Young's Double-Slit Experiment

In 1801 a physician with a taste for heresy took a beam of sunlight, split it in two, and let the halves overlap — and where common sense said "brighter" he found dark bands. Thomas Young's double-slit experiment is the moment light stopped being obviously a stream of particles: only waves can cancel, and cancel they did, in a ruler-straight ladder of bright and dark fringes. The experiment is disarmingly simple. One quasi-monochromatic source lights a barrier with two narrow parallel slits a distance $d$ apart; each slit re-radiates as a secondary source, and because both are carved from the same wavefront they keep a locked phase relationship. On a screen a distance $L$ beyond, the two waves superpose, and the intensity is not $I_1+I_2$ but a pattern that swings from four times a single slit's brightness down to pitch black. Everything hangs on one quantity: the optical path difference between the two routes to a given point on the screen, which for a distant screen is $d\sin\theta$ . Where that difference is a whole number of wavelengths the waves arrive in step and reinforce; where it is a half-integer they arrive opposed and cancel. Sweeping across the screen sweeps the path difference smoothly, and the result — in the small-angle regime that describes almost every real setup — is a set of equally spaced fringes with separation $\Delta y=\lambda L/d$ . Read that spacing off with a ruler and you have measured the wavelength of light, which is exactly what Young did. The same fringes double as the most sensitive coherence meter in optics: their contrast is the degree of coherence of the illuminating light, so the double slit is both the birth certificate of the wave theory and the reference experiment against which coherence is still defined.

Key takeaways

Two slits a distance $d$ apart, cut from one wavefront, act as mutually coherent sources; their overlapping waves interfere into bright and dark fringes on a screen a distance $L$ away.
Bright fringes sit where the path difference $d\sin\theta=m\lambda$ ; dark fringes where it is $(m+\tfrac12)\lambda$ .
In the small-angle limit the fringes are equally spaced by $\Delta y=\lambda L/d$ — so measuring the spacing measures the wavelength.
The intensity of two equal slits is $I=4I_0\cos^2(\pi d\sin\theta/\lambda)$ , swinging from $4I_0$ to $0$ — the plain fingerprint of superposition, with the fringe-averaged power still $2I_0$ .
Fringe contrast equals the degree of coherence $|\gamma_{12}|$ , so the double slit measures both temporal and spatial coherence of a source.
Key idea: two mutually coherent slit-waves superpose with a path difference $d\sin\theta$ that modulates the intensity as $\cos^2$ , producing equally spaced fringes of spacing $\lambda L/d$ that encode both the wavelength and the coherence of the light.

The core insight

Two mutually coherent waves from slits separated by $d$ superpose on a distant screen. Their optical path difference $d\sin\theta$ sets the relative phase $\delta=2\pi d\sin\theta/\lambda$ , so the intensity is $4I_0\cos^2(\delta/2)$ — bright where $d\sin\theta=m\lambda$ , dark where it is $(m+\tfrac12)\lambda$ . In the paraxial limit the fringe spacing is $\Delta y=\lambda L/d$ , which turns a ruler into a wavelength gauge, and the fringe contrast reports the degree of coherence.

Where this lives today

Young's geometry never stopped being the reference experiment. Feed a double slit one photon at a time and the fringes still build up, dot by dot, from individually detected particles — Taylor saw it with faint light in 1909, and it has since been repeated with electrons, neutrons, atoms, and molecules of a thousand atoms, making the double slit the canonical stage for quantum superposition and wave–particle complementarity: learn which slit a particle took and the fringes vanish. In metrology, the same fringe-counting that gave Young a wavelength now runs the other way — a stabilized laser of known wavelength turns fringe counts into displacement measurements of nanometre precision in interferometers on optical benches and semiconductor steppers. And in astronomy the double slit's spatial-coherence lesson scales to the sky: interferometers from Michelson's original stellar apparatus to the VLTI resolve stellar diameters and binary separations by watching fringe contrast fade as the baseline grows.

The physical picture

Drop two stones into a still pond a small distance apart and watch the circular ripples overlap. Along some directions crest always meets crest and the water heaves; along others crest meets trough and the surface lies flat. Those are the bright and dark fringes, frozen into a fan of directions fixed by the wavelength of the ripples and the spacing of the stones. The double slit does exactly this with light: two coherent circular waves, one from each slit, weave an interference pattern whose geometry is set entirely by $d$ , $\lambda$ and the viewing angle.

Figure

Two slits separated by d illuminated by one wavefront; rays to a screen point P differ in path by d·sinθ. Whole-wavelength differences give bright fringes, half-wavelength differences give dark.

The path difference is the whole plot. Stand at a point on the screen and ask how much farther the light from the far slit travelled than from the near one. If that surplus is $0,\lambda,2\lambda,\dots$ the waves are in step and you stand in a bright fringe; if it is $\lambda/2,3\lambda/2,\dots$ they are opposed and you stand in the dark. Because the surplus is $d\sin\theta$ , it grows smoothly with viewing angle, so the brightness runs periodically through bright–dark–bright as you scan. And because $\Delta y=\lambda L/d$ , a shorter wavelength or a wider slit pair crowds the fringes together while a farther screen spreads them out — the same reciprocal, Fourier-like relationship between separation and fringe spacing that runs through all of diffraction. Contrast, meanwhile, is a coherence meter: perfectly coherent light drives the dark fringes to exactly zero (visibility $1$ ); as the source broadens or its lines fatten, the darks fill in and the visibility falls.

Figure

The double-slit intensity I(y) = 4I₀·cos²(πd·y/λL) for λ = 633 nm, d = 0.20 mm, L = 1.5 m: equally spaced fringes of spacing Δy = λL/d = 4.75 mm, running from 4I₀ down to zero.

The punchline figure is the equation drawn to scale. The maxima sit at $y=0,\pm4.75,\pm9.5\,\text{mm}$ — exactly evenly spaced — because $\Delta y=\lambda L/d$ is independent of order; the intensity between them dips all the way to zero, the signature that the two beams are equal and fully coherent. Change any of the three knobs and the whole ladder rescales: halve $d$ and the fringes double their spacing to $9.5$ mm; swap to blue light and they crowd inward. That direct, linear grip of $\lambda$ , $L$ , and $d$ on a spacing you can measure with a millimetre rule is what made the experiment a measurement, not just a demonstration.

The core equations

The bright fringes mark directions where the two paths differ by a whole number of wavelengths. This single condition, read at a distant screen, generates the entire pattern.

d\sin\theta_m = m\lambda,\qquad m=0,\pm1,\pm2,\dots \qquad (1)

where: $d$ — centre-to-centre slit separation; $\theta_m$ — angle from the central axis to the $m$ -th bright fringe; $m$ — fringe order; $\lambda$ — wavelength in the medium. The dark fringes fall exactly halfway between, where the paths differ by a half-integer number of wavelengths, $d\sin\theta=(m+\tfrac12)\lambda$ . Projected onto the screen in the small-angle regime, the equally spaced orders give the fringe spacing that Young turned into a wavelength gauge.

\Delta y = \frac{\lambda L}{d} \qquad (2)

where: $\Delta y$ — linear spacing between adjacent bright (or dark) fringes; $L$ — slit-to-screen distance. Because $\Delta y$ carries no $m$ , the fringes are uniform, and the relation inverts to $\lambda=d\,\Delta y/L$ — three lengths you can measure directly. The full brightness profile follows from adding two equal-amplitude fields with a phase difference set by the path difference. It is the cleanest display of superposition in optics.

I(\theta) = 4I_0\cos^2\!\left(\frac{\pi d\sin\theta}{\lambda}\right),\qquad \delta = \frac{2\pi}{\lambda}\,d\sin\theta \qquad (3)

where: $I_0$ — intensity from one slit alone; $\delta$ — relative phase of the two slit waves; $I$ swings from $4I_0$ (in phase) to $0$ (opposed). Note the peak is four times a single slit, not twice: fields add, then square. Only finitely many orders can appear, because $\sin\theta$ cannot exceed one — a bound worth carrying even when, as here, it is a large number.

m_{max} = \left\lfloor \frac{d}{\lambda} \right\rfloor \qquad (4)

where: $m_{max}$ — highest order with $|\sin\theta_m|\le1$ . For real slits of finite width $a$ , this two-slit pattern is further multiplied by a single-slit diffraction envelope $\mathrm{sinc}^2(\pi a\sin\theta/\lambda)$ , which is developed in the diffraction chapter; the idealization here takes the slits infinitely narrow.

Derivation AFringe spacing from the path difference

The aim is to get from the raw geometry of two point sources and a distant screen to the two headline results — the bright-fringe condition $d\sin\theta=m\lambda$ and the uniform spacing $\Delta y=\lambda L/d$ — making explicit where the far-field and small-angle approximations enter, since almost every real double slit lives in exactly this regime. We take the slits $S_1,S_2$ separated by $d$ and a screen point $P$ at angle $\theta$ , with $L\gg d$ .

Step 1. The two routes to $P$ have lengths $r_1,r_2$ ; the physics is entirely in their difference $\Delta=r_2-r_1$ , since only the relative phase governs interference. Why: a common path length shifts both waves equally and cancels out, so only the surplus travelled by the far slit's light can create fringes.

Step 2. For $L\gg d$ the two rays to $P$ are essentially parallel, both at angle $\theta$ , so the surplus is the leg of a right triangle: $\Delta\approx d\sin\theta$ . Why: dropping a perpendicular from the near slit onto the far ray leaves a small triangle of hypotenuse $d$ and opposite angle $\theta$ — the far-field (Fraunhofer) approximation that makes the path difference a clean $d\sin\theta$ .

Step 3. A bright fringe needs the surplus to be a whole number of wavelengths, $d\sin\theta=m\lambda$ ; a dark fringe a half-integer, $(m+\tfrac12)\lambda$ . Why: whole-wavelength surplus brings the far wave back into step with the near one (crest on crest); half-wavelength surplus sets crest against trough.

Step 4. Project onto the screen: $y_m=L\tan\theta_m\approx L\sin\theta_m=m\lambda L/d$ for small $\theta$ , so adjacent orders are separated by $\Delta y=y_{m+1}-y_m=\lambda L/d$ , independent of $m$ . Why: with $\tan\theta\approx\sin\theta$ the order positions become linear in $m$ , so the fringes are evenly spaced — the property that lets you average many fringes to measure $\Delta y$ precisely. *Verified: $\Delta y=\lambda L/d=4.75$ mm for the punchline geometry, and the back-out $\lambda=d\,\Delta y/L=667$ nm for the second example, in verify_youngs-double-slit-experiment.py.*

Go deeper · Derivation B · 1 minThe cos² intensity and conservation of energy

Derivation BThe cos² intensity and conservation of energy

Here we build the full brightness profile by superposing the two slit fields, and then check that interference has only redistributed light — the fringe-averaged intensity must equal the incoherent sum, or we would be creating energy from geometry. We represent each slit's contribution at $P$ as an equal-amplitude scalar field, with a relative phase $\delta$ set by the path difference.

Step 1. Add the fields, not the intensities: $E_P=E_0(e^{i\phi_1}+e^{i\phi_2})=E_0 e^{i\phi_1}(1+e^{i\delta})$ , with $\delta=\phi_2-\phi_1$ . Why: superposition acts on the fields, and the detector will square the sum — the squaring is where the cross term appears.

Step 2. The intensity is $I\propto|E_P|^2=2E_0^2(1+\cos\delta)$ , i.e. $I=2I_0(1+\cos\delta)$ with $I_0\propto E_0^2$ . The $2I_0\cos\delta$ piece is the interference term; without it $I$ would be a flat $2I_0$ . Why: $|1+e^{i\delta}|^2=2+2\cos\delta$ , and the $\cos\delta$ is precisely the fringe-forming cross term of two-wave interference.

Step 3. Fold to half-angle form using $1+\cos\delta=2\cos^2(\delta/2)$ : $I=4I_0\cos^2(\delta/2)$ , running from $4I_0$ at $\delta=0$ (bright) to $0$ at $\delta=\pi$ (dark). Why: the half-angle identity turns the profile into a manifestly non-negative $\cos^2$ , whose peak of $4I_0$ shows two equal fields adding in amplitude before squaring.

Step 4. Average over one fringe: $\langle\cos^2(\delta/2)\rangle=\tfrac12$ , so $\langle I\rangle=2I_0$ — exactly the incoherent sum. Why: the surplus $2I_0$ at each bright fringe is paid for by a zero at each dark fringe, so interference relocates energy across the screen without creating or destroying any. *Verified: $I(0)=4I_0$ , $I(\Delta y/2)=0$ , and the fringe average $\langle I\rangle=2I_0$ , plus the identity $1+\cos\delta=2\cos^2(\delta/2)$ , in verify_youngs-double-slit-experiment.py.*

Limiting cases

Case	Result	Meaning
paraxial / small angle	$\Delta y\approx\lambda L/d$	$\sin\theta\approx\tan\theta=y/L$ : equally spaced fringes; valid for $d,\lambda\ll L$
wide angle (exact)	$d\sin\theta_m=m\lambda$	fringes crowd together as $\theta\to90^\circ$ ; highest order $m_{max}=\lfloor d/\lambda\rfloor$
finite slit width $a$	$I\propto\mathrm{sinc}^2(\pi a\sin\theta/\lambda)\,\cos^2(\pi d\sin\theta/\lambda)$	a single-slit envelope modulates the fringes; orders vanish when $d/a$ is an integer
partial coherence	$I=2I_0[1+\|\gamma_{12}\|\cos\delta]$ , $V=\|\gamma_{12}\|$	contrast equals the degree of coherence; fringes fade as $\|\gamma_{12}\|\to0$
one slit blocked	$I=I_0\,\mathrm{sinc}^2(\dots)$	no path-difference cross term — the fine fringes disappear, leaving only diffraction

Common trap“The bright fringes are where each beam is individually brightest.”

They are not. The bright fringes fall where the two waves are in phase, and the dark fringes where they are exactly opposed — a location has nothing to do with either beam's own strength there. Two closely related slips reinforce the error. First, do not confuse the two-slit interference spacing $\Delta y=\lambda L/d$ (set by the slit separation $d$ ) with the single-slit diffraction envelope of angular width $\propto\lambda/a$ (set by the slit width $a$ ); the fine fringes and the broad envelope are different physics with different length scales. Second, the tidy condition $d\sin\theta=m\lambda$ assumes the incident wavefront hits both slits in phase — at normal incidence. Tilt the illumination and a constant phase offset shifts the entire pattern sideways, which is not an error but a genuine effect (and the basis of steering fringes).

Worked examples

1 · Fringe spacing of a red-laser double slit (confidence builder)

A helium–neon laser ( $\lambda=633$ nm) illuminates two slits separated by $d=0.20$ mm. The screen sits $L=1.5$ m away. Find the fringe spacing, the positions of the first few maxima, and how many fringes fall within a $\pm3^\circ$ fan.

Step 1. Fringe spacing straight from the paraxial result: $\Delta y=\lambda L/d=(633\times10^{-9}\,\text{m})(1.5\,\text{m})/(0.20\times10^{-3}\,\text{m})=4.75\times10^{-3}\,\text{m}=\mathbf{4.75\ mm}$ . The maxima therefore land at $y=0,\pm4.75,\pm9.5,\dots$ mm — the ladder in the punchline figure.

Step 2. How many orders fit within $\theta=3^\circ$ ? The order there is $m=d\sin3^\circ/\lambda=(0.20\times10^{-3})(0.0523)/(633\times10^{-9})=16.5$ , so orders up to $m=16$ appear on each side. Counting the central fringe, that is $2\times16+1=\mathbf{33}$ bright fringes within the $\pm3^\circ$ fan.

Step 3. The angles are genuinely small — the first-order fringe sits at $\theta_1=\arcsin(\lambda/d)=0.181^\circ$ — which is why the small-angle result is essentially exact here and the fringes are evenly spaced. Δy = 4.75 mm; 33 fringes within ±3°.

2 · Turning fringes into a wavelength (exam level)

In a teaching lab, a student measures a fringe spacing of $\Delta y=2.00$ mm with slits separated by $d=0.50$ mm and a screen at $L=1.50$ m. What wavelength did they use, and roughly what colour is it?

Step 1. Invert the fringe-spacing relation: $\lambda=d\,\Delta y/L$ — the entire measurement is three lengths and a division, which is precisely why Young could do it in 1803 without any modern instrument.

Step 2. Put in the numbers: $\lambda=(0.50\times10^{-3}\,\text{m})(2.00\times10^{-3}\,\text{m})/(1.50\,\text{m})=6.67\times10^{-7}\,\text{m}=\mathbf{667\ nm}$ — deep red, consistent with a red LED or laser.

Step 3. The precision lever is to measure many fringes at once: spanning $N=20$ fringes and dividing by $20$ shrinks the ruler error twentyfold, because the fringes are exactly evenly spaced. This is how a crude ruler yields a wavelength good to a few percent. λ = 667 nm (red).

Physics of the portrait

Young's experiment is remembered as a demonstration, but its enduring importance is as a measurement — the first credible value for the wavelength of light — and as the template for every coherence and matter-wave interference experiment since.

Origins · 1 minThomas Young and how this was discovered

Thomas Young

1773–1829 · London (Royal Institution)

Thomas Young was a physician, linguist, and relentless polymath who read at two, practised medicine in London, and did physics on the side with an amateur's freedom and a genius's reach. Around 1801–1804, lecturing at the Royal Institution, he revived the wave theory of light that Newton's authority had suppressed for a century, and clinched it with interference: light plus light can make darkness, which no particle picture allowed. His double-slit and related experiments let him measure the wavelength of visible light — a few hundred nanometres — the first quantitative optical interferometry. The idea met fierce resistance (Newton was still gospel), and Young's 1817 suggestion to Arago that light's vibrations are transverse — needed to explain why crossed-polarized beams refuse to interfere — was as bold as the interference itself. His curiosity ranged absurdly wide: the elastic modulus that bears his name; the physiological three-receptor theory of colour vision (later Young–Helmholtz); tidal and capillary-action theory; and a decisive share in deciphering the Rosetta Stone's demotic script. He is the rare figure whose name is attached to landmark results in optics, mechanics, physiology, and Egyptology at once.

History

1665

— Grimaldi reports light bending into shadow (diffraction), hinting at wave behaviour

1801

— Young delivers his Bakerian Lecture proposing the interference of light

1803

— Young demonstrates two-source interference fringes and estimates the wavelength of light

1817

— Young proposes to Arago that light vibrations are transverse, explaining polarization

1909

— G. I. Taylor records fringes with light so feeble that photons pass essentially one at a time

1961

— Jönsson demonstrates the same fringe pattern with electrons through a double slit

Going further

Texts. Hecht, Optics, 5th ed., §9.3 (Young's Experiment) · Born & Wolf, Principles of Optics, 7th ed., §7.3 and §10.4 · Saleh & Teich, Fundamentals of Photonics, 3rd ed., §11.1 · Loudon, The Quantum Theory of Light, 3rd ed., ch. 3 · Feynman, Lectures on Physics, Vol. III, ch. 1 (the double slit as the heart of quantum mechanics).

Provenance. The maxima condition $d\sin\theta=m\lambda$ , the spacing $\Delta y=\lambda L/d$ , and the intensity $I=4I_0\cos^2(\pi d\sin\theta/\lambda)$ follow Hecht §9.3 and Born & Wolf §7.3; the visibility–coherence identity $V=|\gamma_{12}|$ from Saleh & Teich ch. 11. Every worked number is recomputed in verify_youngs-double-slit-experiment.py: $\Delta y=4.75$ mm for $\lambda=633$ nm, $d=0.20$ mm, $L=1.5$ m; the wavelength back-out $\lambda=d\,\Delta y/L=667$ nm; $16$ orders within $3^\circ$ (33 fringes); the fringe average $\langle I\rangle=2I_0$ ; and $m_{max}=\lfloor d/\lambda\rfloor=315$ for the punchline geometry. Young's original wavelength estimate (a few hundred nanometres) is standard textbook history.

Prerequisites. Superposition Principle of Electromagnetic Waves · Optical Path Difference and Phase Retardation · Two-Wave Interference and the Interference Term · Conditions for Observable Interference. Leads to. Fringe Visibility and Contrast · Spatial Coherence and the Van Cittert–Zernike Theorem · Temporal Coherence and Coherence Length · Wavefront-Splitting and Amplitude-Splitting Interferometers.

Why is the central bright fringe exactly in the middle, and always bright?

At the point equidistant from the two slits the path difference is zero, so the two waves arrive perfectly in step regardless of wavelength — that is the $m=0$ order, and it is bright for every colour at once. This is why in white light the central fringe is white while the fringes on either side smear into spectra: the higher orders occur at $d\sin\theta=m\lambda$ , which lands at a different angle for each wavelength, but the zeroth order pins them all together at the centre. Finding that white central fringe is the standard way to locate zero path difference in an interferometer.

Does the pattern change if I use fatter slits?

Yes, but only its envelope, not its fringe spacing. The fringe spacing $\Delta y=\lambda L/d$ depends on the slit separation $d$ , so widening each slit leaves the fine fringes where they are. What changes is that each slit now diffracts into a narrower cone (angular width $\sim\lambda/a$ for slit width $a$ ), so a single-slit envelope $\mathrm{sinc}^2$ modulates the fringes — the outer fringes dim, and whole orders can go missing when $d/a$ is an integer (a "missing order"). The idealized $\cos^2$ pattern assumes infinitely narrow slits; real data show the fringes riding under this broader diffraction hump.

Can two separate lasers make a double-slit-like pattern?

Not a stable one to the eye. The double slit works because both beams are carved from a single wavefront, so their phase fluctuations are common and cancel — the essence of the coherence condition. Two independent lasers keep a fixed relative phase only for their coherence time (microseconds at best for free-running lasers), so their fringes drift and a normal detector averages them away. With a fast enough detector, or by phase-locking the two lasers, you can recover real interference — but the point of Young's single-source geometry is that it makes coherence automatic.

How did Young get a wavelength without modern equipment?

By measuring three lengths and dividing. The fringe spacing relation $\lambda=d\,\Delta y/L$ needs only the slit separation $d$ , the fringe spacing $\Delta y$ , and the slit-to-screen distance $L$ — all lengths a careful experimenter can measure with a scale. The trick to precision is to count many fringes across a wide span and divide, since the fringes are exactly evenly spaced; that averages down the ruler error. Young's result, a few hundred nanometres, was the first quantitative value for the wavelength of light and stood as a triumph of the wave theory.

Why is the maximum four times a single slit, not two times?

Because you add the fields and then square. Two slits each contribute a field amplitude $E_0$ ; in phase they sum to $2E_0$ , and the intensity — proportional to the square of the field — is $(2E_0)^2=4E_0^2=4I_0$ . If the beams were incoherent you would add intensities, getting $2I_0$ . The extra $2I_0$ at every bright fringe is exactly balanced by the zeros at the dark fringes, so the fringe-averaged intensity is $2I_0$ and energy is conserved — interference redistributes light, it does not manufacture it.

What happens to the fringes if the screen is moved farther away?

They spread out in direct proportion. Since $\Delta y=\lambda L/d$ , doubling $L$ doubles the fringe spacing, so a distant screen makes the pattern easier to measure but dimmer (the same power is spread over a larger area). The angular structure — the directions $\theta_m$ of the maxima, fixed by $d\sin\theta=m\lambda$ — does not depend on $L$ at all; moving the screen only changes where those fixed directions intercept it. This is why interferometrists quote fringe angles for the source-independent geometry and fringe spacings for a particular screen.