Interference and Diffraction of Light: Why Waves Make Patterns

A soap bubble shimmers with swirling rainbows. The underside of a CD throws bands of colour across the room. A Morpho butterfly's wings blaze electric blue. None of these objects contain any blue or rainbow pigment at all — scratch the butterfly's wing into powder and the blue vanishes. The colours come from something stranger: light waves adding together and cancelling out. That single idea, interference, together with its close cousin diffraction, explains some of the most beautiful phenomena in physics — and it was also the discovery that settled a century-long argument about what light actually is.

What Happens When Two Waves Meet? Interference Explained Simply

Drop two pebbles into a still pond at the same time. Each one sends out rings of ripples, and where the rings cross, something interesting happens. At some spots, a crest from one pebble arrives at the same moment as a crest from the other — the water leaps twice as high. At other spots, a crest from one arrives with a trough from the other — they cancel, and the water barely moves at all.

That is the whole secret:

• Crest + crest → bigger wave. This is called constructive interference.
• Crest + trough → nothing. This is called destructive interference.

The waves don't destroy each other — they pass straight through and keep going. But at each point in the pond, their effects simply add up. Physicists call this the principle of superposition.

Light is a wave too (an electromagnetic one), so light does exactly the same thing. Where two light waves arrive crest-on-crest, you see brightness. Where they arrive crest-on-trough, you see darkness — light plus light makes dark. That sentence sounds absurd until you watch it happen.

Path Difference: The Rule Behind Bright and Dark Fringes

Whether two waves arrive crest-on-crest or crest-on-trough depends on one thing: how far each wave travelled to get there.

Imagine two speakers playing the same pure note, perfectly in step. Stand exactly halfway between them and both sound waves travel the same distance, arrive in step, and the note sounds loud. Now take a few steps to one side. The wave from the far speaker now travels a little farther than the wave from the near one. If it travels exactly half a wavelength farther, its crests arrive where the other wave's troughs are — and the sound nearly disappears.

The extra distance one wave travels compared to the other is called the path difference ($\Delta$), and it controls everything:

• Bright (constructive): $\Delta = m\lambda$ — the path difference is a whole number of wavelengths ($m = 0, 1, 2, \dots$)
• Dark (destructive): $\Delta = \left(m + \tfrac{1}{2}\right)\lambda$ — the path difference is a half-odd number of wavelengths

The simulation below is a virtual ripple tank — two point sources sending out circular waves, exactly like the two pebbles. Press Play to watch the waves spread and overlap, then switch to the Intensity view to see the time-averaged pattern: bright spokes where the waves always reinforce, dark spokes where they always cancel. Try changing the wavelength and the source separation and watch the spokes move.

Every dark spoke in that pattern is a line of points where the path difference to the two sources is a half-odd number of wavelengths. The pattern stands perfectly still even though the waves race through it — which is why it is called a stationary interference pattern.

Young's Double-Slit Experiment: Light Behaving as a Wave

In 1801, Thomas Young did exactly this experiment with light — and changed physics. Newton had argued a century earlier that light was a stream of particles ("corpuscles"). Young let sunlight pass through two narrow slits cut close together and looked at a screen behind them. Particles should have produced two bright stripes, one behind each slit. Instead, Young saw a whole ladder of evenly spaced bright and dark fringes — an interference pattern. Light behaves as a wave.

The geometry is the same as the two speakers. Each slit acts as a source. For a point on the screen at angle $\theta$ from the centre, the wave from the lower slit travels an extra distance $d\sin\theta$, where $d$ is the slit separation. Bright fringes appear wherever that extra distance is a whole number of wavelengths:

$d\sin\theta = m\lambda$

For a screen at distance $L$ (with $L$ much larger than $d$, so $\sin\theta \approx \tan\theta = y/L$), the bright fringes land at positions $y_m = m\lambda L/d$, which means they are evenly spaced with separation:

$\Delta y = \frac{\lambda L}{d}$

This little formula is remarkably powerful. It says red light (large $\lambda$) makes wider fringes than blue light, that moving the screen back magnifies the pattern, and that squeezing the slits together spreads the fringes apart. Young used it to make the first-ever measurement of the wavelength of light — armed with nothing but sunlight, two slits, and a ruler.

Verify each of those claims yourself in the simulation: drag the wavelength from red to violet, then double the slit separation and watch the fringe spacing halve.

The two slits must be lit by the same wave so they stay perfectly in step — physicists say the sources must be coherent. This is also why lasers make such crisp interference patterns: our post on how lasers work explains where that perfect coherence comes from.

The Math of Wave Optics

Everything above can be made precise with surprisingly little machinery. This section derives the actual intensity formulas plotted in the simulations.

Adding Waves with Phasors: Deriving the Double-Slit Intensity

At a point on the screen, the electric fields from the two slits are two oscillations with the same amplitude $E_0$ but a phase difference $\delta$ set by the path difference:

$\delta = \frac{2\pi}{\lambda}\, d\sin\theta$

The total field is $E = E_0\cos(\omega t) + E_0\cos(\omega t + \delta)$. Using the sum-to-product identity (or adding the two as phasors — arrows of length $E_0$ with angle $\delta$ between them):

$E = 2E_0\cos\!\left(\frac{\delta}{2}\right)\cos\!\left(\omega t + \frac{\delta}{2}\right)$

Intensity is proportional to the square of the amplitude, so with $I_0$ the intensity from a single slit:

$I = 4I_0\cos^2\!\left(\frac{\delta}{2}\right) = 4I_0\cos^2\!\left(\frac{\pi d \sin\theta}{\lambda}\right)$

Maxima of $\cos^2$ occur when $\delta/2 = m\pi$, which reproduces $d\sin\theta = m\lambda$. Note the peak intensity is $4I_0$, not $2I_0$ — interference doesn't just add intensities, it adds amplitudes first and squares afterwards. The "missing" energy from the dark fringes is redistributed into the bright ones; energy is conserved overall.

Single-Slit Diffraction: Why a Single Opening Spreads Light

Diffraction is what waves do at edges: they bend around obstacles and spread out from openings. Huygens' principle explains why — every point on a wavefront acts as a tiny source of new wavelets. When a wave squeezes through a slit, only the wavelets inside the opening survive, and they interfere with each other.

Treat the slit of width $a$ as a continuous row of tiny sources and add up (integrate) their contributions at angle $\theta$. Each strip at position $x$ across the slit contributes a phase $kx\sin\theta$, so the total amplitude is:

$E(\theta) \propto \int_{-a/2}^{a/2} e^{\,i k x \sin\theta}\, dx \;=\; a\,\frac{\sin\alpha}{\alpha}, \qquad \alpha = \frac{\pi a \sin\theta}{\lambda}$

Squaring gives the famous single-slit intensity pattern:

$I(\theta) = I_0 \left(\frac{\sin\alpha}{\alpha}\right)^2$

This is the $\mathrm{sinc}^2$ function: a tall, wide central maximum flanked by much weaker side lobes. The intensity falls to zero wherever $\alpha = m\pi$, i.e. at:

$a\sin\theta = m\lambda, \qquad m = \pm 1, \pm 2, \dots$

Here is the counterintuitive part: the central maximum has angular half-width $\sin\theta \approx \lambda/a$ — so making the slit narrower makes the light spread wider. Squeeze the opening and the beam fans out. This inverse relationship between confinement and spread runs deep in physics; the same mathematics reappears as the uncertainty principle in quantum mechanics, and our wave packets post explores it from that angle.

Watch it happen below — drag the slit width down and see the central band balloon outward.

The Real Double-Slit Pattern: Interference Inside a Diffraction Envelope

Real slits have width, so a real double-slit pattern is both effects at once: the fast $\cos^2$ interference fringes from the slit separation $d$, multiplied by the broad $\mathrm{sinc}^2$ diffraction envelope from the slit width $a$:

$I(\theta) = I_0 \cos^2\beta \left(\frac{\sin\alpha}{\alpha}\right)^2, \qquad \beta = \frac{\pi d \sin\theta}{\lambda}, \quad \alpha = \frac{\pi a \sin\theta}{\lambda}$

Here $I_0$ is the on-axis peak intensity of the combined pattern (it absorbs the factor of 4 from the phasor addition, so this form is tidier than writing $4I_{\text{slit}}\cos^2\beta\,(\sin\alpha/\alpha)^2$).

That is exactly what the double-slit simulation plots — the dashed grey curve is the envelope. One elegant consequence: if $d/a$ is a whole number, some interference maxima land exactly on a diffraction zero and vanish. These are called missing orders.

How Diffraction Gratings Split Light into Spectra

What happens with three slits? Ten? Ten thousand?

Each extra slit adds one more wave to the sum. At the special angles where the path difference between neighbouring slits is exactly a whole number of wavelengths, all $N$ waves arrive in step and reinforce. At every other angle, the $N$ contributions point every which way and cancel — and the more slits there are, the more unforgiving that cancellation becomes. Think of a crowd clapping in rhythm: with two people, slightly off-beat still sounds fine; with ten thousand, anything short of perfect unison dissolves into noise. For $N$ equally spaced slits, the phasor sum gives:

$I(\theta) = I_0\left[\frac{\sin(N\gamma)}{N\sin\gamma}\right]^2, \qquad \gamma = \frac{\pi d \sin\theta}{\lambda}$

The bright principal maxima stay in exactly the same places as the double slit — wherever $d\sin\theta = m\lambda$ — but two dramatic things happen as $N$ grows:

1. The peaks get sharper. Each principal maximum has angular width proportional to $1/N$. With thousands of slits, the broad fringes collapse into razor-thin lines.
2. The peaks get brighter. Peak intensity grows as $N^2$ (amplitudes add before squaring), while the background between peaks fades to almost nothing.

A diffraction grating is exactly this: thousands of slits (or reflective grooves) per millimetre. Because the angle of each maximum depends on $\lambda$ through the grating equation $d\sin\theta = m\lambda$, every wavelength is sent in its own direction. Shine white light on a grating and it fans out into a full spectrum — this is how spectrometers read the chemical fingerprints of stars, and why a CD (whose data track is a spiral of pits spaced about 1.6 µm apart) acts as a reflection grating and throws rainbows.

Drag the slit-count slider from 2 up to 20 and watch the double-slit fringes sharpen into spectral lines:

Worked Examples: Interference and Diffraction Calculations

Example 1: Fringe Spacing from a Laser Pointer

A red laser pointer ($\lambda = 650$ nm) shines through two slits separated by $d = 0.25$ mm onto a wall $L = 1.5$ m away. How far apart are the bright fringes?

$\Delta y = \frac{\lambda L}{d} = \frac{(650\times 10^{-9}\ \text{m})(1.5\ \text{m})}{0.25\times 10^{-3}\ \text{m}} = 3.9\ \text{mm}$

Easily visible by eye. Check it in the double-slit simulation: set λ = 650 nm, d = 0.25 mm, L = 1.5 m — the status box reads Δy = 3.9 mm.

Example 2: Width of the Central Diffraction Maximum

Green light ($\lambda = 550$ nm) passes through a single slit of width $a = 80$ µm with the screen at $L = 1.5$ m. How wide is the central bright band?

The central maximum spans between the $m = \pm 1$ minima at $y = \pm\lambda L/a$, so its full width is:

$w = \frac{2\lambda L}{a} = \frac{2(550\times 10^{-9})(1.5)}{80\times 10^{-6}} = 20.6\ \text{mm}$

A 0.08 mm slit produces a 2 cm band of light — diffraction in action. Check it in the single-slit simulation with the default settings.

Example 3: Measuring Wavelength with a Diffraction Grating

Light from an unknown source passes through a grating with 600 lines/mm, and the first-order ($m = 1$) maximum appears at $\theta = 19.5°$. What is the wavelength?

The line spacing is $d = \frac{1}{600}\ \text{mm} = 1.667\ \text{µm}$. From the grating equation:

$\lambda = \frac{d\sin\theta}{m} = (1.667\times 10^{-6})\sin(19.5°) \approx 556\ \text{nm}$

Green light — close to the oxygen emission line that paints aurora green (557.7 nm). This is precisely how spectroscopy measures wavelengths to extraordinary accuracy: a sharper peak (more slits) means a more precise angle, and therefore a more precise wavelength.

Where Interference and Diffraction Appear in the Real World

• Telescope resolution — diffraction at a telescope's circular aperture blurs every star into a tiny disc. Two stars closer than $\theta \approx 1.22\,\lambda/D$ (the Rayleigh criterion) blur into one. This is the fundamental reason astronomers build enormous mirrors.
• X-ray crystallography — the atomic planes in a crystal act as a 3D diffraction grating for X-rays. Rosalind Franklin's "Photo 51" diffraction pattern revealed the double-helix structure of DNA.
• Holograms — a hologram is a recorded interference pattern between light from an object and a reference beam. Re-illuminating it diffracts light into a full 3D reconstruction.
• Anti-reflective coatings — camera lenses and glasses carry a thin transparent layer engineered so reflections from its two surfaces interfere destructively, cancelling glare.
• Radio interferometry — the Event Horizon Telescope combined signals from radio dishes across the Earth, interfering them to act as a planet-sized aperture — sharp enough to photograph a black hole's shadow.
• Structural colour in nature — Morpho butterflies, peacock feathers, and beetle shells get their iridescence from microscopic gratings and thin-film interference, not pigment.

Frequently Asked Questions

What is the difference between interference and diffraction?

They are two faces of the same physics — superposition of waves. Interference usually refers to the pattern from a small number of discrete sources (like two slits), while diffraction refers to the spreading and patterning of a wave from a continuous aperture or edge (like one slit, treated as infinitely many tiny sources via Huygens' principle). In any real experiment both happen at once: the double-slit pattern is interference fringes sitting inside a diffraction envelope.

Why don't two flashlights produce an interference pattern?

Because they are not coherent. Interference fringes require a fixed phase relationship between the two sources. A flashlight emits light as countless independent atomic emissions, each a short wave train with random phase, and the phase relationship between two separate bulbs scrambles billions of times per second. The fringes exist momentarily but shift randomly and far too fast to see — what you observe is the smeared-out average: uniform light. Young solved this by splitting one wavefront into two slits; lasers solve it by emitting intrinsically coherent light.

What is the condition for constructive and destructive interference?

Constructive interference (a bright fringe) occurs when the path difference between the two waves is a whole number of wavelengths: $\Delta = m\lambda$. Destructive interference (a dark fringe) occurs when it is a half-odd number: $\Delta = (m + \tfrac{1}{2})\lambda$. For a double slit the path difference is $d\sin\theta$, which gives bright fringes at $d\sin\theta = m\lambda$.

Why do CDs and DVDs show rainbow colours?

The data on a CD is stamped in a spiral track with neighbouring turns about 1.6 µm apart — comparable to the wavelength of visible light. The track acts as a reflection diffraction grating: each wavelength reflects strongly at its own angle ($d\sin\theta = m\lambda$), so white light fans out into a spectrum. DVDs have finer tracks (0.74 µm), so they spread the colours over even wider angles.

What happens to the interference pattern if the experiment is done in water?

The fringes squeeze closer together. In water, light slows down and its wavelength shrinks to $\lambda/n$, where $n \approx 1.33$ is the refractive index. Since fringe spacing is $\Delta y = \lambda L/d$, the whole pattern contracts by a factor of 1.33. The light's frequency — and its colour as perceived — does not change; only the wavelength does.