Superposition and Interference

Drop two stones into a still pond at the same moment and watch the ripples spread. Where the two sets of rings cross, something remarkable happens: in some places the water heaves up into a taller-than-ever swell, and right beside it lies a stretch of water that is almost dead flat, as if no wave had passed there at all. Two waves, each perfectly able to disturb the water on its own, have somehow added up to nothing in one spot and to double trouble in the next.

That criss-cross pattern of reinforcement and cancellation is interference, and it is one of the surest fingerprints of a wave. It follows from a single, disarmingly simple rule about what happens when waves meet — the principle of superposition. This page builds that rule up carefully, uses it to explain constructive and destructive interference, ties everything to the idea of path difference, and finishes with the experiment that used all of this to prove that light itself is a wave: Young's double slit.

The principle of superposition

When two waves pass through the same point at the same instant, each one is still trying to push the medium exactly as it would on its own. Nature's answer is the most democratic one imaginable: the actual displacement at that point is simply the two individual displacements added together. Displacement is a vector (it has a direction — up or down, forwards or back), so this is a vector sum: an upward push of +3 and a downward push of -3 add to 0.

When two or more waves overlap at a point, the resultant displacement at that point is the vector sum of the displacements each wave would produce there on its own:

y_{\text{total}}(x,t) = y_1(x,t) + y_2(x,t).

Displacements in the same direction add to a larger one.
Displacements in opposite directions partly or wholly cancel.
Afterwards the waves separate and travel on completely unchanged — the overlap is temporary, and neither wave is used up, absorbed or altered by the other.

A one-line worked example. At some instant, wave 1 lifts a point on a rope by y_1 = +4\ \text{mm} while wave 2 pushes the same point down by y_2 = -1\ \text{mm}. Superposition gives the resultant at once:

y_{\text{total}} = y_1 + y_2 = (+4) + (-1) = +3\ \text{mm}.

If instead the second wave had pushed down by the full -4\ \text{mm}, the sum would be (+4)+(-4)=0 — the point would sit perfectly still even while two waves rush through it. Hold on to that idea: a place where two full-blooded waves add to zero is the whole secret of a "dark fringe."

Constructive and destructive interference

Superposition explains the pond straight away. When two identical waves arrive at a point in step — crest landing on crest, trough on trough — their displacements always point the same way, so they add to a wave with a bigger amplitude. We say they are in phase, and the result is constructive interference. Two waves of amplitude A in phase make a resultant of amplitude 2A.

When the two waves arrive exactly out of step — one at a crest just as the other reaches a trough — their displacements point opposite ways and cancel. We say they are in antiphase (a phase difference of 180^\circ, or \pi radians), and the result is destructive interference. Two equal waves in antiphase cancel completely, giving amplitude 0.

In general, if the two waves have amplitude A and a phase difference \phi, superposing the sines gives a neat result for the resultant amplitude:

A_{\text{res}} = 2A\left|\cos\!\left(\tfrac{\phi}{2}\right)\right|.

Check the two extremes: \phi = 0 gives A_{\text{res}} = 2A\cos 0 = 2A (fully constructive), while \phi = 180^\circ gives A_{\text{res}} = 2A\cos 90^\circ = 0 (fully destructive). Everything in between is partial interference.

Path difference: what decides in phase or out of phase

In practice the two waves usually start out identical (from two slits, two speakers, two dippers in a ripple tank) and the phase difference at some point P is caused by the two waves having travelled different distances to get there. That extra distance is the path difference, and it is measured in wavelengths.

If one wave has travelled an extra whole number of wavelengths, it arrives back in step — crest on crest — and you get constructive interference. If it has travelled an extra odd number of half-wavelengths, it arrives exactly out of step and you get destructive interference.

For two coherent sources meeting at a point with path difference \Delta (where n = 0, 1, 2, \dots):

Constructive (bright, maximum): the waves are in phase when \Delta = n\lambda.
Destructive (dark, minimum): the waves are in antiphase when \Delta = \left(n + \tfrac{1}{2}\right)\lambda.

Phase difference and path difference are two ways of saying the same thing, linked by \phi = \dfrac{2\pi}{\lambda}\,\Delta: one whole wavelength of path is a full 2\pi of phase.

Worked example — bright or dark? Two loudspeakers in step play a note of wavelength \lambda = 0.50\ \text{m}. You stand where the sound from one speaker has travelled 4.00\ \text{m} and the other 4.75\ \text{m}. The path difference is

\Delta = 4.75 - 4.00 = 0.75\ \text{m} = \frac{0.75}{0.50}\,\lambda = 1.5\lambda = \left(1 + \tfrac12\right)\lambda.

That is an odd number of half-wavelengths, so the waves arrive in antiphase — you are standing in a quiet spot (destructive interference). Step a little to where the path difference becomes 1.0\ \text{m} = 2\lambda and the sound is suddenly loud again.

Coherence: why the pattern holds still

There is a catch. For the bright-and-dark pattern to stay put long enough to see, the two sources must be coherent: they must have the same frequency and a constant phase difference between them. If the phase relationship keeps jumping about at random, then a point that is bright one microsecond is dark the next, and averaged over the time your eye takes to look, the whole pattern smears out into a dull, uniform glow.

This is exactly why two ordinary light bulbs — or even the two headlamps of a car — never produce interference fringes on a wall, no matter how carefully you arrange them. Each bulb emits light in countless tiny, independent bursts with random phases, so the phase difference between the two sources scrambles billions of times a second. To get coherent light Young had to be clever, as we'll see. (A laser, invented much later, is coherent by design — which is why a laser pointer through a double slit throws up crisp fringes with no trickery at all.)

See it happen: add two waves yourself

Below are two identical waves (wave 1 and wave 2) drawn on the same axis, and beneath them their superposition — the point-by-point sum. Drag the path difference slider, measured in wavelengths, to slide wave 2 relative to wave 1. At 0\lambda, 1\lambda, 2\lambda the two line up and the resultant swells to double the amplitude — constructive. At \tfrac12\lambda, \tfrac32\lambda they are perfectly opposed and the resultant flattens to nothing — destructive. Everywhere in between is partial. Read the live amplitude and compare it with A_{\text{res}} = 2A\,|\cos(\phi/2)|.

Young's double-slit experiment

In 1801 Thomas Young settled a centuries-old argument about the nature of light with a breathtakingly simple setup. He shone light through a single narrow slit first (so that whatever came out was coherent), then let it fall on two close slits side by side. Each slit acts as a source of spreading waves; the two overlap beyond the slits and interfere. On a screen some distance away appears a row of evenly spaced bright and dark fringes.

The bright fringes fall where the path difference from the two slits is 0, \lambda, 2\lambda, \dots (constructive); the dark fringes fall where it is \tfrac12\lambda, \tfrac32\lambda, \dots (destructive). Working through the geometry for slits a small distance s apart and a screen a distance D away (with D \gg s) gives the famous fringe-spacing equation:

The distance between adjacent bright fringes (the fringe spacing w) is

w = \frac{\lambda D}{s},

w — fringe spacing (m)
\lambda — wavelength of the light (m)
D — distance from slits to screen (m)
s — separation of the two slits (m)

Read the equation and it tells a story: longer wavelengths, or a more distant screen, spread the fringes wider apart; slits that are closer together also spread the fringes out, while slits far apart crowd them together. And here is the punch line — only waves interfere. Streams of particles fired through two slits would just make two bright strips. The fact that light instead paints an interference pattern was decisive, historic evidence that light is a wave.

Worked examples with w = \lambda D / s

Example 1 — find the fringe spacing. Red laser light of wavelength \lambda = 600\ \text{nm} = 6.0\times10^{-7}\ \text{m} passes through slits s = 0.50\ \text{mm} = 5.0\times10^{-4}\ \text{m} apart, onto a screen D = 2.0\ \text{m} away. What is the fringe spacing?

w = \frac{\lambda D}{s} = \frac{(6.0\times10^{-7})(2.0)}{5.0\times10^{-4}} = 2.4\times10^{-3}\ \text{m} = 2.4\ \text{mm}.

Comfortably wide enough to see and measure with a ruler.

Example 2 — find the wavelength. A double slit with separation s = 0.40\ \text{mm} and a screen at D = 1.5\ \text{m} gives fringes spaced w = 2.1\ \text{mm} apart. What is the wavelength of the light? Rearrange w = \lambda D / s to make \lambda the subject:

\lambda = \frac{ws}{D} = \frac{(2.1\times10^{-3})(4.0\times10^{-4})}{1.5} = 5.6\times10^{-7}\ \text{m} = 560\ \text{nm}.

That is green-yellow light — measured with nothing but a ruler and this equation.

Example 3 — bright or dark by path difference. Two coherent microwave sources emit \lambda = 3.0\ \text{cm}. At a detector the path difference is \Delta = 7.5\ \text{cm}. Constructive or destructive?

\frac{\Delta}{\lambda} = \frac{7.5}{3.0} = 2.5 = \left(2 + \tfrac12\right)\lambda\text{-worth}.

An odd number of half-wavelengths, so the waves arrive in antiphase — the detector reads a minimum (destructive interference).

Three traps catch almost everyone on this topic:

You need coherent sources, not just two sources. Two ordinary bulbs, or two torches, will not interfere into a stable pattern — their phase difference jitters randomly, washing the fringes out. The sources must share one frequency and hold a constant phase difference (that's why Young split one beam in two, and why lasers work so well).
Superposition is a vector sum, so a crest and a trough can make nothing. Energy isn't destroyed — at a dark fringe it has simply been redirected to the bright fringes, so the total is conserved. But at that one dark point, two real waves genuinely add to zero displacement. "Two waves must make a bigger wave" is the classic wrong instinct.
The waves pass straight through each other and carry on unchanged. The overlap is only while they share the same space; a moment later each wave continues exactly as if the other had never been there. Waves don't collide and bounce off like billiard balls.

Superposition isn't just a textbook curiosity — you may be wearing it. A pair of noise-cancelling headphones has a tiny microphone that listens to the droning sound coming towards your ear (the rumble of a plane engine, say). The headphones instantly generate a second sound wave that is the exact antiphase of that noise — a crest wherever the noise has a trough. By the principle of superposition the two sound waves add to (almost) zero at your eardrum, and the roar melts into near-silence.

The very same idea paints the shimmering colours on a soap bubble or an oil slick on a wet road. Light reflects off both the top and the bottom of the ultra-thin film, and the two reflections interfere. Whether a given colour comes out bright or dark depends on the film's thickness (which sets the path difference) — so as the film thins and swirls, the colour you see shifts, and different wavelengths cancel in different places. Interference, written in rainbow.