
Optional unit · institution-dependent · Interactive lesson
Diffraction & Resolution
A flawless lens still cannot make a point look like a point. The opening that gathers the light also spreads it, and the width of that spread — not the magnification, not the pixel count — decides the finest detail an instrument can ever deliver.
UPII-14.05 · Lesson 5 of 8 in this unit. Position is not a completion score.Optional unit · institution-dependent. Your university may teach this optics or introductory modern-physics material separately. Follow your institution’s required sequence. Optional work is not a required completion gate.
Build the model
Every aperture has a smallest angle it can tell apart.
The key idea
The angular resolution of an instrument is the smallest angle between two point sources whose images it can still show as two, and for a clean circular aperture of diameter D at wavelength λ the Rayleigh criterion places it at θₘᵢₙ ≈ 1.22 λ/D radians.
Example. A 0.20 m telescope at λ = 550 nm has θₘᵢₙ = 1.22 × 550 × 10⁻⁹ ÷ 0.20 = 3.4 × 10⁻⁶ rad, about 0.69 arcsecond: two stars closer together than that arrive as one point of light.
The complete physical model and assumptions
Light that passes through an opening of finite size arrives at the image already spread out. Sum the Huygens wavelets over a circular aperture of diameter D and the image of a distant point is not a point but an Airy pattern: a bright central disk holding about 84 % of the light, ringed by faint maxima, with its first dark ring at sin θ₁ = 1.22 λ/D. The 1.22 is not a fudge factor — it is the first zero of the Bessel function J₁ divided by π, the circle's answer to the slit's 1. Two stars are mutually incoherent, so the detector adds their intensities rather than their amplitudes; push their images together and the dip between the two peaks fades until the pair reads as one blur. Rayleigh drew the line where the peak of one pattern falls on the first dark ring of the other, θₘᵢₙ ≈ 1.22 λ/D in radians, a valley at about 74 % of the peak. That is a convention, not a law, but a useful one, and everything instrumental follows from it: resolution improves with a wider aperture and a shorter wavelength, and with nothing else. The model assumes a clean unobstructed circular opening, incoherent sources and no aberration, so the moment the atmosphere, the pixel grid or the polishing is the larger error, the diffraction limit becomes a floor the instrument never reaches.
1A circular aperture turns a point into a disk
Nothing about a distant star is spread out: its wavefronts arrive flat, and a perfect lens would bring them to a point. The aperture will not allow it. Only the part of each wavefront that passes the opening survives, and summing the Huygens wavelets across a circular opening of diameter D gives an image that is a bright central disk surrounded by faint rings — the Airy pattern. About 84 % of the light lands in the central disk and 7 % in the first ring. The edge of that disk, the first dark ring, sits at sin θ₁ = 1.22 λ/D, and because λ/D is minute for real instruments the sine is almost always dropped: θ₁ ≈ 1.22 λ/D radians. For a 0.20 m telescope at λ = 550 nm that is 3.4 × 10⁻⁶ rad, a blur of angular radius 0.69 arcsecond. The blur is not a flaw to be polished away. It is the price of having a finite opening at all, and it is the single-slit calculation wrapped around a circle.
2Where the 1.22 comes from
A slit of width a puts its first zero at a sin θ = λ, so the number in front of λ/a is exactly 1. A circle gives 1.22, and the extra 22 % is pure geometry. Adding wavelets across a slit is a sum over one coordinate and produces sin u / u, whose first zero is at u = π. Adding them across a circle is a sum over a disc, and near the rim the chords of the opening are short, so the outer parts contribute less amplitude than the centre. The result is 2J₁(u)/u, where J₁ is a Bessel function of the first kind, and its first zero lies at u = 3.8317 rather than at π = 3.1416. Since u = πD sin θ/λ, the zero sits at sin θ₁ = (3.8317/π)(λ/D) = 1.2197 λ/D, which everyone rounds to 1.22. You will not be asked to evaluate a Bessel function. You should know that 1.22 is a property of the circle, and that an aperture of another shape carries a different number.
3Rayleigh's line, and what “just resolved” means
Two stars are mutually incoherent: their phases drift independently, so the interference between them averages to nothing and the detector adds intensities, not amplitudes. Slide the two Airy patterns together and their summed profile keeps a dip between the peaks — until it does not. Rayleigh's convention places the limit where the peak of one pattern falls on the first dark ring of the other, θₘᵢₙ ≈ 1.22 λ/D. At that spacing the valley between the maxima sits at about 74 % of a peak, a dip of 26 %: obvious on a good detector, marginal to the eye. Nothing physical happens at exactly that separation. Push closer and the dip shrinks smoothly to nothing near 0.95 λ/D, the Sparrow limit, where the profile first flattens; with a strong signal and a known point-spread function a separation well inside Rayleigh's line can still be measured. Treat θₘᵢₙ as an honest benchmark, not a wall.
4Only the wavelength and the aperture buy resolution
θₘᵢₙ = 1.22 λ/D carries no magnification, no focal length and no exposure time. It falls when the aperture grows and when the wavelength shrinks, and for no other reason. Take an eye whose pupil is 4.0 mm wide in ordinary room light, at λ = 550 nm: θₘᵢₙ = 1.22 × 550 × 10⁻⁹ ÷ (4.0 × 10⁻³) = 1.7 × 10⁻⁴ rad, or 35 arcseconds. At a reading distance of 0.25 m that is s = L θₘᵢₙ = 42 μm, the right order for the finest print a reader can separate — although a real eye manages only about 1 arcminute, because the spacing of the retinal cones and the lens's own aberrations give up before diffraction does. An angle becomes a length only through the distance, so multiply by L, and remember that 1 rad = 206265 arcseconds. Moving from red light at 700 nm to blue at 400 nm sharpens any instrument by a factor of 1.75 for free, which is why microscopes reach further in the ultraviolet and an electron beam further still.
5The blur that lands on the sensor
A camera does not hand you an angle; it hands you a spot at the focal plane. Multiply the angular diameter of the Airy disk, 2 × 1.22 λ/D, by the focal length f and the spot's diameter is dAiry = 2.44 λ f/D, which depends on the two lengths only through their ratio — the f-number N = f/D. At λ = 550 nm an f/8 lens spreads every point into a disk 2.44 × 550 × 10⁻⁹ × 8 = 10.7 μm wide, and an f/2.8 lens into 3.8 μm. Set that against a sensor of 3.8 μm pixels: at f/8 one blur covers nearly three pixels, so the lens sets the detail and extra pixels buy nothing, while at f/2.8 the blur has shrunk to one pixel and the sensor takes over. This is why stopping a lens down stops helping. A smaller opening tames the aberrations and deepens the depth of field, but it widens every Airy disk in proportion.
6When diffraction is not the limit
The diffraction limit is a floor, and most instruments sit well above it. A ground-based telescope looks through turbulent air that smears a star into a blob about 1 arcsecond across on a good night. A 2.4 m mirror has θₘᵢₙ = 0.058 arcsecond, so the atmosphere throws away a factor of 17 and a 0.3 m telescope on the same site resolves nearly as well — which is the case for putting that mirror in orbit, or for reshaping the wavefront hundreds of times a second with adaptive optics. Polishing errors, focus errors and a central obstruction add blur of their own. Radio astronomy has the opposite trouble: at λ = 6 cm a 25 m dish gives θₘᵢₙ = 2.9 × 10⁻³ rad, about 10 arcminutes, which is why dishes are linked into arrays whose resolution goes as λ/B for a baseline B — 30 km of baseline at that wavelength reaches 0.4 arcsecond. Before quoting 1.22 λ/D, check that diffraction really is the largest error in the instrument.
Key equations
Read the units, assumptions and limits before substituting.
Airy pattern: the first dark ring
sin θ₁ = 1.22 λ/D
Gives the width of the blur a circular opening puts on the image of a point source.
λ the wavelength in m, D the aperture diameter in m, θ₁ the angular radius of the central disk in radRayleigh criterion
θₘᵢₙ ≈ 1.22 λ/D
The smallest angle two incoherent point sources can subtend and still be seen as two.
In radians, valid while θₘᵢₙ ≪ 1 rad; multiply by 206265 to read it in arcseconds.From angle to separation on the object
sₘᵢₙ = L θₘᵢₙ = 1.22 λL/D
Turns the instrument's angular limit into the size of detail it can pick out at range L.
L the distance to the pair in m, sₘᵢₙ the smallest resolvable separation in mWhy the number is 1.22
1.22 = 3.8317/π
A slit of width a gives 1.00 λ/a instead; the extra 22 % belongs to the circular shape alone.
3.8317 is the first zero of the Bessel function J₁, the circle's counterpart of π for a slitAiry disk at the focal plane
dAiry = 2.44 λ f/D
Compare it with the pixel pitch: the larger of the two sets the finest detail a sensor can record.
f the focal length in m; with the f-number N = f/D this is 2.44 λN, a diameter in mResolution of a microscope objective
sₘᵢₙ = 0.61 λ/NA
The same 1.22 halved, for an object at the focus: immersion oil raises n and buys finer detail.
NA = n sin α, the numerical aperture; λ the vacuum wavelength in m, sₘᵢₙ in mInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Hold the gap at 9.0 μrad and widen the aperture: the valley between the peaks sinks until, at D = 75 mm, it touches the dashed 74 % line, the same moment the vertical marker — the first dark ring of the left source — lands on the right source.
Live results
- RAYLEIGH LIMIT θₘᵢₙ
- 11.18 μrad
- GAP IN UNITS OF θₘᵢₙ
- 0.80
- CENTRAL DIP
- 1 %
- SOURCE GAP AT 2.0 km
- 18 mm
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Aperture diameter D: 60 mm; Wavelength λ: 550 nm; True angular gap of the sources: 9.0 μrad
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Resolution is routinely read as a property of the eyepiece or the sensor, and it is neither. Two sources that fall inside one Airy disk have already merged in the light itself, before any lens forms an image, so enlarging that image delivers a bigger blur and nothing more — empty magnification — while a finer pixel grid only samples the same blur more often. The second error is reading D as something other than the diameter of the opening that admits the light: not the focal length, not the radius, not the eyepiece. And θₘᵢₙ is an angle in radians; it becomes a length only after multiplying by the distance to the object.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A camera lens is stopped down to a circular opening 6.0 mm across and records at λ = 600 nm. Find the angular radius of the Airy disk it makes from a distant point source, and the smallest separation of two small lamps 40 m away that the lens could still resolve.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Use θₘᵢₙ = 1.22 λ/D with λ = 600 × 10⁻⁹ m and D = 6.0 × 10⁻³ m. Both must be in metres before dividing.
- λ/D = (6.00 × 10⁻⁷) ÷ (6.0 × 10⁻³) = 1.00 × 10⁻⁴, so θₘᵢₙ = 1.22 × 1.00 × 10⁻⁴ = 1.22 × 10⁻⁴ rad. This one angle is both the angular radius of the Airy disk and the Rayleigh limit, because the criterion puts one peak on the other's first dark ring.
- The angle is far below 1 rad, so the small-angle form s = L θₘᵢₙ applies: s = 40 m × 1.22 × 10⁻⁴ = 4.88 × 10⁻³ m.
- Check the size: 1.22 × 10⁻⁴ rad × 206265 = 25 arcseconds, about 36 times coarser than a 0.20 m telescope at 550 nm, which is what a 6.0 mm opening should give.
Answer. θₘᵢₙ = 1.22 × 10⁻⁴ rad (25 arcseconds), and the lamps must be at least 4.9 mm apart at 40 m.
MediumExample 2
An imaging satellite flies 520 km above the ground carrying a telescope of aperture 0.35 m that works at λ = 520 nm. Find the smallest ground separation it can resolve if diffraction is the only limit, then find the aperture that would be needed to reach 0.10 m from the same altitude and wavelength.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- θₘᵢₙ = 1.22 λ/D = 1.22 × (520 × 10⁻⁹) ÷ 0.35 = 1.81 × 10⁻⁶ rad.
- sₘᵢₙ = L θₘᵢₙ = (5.20 × 10⁵ m) × (1.81 × 10⁻⁶) = 0.94 m.
- For the second part invert sₘᵢₙ = 1.22 λL/D: D = 1.22 λL/sₘᵢₙ = 1.22 × (520 × 10⁻⁹) × (5.20 × 10⁵) ÷ 0.10 = 3.3 m.
- Check the scaling: sₘᵢₙ ∝ 1/D, so improving 0.94 m to 0.10 m by a factor of 9.4 demands 9.4 times the aperture, and 0.35 × 9.4 = 3.3 m agrees. Mirror mass grows roughly as D³, so that is a factor of about 830 in glass.
Answer. θₘᵢₙ = 1.81 × 10⁻⁶ rad, giving sₘᵢₙ ≈ 0.94 m; reaching 0.10 m needs D ≈ 3.3 m.
HardExample 3
An amateur telescope has an aperture of 0.28 m and a focal length of 2.20 m, and images at λ = 550 nm onto a sensor whose pixels are 3.8 μm across. The site's atmosphere blurs a star to 1.4 arcseconds. Find the Airy disk diameter at the focal plane and its width in pixels, compare the diffraction angle with the seeing, and give the width of the real star image in pixels.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Airy disk at the focal plane: dAiry = 2.44 λ f/D = 2.44 × (550 × 10⁻⁹) × 2.20 ÷ 0.28 = 1.05 × 10⁻⁵ m = 10.5 μm, which is 10.5 ÷ 3.8 = 2.8 pixels across. Its radius is 1.4 pixels, a little coarser than the two-pixels-per-Airy-radius sampling rule.
- Diffraction angle: θₘᵢₙ = 1.22 × (550 × 10⁻⁹) ÷ 0.28 = 2.40 × 10⁻⁶ rad. In arcseconds, 2.40 × 10⁻⁶ × 206265 = 0.49 arcsecond.
- Compare: the seeing of 1.4 arcseconds is 2.8 times the diffraction angle, so the atmosphere, not the aperture, sets the blur. The telescope never reaches its own 1.22 λ/D limit on this site.
- Real image width: the seeing dominates, so take the star image as the seeing disk. The seeing angle in radians is 1.4 ÷ 206265 = 6.79 × 10⁻⁶ rad, and multiplying by f gives 6.79 × 10⁻⁶ × 2.20 = 1.49 × 10⁻⁵ m = 14.9 μm, which is 14.9 ÷ 3.8 = 3.9 pixels.
- So the pixels sample the seeing disk adequately. Buying a finer sensor would record the same blob in more pixels; only a better site, or adaptive optics, moves the limit.
Answer. dAiry = 10.5 μm, about 2.8 pixels; θₘᵢₙ = 0.49 arcsecond against 1.4 arcseconds of seeing, so the atmosphere limits, and the star image is about 14.9 μm, near 3.9 pixels wide.
Quick check
Select an answer and explain it before checking.
Review & continue
The angular resolution of an instrument is the smallest angle between two point sources whose images it can still show as two, and for a clean circular aperture of diameter D at wavelength λ the Rayleigh criterion places it at θₘᵢₙ ≈ 1.22 λ/D radians.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.