Young's double-slit experiment

Shine one narrow source of light through two closely spaced slits and something water waves have done for centuries suddenly shows up on a screen of pure light: a striped pattern of bright and dark bands, alternating with almost machine-like regularity. This is Young's double-slit experiment, the single most decisive demonstration that light travels as a wave. Thomas Young performed it in 1801, when the dominant view still held light to be a stream of particles, and the neat fringe pattern he recorded could not be explained any other way. The experiment does more than settle a historical argument — it gives you a precise, testable geometry: two coherent sources, a measurable slit separation, a screen distance, and a fringe spacing that obeys one clean formula. This article builds that geometry from scratch, derives the fringe-width relation, and works two numerical problems in the pattern your exam will ask.

The core insight

Two coherent light sources spaced a distance $d$ apart produce, on a screen a distance $D$ away, alternating bright and dark fringes spaced $\beta=\lambda D/d$ apart, because the path difference $\Delta x = dx/D$ between the two rays sweeps through whole and half wavelengths as you move across the screen.

The core idea

Coherence is the non-negotiable starting condition. Two independent lamps, however identical, flicker in and out of phase with each other randomly, so any interference pattern they might momentarily produce is scrambled thousands of times a second and averages out to uniform brightness. Young's trick was to derive both slits' light from a single source, so $S_1$ and $S_2$ always keep a fixed phase relationship — this alone is what makes a stable, observable pattern possible.

The two wave trains overlap and superpose on the screen. Where crest meets crest (or trough meets trough), the waves reinforce and a bright fringe appears; where crest meets trough, the waves cancel and a dark fringe appears. Because $D\gg d$ in a real setup, the two rays $S_1P$ and $S_2P$ reaching any point $P$ on the screen are very nearly parallel, and simple triangle geometry gives the path difference directly in terms of the angle $\theta$ from the central axis:

\Delta x = d\sin\theta \approx \frac{dx}{D}

where: $d$ is the slit separation, $D$ is the slit-to-screen distance, $x$ is the distance of point $P$ from the central axis, and $\theta\approx x/D$ is the small angle subtended (valid because $D\gg d$ keeps $x$ small compared with $D$ near the centre of the pattern).

Bright and dark fringes follow directly from this path difference. A bright fringe (constructive interference) forms wherever the two rays arrive exactly in step, i.e. an integer number of full wavelengths apart; a dark fringe (destructive interference) forms wherever they arrive exactly out of step, an odd number of half-wavelengths apart:

\Delta x = n\lambda \ \ (\text{bright}), \qquad \Delta x = \left(n+\tfrac12\right)\lambda \ \ (\text{dark}), \qquad n=0,\pm1,\pm2,\ldots

where: $\lambda$ is the wavelength of the light and $n$ is the fringe order, counted outward from the central bright fringe ( $n=0$ ). The figure below sketches the geometry: two slits act as coherent sources, their overlapping wavefronts reinforce along certain directions and cancel along others, and the screen intercepts this pattern as a row of evenly spaced bright and dark bands.

Figure

Two coherent slits sending overlapping wavefronts to a distant screen where they form alternating bright and dark fringes

The derivation

Derivation AFringe width from the path-difference geometry

The goal is to find the spacing $\beta$ between two consecutive bright fringes on the screen, starting only from the path-difference expression $\Delta x = dx/D$ and the bright-fringe condition $\Delta x = n\lambda$ , using nothing beyond algebra — no wave-equation superposition is required for this geometric derivation.

Step 1. Locate the $n$ -th bright fringe. Substituting the bright-fringe condition $\Delta x=n\lambda$ into the path-difference formula $\Delta x = dx/D$ gives $\dfrac{dx_n}{D}=n\lambda$ , so the position of the $n$ -th bright fringe measured from the centre is $x_n=\dfrac{n\lambda D}{d}$ .

Step 2. Locate the next bright fringe, one order higher. Replacing $n$ with $n+1$ in the same relation gives $x_{n+1}=\dfrac{(n+1)\lambda D}{d}$ .

Step 3. Subtract to get the spacing between them. The fringe width is $\beta=x_{n+1}-x_n=\dfrac{(n+1)\lambda D}{d}-\dfrac{n\lambda D}{d}=\dfrac{\lambda D}{d}$ .

Step 4. Note that $n$ has cancelled entirely, so this spacing is the same for every pair of consecutive bright fringes — and, since dark fringes sit exactly midway between bright ones, the same $\beta$ also gives the spacing between consecutive dark fringes. The fringe pattern is therefore uniformly spaced (near the centre, where the small-angle approximation $\theta\approx x/D$ holds):

\beta = \frac{\lambda D}{d}

where: $\lambda$ is the wavelength, $D$ the slit-to-screen distance, and $d$ the slit separation.

Worked example

In a Young's double-slit experiment, the slits are separated by $d=0.5\ \text{mm}$ , the screen is $D=1.5\ \text{m}$ away, and the light used has wavelength $\lambda=600\ \text{nm}$ . Find the fringe width and the position of the 4th bright fringe; then find how far the 2nd dark fringe lies from the centre.

Step 1. Convert to SI: $d=0.5\times10^{-3}\ \text{m}$ , $D=1.5\ \text{m}$ , $\lambda=600\times10^{-9}\ \text{m}$ .

Step 2. Fringe width: $\beta=\dfrac{\lambda D}{d}=\dfrac{600\times10^{-9}\times1.5}{0.5\times10^{-3}}=\dfrac{9\times10^{-7}}{5\times10^{-4}}=1.8\times10^{-3}\ \text{m}=1.8\ \text{mm}$ .

Step 3. Position of the 4th bright fringe ( $n=4$ ): $x_4=n\beta=4\times1.8\ \text{mm}=7.2\ \text{mm}$ from the centre.

Step 4. For the 2nd dark fringe, numbering the first dark fringe adjacent to the centre as $n=0$ , the position is $x=(n+\tfrac12)\beta$ ; the "2nd dark fringe" is $n=1$ , so $x=\left(1+\tfrac12\right)\times1.8\ \text{mm}=1.5\times1.8\ \text{mm}=2.7\ \text{mm}$ .

Answer: the fringe width is 1.8 mm, the 4th bright fringe lies 7.2 mm from the centre, and the 2nd dark fringe lies 2.7 mm from the centre.

Competitive stretch

If the same setup were immersed so that the effective wavelength shortened to $\lambda'=450\ \text{nm}$ (as happens in a denser medium, where $\lambda$ scales down), the fringe width would shrink in direct proportion: $\beta'=\beta\times\dfrac{\lambda'}{\lambda}=1.8\times\dfrac{450}{600}=1.35\ \text{mm}$ — a reminder that $\beta$ depends on $\lambda$ , $D$ and $d$ only, so changing any one of them rescales every fringe position together.

Common traps

Using two separate lamps instead of one coherent source. Even identical bulbs drift in and out of phase randomly, so no stable pattern forms — YDSE works only because both slits are driven by a single source.
Forgetting that $\Delta x\approx dx/D$ needs $D\gg d$ . The small-angle approximation breaks down if the screen sits close to the slits.
Swapping the bright and dark conditions. Bright is $\Delta x=n\lambda$ (whole wavelengths); dark is $\Delta x=(n+\tfrac12)\lambda$ (half-integer wavelengths) — mixing these up is a common slip under time pressure.
Assuming fringe width changes with order $n$ . It does not: $\beta=\lambda D/d$ is constant, so every pair of neighbouring fringes near the centre is equally spaced.

Ask the tutor: want to work through how the fringe pattern changes if the whole setup is immersed in a medium of refractive index other than 1, or try a numerical where you're given the fringe width and asked to find $d$ instead of $\lambda$ ?

Why do we need two slits fed from a single source instead of just two separate lamps close together?

Interference needs the two waves arriving at any point to keep a fixed phase relationship over time. Two independent lamps emit light in unrelated, randomly fluctuating bursts, so the phase difference between them changes millions of times a second — any bright-dark pattern would shift too fast to see and would average out to uniform illumination. Deriving both slits from one source ties their phases together permanently, which is exactly what coherence means and why YDSE always starts from a single source split in two.

Why is the fringe width the same for bright and dark fringes if they come from different conditions

Because bright and dark fringes are not independent patterns — they are evenly interleaved parts of the same continuous path-difference sweep. As you move across the screen, the path difference $\Delta x=dx/D$ increases smoothly; it passes through $n\lambda$ (bright) and then, exactly half a wavelength later, through $(n+\tfrac12)\lambda$ (dark), before reaching the next bright fringe at $(n+1)\lambda$ . Since the spacing between successive bright fringes and between successive dark fringes is derived from the same linear relation, both work out to the identical $\beta=\lambda D/d$ .

Does the fringe pattern stay perfectly evenly spaced all the way to the edges of the screen?

Only approximately, and only near the centre. The formula $\Delta x\approx dx/D$ relies on the small-angle approximation $\theta\approx x/D$ , which holds well when $x$ stays much smaller than $D$ — true for the central portion of the pattern in a typical lab setup where $D$ is a metre or more and $d$ is a fraction of a millimetre. Far out toward the edges of a wide screen, the angle $\theta$ is no longer small, the approximation $\sin\theta\approx\theta$ weakens, and the true fringe spacing very slightly departs from the constant value $\lambda D/d$ . For exam purposes, treat the fringes as uniformly spaced.