Double-Slit Diffraction of Light: Young’s Experiment

The screen in Young’s experiment is far away (L ≫ d), so the rays from the two slits to one point of the screen are almost parallel. Drop a perpendicular from one slit onto the ray from the other — the remaining segment is the path difference: Δ = d·sin θ. At the centre of the screen θ = 0 and Δ = 0, so the waves arrive in phase. The path difference converts into a phase difference: φ = 2πΔ/λ.

In the simulator choose a point P on the graph — on the “path difference” panel the segment Δ is highlighted in red, and next to it you can see how much the waves from S₁ and S₂ are shifted relative to each other.

If Δ = kλ, the crests of the two waves arrive at the point at the same time, the amplitudes add, and the intensity is 4 times larger than from one slit. If Δ = (k + ½)λ, the crest of one wave arrives together with the trough of the other, and the waves cancel completely. The energy does not disappear: it is redistributed — from the dark fringes into the bright ones.

Since sin θ ≤ 1, the order of a maximum cannot exceed d/λ. For d = 5 µm and λ = 532 nm we get k < 9.4, i.e. at most 9 maxima on each side of the centre. Look at the measurements panel: it shows k_max and the angle of the first maximum.

The position of the k-th maximum on the screen is y_k = L·tan θ_k. For small angles tan θ ≈ sin θ = kλ/d, so y_k = kλL/d, and the distance between neighbouring fringes Δy = λL/d does not depend on the fringe number. This is the main formula of Young’s experiment: the fringes are wider for long waves and a distant screen, and narrower for a large slit separation.

This formula is used to measure the wavelength: λ = Δy·d/L. For example, for d = 0.2 mm, L = 2 m and Δy = 5.3 mm we get λ = 530 nm — green light. In the simulator the angles are large (the slits are micrometre-sized), so the far maxima lie slightly farther than λL/d predicts: on the measurements panel compare Δy with the exact y₁ = L·tan θ₁.

Mentally split the slit into two halves. If a·sin θ = λ, every point of the upper half has a point in the lower half whose light arrives with a path difference of λ/2 — they cancel in pairs, and it is dark in this direction. The width of the central maximum of one slit on the screen is 2L·tan θ₁, where sin θ₁ = λ/a: the narrower the slit, the wider the spot.

The two-slit interference fringes “live” inside this envelope. About 2d/a − 1 fringes fit inside the central hump. If d/a is an integer m, the maxima of orders m, 2m… fall on zeros of the envelope and disappear — on the graph they are marked with a red cross. Try d = 5 µm, a = 1 µm.

The simulator uses this formula for the screen and the graph. Two waves of equal amplitude E with a phase shift φ = 2πΔ/λ add up to a wave of amplitude 2E·cos(φ/2); the intensity is proportional to the square of the amplitude, hence 4I₀·cos²(πd·sin θ/λ). At a maximum the brightness is 4 times that of one slit, at a minimum it is zero, and on average over the screen it is exactly twice as much — energy is conserved.

The second factor describes how much light each slit sends at the angle θ. For a very narrow slit (a ≪ λ) it is almost 1, and all fringes are equally bright. For white light the simulator adds up the patterns for 16 wavelengths from 400 to 700 nm — this is how rainbow fringes appear.