The Diffraction Limit

Set the diameter of the opening, the wavelength, and two points of light at some distance. The page renders the image a perfect, clear round opening makes of them, with the two points ray optics would draw as a ghost, and says whether the image has one peak or two.

Worked example — The Diffraction Limit

First dark ring of one point's image, θ1	0.0577″
That angle as a distance at the points	107.4 m
Angle between the two points, α	0.00504″
Their separation as a fraction of θ1	0.087
Wavelengths across the opening	4,363,636

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — The Diffraction Limit

The diffraction limit is the finest angular detail an opening of diameter D can show in light of wavelength λ. A point source seen through a clear round opening makes Airy's pattern, I = I₀[2J₁(v)/v]² with v = πD sin θ/λ, whose first dark ring is at sin θ₁ = 1.21967 λ/D. By Rayleigh's criterion two equal points are just resolved when each sits on the other's first dark ring; the brightness midway then falls to 73.50% of the peak. Sparrow's criterion (1916) takes the separation at which the dip between them just disappears, which for this pattern is 0.947 λ/D.

What each symbol means

D: the diameter of the opening: a pupil, a lens, a mirror
λ: the wavelength of the light
θ_1: the angle from the centre of one point's image to its first dark ring, 1.21967 λ/D
α: the angle between the two points as seen from the opening, s/d
d: the distance from the opening to the two points

Breaks when: Hubble's secondary mirror blocks the middle of its opening, and JWST's mirror is eighteen hexagonal segments, so neither makes Airy's pattern exactly: an obstructed centre narrows the core slightly and moves light out into the rings, and the segment edges add spikes. A telescope on the ground looks through moving air, which usually blurs the image far more than its mirror's diffraction limit unless adaptive optics corrects it.

Hubble cannot make out the Moon landers 🖖

Press Hubble and the lunar module. “The image of a star will not be a point but a bright circle surrounded by a series of bright rings,” George Airy wrote in a paper read to the Cambridge Philosophical Society in November 1834, and he found the first black ring where his variable reached 3.832. In today's terms that ring is at the angle θ₁ = 1.21967 λ/D, where 1.21967 is the first zero of the Bessel function J₁, 3.83171, divided by π; 83.78% of the light lies inside it. For Hubble's 2.4 m mirror, taken as a clear opening, in green 550 nm light, θ₁ is 0.0577″, and at the Moon's 384,400 km that is 107.4 m. Every point on the Moon becomes a blur that size, and points much closer together than that run into one. The landing legs of a lunar module span 9.4 m from footpad to footpad, 0.087 of the distance: both sides of the craft fall inside one blur and make a single peak. Whether their extra light could be noticed at all as a change in that blur's brightness is a question of contrast and noise, which this page does not model; what diffraction rules out is seeing the lander's shape. NASA's own caption for Hubble's 2005 pictures of the Apollo 17 landing region claims features “smaller than a soccer field”, and no more.

Magnifying a blur gives you a bigger blur 🖖

Press Hubble and the lunar module, then The same, magnified 8 times, or drag the Magnify slider. Everything grows together: the blur, its rings and the gap between the two dashed ghosts, so the shape of the image does not change and it still has one peak. A round opening passes no pattern finer than λ/D in angle. Stripes on the Moon with a period below λd/D, 88.09 m for Hubble, reach the image with no contrast at all: exactly zero, not merely faint. The two sides of the lander do not make two peaks: along the line through them, the image differs from that of one point of twice the brightness by at most 0.71% of its peak. Past the magnification at which your eye can make out the blur, more of it shows you nothing new, which is why opticians call it empty magnification.

JWST's bigger mirror is blurrier than Hubble's in the infrared 🖖

The limit depends on D/λ, the opening counted in wavelengths, and not on D alone. JWST's eighteen segments act as a single mirror about 6.5 m across, 2.71 times Hubble's, but its instruments work from 0.6 µm, at the red end of the visible, out to 28.5 µm in the infrared, and STScI's documentation says it gives diffraction-limited images at wavelengths beyond 2 µm. At 2 µm, 6.5 m holds 3,250,000 wavelengths; Hubble's 2.4 m holds 4,363,636 wavelengths of green light. So at 2 µm JWST's first dark ring is at 0.0774″, 1.34 times wider than Hubble's 0.0577″ in green, and the two are equal at 1,490 nm. Press Two points 110 m apart, which Hubble splits, then The same 110 m, JWST at 2 µm, where they merge. The table below runs the same comparison down to your eye: a 3 mm pupil holds 5,455 wavelengths of green light, and its limit is 46.1″.

Problem solved in full

How big a mirror would split the lander? 3 steps
A lunar module's landing legs span 9.4 m. How large a mirror would put its two sides one first dark ring apart, seen from Earth in green 550 nm light? The diameter field stops at 10 m, short of the answer. Press Hubble and the lunar module first: the page gives θ₁ and the 107.4 m.

Open this problem in the tool
1. The two sides are α = s/d apart. Set that equal to θ₁ = 1.21967 λ/D and solve for D:
  
  $α = \frac{s}{d} = 1.21967 \frac{λ}{D} \Rightarrow D = \frac{1 . 21967 λ d}{s} = \frac{1 . 21967 \times 550 \times 1 0 ^{- 9} \times 3 . 844 \times 1 0 ^{8}}{9 . 4} = \frac{257 . 86}{9 . 4} = 27.43 m$
2. The same from the page's own figure: Hubble's limit on the Moon is 107.4 m, 11.4 times 9.4 m, and the limit goes as 1/D:
  
  $D = 2.4 m \times \frac{107 . 4 m}{9 . 4 m} = 2.4 m \times 11.4 \approx 27.4 m$
3. In the infrared at 2 µm, D/λ must stay the same, so the mirror grows in proportion to λ:
  
  $D_{2 μ m} = 27.43 m \times \frac{2000 nm}{550 nm} = 99.75 m$
Answer

About 27.4 m, 11.4 times Hubble's mirror. At 2 µm it would take 99.8 m. Even then the lander would make two overlapping Airy patterns at Rayleigh's separation, not a picture of a spacecraft: showing its shape takes several resolution elements across it.

$D \approx 27.4 m = 11.4 \times D_{Hubble}; D_{2 μ m} \approx 99.8 m$

Learning path

Wave optics

References (3)

Airy's own derivation of the pattern a round object-glass makes of a star: “The image of a star will not be a point but a bright circle surrounded by a series of bright rings”, their sizes “inversely as the aperture”; the central spot at half its light at n = 1.616 and “a black ring, when n = 3,832” (p. 287): G. B. Airy, “On the Diffraction of an Object-glass with Circular Aperture”, Transactions of the Cambridge Philosophical Society, vol. 5, pp. 283–291 (1835; read 24 November 1834).
The Apollo 11 lunar module's size: the descent stage “an octagonal prism 4.2 meters across”, and “the distance between the ends of the footpads on opposite landing legs was 9.4 m”: NASA Space Science Data Coordinated Archive, “Apollo 11 Lunar Module / EASEP”, NSSDCA ID 1969-059C.
JWST's mirror and where it becomes diffraction-limited: the primary is “18 hexagonal segments … which, when properly phased together, act as a single mirror ~6.5 m in diameter”, and “At wavelengths λ > 2 μm, JWST obtains diffraction-limited imaging with a Strehl ratio = 0.8”: Space Telescope Science Institute, JWST User Documentation: “JWST Telescope” and “MIRI Point Spread Functions”.

By the 42Math team · Reviewed against published sources. Corrections welcome.