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Diffraction Simulation

Far-field (Fraunhofer) diffraction from a long single slit and a circular aperture, computed in physical detector coordinates: sinc² for the slit, the Airy function [2J₁(u)/u]² for the circle, with enclosed power, absolute irradiance and a Fresnel-number check of the far-field regime.

Type exact values in the number boxes • Click or use the arrow keys on a plot to move the probe • Share the setup with “Copy link”
↔️ a = 100 µm Slit Width
λ 550 nm Wavelength
📏 1.00 m Screen Distance
📊 Single Slit Aperture Type
◐ 5.50 mm First Minimum y₁
NF 0.0045 Fresnel No. b²/(λL)

2D detector Fraunhofer

Time-averaged intensity on a flat screen at distance L, in millimetres. Dashed lines mark the computed minima. Click or use the arrow keys to move the probe (+).

Line cut through the centre I/I₀

Same data as the detector image along its horizontal axis. y = L tan θ with the exact angle; the far-field formula itself assumes NF ≪ 1.

Enclosed power

Fraction of the transmitted power that lands inside |y| < Y (slit) or r < R (circle). The curve ends at the edge of the field of view, so the power that misses the detector is reported, not renormalized away.

Apparatus schematic, not to scale

Angles are exaggerated. Wavefront brightness after the aperture follows the computed far-field intensity in that direction; the profile drawn on the screen is the line cut. The wavelets are construction lines, not a propagated field.

First minimum y₁
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Small-angle y₁
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Angle θ₁
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Central max width
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Fresnel number
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On-axis peak
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Power in central max
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Power on detector (FOV)
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Probe position
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Probe angle θ
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Probe intensity
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Dark fringes / rings on the positive side (exact flat-screen position and paraxial estimate)
m sin θm ym (mm) paraxial (mm) Δy from previous (mm)
💡 How to Use

Learn with this tool

Learning objectives

  • Predict the position of the diffraction minima from a, D, λ and L, and explain why the angular spread scales as λ/a.
  • Explain the factor 1.22 for a circular aperture from the first zero of J₁, and use it to state a resolution limit.
  • Decide from the Fresnel number whether the Fraunhofer (far-field) model applies to a given setup.

Prerequisites

  • Two-beam interference and phasor addition
  • Double-slit experiment (the single-slit envelope)
  • Huygens’ principle; small-angle approximations; complex exponentials
  • Helpful: Fourier transforms and the Bessel function J₁

Model

A monochromatic plane wave of vacuum wavelength λ illuminates the aperture at normal incidence. In the far field the complex amplitude on a screen at distance L is the Fourier transform of the aperture transmission, evaluated at spatial frequency sin θ/λ. For a flat screen, a point at transverse position y is seen at angle θ with y = L tan θ.

  • Long slit: I(θ) = I₀ sinc²β, β = π a sin θ / λ, minima at a sin θm = mλ.
  • Circle: I(θ) = I₀ [2J₁(u)/u]², u = π D sin θ / λ, dark rings at u = 3.8317, 7.0156, 10.1735, …, so sin θ₁ = 1.22 λ/D.
  • Peak values (paraxial, obliquity neglected): slit I₀ = Iinc a²/(λL); circle I₀ = Iinc A²/(λ²L²) with A = πD²/4. These make ∫I dy = Iinca (per metre of slit) and ∬I dA = IincA.
  • Validity: NF = b²/(λL) ≪ 1 with b = a/2 or D/2 (this tool warns at NF ≥ 0.1), aperture ≫ λ (scalar theory), fully coherent monochromatic illumination.
a, D
slit width, circular aperture diameter (full size)
λ
vacuum wavelength (the medium is air, n ≈ 1)
L
aperture-to-screen distance
y, r
transverse screen coordinate / radius, y = L tan θ
I₀
on-axis intensity (time-averaged irradiance)
Iinc
irradiance of the incident plane wave
NF
Fresnel number b²/(λL), b = half-width or radius
J₀, J₁
Bessel functions of the first kind
Derivation

The Fresnel–Kirchhoff integral sums secondary wavelets from each aperture point x′: U(x) ∝ ∫ t(x′) exp[ik(x − x′)²/(2L)] dx′. Expanding the square gives a phase kx′²/(2L). Its largest value, kb²/(2L) = π NF, is negligible when NF ≪ 1. Only the linear term remains, and U(x) ∝ ∫ t(x′) exp(−2πi x′ sin θ/λ) dx′, the Fourier transform of the aperture.

For a slit, t = 1 on |x′| < a/2, which gives U ∝ a sin β/β. Squaring gives sinc²β, which is zero wherever β = mπ (m ≠ 0). You can also see this with Huygens zones: at a sin θ = λ the slit splits into two halves whose wavelets cancel in pairs.

For a circle, the 2D transform of a disc of radius D/2 in polar coordinates uses ∫₀^{2π} e^{−iρ cos φ} dφ = 2πJ₀(ρ) and ∫ ρJ₀(ρ) dρ = ρJ₁(ρ). This gives U ∝ A · 2J₁(u)/u. Because 2J₁(u)/u → 1 as u → 0, the central value is finite.

Enclosed power follows by integrating the intensity: for the circle d/du[J₀² + J₁²] = −2J₁²/u, so the fraction inside u is 1 − J₀²(u) − J₁²(u) (Rayleigh). For the slit it is (2/π)[Si(2β) − sin²β/β]. At the first zero these fractions are 83.8 % and 90.3 %.

Exercise 1 — Width versus spread

  1. With λ = 550 nm and L = 1 m, where is the first minimum of a 100 µm slit? What happens to it when the width is halved?
  2. Run 100 µm Slit, then Half Width.
  3. Put the probe on the first dark fringe in each case and read y₁. Compare it with the minima table.
  4. Why does a narrower opening spread the light more?
Show answer

y₁ ≈ λL/a = 5.50 mm, and 11.0 mm for a = 50 µm. The far field is the Fourier transform of the aperture, so a feature of size a produces angular widths of order λ/a. This is the reciprocal width relation behind the uncertainty principle for photon momentum.

Exercise 2 — The factor 1.22

  1. A circular hole of diameter D = 100 µm replaces the slit. Is its first dark ring inside or outside 5.50 mm, and by what ratio?
  2. Run Airy Disk. The dashed curve is the slit of the same size.
  3. Read the first dark ring radius and compute the ratio to the slit value. Read the power in the central disk.
  4. Why is a circle of diameter D “effectively narrower” than a slit of width D?
Show answer

r₁ ≈ 6.71 mm, which is 1.2197 × 5.50 mm (= 3.8317/π). A disc has less open area near its edges than a slit of the same width. Its effective width, weighted by open area, is smaller, so the pattern is wider. The Airy disk holds 83.8 % of the power, against 90.3 % between the slit's first zeros.

Exercise 3 — Limiting case: when does “far” stop being far?

  1. Increase a at fixed λ = 500 nm, L = 100 mm. The Fraunhofer y₁ = λL/a shrinks, but the geometric shadow of the slit (half-width a/2) grows. At what width are they equal, and what is NF there?
  2. Set λ = 500 nm, L = 100 mm, a ≈ 316 µm. Then try the Near-Field Limit experiment.
  3. Compare y₁ with a/2, and watch when the Fresnel-number warning appears.
  4. What does a real screen show when NF ≳ 1?
Show answer

λL/a = a/2 gives a = √(2λL) ≈ 316 µm and NF = 1/2. Beyond this, the predicted far-field pattern would be smaller than the shadow of the aperture itself, which is impossible. A real screen shows a Fresnel pattern instead: a bright image of the aperture with edge fringes. As NF → ∞ this becomes the geometric shadow. Explore it numerically in the aperture propagation tool.

Exercise 4 — Where does the power go?

  1. Switch to absolute irradiance and hold the scale. When the circle diameter doubles, by what factor does the on-axis peak change? And the ring radius?
  2. Start from Airy Disk, pick Absolute, tick Hold intensity scale, then change D from 100 µm to 200 µm.
  3. Read the on-axis peak and y₁ before and after.
  4. Reconcile the two factors with energy conservation.
Show answer

The peak rises 16× (I₀ ∝ D⁴) and the ring radius halves. The transmitted power grows as D² (4×) and the disk area shrinks as 1/D² (¼), so the peak irradiance grows by 4 × 4 = 16. The enclosed-power fractions do not change.

Worked example — measuring a hair with a laser pointer

A red laser pointer (λ = 650 nm) illuminates a hair. On a wall L = 1.50 m away, the dark fringes are spaced 12.0 ± 0.5 mm apart near the axis. By Babinet’s principle, an opaque strand of width a gives the same pattern off-axis as a slit of width a.

  1. Small angles: Δy = λL/a, so a = λL/Δy = (650 × 10⁻⁹ m)(1.50 m)/(12.0 × 10⁻³ m) = 81 µm.
  2. Uncertainty: δa/a = δΔy/Δy = 0.5/12 ≈ 4 %, so a = 81 ± 3 µm (λ and L are assumed exact).
  3. Regime: NF = (40.5 µm)²/(650 nm × 1.5 m) ≈ 1.7 × 10⁻³ ≪ 1, so the Fraunhofer model is justified. θ₁ ≈ 8 mrad, so tan θ ≈ sin θ to 3 × 10⁻⁵.
  4. Check it here: set a = 81 µm, λ = 650 nm, L = 1500 mm. The minima table should show Δy ≈ 12.0 mm.

When the model fails

  • Near field: at NF ≳ 0.1 the quadratic aperture phase matters. At NF ≳ 1 the screen shows a Fresnel pattern (including the on-axis dark spot of a circular hole for even zone counts), not sinc² or Airy. Use the numerical propagation tool.
  • Apertures comparable to λ: scalar theory ignores polarization and the boundary conditions at the metal edges. Results for a ≲ 2λ are qualitative only.
  • Partial coherence and bandwidth: a finite source size or a broad spectrum fills in the zeros. The minima here are exact zeros only for a perfectly coherent monochromatic plane wave.
  • Finite slit length, obliquity and 1/r² fall-off: these are neglected. The absolute scale is paraxial, so it overstates the irradiance at large angles.
  • Real detectors: pixel size, noise and dynamic range limit what you can see. The log floor here is a display choice, not a detector model.

References

  • E. Hecht, Optics, 5th ed., §10.2 (Fraunhofer diffraction), §10.3 (Fresnel diffraction).
  • M. Born & E. Wolf, Principles of Optics, 7th ed., §8.5 (Fraunhofer diffraction at apertures; Airy pattern and encircled energy).
  • J. W. Goodman, Introduction to Fourier Optics, 4th ed., ch. 4 (Fresnel number and the Fraunhofer approximation).
  • B. E. A. Saleh & M. C. Teich, Fundamentals of Photonics, 3rd ed., ch. 4.
  • MIT OCW 2.71 Optics lecture notes.

📚 Physics Background

🌊 What Is Diffraction?

Diffraction is the spreading of a wave after it passes through an aperture or around an obstacle. The angular spread is of order λ/a, so it is conspicuous when the aperture is not many thousands of wavelengths wide. Huygens’ principle treats every point of the open wavefront as a source of secondary wavelets. The field at the screen is their coherent sum.

📐 Single-Slit Diffraction

Single-slit intensity

I(θ)=I0(sin⁡ββ)2I(\theta)=I_0\left(\frac{\sin\beta}{\beta}\right)^2
β=πasin⁡θλ\beta=\frac{\pi a\sin\theta}{\lambda}

Dark fringes

asin⁡θm=mλ,ym=Ltan⁡θma\sin\theta_m=m\lambda,\qquad y_m=L\tan\theta_m

Orders m = ±1, ±2, …

Example: a = 100 µm, λ = 550 nm and L = 1 m give the first zero at y₁ ≈ 5.50 mm. At β = 0 the ratio sin β/β has the finite limit 1. The first side lobes carry 4.7 % of the peak intensity.

⭕ Circular Aperture & Airy Disk

Circular-aperture intensity

I(θ)=I0(2J1(u)u)2I(\theta)=I_0\left(\frac{2J_1(u)}{u}\right)^2
u=πDsin⁡θλu=\frac{\pi D\sin\theta}{\lambda}

First dark ring

sin⁡θ1=1.2197λD\sin\theta_1=1.2197\frac{\lambda}{D}

At u = 3.8317, the first zero of J₁.

For D = 100 µm, λ = 550 nm and L = 1 m, the first dark ring has a radius of ≈ 6.71 mm, 1.22× the slit's first zero. The pattern depends only on r, so the 2D detector image is radially symmetric. The Airy disk sets the Rayleigh resolution θR = 1.22 λ/D of telescopes, microscopes and cameras.

🔭 Fraunhofer vs Fresnel

The far-field formulas require NF = b²/(λL) ≪ 1, with b the half-width or radius. This page computes only the Fraunhofer model. It reports NF and warns when NF ≥ 0.1. For example, a = 200 µm, λ = 380 nm and L = 50 mm give NF ≈ 0.53, where the true pattern differs visibly from the curve shown. A lens placed after the aperture produces the Fraunhofer pattern in its focal plane, with L replaced by the focal length f.

🔬 Applications

  • Resolution limits: the Airy disk limits telescopes, microscopes and cameras (Rayleigh criterion).
  • Particle and fibre sizing: laser diffraction measures hairs, wires and droplets (Babinet’s principle).
  • Spectroscopy: the single-slit envelope multiplies grating orders.
  • Beam divergence: any finite beam spreads at an angle of order λ/D.

These applications are described here but not simulated.