Single-Slit Diffraction

Send a wide, clean beam of light at a single narrow slit and geometry predicts a boring result: a bright bar the width of the slit, sharp-edged, sitting in shadow. What you actually see on a distant screen is nothing of the kind. The slit throws a broad central band of light, far wider than itself, flanked by a fading train of weaker bright fringes separated by pitch-black nulls. Narrow the slit and the pattern widens — the opposite of what a shadow should do. This single, humble experiment is the cleanest proof that light is a wave, and the pattern it makes is the mathematical fingerprint of the aperture itself. The mechanism is interference of the slit with itself. Every point across the open strip re-radiates a Huygens wavelet, and on the way to a given point on the screen these wavelets arrive with different phases set by their differing path lengths. In the far field — where the screen is distant enough that rays to a common point are parallel — those path differences are simply linear in position across the slit, and summing the wavelets is a Fourier transform. A uniform slit is a top-hat function, and the Fourier transform of a top-hat is the cardinal-sine, the sinc. The observed intensity is its square: a $\mathrm{sinc}^2$ profile governed by the single phase variable $\beta=(\pi a/\lambda)\sin\theta$ , half the edge-to-edge phase slip across the slit. The dark fringes have an exact, intuitive cause: at those angles the slit divides into an even number of strips that cancel in pairs. The bright central lobe, where every wavelet adds in phase, towers over the rest and carries about nine-tenths of the light. This one pattern is the parent of rectangular-aperture and circular-aperture diffraction, and it fixes the resolution of every slit spectrometer, laser, and antenna.

Key takeaways

A plane wave through a long narrow slit of width $a$ produces, in the far field, an intensity $I(\theta)=I_0\,\mathrm{sinc}^2\beta$ with $\beta=(\pi a/\lambda)\sin\theta$ — the squared Fourier transform of a top-hat.
Dark fringes fall where $a\sin\theta=m\lambda$ ( $m=\pm1,\pm2,\dots$ ); the first minima at $\pm\lambda/a$ bracket a central lobe of full angular width $2\lambda/a$ .
The central maximum is twice as wide as the side lobes and holds about 90.3% of the transmitted energy; the first side lobe carries only 4.72% of the peak, the second 1.65%.
Narrowing the slit widens the pattern ( $\Delta\theta\sim\lambda/a$ ) — the diffraction analogue of the position–momentum uncertainty relation.
Key idea: the far-field amplitude is the Fourier transform of the aperture; a slit is a top-hat, so its transform is a sinc and the intensity is a $\mathrm{sinc}^2$ controlled by the single variable $\beta=(\pi a/\lambda)\sin\theta$ .

The core insight

In the far field the diffracted amplitude is the Fourier transform of the aperture's transmission. A uniform slit is a top-hat, whose Fourier transform is the cardinal-sine $\mathrm{sinc}\,\beta=\sin\beta/\beta$ , so the observed intensity is a $\mathrm{sinc}^2$ pattern. Everything — the dark fringes at $a\sin\theta=m\lambda$ , the dominant central lobe, the rapidly fading side lobes — is read off from this one function of $\beta=(\pi a/\lambda)\sin\theta$ , half the total phase difference between wavelets leaving the two edges of the slit.

Where this lives today

The single slit is the diffraction limit made concrete. In a grating or prism spectrometer the finite entrance slit imposes a $\mathrm{sinc}^2$ instrumental profile on every spectral line: even with a perfect grating, the slit width $a$ sets a floor on resolving power through the central-lobe width $\sim\lambda/a$ , forcing the classic throughput-versus-resolution compromise (a wider slit gathers more light but blurs the line). The same $\lambda/a$ governs the far-field divergence of any beam clipped by an aperture — the spreading of a laser, the beam-width of a radar dish or phased-array antenna, and the pointing budget of a free-space optical link. The pattern's slowly-decaying side lobes are a live engineering problem. High-contrast instruments — stellar coronagraphs hunting exoplanets against a star $10^{10}$ times brighter, and high-dynamic-range spectrographs — replace the hard-edged slit with a graded-transmission (apodized) aperture, trading a slightly broader central lobe for side lobes suppressed by orders of magnitude, exactly the windowing tactic used to tame spectral leakage in digital signal processing.

The physical picture

Split the slit down the middle. Pair each wavelet in the upper half with the one exactly half a slit-width below it. When the extra path travelled by the lower partner is half a wavelength, the two arrive exactly out of phase and cancel — and because every pair cancels together, the whole field vanishes. That condition, $(a/2)\sin\theta=\lambda/2$ , is the first dark fringe $a\sin\theta=\lambda$ , obtained without a single integral.

Figure

The pairing mechanism: split the slit of width a in half; a wavelet in the upper half and its partner (a/2) below travel to the screen with path difference (a/2) sinθ. When that equals λ/2 the pair cancels, giving the first dark fringe at a·sinθ = λ

At $\theta=0$ every wavelet travels an equal path and they add perfectly — the tall central peak. As the angle grows the phasors (arrows representing each wavelet's contribution) fan out and curl into a chain that closes on itself at the minima and forms shrinking loops at the secondary maxima. The whole intensity profile is the Fourier transform of the aperture, plotted below with real numbers; because a hard-edged slit is rich in high spatial frequencies, its side lobes decay only slowly, as $1/\beta^2$ .

Figure

The sinc² intensity for λ = 633 nm, a = 0.10 mm: minima at a·sinθ = mλ (±6.33, ±12.66, ±19.0 mrad), a central lobe holding 90.3% of the light, and side lobes at just 4.72% and 1.65% of the peak

Narrowing the slit spreads the pattern. Confining the wave to a region $\Delta x\sim a$ forces a spread of transverse wavevectors $\Delta k_x\sim2\pi/a$ , hence an angular spread $\Delta\theta\sim\lambda/a$ — the diffraction face of the position–momentum uncertainty relation. Smoothing (apodizing) the slit's hard edges suppresses the slowly-decaying side lobes at the cost of a broader central peak, the same trade-off as windowing in signal processing.

The core equations

The far-field intensity of a uniform slit is set entirely by one dimensionless phase variable, $\beta$ , equal to half the total phase difference between wavelets leaving the two edges. The result is the squared cardinal-sine — the transform of the top-hat aperture — normalised to its on-axis peak.

I(\theta)=I_0\,\mathrm{sinc}^2\beta=I_0\left(\frac{\sin\beta}{\beta}\right)^2,\qquad \beta=\frac{\pi a}{\lambda}\sin\theta \qquad (1)

where: $I(\theta)$ — intensity at diffraction angle $\theta$ ; $I_0$ — on-axis peak intensity; $\mathrm{sinc}\,\beta\equiv\sin\beta/\beta$ (the unnormalised convention, $=1$ at $\beta=0$ ); $a$ — slit width; $\lambda$ — wavelength. Physically, $\beta$ counts the phase fan: at $\beta=0$ all wavelets are in step ( $I=I_0$ ), and each time $\beta$ passes a multiple of $\pi$ the phasor chain closes and the field nulls. The dark fringes follow from setting the numerator of the sinc to zero away from the origin. This reproduces the pairing result and extends it to every order.

a\sin\theta_{\min}=m\lambda,\qquad m=\pm1,\pm2,\dots \qquad (2)

where: $m$ — nonzero integer order ( $m=0$ is the central maximum, not a zero); $\theta_{\min}$ — angle of the $m$ -th dark fringe. The first minima at $\sin\theta=\pm\lambda/a$ set the central-lobe half-width; the geometric shadow of ray optics is recovered only as $a\gg\lambda$ , where this angle collapses to zero. The amplitude behind $(1)$ is the coherent sum of wavelets across the slit — the Fourier transform that the far field (recall Fraunhofer Diffraction Regime) guarantees.

E(\theta)\propto\int_{-a/2}^{a/2}e^{\,ik\xi\sin\theta}\,\mathrm{d}\xi = a\,\frac{\sin\beta}{\beta} \qquad (3)

where: $E(\theta)$ — complex far-field amplitude; $k=2\pi/\lambda$ — wavenumber; $\xi$ — coordinate across the slit. Squaring $|E|^2$ gives $(1)$ . The secondary maxima do not sit halfway between the minima but slightly inward, at the roots of a transcendental equation obtained by differentiating the sinc.

\tan\beta=\beta \quad\Rightarrow\quad \beta_1=4.4934=1.4303\pi,\ \ \beta_2=7.7253,\dots \qquad (4)

where: $\beta_1,\beta_2,\dots$ — successive side-lobe peak positions. Substituting back gives relative intensities $0.0472$ , $0.0165$ , $\dots$ — the side lobes fade fast because $(\sin\beta/\beta)^2$ is bounded by $1/\beta^2$ at the peaks.

Derivation AThe sinc² pattern from the Huygens superposition

The aim of this derivation is to build the entire pattern from first principles by summing the secondary wavelets across the slit, so that the $\mathrm{sinc}^2$ shape, its normalisation, and its exact minima all emerge from a single elementary integral rather than being asserted. We illuminate the strip with a plane wave, assign each point its far-field phase, integrate, and square — the whole content of far-field single-slit diffraction in four steps.

Step 1. A strip at position $\xi$ (with $-a/2\le\xi\le a/2$ ) contributes a wavelet whose path to the screen is advanced by $\xi\sin\theta$ relative to the centre, so $\mathrm{d}E\propto e^{\,ik\xi\sin\theta}\,\mathrm{d}\xi$ . Why: in the far field the rays to a common screen point are parallel, so the only phase difference between strips is the projected path $\xi\sin\theta$ — linear in position.

Step 2. Sum coherently over the open strip: $E(\theta)=C\displaystyle\int_{-a/2}^{a/2}e^{\,ik\xi\sin\theta}\,\mathrm{d}\xi$ , with $C$ the (uniform) amplitude density. Why: coherent illumination means the wavelets add as complex amplitudes, and uniform transmission makes $C$ constant — this integral is exactly the Fourier transform of the top-hat aperture.

Step 3. Evaluate: $E=C\,\dfrac{e^{\,i\beta}-e^{-i\beta}}{ik\sin\theta}=Ca\,\dfrac{\sin\beta}{\beta}$ with $\beta=\tfrac12 ka\sin\theta=(\pi a/\lambda)\sin\theta$ . Square it: $I=|E|^2=I_0(\sin\beta/\beta)^2$ , and $I_0=|Ca|^2$ since $\sin\beta/\beta\to1$ on axis. Why: the symmetric limits turn the exponentials into a sine, and the $\beta\to0$ limit fixes the normalisation so the peak is $I_0$ .

Step 4. Read off the zeros: $\sin\beta=0$ with $\beta\ne0$ gives $\beta=m\pi$ , i.e. $a\sin\theta=m\lambda$ . Why: the amplitude vanishes whenever the phasor chain of wavelets closes into a full loop, which is exactly the pairwise-cancellation condition of the physical picture. *The integral evaluates to $\int_{-a/2}^{a/2}e^{\,ik\xi\sin\theta}\mathrm{d}\xi=a\sin\beta/\beta$ , with minima at $\beta=m\pi$ and on-axis limit $1$ .*

Go deeper · Derivation B · 1 minSide-lobe heights and the energy in the central lobe

Derivation BSide-lobe heights and the energy in the central lobe

This derivation quantifies the two facts that make the single-slit pattern so lopsided: the secondary maxima are startlingly faint, and the central lobe hoards almost all the light. The aim is to locate the side-lobe peaks by extremising the intensity, evaluate their relative heights, and then integrate the pattern to prove that roughly nine-tenths of the transmitted power lands in the central maximum — using the standard Dirichlet integral of the squared sinc.

Step 1. Extremise $I\propto(\sin\beta/\beta)^2$ : setting $\mathrm{d}/\mathrm{d}\beta\,(\sin\beta/\beta)=(\beta\cos\beta-\sin\beta)/\beta^2=0$ gives $\tan\beta=\beta$ . Its first nonzero roots are $\beta_1=4.4934$ ( $=1.4303\pi$ ) and $\beta_2=7.7253$ . Why: the peaks of the side lobes are true stationary points of the intensity, and the transcendental $\tan\beta=\beta$ pulls them slightly inward of the naive midpoints $(m+\tfrac12)\pi$ .

Step 2. Substitute back: $(\sin\beta_1/\beta_1)^2=0.0472$ and $(\sin\beta_2/\beta_2)^2=0.0165$ — the first side lobe carries $4.72\%$ of the central peak ( $-13.3$ dB), the second only $1.65\%$ . Why: at a peak $|\sin\beta|\approx1$ , so the height is close to $1/\beta^2$ , which shrinks quadratically with order — the side lobes vanish quickly.

Step 3. Integrate the power. The total is the Dirichlet integral $\displaystyle\int_{-\infty}^{\infty}(\sin\beta/\beta)^2\,\mathrm{d}\beta=\pi$ ; the central lobe is $\displaystyle\int_{-\pi}^{\pi}(\sin\beta/\beta)^2\,\mathrm{d}\beta\approx2.8362$ . Their ratio is $2.8362/\pi\approx0.9028$ . Why: power is proportional to $\int I\,\mathrm{d}\beta$ , and comparing the central span $|\beta|<\pi$ to the whole line gives the fraction of light in the central maximum directly.

Step 4. Conclude: about $90.3\%$ of the diffracted energy sits in the central lobe, leaving under $10\%$ spread across all side lobes combined. Why: this concentration is why a diffraction-limited spot or spectral line is dominated by its central peak, and why side-lobe suppression (apodization) is worth the wider core in high-contrast instruments. *The $\tan\beta=\beta$ roots are $4.4934$ and $7.7253$ , giving side-lobe heights $0.04719$ and $0.01642$ ; the Dirichlet integral $\int_{-\infty}^{\infty}\mathrm{sinc}^2\,\mathrm{d}\beta=\pi$ yields a central-lobe energy fraction of $0.9028$ .*

Go deeper · Derivation C · 1 minThe minima from half-slit pairing

Derivation CThe minima from half-slit pairing

The Fourier integral gives the minima cleanly, but the pairing argument explains why they land where they do without any calculus — the intuition worth carrying into every diffraction problem. The aim is to reproduce $a\sin\theta=m\lambda$ purely by matching each wavelet to a cancelling partner, and to see why $m=0$ is excluded.

Step 1. Divide the slit into an upper and a lower half and pair a wavelet at $\xi$ with the one at $\xi+a/2$ , exactly half a slit-width away. Their path difference to the screen is $(a/2)\sin\theta$ . Why: pairing points a fixed distance $a/2$ apart gives every pair the same path difference, so they all cancel or reinforce together.

Step 2. Cancellation requires that difference to be half a wavelength: $(a/2)\sin\theta=\lambda/2$ , i.e. $a\sin\theta=\lambda$ — the first dark fringe. Why: two wavelets half a wavelength out of phase are exactly opposite, so each upper wavelet annihilates its lower partner and the total field is zero.

Step 3. For higher orders divide the slit into $2m$ equal strips; adjacent strips cancel in pairs when $a\sin\theta=m\lambda$ . The excluded case $m=0$ is $\theta=0$ , where all paths are equal and the wavelets add — the central maximum. Why: only an even number of equal, oppositely-phased strips cancels completely, so minima require nonzero integer $m$ ; zero path difference is maximal reinforcement, not a null. *The pairing condition $a\sin\theta=m\lambda$ coincides exactly with the sinc zeros $\beta=m\pi$ for all $m\ne0$ .*

Limiting cases

Case	Result	Meaning
Wide slit, $a\gg\lambda$	first minimum $\theta_1\approx\lambda/a\to0$	pattern collapses toward the geometric beam; ray optics recovered
Narrow slit, $a=\lambda$	$\theta_1=90^\circ$	the single lobe fills the whole forward hemisphere
Sub-wavelength, $a<\lambda$	no real minima	$\sin\theta=\lambda/a>1$ has no solution — one broad lobe, near-point source
Small angles	$I\approx I_0\,\mathrm{sinc}^2(\pi a\theta/\lambda)$	minima equally spaced in angle at $\lambda/a$
Central vs first side lobe	$I_1/I_0=0.0472$	the first secondary peak is only 4.7% as bright, near $\beta=1.4303\pi$

Common trapConfusing the single-slit width formula with the double-slit spacing formula

The two look identical and mean opposite things. Single-slit minima obey $a\sin\theta=m\lambda$ , where $a$ is the slit width; double-slit (or grating) maxima obey $d\sin\theta=m\lambda$ , where $d$ is the slit separation. Same algebra, but one locates darkness and the other brightness, and they involve different lengths — a real slit system shows the fine bright grating fringes riding inside the broad single-slit envelope, with envelope minima suppressing the grating orders that fall on them (the "missing orders"). A second trap hides in $m=0$ : it is excluded from the minima because $\theta=0$ is the central maximum, where all wavelets are in phase. Write $a\sin\theta=m\lambda$ and call $m=0$ a dark fringe and you have inverted the entire pattern.

Worked examples

1 · The fringe pattern on a wall (confidence builder)

Red helium–neon light, $\lambda=633$ nm, illuminates a slit of width $a=0.10$ mm. Find the first-minimum angle, the central-lobe angular width, and the width of the central bright band on a screen $L=2$ m away.

Step 1. First dark fringe: $\sin\theta=\lambda/a=633\times10^{-9}/1.0\times10^{-4}=6.33\times10^{-3}$ , so $\theta_1=\arcsin(6.33\times10^{-3})=\mathbf{0.363^\circ}$ ( $6.33$ mrad).

Step 2. The central lobe runs from $-6.33$ to $+6.33$ mrad, an angular full width $\Delta\theta=2\lambda/a=\mathbf{12.66\ \text{mrad}}$ — twice the first-minimum angle, and twice the width of each side lobe.

Step 3. On a screen a distance $L$ away the central band has full width $L\,\Delta\theta=2.0\ \text{m}\times12.66\times10^{-3}=0.0253\ \text{m}=\mathbf{25.3\ \text{mm}}$ . A tenth-of-a-millimetre slit paints a $2.5$ cm band of light.

Step 4. Sanity: the $m=2$ minimum is at $2\lambda/a=12.66$ mrad, so the second dark fringe on the wall is at $25.3$ mm from centre and the first side-lobe peak (at $9.05$ mrad, $18.1$ mm) sits between them at $4.7\%$ brightness. First minimum $0.363^\circ$ ; central width $12.66$ mrad; band $25.3$ mm at $2$ m.

2 · How faint are the side lobes, and where does the energy go? (exam level)

For the same slit, quantify the brightness of the first two secondary maxima relative to the central peak, and find what fraction of the transmitted light lands in the central lobe.

Step 1. Side-lobe positions solve $\tan\beta=\beta$ : the first root is $\beta_1=4.4934=1.4303\pi$ , the second $\beta_2=7.7253=2.459\pi$ . In angle these are $\sin\theta=1.4303\,\lambda/a=9.05$ mrad and $2.459\,\lambda/a=15.6$ mrad.

Step 2. Their heights are $(\sin\beta_1/\beta_1)^2=\mathbf{0.0472}$ and $(\sin\beta_2/\beta_2)^2=\mathbf{0.0165}$ — the first side lobe is $4.72\%$ of the central peak ( $-13.3$ dB), the second $1.65\%$ . The pattern's dynamic range is enormous.

Step 3. Energy split: the total power is the Dirichlet integral $\int_{-\infty}^{\infty}\mathrm{sinc}^2\beta\,\mathrm{d}\beta=\pi$ , the central lobe $\int_{-\pi}^{\pi}\mathrm{sinc}^2\beta\,\mathrm{d}\beta\approx2.836$ , so the central fraction is $2.836/\pi=\mathbf{0.903}$ .

Step 4. So about $90.3\%$ of the light is in the central maximum and under $10\%$ in all side lobes together. That concentration is why a diffraction-limited beam or spectral line is dominated by its core — and why coronagraphs apodize the aperture to bury those $4.7\%$ side lobes. Side lobes $4.72\%$ and $1.65\%$ ; central lobe holds $90.3\%$ .

Physics of the portrait

The single slit's pattern is the square of one specific function — the same shape that governs sampling, filtering, and antenna theory. It deserves a portrait of its own.

Origins · 1 minsinc function and how this was discovered

sinc function

the cardinal sine sinβ/β — Fourier transform of a rectangle · Fourier optics and signal theory

The sinc function, $\mathrm{sinc}\,\beta=\sin\beta/\beta$ , is the Fourier transform of a rectangle, and that single fact places it at the centre of physics and engineering alike. It equals one at the origin (the removable singularity $\sin\beta/\beta\to1$ ), crosses zero at every nonzero multiple of $\pi$ , and oscillates with an envelope falling as $1/\beta$ , so its square — the single-slit intensity — decays as $1/\beta^2$ . Its integral is finite and clean: $\int_{-\infty}^{\infty}\mathrm{sinc}^2\beta\,\mathrm{d}\beta=\pi$ , the Dirichlet result that fixes the 90.3% central-lobe energy. Because any hard-edged aperture, gate, or sampling window is a rectangle in disguise, the sinc surfaces everywhere: as the diffraction pattern of a slit, the impulse response of an ideal low-pass filter, the interpolation kernel of the Nyquist–Shannon sampling theorem, and the far-field beam of a uniformly-illuminated antenna. Its slowly-decaying side lobes are the recurring villain — the reason instruments apodize their apertures and signal processors window their data. Two conventions coexist: the unnormalised $\sin\beta/\beta$ used in optics, and the normalised $\sin(\pi x)/(\pi x)$ favoured in signal processing, whose zeros fall at integers.

History

1665

— Francesco Maria Grimaldi's Physico-mathesis de lumine names the bending of light "diffractio"

1803

— Thomas Young's interference experiments establish the wave picture that explains diffraction

1818

— Augustin-Jean Fresnel's prize memoir puts diffraction on a quantitative wave footing

1821

— Joseph von Fraunhofer analyses the far-field diffraction of slits and gratings

1835

— Friedrich Magnus Schwerd computes single-slit fringe patterns in quantitative detail

1952

— Philip Woodward coins the name "sinc" for the cardinal-sine function in his work on radar and information theory

Going further

Texts. Hecht, Optics, 5th ed., §10.2 (Fraunhofer single-slit diffraction) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., §4.3 (Fourier-transform property) · Loudon, The Quantum Theory of Light, 3rd ed. (diffraction and uncertainty) · Foot, Atomic Physics, §7 (diffraction limits on resolution) · Demtröder, Laser Spectroscopy, Vol. 1 (spectrometer slit functions). Key results: $\int_{-a/2}^{a/2}e^{ik\xi\sin\theta}\mathrm{d}\xi=a\sin\beta/\beta$ ; minima at $\beta=m\pi$ ; side-lobe roots of $\tan\beta=\beta$ at $\beta\approx1.4303\pi$ , relative intensity $\approx0.0472$ ; the Dirichlet integral $\int_{-\infty}^{\infty}(\sin\beta/\beta)^2\mathrm{d}\beta=\pi$ and central-lobe energy fraction $\approx0.9028$ ; worked example $\theta\approx0.363^\circ$ for $\lambda=633$ nm, $a=0.10$ mm.

Prerequisites. Huygens–Fresnel Principle · Fraunhofer Diffraction Regime · Fresnel Number. Leads to. Rectangular Aperture Diffraction · Circular Aperture Diffraction and the Airy Pattern · Rayleigh Resolution Criterion · Fresnel Diffraction Regime.

Why does a narrower slit make a wider pattern — isn't that backwards?

It feels backwards only if you expect a shadow. Diffraction is a Fourier relationship: the angular spread of the light is the Fourier transform of the aperture, and a narrow function in space has a broad transform in angle. Confining the wave to a width $a$ forces a transverse-wavevector spread $\Delta k_x\sim2\pi/a$ , hence an angular spread $\Delta\theta\sim\lambda/a$ that grows as $a$ shrinks. It is the identical mathematics to the position–momentum uncertainty relation and to the fact that a short pulse has a broad spectrum. A sharp shadow is the opposite, large-aperture limit, recovered only when $a\gg\lambda$ .

What is β physically, and why does it use sinθ rather than θ?

$\beta=(\pi a/\lambda)\sin\theta$ is half the total phase difference between wavelets leaving the two edges of the slit. The full edge-to-edge path difference is $a\sin\theta$ — the projected separation of the edges along the viewing direction — which is $2\beta$ in radians of phase. It must be $\sin\theta$ , not $\theta$ , because the path difference is a geometric projection; the approximation $\sin\theta\approx\theta$ is fine for the small angles of a wide slit but fails badly once $a$ approaches $\lambda$ and the minima march toward $90^\circ$ . Keep $\sin\theta$ whenever the slit is only a few wavelengths wide.

Is the sinc here sinβ/β or sin(πx)/(πx)? The two conventions confuse me.

Both name the same shape but scale the argument differently. Optics uses the unnormalised sinc, $\mathrm{sinc}\,\beta=\sin\beta/\beta$ , whose zeros fall at multiples of $\pi$ ; then the single-slit intensity is $\mathrm{sinc}^2\beta$ with $\beta=(\pi a/\lambda)\sin\theta$ . Signal processing prefers the normalised sinc, $\mathrm{sinc}(x)=\sin(\pi x)/(\pi x)$ , whose zeros fall at integers, so the same pattern reads $\mathrm{sinc}^2(a\sin\theta/\lambda)$ . They are identical curves — just check where the argument's zeros land before trusting a formula, since a stray factor of $\pi$ silently misplaces every fringe.

Does the pattern change if the slit is illuminated non-uniformly?

Yes — the far field is always the Fourier transform of the aperture field, so changing the illumination changes the transform. A uniform (top-hat) field gives the sinc with its hard $4.7\%$ side lobes; a field that tapers smoothly to zero at the edges (a Gaussian or cosine profile) transforms into a pattern with a broader central lobe but dramatically weaker side lobes. This is apodization, and it is deliberate in high-contrast optics. The clean $\mathrm{sinc}^2$ assumes uniform, fully coherent illumination; real beams with Gaussian profiles or partial coherence give softened versions of it.

How far away must the screen be for the sinc² to be valid?

Far enough that the Fresnel number is small: $N_F=(a/2)^2/(\lambda L)\ll1$ , i.e. $L\gg a^2/(4\lambda)$ , so that the wavefront curvature across the slit is negligible and the far-field Fourier relation holds (recall Fraunhofer Diffraction Regime and Fresnel Number). For a $0.10$ mm slit at $633$ nm this needs only $L$ of order centimetres, so a metre-scale bench is comfortably in the far field. Closer in, the pattern deforms into the distance-dependent Fresnel (near-field) form, and one must keep the quadratic phase the far-field limit discards.

Why do the side lobes matter if they carry so little energy?

Because contrast, not total energy, is what limits many instruments. A first side lobe at $4.7\%$ of the peak is faint next to the central lobe but blindingly bright next to a signal you are trying to detect at $10^{-6}$ of the peak — an exoplanet beside its star, a weak spectral line beside a strong one, a faint radar return beside a strong one. The side lobes set the dynamic range, so coronagraphs, high-resolution spectrographs, and radar systems spend real effort apodizing or windowing to push them down, accepting a broader central lobe as the price of a darker background.