Interference and the Double Slit

Setting up the experiment

Place an opaque barrier with two narrow slits — call them slit A and slit B — a distance $d$ apart. A monochromatic plane wave of wavelength $\lambda$ illuminates the barrier from the left. A detection screen sits a distance $L$ far to the right, with $L \gg d$ . This is Thomas Young's original 1801 setup, which provided the first decisive evidence for the wave nature of light.

Path-length difference

Consider a point P on the screen at angle $\theta$ above the central axis. Light reaching P from slit A travels a slightly different path length than light from slit B. Because the screen is far away ( $L \gg d$ ), the two rays heading toward P are effectively parallel, and the path-length difference is

$\Delta \ell = d \sin\theta.$

This is the key quantity that determines whether the two contributions add up or cancel.

Constructive and destructive interference

Two waves of equal amplitude arrive at P with a phase difference

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

The total electric field at P is the sum of the two waves. When $\delta$ is a multiple of $2\pi$ (equivalently, when $\Delta\ell$ is an integer multiple of $\lambda$ ), the crests align — constructive interference — and the intensity is maximised:

$\text{bright fringe:} \quad d\sin\theta_m = m\lambda, \quad m = 0, \pm 1, \pm 2, \ldots$

When $\delta$ equals an odd multiple of $\pi$ (i.e., $\Delta\ell$ is a half-integer multiple of $\lambda$ ), crest meets trough — destructive interference — and the intensity falls to zero:

$\text{dark fringe:} \quad d\sin\theta_m = \left(m + \tfrac{1}{2}\right)\lambda, \quad m = 0, \pm 1, \pm 2, \ldots$

The intensity pattern

For two identical, coherent slits, the superposition of the two waves at P gives an electric field $E \propto \cos(\delta/2)$ . Intensity is proportional to $E^2$ , so

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

where $I_0$ is four times the single-slit intensity (both amplitudes add in phase at the centre). The factor of four — not two — is a signature of wave interference: energy is not destroyed at dark fringes but redistributed to the bright ones.

On the screen the pattern appears as a sequence of equally-spaced bright and dark bands. In the small-angle limit $\sin\theta \approx \tan\theta \approx y/L$ , where $y$ is the distance from the centre of the screen, the fringe spacing is

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

So a wider slit separation $d$ crowds the fringes together, and a longer wavelength $\lambda$ spreads them apart.

Why this matters for quantum mechanics

Young's experiment works identically with electrons, neutrons, atoms, and even large molecules — even when the particles are sent through one at a time so that only a single particle is in the apparatus. Individual particles land at seemingly random spots, but after many particles the familiar interference fringes build up. This means the interference is not between two different particles: each particle interferes with itself.

Classical wave optics gives a complete account of the fringe positions (the $d\sin\theta = m\lambda$ condition) but cannot explain what happens particle by particle. That explanation requires quantum mechanics — and the double-slit experiment is the canonical demonstration that quantum particles behave as waves until they are detected.