In short

Send monochromatic light of wavelength \lambda through a single narrow slit of width a. On a screen at distance D behind the slit, the light splays into a bright central maximum flanked by progressively weaker secondary maxima separated by dark minima.

Condition for the n-th minimum (on either side of the centre):

\boxed{\;a\,\sin\theta \;=\; n\lambda,\qquad n = \pm 1,\,\pm 2,\,\pm 3,\,\ldots\;}

(Note: n = 0 is the bright centre, not a minimum.)

Angular half-width of the central maximum (from centre to first dark band):

\sin\theta_1 \;=\; \frac{\lambda}{a} \;\approx\; \theta_1 \quad\text{(for } \lambda \ll a\text{)}

Linear width of the central maximum on a screen at distance D:

\boxed{\;W_0 \;=\; \frac{2\lambda D}{a}\;}

This is exactly twice the width of every other bright fringe. A single slit of width a thus imposes a fundamental angular spread \sim \lambda/a on any beam that passes through it — the spread gets wider as the slit narrows, which is why you cannot focus light to a point. This is the resolution limit for every camera, microscope, telescope, and eye.

Take a red laser pointer from a physics lab drawer. Close a door so the room is dark. Now slide a thin vertical slit — the edge of two razor blades held close together, or the gap in a slit card from the class 12 optics kit — into the beam. On the wall a metre behind the slit you expect a narrow red line. You do not get one.

What you get is a bright central band, several centimetres wide, surrounded by a series of weaker side bands. The side bands get dimmer as you move away from the centre, and between each pair of bright bands is a dark gap — a place on the wall that the laser is, strangely, not reaching. Slide the blades closer together to narrow the slit further, and the pattern spreads out rather than shrinking. Narrow the slit enough and the light seems to fill half the wall.

This is single-slit diffraction. It is what happens when a wave passes through an opening comparable in size to its wavelength, and it is something a particle theory of light cannot produce. A hail of photons flying through a slit should land in a sharp geometric shadow of the slit's shape. Waves do not.

Why a slit spreads the light — the Huygens picture

In the double-slit experiment you treated each of the two slits as a single point source and let the two sources interfere. But a real slit has a width — it is not a point, it is a line segment of length a. Huygens' principle tells you what to do with it: treat every point inside the slit opening as a secondary source of spherical wavelets. When those wavelets reach the screen, they all arrive with slightly different phases because they had slightly different distances to travel. The interference of this whole continuum of sources is what produces the diffraction pattern.

Imagine the slit divided into a large number N of tiny strips. Each strip sends its own wavelet toward a point on the screen. In the direction straight ahead (\theta = 0), every strip's wavelet has travelled the same distance — they are all in phase, they all add constructively, and the amplitude at the centre is huge. This is the central bright maximum.

Now aim slightly off-axis. The wavelets from different strips arrive with different phases. Some of them reinforce each other, some partially cancel. For most angles the cancellation is partial, producing a dim glow. But at a special set of angles the cancellation is complete — every strip's contribution is exactly cancelled by another strip's contribution, and the screen at that angle sees zero light. Those are the dark minima.

Finding the directions where total cancellation happens is all the single-slit problem is about.

Deriving the minima — the pairing argument

Consider the slit of width a with monochromatic plane-wave light falling on it straight ahead. Look at the light travelling at angle \theta to the forward direction, aimed at some point P on a distant screen.

Single-slit geometry: path difference between wavelets from top and bottom of the slitA vertical slit of width a on the left. Plane waves arrive from the left. Rays leave the slit at angle theta toward a distant point P on the right. The ray from the top of the slit travels a distance a sin theta more than the ray from the bottom, marked as the path-difference segment.$a$topbottomincoming plane waves$P$$a\sin\theta$$\theta$
A slit of width $a$ with plane waves arriving from the left. Consider two wavelets leaving at angle $\theta$ — one from the top of the slit and one from the bottom. The path from the top is longer than the path from the bottom by exactly $a\sin\theta$.

The ray from the top of the slit has to travel a bit further to reach P than the ray from the bottom does. The extra distance is the short perpendicular segment you can drop from the bottom onto the top ray: a right triangle of hypotenuse a and opposite side a\sin\theta. So the path difference across the full slit is

\Delta = a\sin\theta.

That is the only geometric fact you need.

The pairing trick — pair strip 1 with strip (N/2 + 1)

Divide the slit into N equal thin strips, N even and large. Label them 1, 2, 3, \ldots, N from bottom to top.

Pair strip 1 (at the bottom) with strip N/2 + 1 (just above the middle). These two strips are separated by exactly a/2 — half the slit width. Their path difference to P is therefore (a/2)\sin\theta.

Claim: these two wavelets cancel at P when (a/2)\sin\theta = \lambda/2, that is, when a\sin\theta = \lambda.

Why: a path difference of \lambda/2 means the two waves arrive exactly out of phase — one is at a crest while the other is at a trough. Their amplitudes are equal (each strip is the same size, equally illuminated, sends the same amplitude), so they cancel exactly. Two equal antiphase amplitudes sum to zero.

Now pair strip 2 with strip N/2 + 2. They are also separated by a/2. Same argument: they cancel.

Pair strip 3 with strip N/2 + 3. Separation a/2. Cancel.

Keep going. Every strip in the bottom half pairs with exactly one strip in the top half, at separation a/2, and the two cancel. Since every strip is paired, and every pair sums to zero, the total amplitude at P is exactly zero.

This happens whenever

a\sin\theta = \lambda.

That is the first minimum, the first dark band on either side of the bright centre.

Higher-order minima — divide into more pieces

For the second minimum, split the slit into four equal quarters instead of two halves. Pair each quarter with the next one at separation a/4. The path difference across one quarter is (a/4)\sin\theta. Complete cancellation happens when this difference equals \lambda/2 — that is, when

\tfrac{a}{4}\sin\theta = \tfrac{\lambda}{2} \quad\Longrightarrow\quad a\sin\theta = 2\lambda.

Why: within the first quarter, pair each strip with the corresponding strip in the second quarter — separation a/4, path difference \lambda/2, cancel. Do the same for the third quarter paired with the fourth. All four quarters cancel in pairs, so the total is zero.

For the third minimum, split into six pieces. Same argument gives a\sin\theta = 3\lambda. In general, splitting the slit into 2n equal pieces gives cancellation whenever

\boxed{\;a\sin\theta = n\lambda,\qquad n = 1, 2, 3, \ldots\;}

The negative integers (n = -1, -2, \ldots) give the symmetric dark bands on the other side of the centre.

Why n = 0 is not a minimum

Plug n = 0 into a\sin\theta = n\lambda and you get \theta = 0 — straight ahead. But straight ahead is the brightest point on the screen, not the dimmest. The pairing argument breaks down at \theta = 0 because the path difference across every strip is zero, every wavelet arrives in phase, and they all add up. So n = 0 is excluded from the minimum condition — it corresponds to the central maximum.

The central maximum is twice as wide as every other

Look at where the first minimum sits: at \sin\theta_1 = \lambda/a, on either side of the centre. The angular width of the central bright band, from its first dark edge on one side to its first dark edge on the other, is

\Delta\theta_0 = 2\theta_1 = \frac{2\lambda}{a}.

Now the second minimum is at \sin\theta_2 = 2\lambda/a. The first secondary maximum sits roughly halfway between \theta_1 and \theta_2. The angular width of that secondary maximum — from minimum n=1 to minimum n=2 — is

\theta_2 - \theta_1 = \frac{2\lambda}{a} - \frac{\lambda}{a} = \frac{\lambda}{a}.

That is exactly half the angular width of the central maximum. Every secondary maximum has this same angular width \lambda/a. Only the central one is doubled.

On a screen at distance D (with D \gg a), the angles are small, so \sin\theta \approx \tan\theta \approx y/D where y is the distance from the screen centre. The linear positions of the minima are

y_n = \frac{n\lambda D}{a},

and the linear width of the central maximum on the screen is

\boxed{\;W_0 = 2y_1 = \frac{2\lambda D}{a}.\;}

This is a useful number to remember: double the slit width and the central fringe shrinks in half; halve the slit width and the pattern doubles in spread.

Single-slit diffraction intensity envelopeIntensity on the screen plotted against angle. A tall central peak flanked by much smaller secondary peaks on either side. Dashed lines mark the minima at a sin theta equals plus-or-minus lambda, plus-or-minus 2 lambda, plus-or-minus 3 lambda. The central peak is twice as wide as each secondary peak.$a\sin\theta / \lambda$$I/I_0$0$-3$$-2$$-1$$1$$2$$3$central max2nd max2nd max
Intensity of the single-slit diffraction pattern. The central peak is at $a\sin\theta = 0$ and is flanked by zeros at $a\sin\theta = \pm\lambda, \pm 2\lambda, \pm 3\lambda, \ldots$. The central peak has angular width $2\lambda/a$ — exactly twice the width of every secondary peak.

Explore the slit width yourself

The formula W_0 = 2\lambda D/a says something counter-intuitive: the narrower the slit, the wider the pattern. The interactive figure below lets you drag the slit width a to set a value between 50 μm and 500 μm, then plots the full diffraction intensity pattern I/I_0 = (\sin\alpha/\alpha)^2 that the slit would produce on a screen at D = 1 m with wavelength \lambda = 600 nm.

Interactive: how the diffraction envelope depends on slit widthIntensity envelope sin(alpha)/alpha squared plotted against screen position in millimetres, with a draggable slit-width parameter. As the slit narrows, the central peak widens; as the slit widens, the central peak narrows.screen position $y$ (mm)intensity $I/I_0$-20-100102000.51slit width $a$ →drag the red point below the axis
Drag the red control along the slider row at the bottom to change the slit width $a$ from 50 μm to 500 μm. The diffraction envelope above redraws in real time. Readout shows the central-maximum width $W_0 = 2\lambda D/a$ (with $\lambda = 600$ nm, $D = 1$ m, so $W_0$ in mm equals $1200/a$ when $a$ is in μm).

The narrower the slit, the more the light spreads. A slit that is extremely wide compared to the wavelength (a \gg \lambda) produces a central peak so narrow that you cannot see any of it fringing — the light looks, to your eye, like a straight beam with geometric edges. This is why you do not normally notice diffraction: ordinary doorways and windows are millions of wavelengths wide, and 2\lambda D/a is a number of micrometres that disappears inside the blur of your own eye.

Single slit versus double slit — the two patterns are not the same

It is easy to confuse single-slit diffraction with double-slit interference. They share the wave idea but produce visibly different patterns.

In the double-slit experiment, two narrow coherent sources are separated by a distance d. Their interference gives equally spaced, equally bright fringes, with the bright fringes at d\sin\theta = n\lambda for n = 0, 1, 2, \ldots. The central fringe is not special — every fringe looks the same.

In the single-slit experiment, there is one slit of finite width a. You get a bright central maximum that is twice as wide as the others, flanked by progressively weaker secondary maxima (the second bright band is roughly 4.7% of the central intensity; the third, about 1.7%; the fourth, 0.8%).

Feature Double slit (interference) Single slit (diffraction)
Bright fringe position d\sin\theta = n\lambda, n = 0, 1, 2, \ldots Central peak at \theta = 0; weak secondary peaks between minima
Dark fringe position d\sin\theta = (n+\tfrac{1}{2})\lambda a\sin\theta = n\lambda, n = 1, 2, 3, \ldots
Fringe spacing All bright fringes equally spaced (\beta = \lambda D/d) Central max is twice as wide; secondaries equally spaced
Brightness Every bright fringe equally bright (in the limit) Intensity drops sharply away from the centre
Characteristic length Slit separation d (hundreds of μm) Slit width a (tens to hundreds of μm)

In a real double-slit experiment each slit also has width — so the fringe pattern you actually see is the double-slit interference modulated by the single-slit diffraction envelope. Some high-order interference fringes sit where the single-slit envelope is zero, and those fringes are missing — they are called missing orders. This happens whenever d = ma for some integer m (the m-th diffraction minimum kills the m-th interference maximum beyond the centre).

Worked examples

Example 1: The class-12 lab laser demonstration

A standard class-12 physics lab experiment: a He-Ne laser (\lambda = 632.8 nm) is shone through a single slit of width a = 0.10 mm onto a screen at D = 2.0 m. Find the width W_0 of the central maximum on the screen and the position of the first minimum.

Layout for class-12 single-slit laser experimentA red laser on the left aimed at a vertical slit of width a. Light spreads and forms a diffraction pattern on a screen 2 metres to the right, with a bright central band flanked by weaker side bands.laserslit, $a$ = 0.1 mmscreen, $D$ = 2.0 m$D$
The laser beam, the slit, and the screen. The central bright band and its dimmer companions are roughly sketched on the screen.

Step 1. List the known quantities in SI units.

\lambda = 632.8 \text{ nm} = 6.328 \times 10^{-7} m, a = 0.10 \text{ mm} = 1.0 \times 10^{-4} m, D = 2.0 m.

Why: wave-optics numbers span many orders of magnitude. Always convert to SI before you substitute, or you will drop a factor of 10^3 somewhere.

Step 2. Find the angle to the first minimum from a\sin\theta_1 = \lambda.

\sin\theta_1 = \frac{\lambda}{a} = \frac{6.328 \times 10^{-7}}{1.0 \times 10^{-4}} = 6.328 \times 10^{-3}

Why: the ratio \lambda/a is a tiny number. Its sine is essentially equal to itself in radians, so you can also read this as \theta_1 \approx 6.3 \times 10^{-3} rad. The small-angle approximation is excellent here.

Step 3. Find the linear position y_1 of the first minimum on the screen.

y_1 = D\tan\theta_1 \approx D\sin\theta_1 = 2.0 \times 6.328 \times 10^{-3} = 1.27 \times 10^{-2} \text{ m} = 12.7 \text{ mm}

Why: the first minimum on either side of the centre sits 12.7 mm from the axis — a distance you can comfortably measure with a ruler. This is why the demonstration works in a classroom: the pattern has centimetre-scale features.

Step 4. Find the width of the central maximum.

W_0 = 2y_1 = 2 \times 12.7 = 25.4 \text{ mm}

Or directly from the formula:

W_0 = \frac{2\lambda D}{a} = \frac{2 \times 6.328 \times 10^{-7} \times 2.0}{1.0 \times 10^{-4}} = 2.53 \times 10^{-2} \text{ m} = 25.3 \text{ mm}.

Why: the two methods agree to within rounding. A central maximum of ~2.5 cm is what students actually measure with a metre rule on the wall — and the measurement is how you extract the slit width (or, equivalently, test that the theory works).

Result: The first minimum is at y_1 \approx 12.7 mm from the centre. The central maximum is W_0 \approx 25.4 mm wide.

What this shows: A hundredth-of-a-millimetre slit spreads a laser across a 2.5-cm-wide band on a wall two metres away. The spreading is a direct measurement of \lambda/a. If you did not know the wavelength, you could solve \lambda = W_0 \cdot a / (2D) and measure it — which is exactly how spectroscopists once pinned down the wavelength of sodium's yellow line.

Example 2: Reading the grooves on a CD

Shine a red laser (\lambda = 650 nm) through a single slit and you see a diffraction pattern spread over the wall. Now cover the slit and bounce the same laser off the playing surface of a compact disc held vertically. You see almost the same thing — a set of bright spots spread across the wall. The CD's spiral data track, which is almost perfectly regular at around 1.6 μm between grooves, behaves like a very fine diffraction element. If the central (zero-order) reflected spot sits straight back at you and the first-order bright spots sit at angles \theta = \pm 23.9° from the normal, estimate the groove spacing.

This is technically a grating problem (chapter on diffraction grating), but the single-slit idea is the parent. The same path-difference argument applies: neighbouring grooves differ by a path of d\sin\theta, and constructive interference happens when this equals n\lambda.

Diffraction from the grooved surface of a CDA red laser beam incident on a CD surface reflects into several ordered bright spots. The central zero-order spot goes straight back. First-order spots come off at plus or minus 23.9 degrees from the normal.CD grooves (spacing $d \approx 1.6$ μm)incident lasernormal$+1$ order$-1$ order
A CD acts as a reflection grating. The first-order bright spots at $\theta = \pm 23.9°$ come from path differences of exactly one wavelength between neighbouring grooves.

Step 1. Use the bright-fringe condition for a grating (which is the same algebra as the single-slit minimum condition, applied to the spacing d instead of the width a).

d\sin\theta = n\lambda

For n = 1 and \theta = 23.9°:

d = \frac{\lambda}{\sin\theta} = \frac{650 \times 10^{-9}}{\sin 23.9°}

Why: the neighbouring grooves act like the top and bottom of the slit in the earlier argument — they are the two ends of a path-difference triangle of length d\sin\theta. When that path difference is one full wavelength, every groove adds in phase, producing a bright spot.

Step 2. Compute.

\sin 23.9° = 0.405
d = \frac{650 \times 10^{-9}}{0.405} = 1.60 \times 10^{-6} \text{ m} = 1.60 \text{ μm}.

Why: the CD specification is exactly this — the data track pitch is 1.6 μm. The laser diffraction measurement gives you the spacing without ever touching the disc.

Result: d \approx 1.6 μm.

What this shows: Diffraction is a ruler. Shine light of known wavelength through (or off) a regular structure, measure the angles where the bright spots land, and read off the spacing. The same principle scales from CDs down to crystals, where the spacing is the size of an atom and the "light" is X-rays — which is how crystal structures are solved.

Common confusions

If you wanted to know how diffraction comes from, where the minimum condition comes from, and why the central maximum is twice as wide, you have what you need. What follows is the full amplitude calculation (the \text{sinc}^2 intensity profile), an exact derivation of the secondary-peak positions, and a note on what happens when the slit is not illuminated by a plane wave.

The full intensity pattern — deriving I(\theta) = I_0\left(\frac{\sin\alpha}{\alpha}\right)^2

The pairing argument gives you the minima, but not the shape of the whole pattern. To get that, add up the contributions from every strip as a continuous integral.

Divide the slit into N strips of width \Delta a = a/N, each sending a wavelet of amplitude E_0 / N (equal, because the strips are equally illuminated and equally sized). The strip at position y inside the slit (measured from the centre, so y runs from -a/2 to +a/2) contributes a wavelet to the point on the screen at angle \theta. Its extra path compared to the centre of the slit is y\sin\theta, so its extra phase is

\delta(y) = \frac{2\pi}{\lambda}\,y\sin\theta.

The total complex amplitude at angle \theta is

E(\theta) = \frac{E_0}{N}\sum_{k=1}^{N} e^{i\delta(y_k)}.

Taking N \to \infty and replacing the sum by an integral,

E(\theta) = \frac{E_0}{a}\int_{-a/2}^{+a/2} e^{i(2\pi/\lambda)y\sin\theta}\,dy.

Let \alpha \equiv \pi a\sin\theta/\lambda. The integral evaluates to

E(\theta) = E_0 \cdot \frac{\sin\alpha}{\alpha}.

Why: the integral of a complex exponential e^{iky} from -a/2 to a/2 is \frac{2\sin(ka/2)}{k}. With k = 2\pi\sin\theta/\lambda, the prefactor becomes a \cdot \sin\alpha / \alpha after simplification.

The intensity is the square of the amplitude:

\boxed{\;I(\theta) = I_0\left(\frac{\sin\alpha}{\alpha}\right)^2, \quad \alpha = \frac{\pi a\sin\theta}{\lambda}.\;}

This is the sinc-squared function. You can read everything about the pattern off this one formula:

  • At \theta = 0: \alpha = 0. The limit (\sin\alpha/\alpha)|_{\alpha\to 0} = 1, so I(0) = I_0. That is the central peak.
  • Zeros at \sin\alpha = 0 (but \alpha \neq 0): \alpha = n\pi for n = \pm 1, \pm 2, \ldots. That is a\sin\theta = n\lambda — the minima we derived.
  • Secondary maxima between zeros.

Exact positions of the secondary maxima

At a secondary maximum, dI/d\alpha = 0. Differentiating (\sin\alpha/\alpha)^2:

\frac{d}{d\alpha}\left(\frac{\sin\alpha}{\alpha}\right)^2 = 2\,\frac{\sin\alpha}{\alpha}\,\frac{\alpha\cos\alpha - \sin\alpha}{\alpha^2} = 0.

The non-trivial solutions come from \alpha\cos\alpha = \sin\alpha, i.e.

\tan\alpha = \alpha.

This transcendental equation has no closed-form solution but can be solved numerically. The first few roots are

\alpha_1 \approx 4.4934, \qquad \alpha_2 \approx 7.7253, \qquad \alpha_3 \approx 10.9041, \ldots

Compare these to the naive "halfway between zeros" estimates \alpha_n \approx (n + 1/2)\pi: 4.712, 7.854, 10.996, \ldots. The actual peaks fall slightly before the half-integer multiples — by a few percent.

The intensity at the first secondary peak is

\left(\frac{\sin 4.4934}{4.4934}\right)^2 \approx \left(\frac{-0.2172}{4.4934}\right)^2 \approx 0.0472.

So the first secondary maximum carries only about 4.7% of the central intensity. The second is at about 1.65%, the third at 0.83%, and so on.

Fraunhofer vs Fresnel — when is the formula valid?

The derivation above assumes two things:

  1. The light arrives at the slit as a plane wave (parallel rays, equivalent to a point source at infinity).
  2. The screen is at infinite distance (so the diffracted rays leaving the slit can be treated as parallel, all reaching the same point on the screen at angle \theta).

This limit is called Fraunhofer diffraction and is what produces the \text{sinc}^2 pattern. In a classroom experiment you do not have an infinite source or infinite screen, but you can reach the Fraunhofer regime by passing the light through a lens that focuses the parallel rays to a point — this is how a real optics-kit apparatus is built.

If the slit, source, and screen are all at finite distances comparable to a^2/\lambda, the calculation becomes much more complicated (the rays from different parts of the slit reach the screen point at different angles). That regime is called Fresnel diffraction, and the pattern it produces near a slit or edge contains the famous Fresnel fringes that snake along the edges of a geometric shadow. The key number is the Fresnel number

F = \frac{a^2}{\lambda D}.

If F \ll 1, you are in the Fraunhofer regime and the \text{sinc}^2 pattern applies. If F \sim 1 or larger, you see Fresnel fringes, not the clean pattern of this article.

For the class-12 laser setup (a = 0.1 mm, D = 2 m, \lambda = 633 nm): F = (10^{-4})^2 / (633 \times 10^{-9} \times 2) = 10^{-8}/(1.27 \times 10^{-6}) \approx 0.008. That is safely in the Fraunhofer regime — good, because that is what the formula assumes.

Where diffraction sets the resolution limit

The angular width of the central peak, 2\lambda/a, is more than just a feature of the diffraction pattern — it is the minimum angular size of a spot that an aperture of width a can produce. This means two point sources of light whose angular separation is less than about \lambda/a cannot be distinguished by an aperture of width a. Their diffraction patterns overlap and blur into one.

This is the origin of the Rayleigh criterion for optical resolution: two stars are "just resolved" when the central peak of one falls on the first minimum of the other — that is, when they are separated by the angle \lambda/a. A telescope of aperture 1 metre can resolve two stars separated by 5.5 \times 10^{-7}/1 \approx 0.1 arc-seconds at visible wavelengths. Bigger mirrors see finer detail — that is why astronomers keep building bigger telescopes.

The same argument in reverse tells you why your phone camera cannot make arbitrarily sharp photographs. A lens of diameter D_L imaged at distance f (the focal length) focuses a distant point to a spot of diameter about 1.22\,f\lambda/D_L. Smaller lenses produce fuzzier images; this is the fundamental limit no clever engineering can beat. Optical microscopes hit this limit at about \lambda/2, below which you cannot resolve anything — a wall physicists labelled the Abbe limit and which pushed biology toward electron microscopy (electrons have much shorter de Broglie wavelengths) and super-resolution fluorescence techniques (which beat the limit by clever tricks about where single molecules sit, not by making a sharper image).

Where this leads next