoint; both centers have total 1. Each first zero reaches the other center. — Two equal incoherent Airy point images at Rayleigh separation: each maximum falls on the first minimum of the other. Curves show intensities and their sum, not coherent field amplitudes. For 550 nm and D = 0.200 m, approximately θmin ≈ 3.355 µrad.

Rayleigh criterion calculator

Find the smallest resolvable angle of a round aperture such as a telescope, camera or eye from wavelength and diameter, in the ideal diffraction-limited case.

Calculation model

Find

Wavelength λ

Wavelength of the light, e.g. 550 nm for green light in the middle of the visible range.

Aperture diameter D

Diameter of the clear aperture, not the radius, e.g. 20 cm for a telescope.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for θR, substitute, evaluate

1. Rearrange for θR
$\theta_R\approx1.22\frac\lambda D$
2. Substitute values (in SI units)
- λ = 550 nm = 5.5 · 10⁻⁷ m
- D = 20 cm = 0.2 m
$\frac{1.22\cdot 5.5\times10^{-7}}{0.2}$
3. Result
θR = 3.355 μrad

Result: Smallest resolvable angle θ

θR = 3.355 μrad

With λ = 550 nm and D = 20 cm, θR = 3.355 μrad.

Quick tips

Twice the aperture diameter, half the angle: θ is proportional to λ ÷ D. That is why blue light resolves finer detail than red light.
Use the diameter, not the radius: the radius would make the angle twice as large. Convert nanometres to metres first: 550 nm = 5.5 · 10⁻⁷ m.

Assumptions and limits

Ideal diffraction-limited paraxial Fraunhofer imaging of equal-brightness incoherent monochromatic point sources through an unobstructed uniformly illuminated circular aperture. Wavelength in the medium. Rounded factor 1.22; the result is approximate. Deliberate technical limit of this approximation calculator: θR ≤ 0.1 rad, not a universal Rayleigh validity threshold. No guarantee of actual visibility or sensor resolution.

Derivation & background

A circular pupil gives intensity factor $\left[\frac{2J_1(u)}u\right]^2$ . Its first Bessel zero yields the small-angle coefficient 1.22. The criterion concerns incoherent point images, not a universal hard detection limit. Focal-plane first-minimum radius is $r_1\approx f\theta_{\min}$ ; Airy diameter is 2 r₁. Different pupils, strong brightness contrast, atmosphere, aberrations and sampling require further assessment.

Exam blueprint

Validity range

Diffraction-limited imaging of equal incoherent point sources through a clear uniformly illuminated circular aperture in the Fraunhofer/paraxial model. λ is wavelength in the propagation medium; the result is in radians. Obstruction, apodization, coherence, unequal brightness, aberrations, atmosphere and detector properties alter real resolution or detectability.

Derivation steps

Find the circular-pupil pattern’s first minimum and use it as Rayleigh separation of incoherent points.

1Model a clear uniformly illuminated circular aperture of diameter D and monochromatic wavelength λ in the medium.
2Fraunhofer field amplitude is the two-dimensional Fourier transform of that opening.
3Circular symmetry gives $\frac{I(\theta)}{I(0)}=\left[\frac{2J_1(u)}u\right]^2$ with $u=\frac{\pi D\sin\theta}{\lambda}$ .
4The first positive zero is u≈3.831706, hence sinθ₁≈1.219670 $\frac\lambda D$ .
5For small angles in radians $\sin\theta\approx\theta$ . Round the factor to 1.22.
6Rayleigh places one incoherent point-image maximum on the other’s first minimum: $\theta_{\min}\approx1.22\frac\lambda D$ . This is a resolution convention, not a universal information limit.

Rearrangements

Aperture diameter

D\approx1.22\frac{\lambda}{\theta_{\min}}

Use θmin in radians.

Task variant

A clear uniformly illuminated circular aperture has D = 0.20 m; monochromatic light has λ = 550 nm. Find ideal Rayleigh angle in radians and arcseconds and, for f = 1.20 m, first-minimum radius and Airy diameter. What diameter is needed for 0.80 arcseconds at the same wavelength?

Approach: $\theta_R=1.22\frac\lambda D$ .
This gives 3.355 × 10⁻⁶ rad ≈ 0.692018 arcseconds.
In the focal plane $r_1\approx f\theta_R=4.026\,\mathrm{\mu m}$ ; central diameter is 8.052 μm.
Convert the required 0.80 arcseconds to radians before using $D\approx1.22\frac\lambda{\theta_R}=0.1730046\,\mathrm m$ : ideally at least about 17.30 cm.
Noise or aberrations can still prevent actual separation.

Common mistakes

Ignoring units or assumptions stated for the formula: λ,D in the same length unit, preferably meters; θmin in radians.1 rad≈206264.806 arcseconds.

Convert units before substitution and verify the validity range; round only at the end.

Exam context

Analyze telescope and camera resolution, focal-plane sampling and diffraction reference limits; describe microscope objects with the appropriate numerical aperture.

Formula cluster

Diffraction and resolution

Related relations: beugungsgitter-maxima, linsengleichung, doppelspalt-interferenz.

d sin θ_m = mλ

Diffraction grating: principal maxima

1/f = 1/g + 1/b

Thin Lens Equation

d·sin(α_k) = k·λ

Double-Slit Interference

Worked example

λ=550 nm and D=0.20 m gives θmin=3.355 × 10⁻⁶rad≈0.6920 arcseconds. For f=1.20 m, r₁=4.026μm and Airy diameter 8.052μm. At that wavelength,0.80 arcseconds ideally requires at least 17.30 cm clear aperture.

Applications

Analyze telescope and camera resolution, focal-plane sampling and diffraction reference limits; describe microscope objects with the appropriate numerical aperture.

How do you calculate with Rayleigh criterion for a circular aperture?

Here is how to work through a typical Rayleigh criterion for a circular aperture (θmin ≈ 1.22 λ/D) task step by step:

1
Task
A clear uniformly illuminated circular aperture has D = 0.20 m; monochromatic light has λ = 550 nm. Find ideal Rayleigh angle in radians and arcseconds and, for f = 1.20 m, first-minimum radius and Airy diameter. What diameter is needed for 0.80 arcseconds at the same wavelength?
Solution path
1. Approach: $\theta_R=1.22\frac\lambda D$ .
2. This gives 3.355 × 10⁻⁶ rad ≈ 0.692018 arcseconds.
3. In the focal plane $r_1\approx f\theta_R=4.026\,\mathrm{\mu m}$ ; central diameter is 8.052 μm.
4. Convert the required 0.80 arcseconds to radians before using $D\approx1.22\frac\lambda{\theta_R}=0.1730046\,\mathrm m$ : ideally at least about 17.30 cm.
5. Noise or aberrations can still prevent actual separation.

Understand and apply the equation

Two distant light points can merge into one spot even when the lens is focused. Diffraction through a finite aperture spreads each point image. Rayleigh’s criterion defines one comparison spacing for that spread. First distinguish pattern radius from diameter, then calculate angle and image separation and examine what a larger aperture actually improves.

Before calculating

You need wavelength conversion, small angles in radians and the idea of a point image. An ideal object point becomes a finite-width optical pattern. Two points therefore form two overlapping patterns. For incoherent sources add intensities without assuming a fixed relative field phase.[1][2]

The comparison uses equal-brightness incoherent point sources and an unobstructed circular uniformly weighted aperture in a diffraction-limited model. D is diameter, not radius. The first dark ring bounds the central Airy disk. This criterion describes that optical model, not automatically eyesight, sensor resolution or detection of an arbitrarily faint companion.[1][3]

A minimum separation under a specified convention

\theta_{\min}\approx1.22\frac\lambda D

D is the clear diameter of a uniformly illuminated unobstructed circular aperture. λ is wavelength in the propagation medium; θ is calculated in radians.[1]

One point gives an Airy pattern with a central maximum and rings. Two equally bright incoherent points meet Rayleigh’s criterion when one maximum lies at the first minimum of the other. Add their intensity patterns, not their field amplitudes. For coherent sources a fixed relative phase can change the combined image.[3]

θmin is source angular separation and also the angular radius to the first minimum of one point image. It is not the Airy-disk diameter or the full spread angle of an arbitrary source. The central pattern’s diameter between first minima is twice this radius.[1]

Two point images at Rayleigh spacing

QUANTA STUDYquanta-study.de

A cut through the circular-aperture image, not the entire two-dimensional Airy disk. The x scale uses multiples of the first-minimum angle; the combined curve adds intensities, not amplitudes.[1][2]

Free with a Quanta account. Your download stays ready after registration.

Why a circular aperture gives the factor 1.22

In the Fraunhofer model, field amplitude is proportional to the two-dimensional Fourier transform of the aperture. For a uniformly illuminated circle, integrate over azimuth and then over radius to obtain a Bessel expression. The result is rotationally symmetric. Its zero depends on aperture shape; a slit or rectangular opening has a different pattern.[2][3]

\begin{aligned}&\frac{I(\theta)}{I(0)}=\left[\frac{2J_1(u)}u\right]^2\\&u=\frac{\pi D}{\lambda}\sin\theta\end{aligned}

The continuous central limit is one. The first positive minimum occurs where $J_1(u)=0$ .[3]

\begin{aligned}&u_1\approx3.831706\\&\sin\theta_1=\frac{u_1}\pi\frac\lambda D\approx1.219670\frac\lambda D\end{aligned}

For small angles $\sin\theta\approx\theta$ in radians; rounding the coefficient to 1.22 gives the familiar relation.[3][1]

A single-point pattern’s first minimum is a diffraction property. Choosing that separation as just resolved is a resolution convention. It does not mean that all information vanishes below that spacing. Measurements with known source models and high signal quality can have different estimation limits.[1]

From radians to arcseconds and image separation

The length ratio first gives an angle in radians. A smaller value means a smaller just-resolved separation and better ideal angular resolution. Focal length enters only when converting that angle into separation on a sensor or screen.[1]

\begin{aligned}&1\,\mathrm{rad}=\frac{180}{\pi}\cdot3600\,\mathrm{arcsec}\approx206264.806\,\mathrm{arcsec}\\&r_1\approx f\theta_{\min}\end{aligned}

r₁ is the first-minimum radius in a paraxial focal plane, not its diameter.[1]

λ and D must use the same length unit. 550 nm with D=0.20 m requires λ=550 × 10⁻⁹m first. The ratio is dimensionless but interpreted as an angle in radians. Degrees or arcseconds require conversion. For distant points at range L, corresponding transverse separation is approximately $s\approx L\theta_{\min}$ .[1]

In air, vacuum wavelength is often a sufficient approximation for λ. In a medium with relevant index n, $\lambda=\frac{\lambda_0}{n}$ . Object-side microscope resolution instead uses the appropriate numerical aperture and vacuum wavelength: $\delta\approx0.61\frac{\lambda_0}{\mathrm{NA}}$ . A physical diameter alone does not fully describe high-aperture collection geometry.[1]

A 20-cm telescope in angle and focal plane

Two equally bright incoherent stars in an ideal model

Given: D=0.200 m; λ=550 nm in air; f=1.20 m; clear circular aperture, diffraction limited.
Find: Find Rayleigh angle, arcseconds and focal-plane separation.

Approach: For equal incoherent point images through a clear circular aperture, find Rayleigh angle and then focal-plane separation.
$\begin{aligned}\theta_{\min}&\approx1.22\frac\lambda D\\r_1&\approx f\theta_{\min}\end{aligned}$
Convert wavelength to meters and calculate angle.
$\theta_{\min}=1.22\frac{550\times10^{-9}}{0.200}=3.355\times10^{-6}\,\mathrm{rad}$
Convert to arcseconds.
$\theta_{\min}=0.692018\,\mathrm{arcsec}$
Use focal length to convert to the focal plane.
$r_1=f\theta_{\min}=4.026\,\mathrm{\mu m}$

Result: Rayleigh separation is about 0.692 arcseconds; point-image centers are 4.026μm apart. The Airy diameter between first minima is 8.052μm.

Check: Units check: $\frac\lambda D$ is dimensionless and fθ is a length. Doubling f enlarges sensor pattern and separation but does not change angular resolution when D stays fixed.

Convert a target angular separation into aperture

0.80 arcseconds at 550 nm

Given: Target separation 0.80 arcsec; λ=550 nm; equal incoherent points and ideal clear circular aperture.
Find: Find minimum diameter under the Rayleigh model.

Approach: Express target separation in radians first, then solve the circular-aperture criterion for minimum diameter.
$\begin{aligned}\theta_{\mathrm{target,rad}}&=\theta_{\mathrm{target,arcsec}}\frac{\pi}{180\cdot3600}\\D&\ge\frac{1.22\lambda}{\theta_{\mathrm{target,rad}}}\end{aligned}$
Convert arcseconds into radians first.
$\theta=\frac{0.80}{206264.806}=3.87851\times10^{-6}\,\mathrm{rad}$
Rearrange the resolution relation for D.
$D\ge\frac{1.22\lambda}{\theta}=0.173005\,\mathrm m$

Result: At least about 17.30 cm clear aperture in the ideal model. This is a diffraction requirement, not a guarantee of a real observation.

Check: Back-substitution gives 0.80 arcsec. A 20-cm system gives 0.692 arcsec and meets the ideal condition;15 cm gives 0.923 arcsec and misses it.

Explore further: f-number and sensor pixels

At fixed λ, doubling D halves Rayleigh angle. Doubling focal length without changing D only increases image scale. The sensor may sample that larger pattern more easily, but optical angular information available before sampling is unchanged. Digital enlargement likewise does not create missing optical detail.[1]

For a camera, $r_1\approx1.22\lambda N$ where $N=\frac fD$ is f-number. This is ideal focal-plane Airy radius. Shorter wavelength gives a smaller pattern for unchanged aperture. Stopping down often reduces aberrations and increases depth of field while enlarging diffraction blur; stopping down indefinitely does not always improve sharpness.[3]

Compare diffraction disk with pixel size

Given: λ=550 nm in air; ideal camera lens at N=8; pixel pitch p=3.00 μm. Second comparison at N=16.
Find: Find Airy radius, diameter and diameter in pixel pitches.

Approach: use the focal-plane expression with f-number. Calculate radius first and double it for diameter.
$\begin{aligned}r_1&\approx1.22\lambda N\\d_{\mathrm{Airy}}&=2r_1\\q&=\frac{d_{\mathrm{Airy}}}{p}\end{aligned}$
Convert nanometres to micrometres and substitute N=8.
$\begin{aligned}r_1&=1.22(0.550\,\mathrm{\mu m})8=5.368\,\mathrm{\mu m}\\d_{\mathrm{Airy}}&=10.736\,\mathrm{\mu m}\\q&=\frac{10.736\,\mathrm{\mu m}}{3.00\,\mathrm{\mu m}}=3.57867\end{aligned}$
At N=16 the geometric pattern dimensions double.
$\begin{aligned}r_{1,16}&=10.736\,\mathrm{\mu m}\\d_{\mathrm{Airy},16}&=21.472\,\mathrm{\mu m}\\q_{16}&=7.15733\end{aligned}$

Result: At N=8 the central disk spans about 3.58 pixel pitches; at N=16 it spans 7.16. This is diameter between the first minima.

Check: Doubling N doubles Airy diameter at fixed wavelength. Pixel counts are only a scale comparison, not proof of a general sampling criterion or actual resolution under noise and aberrations.

The thin-lens equation adds geometric image position and scale under suitable imaging assumptions. It does not automatically determine the diffraction separation of two point images; clear aperture, wavelength and the chosen resolution criterion remain essential.[4][1]

Brightness contrast, atmosphere and different pupils

The coefficient 1.22 belongs to the specified unobstructed uniformly weighted circular pupil. Central obstruction, apodization or rectangular apertures alter the central lobe and rings. Broadband light superposes patterns of different sizes. A faint companion can disappear in a bright object’s diffraction rings despite adequate geometric separation; D and λ alone cannot decide detectability.[3]

Atmospheric turbulence, defocus, aberrations, motion, detector noise and sampling further degrade real images. Label the result as an optical reference criterion, not measured instrument resolution. The paraxial far-field form uses small angles; large-angle diffraction geometry retains the sine.[1]

The wave relation for speed, wavelength and frequency helps assign wavelength to its medium: $\lambda=v/f$ . At the same specified frequency, a different phase speed also means a different wavelength. This assignment replaces neither the circular-aperture assumption nor an assessment of real imaging errors.[5][1]

Five exercises checking angle, color and distance

Calculate · Foundation

D=0.30 m, λ=600 nm: find θmin in radians and arcseconds.

Show hint

Convert nanometers to meters.

Learning goal: Calculate Rayleigh angle and convert angular units.

Show solution and calculation

Approach: convert wavelength to metres, insert clear diameter in Rayleigh’s criterion and then convert from radians.
The result is 2.44 microradians or 0.503286 arcseconds.

Check: both values describe the same angle. Reporting the initial result directly as degrees would confuse angular units even though the length ratio is dimensionless.

\begin{aligned}\theta_R&=1.22\frac{600\times10^{-9}\,\mathrm m}{0.30\,\mathrm m}=2.44\times10^{-6}\,\mathrm{rad}\\\theta_R&\approx0.503286\,\mathrm{arcsec}\end{aligned}

Calculate · Application

A 15-cm system observes at 500 nm and 650 nm. How do Rayleigh angles change?

Show hint

For fixed D, θ is proportional to λ.

Learning goal: Check wavelength scaling.

Show solution and calculation

Approach: at fixed diameter Rayleigh angle is proportional to wavelength.
At 500 nm it is 4.06667 microradians or 0.838810 arcseconds; at 650 nm it is 5.28667 microradians or 1.090453 arcseconds.

Check: their ratio 1.30 equals the wavelength ratio. The longer red wavelength gives the larger ideal separation threshold.

\frac{\theta_{650}}{\theta_{500}}=\frac{650}{500}=1.30

Calculate · Application

D=20 cm, λ=550 nm; two points lie 2000 m away. What transverse separation matches ideal Rayleigh angle?

Show hint

Use s≈Lθ with θ in radians.

Learning goal: Convert angular resolution to transverse separation.

Show solution and calculation

Approach: first calculate the circular aperture’s Rayleigh angle, then use viewing distance for small transverse separation.
This gives s≈6.710 mm.

Check: $\frac{s}{L}=3.355\times10^{-6}$ recovers the angle in radians. The small-angle approximation is excellent here; contrast and air turbulence remain additional practical limits.

\begin{aligned}\theta_R&=1.22\frac{550\times10^{-9}\,\mathrm m}{0.20\,\mathrm m}=3.355\times10^{-6}\,\mathrm{rad}\\s&\approx L\theta_R=(2000\,\mathrm m)(3.355\times10^{-6})\\&=6.710\,\mathrm{mm}\end{aligned}

Interpret · Application

Two equally bright incoherent point sources are separated by 0.8 θR in an ideal circular-aperture image. A second image uses twice the diameter at the same wavelength. Classify both relative to Rayleigh. Does the second guarantee detection under arbitrary noise?

Show hint

Source separation stays fixed while the diffraction limit halves.

Learning goal: Distinguish diffraction limits from measurement guarantees.

Show solution and calculation

Approach: Rayleigh separation is inversely proportional to diameter.
The first image lies below the criterion; the second lies at 1.6 times its new threshold.
The second meets the ideal criterion.
Arbitrarily high noise, aberrations or changed contrast can still prevent detection; Rayleigh is not a universal measurement guarantee.

\begin{aligned}\theta_{R,2}&=\frac{\theta_{R,1}}2\\{}\frac{\Delta\theta}{\theta_{R,2}}&=\frac{0.8\theta_{R,1}}{0.5\theta_{R,1}}=1.6\end{aligned}

[1]

Error analysis · Application

A circular clear aperture has radius 0.10 m. At λ=500 nm someone inserts D=0.10 m into Rayleigh’s criterion. Find correct and incorrect limits and the error factor; assume ideal optics.

Show hint

D denotes diameter.

Learning goal: Distinguish aperture radius and diameter.

Show solution and calculation

Approach: Double the given radius first.
With the correct 0.20 m diameter, the threshold is 3.05 microradians; the incorrect substitution gives 6.10 microradians.
It predicts twice as poor angular resolution.
Both calculations have correct units, so dimensional checks alone cannot detect the radius-diameter confusion.

\begin{aligned}D&=2r=0.20\,\mathrm m\\{}\theta_R&=1.22\frac{500\times10^{-9}\,\mathrm m}{0.20\,\mathrm m}=3.05\times10^{-6}\,\mathrm{rad}\\{}\theta_{\mathrm{wrong}}&=2\theta_R\end{aligned}

[1]

What to take away

Use the same length unit for wavelength and clear diameter and interpret the ratio in radians. Rayleigh separation is a first-minimum radius, not Airy diameter. A larger aperture improves the ideal angular comparison; longer focal length alone enlarges the image. Actual resolution additionally needs contrast, sound imaging and appropriate collection.[1][2]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 2, 2026Added relevant connections; preserved existing derivations, calculations, exercises, figures and model limits.
October 1, 2026Explained Rayleigh interpretation and Airy derivation; visualized two incoherent point images; added a pixel comparison and model summary.
October 1, 2026Added prerequisites and model assumptions; clarified derivation steps and arithmetic checks; included an additional original example, diagram and summary.

Cite this reference page

Quanta. Rayleigh criterion for a circular aperture. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/rayleigh-kriterium-kreisblende

When citing a specific scientific statement, also cite the specialist source linked to that statement.

Suggest a correction with a reference

Frequently asked questions about Rayleigh criterion for a circular aperture

What does the Rayleigh criterion for a circular aperture describe?+

A point produces an Airy pattern through a clear circular aperture, with central maximum and rings. Rayleigh labels two equally bright incoherent point sources just resolved when one maximum lies at the first minimum of the other pattern. The small angular separation is approximately 1.22 times wavelength divided by aperture diameter. This is first-minimum radius, not central-disk diameter. It is a particular resolution convention, not a universal boundary below which all information disappears. Brightness contrast, coherence and noise change actual distinguishability. The calculation also assumes a uniformly illuminated unobstructed circular pupil and diffraction-limited paraxial imaging.

How do I check Rayleigh angle and Airy-disk size?+

Convert wavelength and diameter into the same length unit first.550 nm with 0.20 m gives 3.355 × 10⁻⁶rad, or 0.692018 arcseconds. Check scaling: twice the aperture halves angle, twice the wavelength doubles it. Focal length does not enter that angular calculation. For f=1.20 m, multiplying angle by f gives first-minimum focal-plane radius 4.026μm; central Airy diameter is 8.052μm. When solving a diameter requirement from angular separation, substitute the computed diameter back into the original relation. This reveals missed radian conversions and radius-versus-diameter mistakes. It checks the ideal diffraction model, not a complete real instrument.

What assumptions and limits apply to the Rayleigh criterion?+

The coefficient 1.22 belongs to a clear uniformly illuminated circular pupil and monochromatic Fraunhofer pattern. Its Rayleigh interpretation uses incoherent point images of comparable brightness. A rectangular aperture, central obstruction or apodization gives different patterns. Broadband light superposes different pattern scales; coherent sources may also interfere. Atmospheric turbulence, defocus, aberrations, motion, detector noise and sampling limit real instruments. A faint companion can remain hidden in a bright star’s diffraction rings even when separation meets the criterion. Treat the result as an optical reference, not guaranteed visibility. The small-angle approximation must also fit the geometry being modeled.

How do I check units and scale for the Rayleigh criterion?+

Wavelength and aperture diameter form a dimensionless ratio interpreted as a small angle in radians, not directly degrees. One radian is approximately 206264.806 arcseconds.600 nm with 0.30 m gives 2.44 × 10⁻⁶rad or 0.503286 arcseconds. Confusing meters and nanometers changes the answer by nine powers of ten. Also check whether λ denotes medium wavelength or vacuum wavelength. The difference is often neglected in air but matters in a higher-index medium. A requested image separation rather than angle additionally requires focal length or viewing distance and appropriate small-angle geometry. Keep those quantities separate from aperture diameter throughout the calculation.

When do I additionally need focal length or viewing distance?+

For focal-plane separation combine the angle criterion with $r_1\approx f\theta_{\min}$ ; for distant point objects use transverse separation $s\approx L\theta_{\min}$ . The 20-cm 550 nm example corresponds to 6.710 mm separation at 2000 m in the ideal model. Aperture design rearranges the relation for D;0.80 arcseconds ideally requires at least 17.30 cm at that wavelength. Microscope object resolution instead uses the appropriate numerical aperture and vacuum wavelength. The lens equation can add image position and magnification but does not automatically alter diffraction limits. Longer focal length enlarges a sensor pattern; at fixed λ, better angular resolution requires a larger appropriate clear aperture.

Quanta exam set

Curated exam set for "Rayleigh criterion for a circular aperture":

Question (front)

Which formula describes Rayleigh criterion for a circular aperture?

Answer in your set

Question (front)

How do you rearrange Rayleigh criterion for a circular aperture for Aperture diameter?

Answer in your set

Question (front)

Which common mistake happens with Rayleigh criterion for a circular aperture?

Answer in your set

+ 7 more cards: units, variables, derivation, example, exam task

The formula set contains 10 cards. After importing, review them and start practicing; your ratings determine later review intervals.

Create flashcards or a quiz I already have an account

Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

Related formulas

a\sin\theta_m=m\lambda,\quad m=\pm1,\pm2,\ldots

Minima of single-slit diffraction

More formulas: Optics

n_1 \cdot \sin(\alpha_1) = n_2 \cdot \sin(\alpha_2)

Snell's Law of Refraction

I=I_0\cos^2\theta

Malus’s law

\sin\theta_c=\frac{n_2}{n_1}\quad(n_1>n_2)

Critical angle for total internal reflection

\tan\theta_B=\frac{n_2}{n_1}

Brewster angle for polarization

Spaced-repetition apps

Compare scheduling workflows

Study evidence

Read methods, results and limits

Compare flashcard apps

FSRS support and study features

Create a formula exam set

From Essential in Quanta

Retain Rayleigh criterion for a circular aperture for exams

Create a curated FSRS exam set for θmin ≈ 1.22 λ/D: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Create flashcards or a quiz Already registered? Continue directly

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Quanta Study editorial team

Rayleigh criterion for a circular aperture

Rayleigh’s criterion sets two incoherent point-image separation to the first-minimum radius of a circular-pupil Airy pattern.

AdvancedUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

θmin

Rayleigh separation of the specified incoherent point sources, in radians

rad

λ

Wavelength in propagation medium; approximately vacuum wavelength in air

D

Clear diameter of the uniformly illuminated unobstructed circular pupil

Derivation & background

Exam blueprint

Validity range

Derivation steps

Find the circular-pupil pattern’s first minimum and use it as Rayleigh separation of incoherent points.

1Model a clear uniformly illuminated circular aperture of diameter D and monochromatic wavelength λ in the medium.
2Fraunhofer field amplitude is the two-dimensional Fourier transform of that opening.
3Circular symmetry gives $\frac{I(\theta)}{I(0)}=\left[\frac{2J_1(u)}u\right]^2$ with $u=\frac{\pi D\sin\theta}{\lambda}$ .
4The first positive zero is u≈3.831706, hence sinθ₁≈1.219670 $\frac\lambda D$ .
5For small angles in radians $\sin\theta\approx\theta$ . Round the factor to 1.22.
6Rayleigh places one incoherent point-image maximum on the other’s first minimum: $\theta_{\min}\approx1.22\frac\lambda D$ . This is a resolution convention, not a universal information limit.

How do you calculate with Rayleigh criterion for a circular aperture?

Here is how to work through a typical Rayleigh criterion for a circular aperture (θmin ≈ 1.22 λ/D) task step by step:

Task

Solution path

Approach: $\theta_R=1.22\frac\lambda D$ .
This gives 3.355 × 10⁻⁶ rad ≈ 0.692018 arcseconds.
In the focal plane $r_1\approx f\theta_R=4.026\,\mathrm{\mu m}$ ; central diameter is 8.052 μm.
Convert the required 0.80 arcseconds to radians before using $D\approx1.22\frac\lambda{\theta_R}=0.1730046\,\mathrm m$ : ideally at least about 17.30 cm.
Noise or aberrations can still prevent actual separation.

Understand and apply the equation

Before calculating

A minimum separation under a specified convention

\theta_{\min}\approx1.22\frac\lambda D

D is the clear diameter of a uniformly illuminated unobstructed circular aperture. λ is wavelength in the propagation medium; θ is calculated in radians.[1]

Two point images at Rayleigh spacing

QUANTA STUDYquanta-study.de

A cut through the circular-aperture image, not the entire two-dimensional Airy disk. The x scale uses multiples of the first-minimum angle; the combined curve adds intensities, not amplitudes.[1][2]

Free with a Quanta account. Your download stays ready after registration.

Why a circular aperture gives the factor 1.22

\begin{aligned}&\frac{I(\theta)}{I(0)}=\left[\frac{2J_1(u)}u\right]^2\\&u=\frac{\pi D}{\lambda}\sin\theta\end{aligned}

The continuous central limit is one. The first positive minimum occurs where $J_1(u)=0$ .[3]

\begin{aligned}&u_1\approx3.831706\\&\sin\theta_1=\frac{u_1}\pi\frac\lambda D\approx1.219670\frac\lambda D\end{aligned}

For small angles $\sin\theta\approx\theta$ in radians; rounding the coefficient to 1.22 gives the familiar relation.[3][1]

From radians to arcseconds and image separation

\begin{aligned}&1\,\mathrm{rad}=\frac{180}{\pi}\cdot3600\,\mathrm{arcsec}\approx206264.806\,\mathrm{arcsec}\\&r_1\approx f\theta_{\min}\end{aligned}

r₁ is the first-minimum radius in a paraxial focal plane, not its diameter.[1]

A 20-cm telescope in angle and focal plane

Two equally bright incoherent stars in an ideal model

Given: D=0.200 m; λ=550 nm in air; f=1.20 m; clear circular aperture, diffraction limited.
Find: Find Rayleigh angle, arcseconds and focal-plane separation.

Approach: For equal incoherent point images through a clear circular aperture, find Rayleigh angle and then focal-plane separation.
$\begin{aligned}\theta_{\min}&\approx1.22\frac\lambda D\\r_1&\approx f\theta_{\min}\end{aligned}$
Convert wavelength to meters and calculate angle.
$\theta_{\min}=1.22\frac{550\times10^{-9}}{0.200}=3.355\times10^{-6}\,\mathrm{rad}$
Convert to arcseconds.
$\theta_{\min}=0.692018\,\mathrm{arcsec}$
Use focal length to convert to the focal plane.
$r_1=f\theta_{\min}=4.026\,\mathrm{\mu m}$

Result: Rayleigh separation is about 0.692 arcseconds; point-image centers are 4.026μm apart. The Airy diameter between first minima is 8.052μm.

Check: Units check: $\frac\lambda D$ is dimensionless and fθ is a length. Doubling f enlarges sensor pattern and separation but does not change angular resolution when D stays fixed.

Convert a target angular separation into aperture

0.80 arcseconds at 550 nm

Given: Target separation 0.80 arcsec; λ=550 nm; equal incoherent points and ideal clear circular aperture.
Find: Find minimum diameter under the Rayleigh model.

Approach: Express target separation in radians first, then solve the circular-aperture criterion for minimum diameter.
$\begin{aligned}\theta_{\mathrm{target,rad}}&=\theta_{\mathrm{target,arcsec}}\frac{\pi}{180\cdot3600}\\D&\ge\frac{1.22\lambda}{\theta_{\mathrm{target,rad}}}\end{aligned}$
Convert arcseconds into radians first.
$\theta=\frac{0.80}{206264.806}=3.87851\times10^{-6}\,\mathrm{rad}$
Rearrange the resolution relation for D.
$D\ge\frac{1.22\lambda}{\theta}=0.173005\,\mathrm m$

Result: At least about 17.30 cm clear aperture in the ideal model. This is a diffraction requirement, not a guarantee of a real observation.

Check: Back-substitution gives 0.80 arcsec. A 20-cm system gives 0.692 arcsec and meets the ideal condition;15 cm gives 0.923 arcsec and misses it.

Explore further: f-number and sensor pixels

Compare diffraction disk with pixel size

Given: λ=550 nm in air; ideal camera lens at N=8; pixel pitch p=3.00 μm. Second comparison at N=16.
Find: Find Airy radius, diameter and diameter in pixel pitches.

Approach: use the focal-plane expression with f-number. Calculate radius first and double it for diameter.
$\begin{aligned}r_1&\approx1.22\lambda N\\d_{\mathrm{Airy}}&=2r_1\\q&=\frac{d_{\mathrm{Airy}}}{p}\end{aligned}$
Convert nanometres to micrometres and substitute N=8.
$\begin{aligned}r_1&=1.22(0.550\,\mathrm{\mu m})8=5.368\,\mathrm{\mu m}\\d_{\mathrm{Airy}}&=10.736\,\mathrm{\mu m}\\q&=\frac{10.736\,\mathrm{\mu m}}{3.00\,\mathrm{\mu m}}=3.57867\end{aligned}$
At N=16 the geometric pattern dimensions double.
$\begin{aligned}r_{1,16}&=10.736\,\mathrm{\mu m}\\d_{\mathrm{Airy},16}&=21.472\,\mathrm{\mu m}\\q_{16}&=7.15733\end{aligned}$

Result: At N=8 the central disk spans about 3.58 pixel pitches; at N=16 it spans 7.16. This is diameter between the first minima.

Check: Doubling N doubles Airy diameter at fixed wavelength. Pixel counts are only a scale comparison, not proof of a general sampling criterion or actual resolution under noise and aberrations.

Brightness contrast, atmosphere and different pupils

Five exercises checking angle, color and distance

Calculate · Foundation

D=0.30 m, λ=600 nm: find θmin in radians and arcseconds.

Show hint

Convert nanometers to meters.

Learning goal: Calculate Rayleigh angle and convert angular units.

Show solution and calculation

Approach: convert wavelength to metres, insert clear diameter in Rayleigh’s criterion and then convert from radians.
The result is 2.44 microradians or 0.503286 arcseconds.

Check: both values describe the same angle. Reporting the initial result directly as degrees would confuse angular units even though the length ratio is dimensionless.

\begin{aligned}\theta_R&=1.22\frac{600\times10^{-9}\,\mathrm m}{0.30\,\mathrm m}=2.44\times10^{-6}\,\mathrm{rad}\\\theta_R&\approx0.503286\,\mathrm{arcsec}\end{aligned}

Calculate · Application

A 15-cm system observes at 500 nm and 650 nm. How do Rayleigh angles change?

Show hint

For fixed D, θ is proportional to λ.

Learning goal: Check wavelength scaling.

Show solution and calculation

Approach: at fixed diameter Rayleigh angle is proportional to wavelength.
At 500 nm it is 4.06667 microradians or 0.838810 arcseconds; at 650 nm it is 5.28667 microradians or 1.090453 arcseconds.

Check: their ratio 1.30 equals the wavelength ratio. The longer red wavelength gives the larger ideal separation threshold.

\frac{\theta_{650}}{\theta_{500}}=\frac{650}{500}=1.30

Calculate · Application

D=20 cm, λ=550 nm; two points lie 2000 m away. What transverse separation matches ideal Rayleigh angle?

Show hint

Use s≈Lθ with θ in radians.

Learning goal: Convert angular resolution to transverse separation.

Show solution and calculation

Approach: first calculate the circular aperture’s Rayleigh angle, then use viewing distance for small transverse separation.
This gives s≈6.710 mm.

Check: $\frac{s}{L}=3.355\times10^{-6}$ recovers the angle in radians. The small-angle approximation is excellent here; contrast and air turbulence remain additional practical limits.

\begin{aligned}\theta_R&=1.22\frac{550\times10^{-9}\,\mathrm m}{0.20\,\mathrm m}=3.355\times10^{-6}\,\mathrm{rad}\\s&\approx L\theta_R=(2000\,\mathrm m)(3.355\times10^{-6})\\&=6.710\,\mathrm{mm}\end{aligned}

Interpret · Application

Two equally bright incoherent point sources are separated by 0.8 θR in an ideal circular-aperture image. A second image uses twice the diameter at the same wavelength. Classify both relative to Rayleigh. Does the second guarantee detection under arbitrary noise?

Show hint

Source separation stays fixed while the diffraction limit halves.

Learning goal: Distinguish diffraction limits from measurement guarantees.

Show solution and calculation

Approach: Rayleigh separation is inversely proportional to diameter.
The first image lies below the criterion; the second lies at 1.6 times its new threshold.
The second meets the ideal criterion.
Arbitrarily high noise, aberrations or changed contrast can still prevent detection; Rayleigh is not a universal measurement guarantee.

\begin{aligned}\theta_{R,2}&=\frac{\theta_{R,1}}2\\{}\frac{\Delta\theta}{\theta_{R,2}}&=\frac{0.8\theta_{R,1}}{0.5\theta_{R,1}}=1.6\end{aligned}

[1]

Error analysis · Application

A circular clear aperture has radius 0.10 m. At λ=500 nm someone inserts D=0.10 m into Rayleigh’s criterion. Find correct and incorrect limits and the error factor; assume ideal optics.

Show hint

D denotes diameter.

Learning goal: Distinguish aperture radius and diameter.

Show solution and calculation

Approach: Double the given radius first.
With the correct 0.20 m diameter, the threshold is 3.05 microradians; the incorrect substitution gives 6.10 microradians.
It predicts twice as poor angular resolution.
Both calculations have correct units, so dimensional checks alone cannot detect the radius-diameter confusion.

\begin{aligned}D&=2r=0.20\,\mathrm m\\{}\theta_R&=1.22\frac{500\times10^{-9}\,\mathrm m}{0.20\,\mathrm m}=3.05\times10^{-6}\,\mathrm{rad}\\{}\theta_{\mathrm{wrong}}&=2\theta_R\end{aligned}

[1]

What to take away

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 2, 2026Added relevant connections; preserved existing derivations, calculations, exercises, figures and model limits.
October 1, 2026Explained Rayleigh interpretation and Airy derivation; visualized two incoherent point images; added a pixel comparison and model summary.
October 1, 2026Added prerequisites and model assumptions; clarified derivation steps and arithmetic checks; included an additional original example, diagram and summary.

Cite this reference page

Quanta. Rayleigh criterion for a circular aperture. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/rayleigh-kriterium-kreisblende

When citing a specific scientific statement, also cite the specialist source linked to that statement.

Suggest a correction with a reference

Frequently asked questions about Rayleigh criterion for a circular aperture

What does the Rayleigh criterion for a circular aperture describe?+

How do I check Rayleigh angle and Airy-disk size?+

What assumptions and limits apply to the Rayleigh criterion?+

How do I check units and scale for the Rayleigh criterion?+

When do I additionally need focal length or viewing distance?+

Retain Rayleigh criterion for a circular aperture for exams

Create a curated FSRS exam set for θmin ≈ 1.22 λ/D: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential