The double-slit experiment

Shine light of one colour through two narrow slits and the screen does not show two bright lines, but a whole row of bright and dark stripes. That is interference, and it shows that light behaves as a wave.

Under 15Drop two pebbles in a pond and the ripples cross. Where two bumps meet they make a bigger bump; where a bump meets a dip they cancel out and the water stays flat. Light does the same thing! Shine it through two tiny slits and you get stripes: bright where the waves add up, dark where they cancel. Try changing the colour slider: red light makes wider stripes than blue.
20+ · going deeperThe pattern is the squared modulus of the sum of two complex amplitudes, I ∝ |e^{ikr₁} + e^{ikr₂}|², times the Fraunhofer single-slit envelope sinc²(πa sinθ/λ). The fringe visibility V = (Imax − Imin)/(Imax + Imin) = 2√(I₁I₂)/(I₁ + I₂) for unequal slits (try the brightness slider), and it falls to zero if the source's coherence length is shorter than the path difference. In the quantum version, any which-path information, even if never read out, destroys the interference term.
1 · Discover2 · Visualize & experiment3 · Understand4 · Predict5 · Single slit6 · Photons7 · Test yourself

1Discover

In 1801 Thomas Young let sunlight through two pinholes and saw coloured bands. Each slit acts as a new source of waves. Where a crest from one slit meets a crest from the other, they add (constructive interference, a bright fringe). Where a crest meets a trough, they cancel (destructive interference, a dark fringe). Which one happens depends on the path difference: how much farther the light travels from one slit than from the other.

2Visualize & experiment

Top view, not to scale: wavelength and slit gap are hugely enlarged so you can see the waves. Click the waves to inspect a point.
Fringe spacing Δy = λL/d
Bright fringes seen

3Understand

A point on the screen at height y sees light from the two slits with a path difference

Δ = d sin θ ≈ d y / L

Bright fringes appear where Δ is a whole number of wavelengths (Δ = mλ), dark fringes where it is a half-number ((m + ½)λ). So the bright fringes sit at

ym ≈ m λ L / d

and neighbouring bright fringes are separated by

Δy ≈ λ L / d

Right now:

The full intensity on the screen, with both slits equally bright, is

I(θ) = 4I₀ cos²(π d sin θ / λ) · [sin β / β]²,  β = π a sin θ / λ

The cos² part is the two-slit interference; the [sin β/β]² part is the diffraction envelope from each slit's width a. The formulas ym ≈ mλL/d and Δy ≈ λL/d are small-angle, far-field approximations (L ≫ d and y ≪ L); the graph above uses the full sin θ form.

4Predict, then test

Choose an answer first, then press Try it to change the experiment and check.

5Single-slit comparison

Choose Single wide slit above and move the slit width slider. One slit on its own does not give evenly spaced fringes. Instead there is a broad bright centre, with weaker side bands, because waves from different parts of the same slit interfere. The first dark band is at sin θ = λ / a, so a narrower slit spreads the light wider. Diffraction is this spreading from one opening. Interference is the stripe pattern from two (or more) separate sources. In a real double slit you see both: fine two-slit fringes inside the single-slit envelope.

6One photon at a time Advanced · conceptual

Turn the light down until only one photon at a time passes through. Each photon lands at a single random spot, yet after many photons the same fringes appear. The probability of landing at each point follows the interference pattern, as long as nothing records which slit the photon went through. If which-path information is available, the fringes disappear: try Top only above to see a pattern with no interference stripes.

0 photons

This is a conceptual model: hits are drawn at random from the calculated intensity. It does not simulate the full quantum measurement process, and an ordinary wave animation does not by itself explain quantum mechanics.

7Test yourself