Optics

The Double-Slit Pattern, Measured

By Lucian — builder & engineer, LK Forge

Two narrow slits turn a single beam of light into a row of evenly spaced bright bands. The spacing follows a formula so simple it fits on a fingernail — and it behaves backwards from what you would guess. Then, hiding in the same pattern, whole fringes go missing. We computed the interference from the double-slit simulator’s own optics and measured all of it.

 ·  6 min read  ·  exact Fraunhofer intensity, every number reproducible

5.50 mm
fringe spacing (550 nm · 0.1 mm slits · 1 m)
1 / d
closer slits make WIDER fringes
1.75×
red fringes vs violet (700 vs 400 nm)
±5, ±10
fringe orders missing — killed by the envelope

The pattern itself

The intensity on the screen is the product of two things: the fast cos² interference fringes from the two slits, and a slow sinc² diffraction envelope from each slit’s finite width. Here they are together, at the simulator’s default settings. Count the bright fringes in the central hump: nine. The tenth and eleventh — orders ±5 — should be next, but the envelope is exactly zero there, so they are simply gone.

order ±5 missingorder ±5 missing -30-20-100102030 position on screen (mm) →

Intensity = interference fringes under the dashed diffraction envelope. The gaps at ±27.5 mm are the missing 5th orders.

Closer slits, wider fringes

The spacing is Δy = λL/d, and the d in the denominator is the surprise: halve the gap between the slits and the fringes get twice as wide. It is why the effect is visible at all — you need very fine, very close slits to spread the fringes far enough to see. Below, the measured spacing against slit separation traces the hyperbola exactly.

0 3 6 9 12 11 5.5 3.667 2.75 1.833 0.050.100.150.200.250.30 slit separation d (mm) → Δy (mm)

Fringe spacing against slit separation at 550 nm, 1 m screen — a clean 1/d hyperbola. Wavelength scales it too: red (700 nm) fringes run 7.0 mm against violet’s (400 nm) 4.0 mm.

The missing orders

The interference fringes are spaced by the slit separation d; the diffraction envelope is set by the slit width a. Their ratio decides everything about which fringes survive. A bright interference order sits at d·sinθ = nλ; an envelope zero sits at a·sinθ = mλ. When those coincide — when n = m·(d/a) — the fringe falls into a dark gap and disappears. Since d/a = 5 here, every 5th order is a missing order: ±5, ±10, ±15. It is a real, testable fingerprint of the width-to-separation ratio, not a rendering quirk.

One last subtlety the measurement turned up: the envelope also nudges the surviving fringes. A fringe’s brightness peak is pulled slightly toward the centre from its exact interference maximum, because the envelope is sloping across it — we measured the first-order peak at 5.43 mm against the 5.50 mm maximum, about 1.35% inward. The spacing law Δy = λL/d is a statement about the interference maxima; the visible peaks carry this small envelope-induced shift on top.

Reproduce it yourself

The whole pattern is one line of intensity, and the two headline results fall straight out of it.

const d = 1e-4, a = 2e-5, lambda = 550e-9, L = 1; // metres

// Fraunhofer double-slit intensity at screen angle theta
function I(theta) {
  const s = Math.sin(theta);
  const beta  = Math.PI * a * s / lambda;   // single-slit
  const gamma = Math.PI * d * s / lambda;   // two-slit
  const env = beta === 0 ? 1 : (Math.sin(beta) / beta) ** 2;
  return env * Math.cos(gamma) ** 2;
}

const fringeSpacing = lambda * L / d;   // 5.5e-3 m = 5.50 mm
const missingOrders = n => n % (d / a) === 0;   // d/a = 5 -> 5, 10, 15 ...

The I(θ) and Δy = λL/d are lifted from the double-slit solver; the sweeps and peak-finding are the small script behind this post.

Make some fringes

Open the simulator, slide the slits closer, change the colour, and watch the fringes widen — then narrow the slit width and watch orders drop out.

Open the double-slit lab →
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Common questions

What sets the spacing of double-slit fringes?

The bright fringes are evenly spaced by Δy = λL/d, where λ is the wavelength, L the slit-to-screen distance and d the slit separation. For 550 nm light through slits 0.1 mm apart on a screen 1 m away, that is 5.50 mm, and the measured interference maxima confirm it to within 0.05%. Two consequences run against intuition: moving the slits closer together makes the fringes wider (Δy ∝ 1/d), and red light produces fringes about 1.75× wider than violet because its wavelength is longer.

Why do some fringes go missing in a double-slit pattern?

Each slit has a finite width a, which imposes a single-slit diffraction envelope on top of the interference fringes. Where a bright interference order lands exactly on a zero of that envelope, the fringe disappears — a "missing order". It happens whenever the slit separation d is a whole-number multiple of the slit width a: the order n = d/a and its multiples vanish. In this simulation d/a = 5, so orders ±5, ±10 and ±15 are missing, leaving 9 bright fringes under the central diffraction peak.

Does the diffraction envelope move the fringe positions?

Slightly, yes. The interference maxima sit exactly at d·sinθ = nλ, but the observed brightness peak of each fringe is pulled a little toward the centre because the envelope is falling across it. We measured the first-order brightness peak at 5.43 mm against the 5.50 mm interference maximum — about 1.35% inward. It is a second-order effect: the fringe spacing law Δy = λL/d describes the interference maxima, which is what "fringe spacing" conventionally means.