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Double-Slit Experiment

Young's double-slit experiment with physical units. Set the slit width, slit separation, wavelength and screen distance, then measure the interference fringes and the diffraction envelope on a calibrated detector, as a smooth intensity or as individual photon detections.

Fraunhofer model • lengths in nm / µm / mm / m • type exact values in the number boxes • arrow keys move the plot cursor
↕️ 2.00 mm Fringe Spacing λL/d
◌ 10.0 mm Envelope First Zero
# ±7 Orders on Screen
📊 1.00 Fringe Visibility

Apparatus schematic · not to scale

Instantaneous signed field Re ψ (diverging colours, arbitrary units) from a plane wave and two cylindrical wavelets. Only the a : d ratio is to scale; the drawn wavelength and L are illustrative. Partial coherence cannot be shown in one snapshot.

Detector screen time-averaged I

Screen seen face-on: horizontal axis is the physical screen coordinate y; the vertical direction runs along the slits, where the pattern is uniform. Colour bar gives the normalisation.

Intensity versus screen position Fraunhofer · exact angle

Solid: I(y) from one sampled array (the detector, readouts and CSV use the same samples). Dashed/dotted: upper and lower fringe envelopes. Ticks along the top mark interference orders m (orange with × = missing order on an envelope zero). Hover, drag or focus the plot and use ←/→ (Shift = ×10) to move the cursor.

Interference maxima on the screen (φ-shifted; |m| ≤ 10 listed)
m sin θ y exact (mm) y paraxial mλL/d (mm) paraxial error envelope sinc²
💡 How to use

Learn with this tool

Learning objectives

  • Predict fringe spacing, envelope zeros, missing orders and the number of real orders from λ, a, d and L, and measure them on a calibrated screen.
  • Separate the two-slit interference factor from the single-slit diffraction factor, and explain how unequal amplitudes, relative phase and partial coherence change the visibility but not the envelope.
  • Decide when the far-field (Fraunhofer) and paraxial approximations are valid, using the Fresnel number and the observation angle.

Prerequisites

  • Two-beam interference: phasor addition, I = I₁ + I₂ + 2√(I₁I₂) cos Δφ.
  • Single-slit diffraction: the sinc² pattern and its zeros a sinθ = nλ.
  • Complex exponentials and the small-angle approximation sinθ ≈ tanθ ≈ θ.

Model

Two slits of width a, centres ±d/2, lit by a normally incident monochromatic plane wave. In the far field the screen field is the Fourier transform of the aperture, so at observation angle θ = atan(y/L)

Two-slit intensity

I(θ)=I1 sinc⁡2 ⁣β[A12+A22+2A1A2∣γ∣cos⁡(δ−φ)]I(\theta)=I_1\,\operatorname{sinc}^2\!\beta\left[A_1^2+A_2^2+2A_1A_2|\gamma|\cos(\delta-\varphi)\right]
β=πasin⁡θλ,δ=2πdsin⁡θλ,sinc⁡u=sin⁡uu\beta=\frac{\pi a\sin\theta}{\lambda},\qquad\delta=\frac{2\pi d\sin\theta}{\lambda},\qquad\operatorname{sinc}u=\frac{\sin u}{u}

Equal, coherent, in-phase slits

II0=sinc⁡2 ⁣(πasin⁡θλ)cos⁡2 ⁣(πdsin⁡θλ)\frac{I}{I_0}=\operatorname{sinc}^2\!\left(\frac{\pi a\sin\theta}{\lambda}\right)\cos^2\!\left(\frac{\pi d\sin\theta}{\lambda}\right)

The peak reference intensity is I₀ = 4I₁.

λ
vacuum wavelength (the space between slits and screen is taken as vacuum/air, n = 1)
a, d
slit width; centre-to-centre separation (a < d)
L, y
slit-to-screen distance; screen coordinate (y > 0 toward the top slit)
A₁, A₂
field amplitudes of top/bottom slit, A₁ = 1, A₂ = √(I₂/I₁)
φ
phase lead of the bottom slit field relative to the top (φ > 0 shifts the fringes toward the top slit)
|γ|
modulus of the complex degree of mutual coherence of the fields at the two slits
I₀
on-axis intensity of two equal coherent in-phase slits (fixed-scale reference); I₁ = I₀/4 for one slit
NF
Fresnel number (D/2)²/(λL), D = d + a; the model needs NF ≪ 1

Assumptions and validity: scalar field; monochromatic, uniformly illuminated slits; infinitely thin opaque screen; Fraunhofer propagation (NF ≲ 0.1). The exact angle θ = atan(y/L) is used, but obliquity and 1/r fall-off are ignored, so the model is not a near-field solver. The paraxial formulas λL/d and λL/a hold only for |θ| ≲ 10°. Partial coherence enters only through the constant |γ|; spectral bandwidth and finite source size are not modelled.

Derivation

In the Fraunhofer limit the field at direction θ is E(θ) ∝ ∫ t(x) e−ikx sinθ dx, where t(x) is the aperture transmission. For one slit centred at x₀, ∫x₀−a/2x₀+a/2 e−ikx sinθ dx = a sinc(β) e−ikx₀ sinθ with β = ka sinθ/2 = π a sinθ/λ.

Two slits at x₀ = ±d/2 with amplitudes A₁, A₂e−iφ (bottom field leading by φ in the e−iωt convention) give E ∝ a sinc(β) [A₁e−iδ/2 + A₂e−iφe+iδ/2], δ = kd sinθ. Then |E|² ∝ sinc²(β)[A₁² + A₂² + 2A₁A₂ cos(δ − φ)]: the pattern shifts by φ/2π fringes toward the top slit (y > 0, at +d/2).

If the two slit fields are only partially correlated, the time average of the cross term is multiplied by |γ| (Born & Wolf §10.3), giving the model above. For A₁ = A₂ = 1, γ = 1, φ = 0: 2 + 2cos δ = 4cos²(δ/2), so I/I₀ = sinc²β cos²(δ/2).

Maxima of the cos² factor: d sinθ = mλ. Because |sinθ| < 1, only |m| < d/λ exist. Zeros of the envelope: a sinθ = nλ. When d/a is an integer p, orders m = ±p, ±2p… fall on envelope zeros and are missing. Small angles: y ≈ L sinθ gives Δy = λL/d and yzero = λL/a.

Visibility: Imax,min ∝ A₁² + A₂² ± 2A₁A₂|γ|, so V = 2A₁A₂|γ|/(A₁² + A₂²).

Exercise 1: measure a laser wavelength

  1. Set λ = 632.8 nm, d = 0.500 mm, a = 0.100 mm, L = 1.50 m. Predict the fringe spacing and the positions of the first envelope zeros.
  2. Type the values into the number boxes, set the half-width to 12 mm and look at the plot.
  3. Put the cursor on the 5th dark fringe on each side of the centre and read both y values. Divide their distance by the number of fringe spacings between them, then compute λ = dΔy/L.
  4. Why is measuring across many fringes better than measuring one? Which order is missing, and why?
Show answer

Δy = λL/d = 632.8 nm × 1.5 m / 0.5 mm = 1.898 mm; envelope zeros at λL/a = ±9.49 mm. The 5th dark fringe on each side sits at d sinθ = 4.5λ, i.e. y = ±4.5Δy = ±8.54 mm, so the two are 9 fringe spacings (17.08 mm) apart: Δy = 17.08/9 = 1.898 mm and λ = 0.5 mm × 1.898 mm / 1.5 m = 633 nm. The cursor error (about one screen pixel) is shared over nine intervals, so the relative error falls ninefold. d/a = 5, so order m = ±5 falls on the envelope zero and is missing.

Exercise 2: visibility from unequal illumination and coherence

  1. Start from Textbook. Predict the fringe visibility for I₂/I₁ = 25 % with |γ| = 1, and then with |γ| = 0.6.
  2. Set the illumination ratio and coherence sliders; switch the intensity scale to Fixed so the curves are comparable.
  3. Read the measured visibility and the peak I/I₀. Do the fringe positions or the envelope move?
  4. Explain why coherence changes the contrast but not the envelope or the fringe positions.
Show answer

A₂ = √0.25 = 0.5. V = 2(1)(0.5)(1)/(1 + 0.25) = 0.80; with |γ| = 0.6, V = 0.48. The on-axis peak is (1 + 0.25 + 2·0.5·|γ|)/4 = 0.5625 I₀ for |γ| = 1. The envelope is a property of each slit alone (sinc² factor), and the fringe phase δ − φ does not involve A or γ, so only the amplitude of the cos term changes. The readout differs slightly from theory because the envelope slopes across the central fringe.

Exercise 3 (limiting cases): how many bright fringes can exist?

  1. Load “d < λ” (λ = 500 nm, d = 0.4 µm). How many intensity maxima exist over the whole half-space? What happens if a → 0 with d fixed, or if one slit is blocked?
  2. Increase the screen half-width to its maximum (the screen then spans almost ±90°). Then block a slit.
  3. Read “Orders on screen”, the table, and the angle of the half-maximum from the cursor.
  4. Why can no screen, however large, show an m = 1 fringe here?
Show answer

A maximum needs sinθ = mλ/d; with λ/d = 1.25 only m = 0 satisfies |sinθ| < 1, so there is exactly one broad maximum. The paraxial formula λL/d would still predict a “spacing”, but it describes no real direction. As a → 0 the envelope becomes flat (sinc → 1) and the pattern tends to pure cos²(π d sinθ/λ). Blocking a slit removes the cross term, leaving the single-slit envelope at one quarter of I₀. In this regime a and d are below λ, so the scalar model is only qualitative (see the warning).

Exercise 4: photons one at a time

  1. Load “Photon build-up” (N ≈ 300). Predict how many detections are needed before the fringes are visible, if the relative scatter of a bin with n counts is 1/√n.
  2. Raise N step by step to 10⁵; press “New random sample” at each N.
  3. Compare the histogram with the dashed expected curve and note the error bars (±√n).
  4. What does the simulator assume about individual detections, and what does it not model?
Show answer

With 60 bins across ±15 mm and about 15 fringes, a bright-fringe bin needs roughly n ≳ 25 counts (20 % scatter) before bright and dark bins are clearly distinct, i.e. N of a few thousand. Each hit is drawn independently from the normalised intensity (Born rule for the screen position, conditioned on reaching the screen). Detector efficiency, dark counts, dead time and which-path markers are not modelled.

Worked example

Problem. A He–Ne laser (λ = 632.8 nm) illuminates two slits with a = 40 µm and d = 0.20 mm. The screen is at L = 2.0 m. (a) Find the fringe spacing and the first envelope zero. (b) How many bright fringes lie inside the central envelope lobe? (c) Is the Fraunhofer model justified? (d) The laser is replaced by a filtered lamp that gives |γ| = 0.35 at the slits. What visibility do you expect?

Solution. (a) Δy = λL/d = 632.8×10⁻⁹ × 2.0 / 2.0×10⁻⁴ = 6.33 mm; yzero = λL/a = 31.6 mm (θ = 0.91°, so the paraxial forms are accurate to 10⁻⁴). (b) d/a = 5, so orders m = ±5 are missing; the central lobe holds m = −4…4, i.e. 9 bright fringes. (c) D = d + a = 0.24 mm, NF = (0.12 mm)²/(632.8 nm × 2 m) = 0.011 ≪ 1: far field is fine (it would need L ≳ 0.23 m for NF ≤ 0.1). (d) Equal illumination, so V = |γ| = 0.35: minima at 0.65/1.35 ≈ 0.48 of the maxima. The fringe positions and envelope are unchanged. Load these numbers into the tool and check each answer against the readouts and the table.

When the model fails

  • Near field. For NF ≳ 0.1 the screen pattern is a Fresnel pattern of each slit, not a scaled Fourier transform. Use aperture propagation for a numerical near/far-field solution.
  • Slits comparable to λ. Scalar theory ignores polarisation and the boundary conditions at real (metal, finite-thickness) slit edges; for a ≲ λ transmission becomes polarisation dependent and the envelope shape changes.
  • Large angles. Beyond ~10° the paraxial formulas fail (the tool places orders exactly), and the neglected obliquity factor and 1/r² fall-off make real outer fringes dimmer than shown.
  • Finite bandwidth and source size. A spectral width Δλ washes out orders beyond m ≈ λ/Δλ; an extended source reduces |γ| depending on d. Here |γ| is a single constant, not computed from a source.
  • Detectors and quanta. The photon view samples positions from the classical intensity; it does not model detector efficiency, noise, or the which-path measurement apparatus.

References

  • E. Hecht, Optics, 5th ed., Pearson (2017), §9.3 (Young's experiment) and §10.2.4 (double slit).
  • M. Born and E. Wolf, Principles of Optics, 7th ed., Cambridge (1999), §7.3 and §10.3–10.4 (partial coherence, visibility).
  • B. E. A. Saleh and M. C. Teich, Fundamentals of Photonics, 3rd ed., Wiley (2019), Ch. 2, 4 and 12.
  • R. P. Feynman, The Feynman Lectures on Physics, Vol. III, Ch. 1 (quantum behaviour).
  • MIT OCW 2.71 Optics, lecture notes on Fraunhofer diffraction.

📚 Physics background

🔬 Young's double-slit experiment

In 1801 Thomas Young passed light through two closely spaced openings and observed alternating bright and dark bands, evidence that light behaves as a wave. A typical bench version uses slits a few tens of micrometres wide, a fraction of a millimetre apart and a screen about a metre away, which gives fringes a few millimetres apart.

📏 Fringe spacing and allowed orders

  • Bright fringes: d sinθ = mλ, m = 0, ±1, ±2, …; dark fringes: d sinθ = (m + ½)λ.
  • Envelope zeros: a sinθ = nλ; an order on a zero is a missing order (every 5th when d = 5a).
  • Small angles: Δy = λL/d and yzero = λL/a.

Because sinθ cannot exceed 1, only orders with |m|λ/d < 1 exist. When d < λ there is only the central order. At large angles the simulator places orders at y = L tan(asin(mλ/d)) and the table lists the paraxial error.

🌊 Visibility and coherence

Fringe visibility

V=Imax⁡−Imin⁡Imax⁡+Imin⁡=2A1A2∣γ∣A12+A22V=\frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}}=\frac{2A_1A_2|\gamma|}{A_1^2+A_2^2}

Equal, fully coherent illumination gives V = 1. Unequal amplitudes leave a nonzero minimum, and |γ| < 1 adds an unmodulated background; with |γ| = 0 the slit intensities simply add. The tool measures V from the sampled pattern over the central fringe period, which also includes the slight slope of the envelope.

🔮 Quantum interpretation

With single photons or electrons each quantum is detected at one point, yet the accumulated hits reproduce the fringes. Quantum mechanics does not say the particle went through one slit or the other: it assigns a probability amplitude to each path, and the detection probability is |ψ₁ + ψ₂|² = |ψ₁|² + |ψ₂|² + 2 Re(ψ₁*ψ₂). The cross term is the interference.

If anything records which path was taken (a detector, a polarisation tag, a recoiling slit), the cross term is multiplied by the overlap of the marker states. Full path information makes that overlap zero and the fringes vanish; partial information reduces the visibility. This is loss of coherence through entanglement with the marker, not a disturbance caused merely by “looking”. The |γ| slider plays the same mathematical role.

Scope: this page computes a classical scalar-optics intensity. The photon view draws detection positions from that intensity with a seeded random generator; it does not model the measurement apparatus or detector physics.

🔬 Applications (not simulated here)

  • Wavelength measurement: λ = dΔy/L.
  • Coherence testing: fringe visibility measures the mutual coherence of the light at the two slits.
  • Stellar interferometry: visibility versus baseline d gives the angular size of a source.
  • Gratings: many slits sharpen the same orders into narrow spectral lines.