Interference and Coherence

Drop two pebbles into a still pond, side by side, and watch the ripples spread. Where two crests meet they pile up into a taller crest; where a crest lands on a trough the water barely stirs. The surface fills with a fixed lattice of choppy lanes and glassy-calm lanes fanning out from the pair of splashes. Nothing was added to the water — the same two ripple trains are simply overlapping, adding where they agree and cancelling where they disagree. That is interference, and it is one of the surest fingerprints that something is a wave.

The very same thing happens with sound (the "dead spots" as you walk between two speakers playing the same tone) and, most spectacularly, with light: shine light through two fine slits and a screen beyond lights up not with two bright blobs but with a whole striped pattern of bright and dark bands. This page is about that pattern — where the bright and dark bands land, the single number that decides it (the path difference), and the quiet condition the two sources must satisfy for any pattern to appear at all: coherence. It builds directly on the travelling waves that add up to make it happen.

Two waves, added: it all comes down to path difference

Take two sources, S_1 and S_2, sending out identical waves of the same wavelength \lambda. Pick any point P in the region where the two waves overlap. Each wave had to travel some distance to reach P — call them r_1 and r_2. The waves left their sources marching in step (we will insist on that in a moment), so the only thing that can throw them out of step by the time they arrive is the difference in how far they travelled. That difference is the star of the whole show, the path difference:

\Delta = \lvert r_1 - r_2 \rvert.

Here is the crucial move: what matters is not r_1 or r_2 on their own, but how the path difference compares to the wavelength. Slide one wave forward by exactly one whole wavelength and it looks identical — crest still on crest. So:

You can feel this in the diagram below. Two identical waves are drawn faintly; drag the phase difference \varphi between them and watch their sum (the bold curve) swell to double height at \varphi = 0 and flatten to nothing at \varphi = \pi.

The canonical example: Young's double slit

In 1801 Thomas Young sent a single beam of light through two closely spaced slits and let it fall on a screen. If light were simply a stream of particles you would expect two bright lines. Instead he saw a row of evenly spaced bright and dark fringes — interference, and the first hard proof that light is a wave. It remains the cleanest way to see the path-difference rules in action.

The two slits, a distance d apart, act as our two in-step sources. For a screen far away, the two rays heading toward a point at angle \theta from the straight-ahead direction are very nearly parallel, and a neat right triangle at the slits shows the extra distance the lower ray must cover: the path difference is \Delta = d\sin\theta. Reveal the figure step by step to build that triangle.

Feeding \Delta = d\sin\theta into the constructive condition \Delta = m\lambda gives the angles of the bright fringes:

d\sin\theta = m\lambda, \qquad m = 0, \pm 1, \pm 2, \dots

The m = 0 fringe sits dead ahead (\theta = 0, zero path difference — always bright), with the higher orders fanning out symmetrically on either side. The dark fringes fall between them, where d\sin\theta = \left(m + \tfrac12\right)\lambda.

On a screen a distance L away, a fringe at angle \theta lands at height y = L\tan\theta. Because the fringe angles are tiny (\lambda is far smaller than d), we can use \sin\theta \approx \tan\theta \approx \theta, so y_m \approx mL\lambda/d. Consecutive bright fringes are therefore evenly spaced, and the gap between them — the fringe spacing — is the number you actually measure on the screen:

Worked examples

Example 1 — fringe spacing. Green laser light, \lambda = 550\ \text{nm}, passes through two slits d = 0.20\ \text{mm} apart onto a screen L = 1.5\ \text{m} away. How far apart are the bright fringes?

\Delta y = \frac{\lambda L}{d} = \frac{(550\times 10^{-9}\ \text{m})(1.5\ \text{m})}{0.20\times 10^{-3}\ \text{m}} = 4.1\times 10^{-3}\ \text{m} \approx 4.1\ \text{mm}.

A few millimetres — comfortably visible. Notice the answer barely depends on the messy powers of ten once you keep the units straight: 550\times 1.5 / 0.2 = 4125, then shift the exponents.

Example 2 — the angle of a fringe. For that same setup, at what angle does the first-order (m = 1) bright fringe appear?

\sin\theta = \frac{m\lambda}{d} = \frac{(1)(550\times 10^{-9})}{0.20\times 10^{-3}} = 2.75\times 10^{-3},\qquad \theta \approx 0.16^\circ.

A tiny angle — which is exactly why the small-angle approximation behind \Delta y = \lambda L/d is so safe here.

Example 3 — reading a path difference. At some point on the screen the two waves arrive with a path difference of \Delta = 2.5\lambda. Bright or dark? Since 2.5 = 2 + \tfrac12 is a half-integer, this is \left(m + \tfrac12\right)\lambda with m = 2: destructive — a dark fringe. Had it been \Delta = 3\lambda (a whole number), it would be bright.

Example 4 — a diffraction grating angle. Interference is not confined to tiny angles. A grating with lines just d = 2.0\ \mu\text{m} apart is lit by \lambda = 500\ \text{nm} light. The first-order beam comes off at:

\sin\theta = \frac{m\lambda}{d} = \frac{500\times 10^{-9}}{2.0\times 10^{-6}} = 0.25,\qquad \theta \approx 14.5^\circ.

When d is only a few wavelengths, the orders splay out widely — which is how a grating fans white light into a rainbow.

Watch out — this is the trap that makes interference sound like magic (or like it breaks the conservation of energy). It does neither. Destructive interference does not destroy energy; it redistributes it. The light (or sound, or water) that is missing from a dark fringe is not gone — it has been shovelled into the bright fringes, which are correspondingly brighter than either wave alone.

Add up the intensity over the whole screen and you get exactly the total energy the two sources put out — the same as if there were no interference, just smeared evenly. Two waves of amplitude a reaching a bright fringe make amplitude 2a, and because intensity goes as amplitude squared, that fringe carries 4 times the intensity of one wave — not 2. The extra factor is precisely the energy borrowed from the dark fringes. Interference is a bookkeeping of where the energy lands, never a leak in the ledger.

Coherence: why the pattern needs a steady handshake

We have quietly been assuming the two sources leave "in step". That assumption has a name — coherence — and it is the make-or-break condition for seeing any interference at all. Two sources are coherent when they keep a stable phase relationship over time (which requires, among other things, the same frequency). If source 2 is always, say, a quarter-cycle behind source 1, the bright and dark fringes sit still and you see a crisp pattern. If instead the relationship jitters randomly, the fringes jump around faster than any eye or detector can follow, and everything blurs into a uniform, boring glow.

This is exactly why two separate light bulbs never make interference fringes. Each bulb emits countless independent little bursts of light from countless jostling atoms, and the overall phase reshuffles randomly billions of times a second. There is an interference pattern at any given instant — it just relocates so fast that all you ever record is the average: plain brightness. The two sources are incoherent.

A laser is the opposite: its atoms are marshalled into emitting in lockstep, so its light stays in phase with itself over a long distance. That is why Young's experiment is trivial with a laser — and why the honest way to get two coherent sources is to take one source and split it in two (send one beam through two slits, or through a half-silvered mirror). Both halves inherit the same phase history, so whatever random wobbles the source has, they wobble together and cancel out of the path difference.

Even a laser is not perfectly coherent forever. The distance over which a wave keeps a predictable phase is its coherence length, L_c. Real light comes in finite wave-trains with a small spread of wavelengths \Delta\lambda, and the two are tied together by L_c \approx \lambda^2/\Delta\lambda: the purer the colour (smaller \Delta\lambda), the longer the coherence length.

A tungsten bulb manages only a micron or so — a couple of wavelengths — which is why you must use almost-touching slits to see anything. A good laser can stay coherent over metres or even kilometres. The rule of thumb for the lab: interference stays visible only while the path difference \Delta is smaller than the coherence length. Overrun L_c and the two waves at P come from parts of the beam that no longer remember each other's phase — and the fringes wash out.