Single-Slit Diffraction

Young's double slit needed two openings to make its fringes. But send light through a single narrow slit and something just as striking happens: instead of a sharp bright line the shape of the slit, you get a broad, bright central band with dimmer bands fading away on either side. The wave has bent around the edges of the opening and spread — this is diffraction, the same edge-bending that let the wave "leak" sideways in Huygens' picture, now made into a measurable pattern. Narrow the slit and the pattern spreads wider; that counter-intuitive inverse relationship is the heart of the topic and the reason a camera stopped down too far actually loses sharpness. The mechanism is pure Huygens. Every point across the open slit is a source of secondary wavelets, and at any angle off-axis those wavelets arrive with a spread of path differences. Straight ahead they are all in step and build the bright central maximum. But at the special angle where the wavelets from the top edge and the bottom edge of the slit are exactly one wavelength out of step, the slit can be split into pairs that each cancel — and the light goes dark. That condition, $a\sin\theta = m\lambda$ , marks the minima, and it is the mirror image of the double-slit rule: where two slits gave interference maxima at $d\sin\theta = n\lambda$ , one slit gives diffraction minima at $a\sin\theta = m\lambda$ . The single most-tested trap in this chapter is keeping those two straight. The central maximum is special: it runs from $-\lambda/a$ to $+\lambda/a$ , an angular width $2\lambda/a$ — twice as wide as every side band and far brighter, carrying most of the light. This article derives the minima by pairing wavelets, pins down why the central band is double-width, and works the numbers: a tenth-of-a-millimetre slit throws a $12\ \mathrm{mm}$ central band on a screen a metre away, and halving the slit doubles it.

Key takeaways

A wave through a single slit of width $a$ spreads into a broad, bright central maximum flanked by dimmer side maxima — diffraction.
Minima occur at $a\sin\theta = m\lambda$ for $m = \pm1, \pm2, \dots$ ; note $m = 0$ is not a minimum — it is the bright centre.
The central maximum spans $-\lambda/a$ to $+\lambda/a$ : angular width $2\lambda/a$ , linear width $2\lambda D/a$ on a screen a distance $D$ away.
The central band is twice as wide as each side band, and much brighter — most of the energy is in it.
A narrower slit gives a wider pattern (more spreading); halving $a$ doubles the central width.
Contrast with the double slit: single-slit $a\sin\theta = m\lambda$ gives diffraction minima; double-slit $d\sin\theta = n\lambda$ gives interference maxima.
Key idea: splitting a single slit into wavelet pairs shows the light goes dark at $a\sin\theta = m\lambda$ , leaving a central maximum of angular width $2\lambda/a$ that widens as the slit narrows.

The core insight

Wavelets from across a single slit of width $a$ cancel in pairs at $a\sin\theta = m\lambda$ (the minima, with $m = 0$ the bright centre, not a minimum), producing a central maximum of angular width $2\lambda/a$ — twice each side band and brightest — that spreads wider as the slit narrows.

Where this lives today

Single-slit diffraction sets the ultimate sharpness of every optical instrument. Because any lens or mirror is a finite aperture, the light it collects diffracts, and the smallest detail it can resolve is fixed by exactly this spreading — the reason astronomers build ever-larger telescope mirrors and microscopists reach for shorter wavelengths. The same physics is why a laser beam, however tightly you aim it, unavoidably fans out over long distances (its divergence is diffraction from the beam's own width), and why a satellite dish or a phone antenna has a beam width set by its size relative to the wavelength. Photographers meet it as the "diffraction limit": stop a lens down too far and the shrinking aperture widens each point's diffraction blur until the image softens. Diffraction gratings — Fraunhofer's descendants — spread light into spectra to identify the elements in a flame, a star, or a chemical sample, and the pits on a CD act as a reflection grating, which is why the disc shimmers with rainbow colours. The next article turns this spreading into a precise limit on resolution.

The physical picture

Take a single slit of width $a$ , illuminated by parallel monochromatic light — plane wavefronts marching straight at the opening. By Huygens' principle, every point across the slit becomes a source of secondary wavelets that spread forward into the space beyond. On a distant screen (or in the focal plane of a lens, which collects the parallel rays that leave the slit at each angle), those wavelets from all across the slit combine. This is the Fraunhofer, or far-field, regime: the rays leaving the slit at a given angle $\theta$ are treated as parallel, and their interference builds the diffraction pattern. Look straight ahead, $\theta = 0$ . Every wavelet has travelled the same distance to the screen, so they all arrive in step and add to a bright maximum — the central maximum. Now tip the viewing angle up. The wavelet from the top of the slit now has a little farther to go than the one from the bottom; across the slit there is a spread of path differences from zero (at one edge) up to $a\sin\theta$ (at the other). For small angles the wavelets are still roughly in step and the light stays bright, but as $\theta$ grows the spread widens until, at one special angle, the contributions cancel exactly and the screen goes dark. Increase the angle further and a weaker side maximum appears, then another minimum, and so on — a bright centre with a train of fading side bands. The whole pattern is the interference of a continuum of wavelets from one opening, not two separate beams.

Figure

Single-slit diffraction by wavelet pairing — at the angle where the top and bottom edges are one wavelength out of step, the slit splits into pairs that cancel, giving the first minimum

The minima: pairing wavelets across the slit

The condition for darkness comes from a beautiful pairing argument. Split the slit down the middle into a top half and a bottom half, each of width $a/2$ . Pair every point in the top half with the point exactly $a/2$ below it in the bottom half. The path difference between the two members of a pair, at angle $\theta$ , is $(a/2)\sin\theta$ . If that pair-path is half a wavelength, the two wavelets arrive exactly out of step and cancel — and since every pair has the same path difference, the whole slit cancels. Setting $(a/2)\sin\theta = \lambda/2$ gives the first minimum at $a\sin\theta = \lambda$ . Dividing the slit into $2m$ equal strips and pairing within them generalises this to

a\sin\theta = m\lambda, \qquad m = \pm1, \pm2, \pm3, \dots \qquad (1)

where: $a$ is the slit width, $\theta$ the angle from the forward direction, $\lambda$ the wavelength, and $m$ the order of the minimum. The physical reading — and the biggest trap in the chapter — is that equation (1) locates the dark fringes, not the bright ones, and that $m = 0$ is deliberately excluded: at $\theta = 0$ all wavelets are in step, so the centre is the brightest point, not a minimum. Contrast this sharply with the double slit, where $d\sin\theta = n\lambda$ located the bright maxima: two point-like slits interfering give maxima at whole wavelengths, whereas one extended slit self-cancelling gives minima at whole wavelengths. Put in numbers: slit width $a = 0.1\ \mathrm{mm} = 1\times10^{-4}\ \mathrm{m}$ , wavelength $\lambda = 600\ \mathrm{nm}$ . The first minimum is at

\sin\theta = \frac{\lambda}{a} = \frac{600\times10^{-9}}{1\times10^{-4}} = 6\times10^{-3}, \qquad \theta = 0.344^\circ.

The angle is tiny — diffraction from a slit far wider than the wavelength is a gentle spreading — but on a screen a distance $D = 1\ \mathrm{m}$ away it lands at a very readable position, $y_1 = D\tan\theta \approx D\lambda/a = 6\ \mathrm{mm}$ from the centre.

The central maximum: twice as wide, and brightest

The central maximum stretches from the first minimum on one side to the first minimum on the other — from $\sin\theta = -\lambda/a$ to $\sin\theta = +\lambda/a$ . Its angular width (small angles) is therefore

\Delta\theta_{\text{central}} = \frac{2\lambda}{a} \qquad (2)

and its linear width on a screen a distance $D$ away is

w_{\text{central}} = 2D\tan\theta \approx \frac{2\lambda D}{a} \qquad (3)

where: the factor of two is because the central band spans first minima on both sides. The side maxima, by contrast, each lie between two consecutive minima — between $\sin\theta = \lambda/a$ and $2\lambda/a$ , for example — so each side band is only $\lambda/a$ wide in angle, exactly half the central width. The central maximum is thus twice as wide as every side band, and it also carries the great majority of the light; the first side maximum peaks at only about $4.5\%$ of the central intensity, the value $(2/3\pi)^2$ quoted from the full intensity envelope (the sinc-squared function, whose closed-form derivation is a university-track calculation we do not perform here). For the standard numbers, $a = 0.1\ \mathrm{mm}$ , $\lambda = 600\ \mathrm{nm}$ , $D = 1\ \mathrm{m}$ :

w_{\text{central}} = \frac{2\lambda D}{a} = \frac{2(600\times10^{-9})(1)}{1\times10^{-4}} = 12\times10^{-3}\ \mathrm{m} = 12\ \mathrm{mm}, \qquad \Delta\theta_{\text{central}} = \frac{2\lambda}{a} = 0.012\ \mathrm{rad}.

Now the counter-intuitive part. Because $a$ is in the denominator, a narrower slit gives a wider pattern. Halve the slit to $a/2 = 0.05\ \mathrm{mm}$ and the central width doubles to $24\ \mathrm{mm}$ : squeezing the light through a smaller hole makes it spread more, not less. This is diffraction's fundamental trade-off — you cannot confine a wave in space without spreading its directions — and it is why a very small aperture, far from sharpening an image, blurs it.

Figure

The single-slit intensity envelope — a tall central maximum of angular width 2 lambda over a, flanked by side maxima only half as wide and about 4.5 percent as tall

Derivation AThe single-slit minima from pairing wavelets

The goal is to derive the minimum condition $a\sin\theta = m\lambda$ from Huygens' wavelets alone, by the pairing (or "divide into strips") argument, and to see clearly why $m = 0$ is excluded. This is the standard high-school and competitive-exam derivation; the full intensity envelope is a university-track result we do not need. Only geometry and the half-wavelength cancellation rule are used, no calculus.

Step 1. Set up the wavelets. Parallel light fills the slit of width $a$ ; by Huygens each point across the slit emits a secondary wavelet. Consider the wavelets leaving at angle $\theta$ to the forward direction. The path difference between the wavelet from one edge and the wavelet from the other edge is $a\sin\theta$ (the extra distance the far edge's ray must cover), and points in between have path differences spread evenly from $0$ to $a\sin\theta$ .

Step 2. Pair the wavelets for the first minimum. Divide the slit into two halves, top and bottom, each of width $a/2$ . Pair each point in the top half with the point $a/2$ directly below it. Every such pair has the same path difference, $(a/2)\sin\theta$ , because the two members are always $a/2$ apart across the slit.

Step 3. Impose cancellation. If $(a/2)\sin\theta = \lambda/2$ , the two members of every pair are exactly half a wavelength out of step and cancel. Since this holds for all pairs simultaneously, the entire slit's contribution vanishes: total darkness. That first minimum is at $a\sin\theta = \lambda$ , i.e. $m = 1$ .

Step 4. Generalise and exclude $m = 0$ . Dividing the slit into $2m$ equal strips and pairing within adjacent strips gives cancellation whenever $a\sin\theta = m\lambda$ for $m = \pm1, \pm2, \dots$ — equation (1). The value $m = 0$ ( $\theta = 0$ ) cannot be a minimum: there all wavelets travel equal paths, arrive in step, and add to the brightest point, the central maximum. With $a = 0.1\ \mathrm{mm}$ , $\lambda = 600\ \mathrm{nm}$ : first minimum at $\sin\theta = \lambda/a = 6\times10^{-3}$ , $\theta = 0.344^\circ$ , landing $D\lambda/a = 6\ \mathrm{mm}$ from centre on a screen $1\ \mathrm{m}$ away. Verified: $\sin\theta_1 = \lambda/a = 6\times10^{-3}$ , $\theta_1 = 0.344^\circ$ , $y_1 = D\lambda/a = 6\ \mathrm{mm}$ — recomputed in verify_hs2-single-slit-diffraction.py.

Go deeper · Derivation B · 1 minWhy the central maximum is twice as wide as the side bands

Derivation BWhy the central maximum is twice as wide as the side bands

The goal is to show that the central maximum has angular width $2\lambda/a$ — exactly twice the $\lambda/a$ width of each side band — and to compute its linear width on a screen, then use it to explain why a narrower slit spreads the light more. The reasoning is bookkeeping of the minima positions from equation (1); no calculus is needed.

Step 1. Locate the minima that bound the central maximum. The nearest minima to the centre are the $m = \pm1$ minima at $\sin\theta = \pm\lambda/a$ . The central maximum is the bright region between them, so it runs from $-\lambda/a$ to $+\lambda/a$ in $\sin\theta$ .

Step 2. Read off the angular width. For small angles $\sin\theta \approx \theta$ , so the central maximum spans from $\theta = -\lambda/a$ to $\theta = +\lambda/a$ , an angular width $\Delta\theta_{\text{central}} = 2\lambda/a$ — equation (2). On a screen a distance $D$ away, the linear half-width is $D\lambda/a$ , so the full width is $w = 2\lambda D/a$ .

Step 3. Compare with a side band. A side maximum lives between two consecutive minima — for the first one, between $\sin\theta = \lambda/a$ ( $m=1$ ) and $2\lambda/a$ ( $m=2$ ). That interval is only $\lambda/a$ wide in angle, so each side band is exactly half the width of the central band. The central maximum is therefore twice as wide as the side maxima, as well as far brighter (the first side peak is about $4.5\%$ of the centre).

Step 4. Compute and interpret. For $a = 0.1\ \mathrm{mm}$ , $\lambda = 600\ \mathrm{nm}$ , $D = 1\ \mathrm{m}$ : $w = 2\lambda D/a = 12\ \mathrm{mm}$ and $\Delta\theta = 2\lambda/a = 0.012\ \mathrm{rad}$ . Because $a$ is in the denominator, halving the slit to $0.05\ \mathrm{mm}$ doubles the central width to $24\ \mathrm{mm}$ : a narrower slit spreads the light more. Verified: central width $2\lambda D/a = 12\ \mathrm{mm}$ , angular width $2\lambda/a = 0.012\ \mathrm{rad}$ , and halving $a$ multiplies the width by $2$ — recomputed in verify_hs2-single-slit-diffraction.py.

Limiting cases

case	result	meaning
$\theta = 0$ ( $m = 0$ )	brightest point	all wavelets in step — central maximum, not a minimum
first minima $\sin\theta = \pm\lambda/a$	edges of the central band	central angular width $2\lambda/a$
central band width	$2\lambda D/a$	twice each side band, carries most of the light
halve the slit width $a$	pattern width doubles	narrower slit spreads light more
$a \gg \lambda$	tiny spread, near geometric shadow	wide slit barely diffracts
$a \to \lambda$	huge spread, first minimum near $90^\circ$	slit as narrow as a wavelength floods the screen
first side maximum	$\approx 4.5\%$ of central	side bands are dim; energy is in the centre

Common trapa sinθ = mλ is a minimum, and a is the slit width, not the slit separation

Three confusions dominate single-slit problems. First and worst: $a\sin\theta = m\lambda$ gives the minima (dark bands), not the maxima. This looks identical in form to the double-slit maximum condition $d\sin\theta = n\lambda$ , but the physics is opposite — two slits give bright fringes at whole wavelengths, one slit self-cancels to dark fringes at whole wavelengths. Second, $m = 0$ is not a dark fringe; it is the bright central maximum, which is why $m$ starts at $\pm1$ . Third, the length in the formula is the slit width $a$ , not the slit separation $d$ of the double slit — putting $d$ here (or $a$ in the double-slit formula) is a classic mix-up. And remember the direction of the spreading: narrower slit, wider pattern — the opposite of most students' first guess.

Worked examples

Board level — first minimum and central width (CBSE and NEET pattern)

Monochromatic light of wavelength $600\ \mathrm{nm}$ passes through a single slit of width $0.1\ \mathrm{mm}$ , and the pattern is observed on a screen $1\ \mathrm{m}$ away. Find (a) the angle of the first minimum, (b) its distance from the centre on the screen, and (c) the width of the central maximum.

Step 1. First minimum from equation (1) with $m = 1$ : $\sin\theta = \lambda/a = (600\times10^{-9})/(1\times10^{-4}) = 6\times10^{-3}$ , so $\theta = \boxed{0.344^\circ}$ — a small angle, so $\sin\theta \approx \theta$ throughout.

Step 2. Its position on the screen: $y_1 = D\tan\theta \approx D\lambda/a = (1)(600\times10^{-9})/(1\times10^{-4}) = 6\times10^{-3}\ \mathrm{m} = \boxed{6\ \mathrm{mm}}$ from the centre.

Step 3. The central maximum runs from the first minimum on one side to the first minimum on the other, so its width is $2y_1 = 2\times6 = \boxed{12\ \mathrm{mm}}$ , equivalently $2\lambda D/a$ .

Step 4. Sense-check against the side bands: each side maximum is only $\lambda D/a = 6\ \mathrm{mm}$ wide, half the central band, and peaks at about $4.5\%$ of the central brightness. The central band dominates the pattern.

Competitive level — narrowing the slit (JEE Advanced pattern)

In the setup above the central maximum is $12\ \mathrm{mm}$ wide. The slit is now replaced by one half as wide, $0.05\ \mathrm{mm}$ , everything else unchanged. Find the new central width, and state the general rule relating slit width to pattern width. Would a wider slit sharpen or blur a projected image?

Step 1. Central width scales as $2\lambda D/a$ , inversely with $a$ . Halving $a$ from $0.1\ \mathrm{mm}$ to $0.05\ \mathrm{mm}$ therefore doubles the width: $w_{\text{new}} = 2\times12 = \boxed{24\ \mathrm{mm}}$ .

Step 2. Check directly: $w_{\text{new}} = 2\lambda D/a_{\text{new}} = 2(600\times10^{-9})(1)/(0.05\times10^{-3}) = 24\times10^{-3}\ \mathrm{m} = 24\ \mathrm{mm}$ . Consistent.

Step 3. The rule: pattern width $\propto 1/a$ . A narrower slit spreads the light more (wider central band); a wider slit spreads it less. Squeezing a wave in space forces it to spread in angle.

Step 4. For a projected image: a wider slit or aperture reduces diffraction spreading, so up to a point it sharpens the image — but open it too wide and geometric blur and aberrations take over. This diffraction-versus-aperture trade-off is why every lens has an optimum stop, and why a pinhole camera's hole cannot be made arbitrarily small. Timing note: about 60 seconds. The trap is the direction of the effect — write pattern width $\propto 1/a$ at the top so you never accidentally make a smaller slit give a narrower pattern.

Exam traps

$a\sin\theta = m\lambda$ is a minimum. For a single slit this locates the dark bands. Do not read it as a maximum condition just because it looks like the double-slit maximum $d\sin\theta = n\lambda$ . Label it "single slit → minima" in your working.

$m = 0$ is the bright centre. The order $m$ in the single-slit minima runs $\pm1, \pm2, \dots$ ; there is no $m = 0$ minimum, because straight ahead the wavelets are all in step. Starting the count at $m = 0$ puts a dark fringe where the brightest point is.

Slit width $a$ , not separation $d$ . Diffraction uses the width of the one slit; interference uses the gap between two slits. Mixing them swaps the length in your formula. Keep $a$ for diffraction and $d$ for the double slit.

The central maximum is twice as wide. It spans two first-minima and so is $2\lambda/a$ wide, double each $\lambda/a$ side band. A common error is treating all the bands as equal width.

Narrower slit, wider pattern. Because $a$ is in the denominator, reducing the slit width increases the spread. If your answer made a smaller slit give a narrower central band, you inverted the relationship.

Board scope. Single-slit diffraction and the central-maximum width are assessed on CBSE, ICSE, JEE, NEET, AP Physics 2, IB Higher Level, and A-Level (CIE 9702). In IB it is HL only — do not tell an IB Standard Level student it is on their syllabus — and IGCSE has no wave optics at all. The circular-aperture and resolving-power extensions (the $1.22$ factor, Rayleigh's criterion) belong to the next article.

Origins · 1 minJoseph von Fraunhofer and how this was discovered

Joseph von Fraunhofer

1787–1826 · Munich

Joseph von Fraunhofer rose from an orphaned glassworker's apprentice — he was pulled alive from a collapsed workshop as a boy — to the greatest optician of his era, and the far-field diffraction regime this article works in bears his name. In the 1810s he mapped hundreds of dark lines crossing the solar spectrum (the Fraunhofer lines), and to measure their wavelengths precisely he built the first high-quality diffraction gratings, ruling thousands of fine lines with a machine of his own design. His study of diffraction by a single slit and by gratings, with the light treated as parallel rays (the far-field, or Fraunhofer, condition), is exactly the setting of $a\sin\theta = m\lambda$ . His optical glass was so superior that it gave German instrument-making a decades-long lead, and his spectral lines later became the key that unlocked the chemical composition of the stars. He died at 39, the same age as Fresnel, of tuberculosis — the respiratory illness that had dogged him since his years hunched over the glass furnace.

History

1665

— Francesco Maria Grimaldi's posthumous work describes and names the diffraction of light.

1801

— Thomas Young attributes diffraction fringes to the interference of light.

1818

— Augustin-Jean Fresnel's prize memoir establishes the wave theory of diffraction, confirmed by the Poisson–Arago spot.

1821

— Joseph von Fraunhofer studies far-field diffraction by slits and rules the first precision diffraction gratings.

1835

— George Biddell Airy computes the diffraction pattern of a circular aperture, extending the theory to lenses and apertures.

Going further

Texts. NCERT Physics Class 12, Chapter 10 ("Wave Optics"), the single-slit diffraction section and the width of the central maximum · H. C. Verma, Concepts of Physics Vol. 1, "Light Waves", diffraction at a single slit · Cambridge A-Level 9702, the diffraction topic.

Provenance. Standard results. Recompute: first minimum $\sin\theta = \lambda/a = (600\ \mathrm{nm})/(0.1\ \mathrm{mm}) = 6\times10^{-3}$ , $\theta = 0.344^\circ$ ; screen position $y_1 = D\lambda/a = 6\ \mathrm{mm}$ ; central width $2\lambda D/a = 12\ \mathrm{mm}$ ; central angular width $2\lambda/a = 0.012\ \mathrm{rad}$ ; halving $a$ doubles the width to $24\ \mathrm{mm}$ ; first side maximum $\approx (2/3\pi)^2 = 4.5\%$ of the central intensity.

Prerequisites. Huygens' Principle & Wavefronts — the secondary wavelets that the slit re-radiates · Young's Double-Slit Experiment — the interference the diffraction envelope modulates. Leads to. Resolving Power, the Rayleigh Criterion & Gratings-Lite — where diffraction becomes the limit on resolution · ARCHETYPE: Diffraction, Resolution & Grating Numericals — the numerical drills · and, for enrichment, the university seam Single-Slit Diffraction (go-deeper only, never an exam dependency here).

Why does a narrower slit produce a wider pattern?

It seems backwards, but it is the essence of wave behaviour. The width of the central band is $2\lambda D/a$ , with the slit width $a$ in the denominator, so shrinking $a$ increases the spread. Physically, forcing the wave through a smaller opening confines it more tightly in space, and a wave confined in space must spread in direction — you cannot pin down both. When the slit is far wider than the wavelength, the wave barely notices the edges and travels nearly straight (a sharp geometric shadow); as the slit narrows toward the wavelength, the edge-bending becomes dramatic and the light floods out over a wide angle. This inverse relationship is the same trade-off that limits how tightly any beam or image can be focused.

How is single-slit diffraction different from double-slit interference?

Both come from waves combining, but the geometry and the result differ. The double slit has two narrow openings a distance $d$ apart, and their two beams interfere to give equally spaced bright fringes at $d\sin\theta = n\lambda$ . The single slit is one opening of width $a$ , and the continuum of wavelets across it self-cancels to give dark fringes at $a\sin\theta = m\lambda$ , with a broad bright centre. So $a\sin\theta = m\lambda$ and $d\sin\theta = n\lambda$ look identical but mean opposite things — minima versus maxima — and use different lengths, the width $a$ versus the separation $d$ . In a real double-slit experiment both act together: the single-slit diffraction envelope from each slit's width modulates the brightness of the double-slit interference fringes, which is why some fringes are missing where a diffraction minimum falls.

Why is m = 0 excluded from the minima?

Because at $m = 0$ the angle is zero — you are looking straight ahead — and there every wavelet from across the slit has travelled the same distance, so they all arrive perfectly in step and add up. That is the brightest point in the whole pattern, the central maximum, not a dark fringe. The minimum condition $a\sin\theta = m\lambda$ describes the angles at which the wavelets cancel, and cancellation first happens at $m = \pm1$ (edges one wavelength out of step). Plugging $m = 0$ into the formula gives $\sin\theta = 0$ , the centre, but interpreting that as darkness is exactly wrong. So the minima are numbered $m = \pm1, \pm2, \dots$ , deliberately skipping zero, and the central bright band sits where the formula's $m = 0$ would be.

Why is the central maximum twice as wide as the other bands?

Because of how the minima are spaced. The dark fringes fall at $\sin\theta = \lambda/a, 2\lambda/a, 3\lambda/a, \dots$ on each side, evenly spaced by $\lambda/a$ . Every side maximum sits in one such interval, between two consecutive minima, so it is $\lambda/a$ wide. But the central maximum straddles the centre, running from the first minimum on the left ( $-\lambda/a$ ) all the way to the first minimum on the right ( $+\lambda/a$ ) — that is two intervals, so it is $2\lambda/a$ wide, twice any side band. The central band is also much brighter (the first side peak is only about $4.5\%$ of it), because the wavelets are most nearly in step near the centre. Most of the light energy is concentrated in that wide central band.

What sets the "4.5%" brightness of the first side maximum?

The exact intensity across a single-slit pattern follows a specific mathematical curve, the sinc-squared function, which describes how the continuum of wavelets adds up at each angle. Its central peak is tall, and the successive side peaks fall off rapidly: the first side maximum reaches only about $4.5\%$ of the central intensity, the value $(2/3\pi)^2$ , the second even less. The derivation of that envelope in closed form is a university-track calculation, so at this level we quote the $4.5\%$ figure rather than derive it — what you need is the qualitative fact that the side bands are dim and the centre dominates. This rapid fall-off is why, to the eye, a single-slit pattern looks like one bright band with a few faint companions, not a set of equal fringes.

Does diffraction happen with white light, and what would it look like?

Yes. Each wavelength diffracts by its own amount, since the pattern width scales with $\lambda$ (the minima are at $\sin\theta = m\lambda/a$ ). So in white light the central maximum stays white at its very centre, where all colours are in step, but its edges and the side bands fan out into colour: red, with the longest wavelength, spreads most and sits farthest out, while violet spreads least and stays closest to the centre. You see a white centre grading into coloured fringes, the reverse colour order from a prism's spectrum (a prism bends violet most; diffraction spreads red most). This wavelength-dependence is exactly what a diffraction grating exploits to split light into a spectrum, and it is why fine slits and gratings show rainbow effects in ordinary light.