Shine a laser pointer through a narrow slit and catch the light on a wall across the room. You do not get a bright line the shape of the slit. You get a row of dashes: a broad blazing central band, then darkness, then a fainter band, then darkness, then fainter still — a striped pattern far wider than the slit that made it. The slit was a fraction of a millimetre; the pattern is tens of centimetres. Where does the structure come from, and why those spacings?

The answer is one of the most beautiful facts in all of physics, and it is the single idea of this page: the far-field diffraction pattern of an aperture is the Fourier transform of the aperture. The shape of the hole and the shape of the light on the distant wall are a Fourier transform pair — exactly as a musical chord and its spectrum are. A slit is a rectangle in space, so its diffraction pattern is the Fourier transform of a rectangle: the sinc function. This single sentence turns optics into a branch of Fourier analysis, and it is the foundation of Fourier optics, X-ray crystallography, and the theory of the microscope.

The far field: why "Fraunhofer"

Right up against the slit the light is a mess — a near-field jumble called Fresnel diffraction, where the distance to the screen still matters. But travel far enough away (or, equivalently, put a lens after the slit and look in its focal plane) and something clean happens: the rays reaching any one point on a distant screen are effectively parallel, all leaving the slit at the same angle \theta. This is the Fraunhofer, or far-field, regime — named for Joseph von Fraunhofer, who built the first diffraction gratings. Parallel rays are what make the maths simple, because now the only thing that distinguishes one point on the wall from another is a single angle.

In the Fraunhofer regime the field at angle \theta is a sum — an integral — over the aperture, adding up the contribution of every point of the wavefront (this is Huygens' principle again) with the phase each one has picked up by the time it reaches that far point. Write the aperture as a transmission function A(x) (here 1 inside the slit, 0 outside). The field in the far field is

Look at that integral. With the spatial frequency u = k\sin\theta = (2\pi/\lambda)\sin\theta, it is precisely the Fourier transform \tilde{A}(u)=\int A(x)e^{iux}dx of the aperture. The screen is displaying the spectrum of the hole.

The single slit, ray by ray

Before trusting the integral, build the first dark fringe by hand — it is a lovely argument. Divide the slit of width a into a top half and a bottom half. Consider light leaving at angle \theta. A ray from the very top of the slit and its partner from the middle are separated by a/2, so their extra path difference is (a/2)\sin\theta. If that equals half a wavelength, \lambda/2, the two rays arrive exactly out of phase and cancel. And every ray in the top half has such a partner in the bottom half. The whole slit annihilates itself. So the first minimum sits where

Repeating the pairing argument by cutting the slit into 4, 6, … equal parts gives the full family of minima:

The sinc pattern

Doing the integral for a rectangular slit (a constant A(x) between -a/2 and +a/2) is a one-line calculation, and out drops the Fourier transform of a rectangle — the sinc function. With \beta = \dfrac{\pi a}{\lambda}\sin\theta, the intensity on the screen is

The zeros of \sin\beta at \beta = m\pi give back exactly a\sin\theta = m\lambda — the ray argument and the integral agree. Drag the slider to change the slit width a (measured in wavelengths) and watch the pattern breathe: a wide slit gives a tight central spike with many packed side lobes; a narrow slit throws the light out into a broad, gently rippled glow.

Notice how quickly the side lobes fade: the first one peaks at only about 4.7\% of the central brightness, the next at 1.6\%. Almost all the energy is in the central band — which is why a slit still makes a recognisable bright stripe, just with a decorative fringe of echoes around it.

Worked examples

Example 1 — where is the first dark fringe? Red light of wavelength \lambda = 600\ \text{nm} passes through a slit of width a = 0.12\ \text{mm} = 1.2\times10^{-4}\ \text{m}. The first minimum is at

so \theta \approx 0.29^\circ — a tiny angle, but on a wall 3\ \text{m} away the central band is already 2\times(3)(5.0\times10^{-3}) = 3\ \text{cm} wide.

Example 2 — resolving two stars. The same maths with a circular aperture (the sinc becomes a Bessel function, and the first zero shifts to \sin\theta = 1.22\lambda/D) sets the diffraction limit of every telescope and eye. Two stars closer together than this angle blur into one. That factor of 1.22 is the whole reason astronomers crave bigger mirrors: doubling the diameter D halves the smallest angle you can split.

Example 3 — reading the pattern backwards. A slit produces a central band whose half-angular-width (centre to first dark fringe) is measured as \sin\theta = 0.0040 with \lambda = 500\ \text{nm}. Then a = \lambda/\sin\theta = 5.0\times10^{-7}/0.0040 = 1.25\times10^{-4}\ \text{m} — you have measured a slit a tenth of a millimetre wide just from the shape of its shadow. This is exactly how crystallographers measure atomic spacings they can never see directly.