Fraunhofer Diffraction from a Single Slit

Take a slit of width a, lit at normal incidence by a plane wave, and ask for the field far away in a direction θ from the axis. Huygens and Fresnel say that every element dx of the slit radiates a wavelet, and that the far field is the sum of those wavelets with their phases kept. Far away the rays to the observation point are parallel, so an element at position x, measured from the slit centre, is ahead of the centre element by a path x sin θ and a phase kx sin θ with k = 2π/λ. Put numbers on it: a 0.10 mm slit under 633 nm light at θ = 0.20° has an edge-to-edge path difference a sin θ = 0.349 μm, which is 0.55λ, so the two extreme wavelets arrive 2π × 0.55 = 3.46 rad apart. The sum therefore runs over a continuum of equal-amplitude contributions whose phase grows linearly across the aperture, and a sum over a continuum with a linear phase is a definite integral of e^{(ikx sin θ)} dx from −a/2 to a/2.

Write the on-axis amplitude as E₀ and share it evenly, E₀ dx/a per element. Then E(θ) = (E₀/a) ∫ e^{(ikx sin θ)} dx over −a/2 ≤ x ≤ a/2. The antiderivative is e^{(ikx sin θ)}/(ik sin θ); between the limits it gives 2 sin(½ka sin θ)/(k sin θ), and dividing by a leaves sin β/β with β = ½ka sin θ = πa sin θ/λ. Check it. β is a length over a length, so it is dimensionless, as the argument of a sine must be. As θ → 0, sin β/β → 1 and E → E₀, which is the field when every wavelet is in step. Intensity is proportional to the square of the amplitude, so I(θ) = I₀ (sin β/β)²: the sinc-squared pattern. For the 0.20° example above, β is half the total phase, 1.732 rad, so I/I₀ = (sin 1.732/1.732)² = 0.325, and the point already has less than a third of the axial brightness.

Draw the same integral. Each element contributes a small phasor of equal length, rotated by k dx sin θ from its neighbour, so head to tail they form an arc of fixed length E₀ that bends through the total angle 2β = ka sin θ. The resultant is the chord, and the chord of an arc of length E₀ subtending 2β is E₀ sin β/β, the same answer by geometry. The chord vanishes when the arc closes on itself, which needs 2β = 2πm, so a sin θ = mλ with m = ±1, ±2, …. That is the pairing argument in disguise: at the first minimum an element at x and its partner at x + a/2 differ in path by (a/2) sin θ = λ/2 and cancel, and every element has such a partner. Between π and 2π the arc has curled past a full turn, and the chord points backwards: light in the first side lobe is π out of phase with the centre, which is why the amplitude there is written as negative. At m = 0 there is no closure at all; the phasors are parallel and the chord is the arc itself, the largest resultant the slit can give.

The zeros of sin β/β are at β = ±π, ±2π, …, so the central lobe spans 2π in β while every other lobe spans π: the central band is twice as wide as any side band, a signature that separates single-slit diffraction from the equally spaced fringes of two slits. On a screen at distance L the first minimum lies at y₁ = L tan θ₁ ≈ λL/a, and the central band has width w = 2λL/a. With a = 0.10 mm, λ = 633 nm and L = 1.00 m that is 12.7 mm, against 6.3 mm for each side band. The bright directions between the zeros are where dI/dβ = 0, which gives β cos β = sin β, that is tan β = β. Its roots are 4.493, 7.725 and 10.904, each a little inside the half-integer midpoints 4.712, 7.854 and 10.996, because the 1/β² envelope pulls every maximum towards the centre. The intensities there are 4.72%, 1.65% and 0.83% of I₀. Integrating sinc² over the central lobe shows it carries 90.3% of the diffracted power, and each first side lobe 2.4%.

When a < λ, the equation sin θ₁ = λ/a has no solution: the pattern has no minima at all, and as a/λ → 0 the argument β stays small for every θ, so the slit radiates almost uniformly, like a line source sending out a cylindrical wave. A 2.0 μm slit under 500 nm light is on the way there, with dark directions at sin θ = 0.25, 0.50, 0.75 and 1.00, that is at 14.5°, 30°, 48.6° and 90°. In the other limit, a ≫ λ, the central lobe narrows as 2λ/a and the light goes straight through: a 1.0 mm slit at 500 nm has a half-width of only 0.5 mrad, and geometric optics is recovered as the a/λ → ∞ limit. One caution belongs here. In the Fraunhofer regime the central band is always much wider than the slit, because w/a = 2λL/a² = 2/N_F ≫ 1; that 1.0 mm slit needs L ≫ a²/λ = 2.0 m, and at L = 20 m its central band is 20 mm wide, twenty times the slit. The sharp geometric shadow that the wide-slit limit suggests is a near-field object, and the far-field model never shows it.

The integral assumed that the phase varies linearly across the aperture. For a point at distance L from the slit, the exact distance from element x contains a quadratic correction, and the extra phase at the slit edge x = a/2 is k(a/2)²/(2L) = πa²/(4λL) = (π/4) N_F, where N_F = a²/(λL) is the Fresnel number. The linear-phase integral is trustworthy only while that correction is a small fraction of a radian, which is the statement N_F ≪ 1: the wavefront curvature across the aperture is then negligible. A 0.10 mm slit at 633 nm has a²/λ = 15.8 mm, so a screen 1.00 m away gives N_F = 0.016 and an edge phase error of 0.012 rad, and the sinc-squared form is safe. A 1.0 mm slit has a²/λ = 1.58 m and needs tens of metres, or a converging lens, which delivers the far-field pattern in its focal plane by cancelling the quadratic phase. Two further approximations sit underneath: the light is treated as a scalar, and the illumination as monochromatic and coherent across the whole slit. Where N_F is not small, the Fresnel integrals of the next topic replace this one.