Young's Double Slit Experiment

In short

Shine monochromatic light through two narrow parallel slits separated by distance d, and look at a screen at distance D beyond them (D \gg d). The two slits act as coherent sources. Waves from them travel different distances to any point on the screen — a path difference \Delta. Where \Delta is an integer number of wavelengths, the two waves arrive in step and reinforce (bright fringe). Where \Delta is an odd number of half-wavelengths, they arrive out of step and cancel (dark fringe).

Path difference at screen distance y from the central axis (small-angle approximation):

\boxed{\;\Delta \;=\; \frac{y d}{D}\;}

Bright fringe condition (constructive):

\Delta \;=\; n\lambda \qquad (n = 0, \pm 1, \pm 2, \ldots)

Dark fringe condition (destructive):

\Delta \;=\; \left(n + \tfrac{1}{2}\right)\lambda \qquad (n = 0, \pm 1, \pm 2, \ldots)

Fringe width — the distance between consecutive bright fringes (or consecutive dark fringes):

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

Why this proved light is a wave. Interference — the cancellation of light by light — cannot happen in a particle model. Two rivers of bullets cannot produce regions empty of bullets. Only waves can produce dark bands by destructive cancellation of their amplitudes. The experiment Thomas Young performed in 1801 is the single most compelling demonstration of the wave nature of light in all of physics, and its modern quantum-mechanical echo (single-electron interference) is the single most compelling demonstration of the wave nature of matter.

Scratch two parallel slits into a piece of black paper with a razor blade. Hold a red laser pointer a metre behind the paper and aim the beam at the slits. On the wall another metre beyond, a set of bright and dark vertical stripes spreads across a span of several centimetres. Move the slits closer together — the stripes spread farther apart. Move the slits farther apart — the stripes cram together. Swap the red laser for a green one — every stripe slides across the wall as the pattern reshuffles. Cover one slit with your finger — the stripes vanish and you see a single diffuse blob of light.

This is the double-slit experiment. Any Class 12 physics lab in India will do it for you with a sodium lamp and a pair of slits on a mounted card; you will see sodium-yellow stripes on a screen and count their spacing with a travelling microscope. Every number the experiment produces — the stripe spacing, the positions of the dark bands, the way the pattern changes when you swap wavelengths — falls out of one idea: light is a wave, and waves interfere.

The setup and what you observe

Apparatus. A monochromatic light source (a sodium lamp, a laser pointer, or — in Young's original experiment — sunlight passed through a single pinhole). A first slit S_0 that selects a narrow coherent source. Two closely spaced parallel slits S_1 and S_2, separation d (typically a fraction of a millimetre). A screen at distance D beyond the two slits (typically \sim 1 m). Everything is in a dark room.

The double-slit setup. Monochromatic light is selected through the single slit $S_0$ to produce a coherent wavefront. This wavefront passes through the two closely spaced slits $S_1, S_2$, which act as new coherent secondary sources. The circular waves from $S_1$ and $S_2$ overlap on the screen a distance $D$ beyond, producing a pattern of alternating bright and dark fringes.

What you see on the screen. A set of parallel bright and dark fringes, spaced evenly, centred on the geometric midpoint between the two slits. The central fringe (labelled n = 0) is bright. On either side, alternating dark and bright fringes extend outward, all parallel, all the same spacing — until the intensity fades into the background far from the centre (an effect of single-slit diffraction covered in Diffraction at a Single Slit).

What changes when you change the setup.

Smaller slit separation d → fringes spread farther apart.
Larger screen distance D → fringes spread farther apart.
Shorter wavelength (bluer light) → fringes crowd together.
Longer wavelength (redder light) → fringes spread apart.
Covering one slit → the pattern disappears; only the diffuse diffraction blob from the remaining slit survives.

These observations rule out every particle-based explanation of light. A stream of corpuscles cannot produce dark bands by cancellation with another stream of corpuscles. A wave theory — and only a wave theory — produces interference.

The path-difference geometry

Why does any particular point on the screen receive constructive or destructive interference? Because the two waves, one from S_1 and one from S_2, arrive at the same point having travelled slightly different distances. The difference in path is called the path difference \Delta.

Path difference geometry. The ray from $S_2$ must travel an extra distance $\Delta = d\sin\theta$ compared with the ray from $S_1$ to reach the point $P$ at height $y$ on the screen. For small angles, $\sin\theta \approx y/D$, so $\Delta \approx yd/D$.

Step 1. Let P be a point on the screen at height y above the central axis. The line from P back to the midpoint of the two slits makes an angle \theta with the axis, where \tan\theta = y/D.

Step 2. Drop a perpendicular from S_1 to the line joining S_2 to P. The extra path travelled by the ray from S_2 (compared to the ray from S_1) is the hypotenuse of a tiny right triangle whose legs are d and the offset along the ray.

For distances D \gg d (which is always true — D is a metre, d is half a millimetre), the two rays from S_1 and S_2 to P are almost parallel, and the extra path is

\Delta \;=\; d \sin\theta

Why: in the "far-field" limit, the angle \theta subtended by P at each slit is essentially the same. The geometric construction — drop a perpendicular from S_1 to the S_2-to-P line — gives a right triangle with hypotenuse d and the side opposite \theta equal to d\sin\theta. This is the extra distance S_2's wave has to cover.

Step 3. For small \theta (every fringe of interest lies within a few degrees of the axis), \sin\theta \approx \tan\theta = y/D. Substitute:

\boxed{\;\Delta \;=\; d\sin\theta \;\approx\; \frac{yd}{D}\;}

Why: the small-angle approximation. Close to the central axis, \sin\theta, \tan\theta, and \theta itself (in radians) are all essentially equal. For a screen 1 m away and fringes a centimetre off-axis, \theta \approx 0.01 rad and the approximation is good to one part in 10^4.

This is the master formula of the double-slit experiment. Every fringe position follows from it.

Bright and dark fringes

At any point P on the screen, the total electric-field amplitude is the sum of contributions from S_1 and S_2. Write them as two sinusoidal waves of the same angular frequency \omega and wavelength \lambda, same amplitude E_0, but with different path lengths to P:

E_1 \;=\; E_0 \cos(\omega t - k r_1)

E_2 \;=\; E_0 \cos(\omega t - k r_2)

where k = 2\pi/\lambda and r_1, r_2 are the distances from S_1, S_2 to P. The total field is E_1 + E_2, and the intensity you see on the screen is the time-averaged square of the amplitude.

Constructive interference — bright fringes. If the path difference \Delta = r_2 - r_1 is an integer multiple of the wavelength:

\Delta \;=\; n\lambda \qquad (n = 0, \pm 1, \pm 2, \ldots)

then the two waves reach P with the same phase — crests on top of crests, troughs on top of troughs. The amplitudes add: E_1 + E_2 = 2E_0\cos(\omega t - k r_1). The intensity is 4\,I_0, where I_0 is the intensity one slit would give alone.

Destructive interference — dark fringes. If the path difference is an odd multiple of a half-wavelength:

\Delta \;=\; \left(n + \tfrac{1}{2}\right)\lambda \qquad (n = 0, \pm 1, \pm 2, \ldots)

then the waves arrive exactly out of phase — crests on troughs, troughs on crests. They cancel: E_1 + E_2 = 0. Intensity is zero. This is the remarkable feature: two real waves, both carrying energy, add up to no light at all at this point.

Positions of the fringes

Substitute \Delta = yd/D into the conditions.

Bright fringe at height y_n^{\text{bright}}:

\frac{y_n^{\text{bright}} d}{D} \;=\; n\lambda \;\Rightarrow\; y_n^{\text{bright}} \;=\; \frac{n\lambda D}{d}

Dark fringe at height y_n^{\text{dark}}:

y_n^{\text{dark}} \;=\; \left(n + \tfrac{1}{2}\right)\frac{\lambda D}{d}

Notice how evenly they're spaced. The distance between the n-th and (n+1)-th bright fringe is

\beta \;=\; y_{n+1}^{\text{bright}} - y_n^{\text{bright}} \;=\; \frac{(n+1)\lambda D}{d} - \frac{n\lambda D}{d} \;=\; \frac{\lambda D}{d}

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

Why: the spacing \beta is the "fringe width." It is the same between any two consecutive bright (or consecutive dark) fringes because the condition \Delta = n\lambda places them at equal intervals in y. Note \beta does not depend on which fringe n you pick.

This single formula is the working tool of Young's experiment. Change \lambda → fringes change spacing. Change D → fringes change spacing. Change d → fringes change spacing. Count \beta and D and d on a lab bench, and you get \lambda — the wavelength of light. That is how the wavelengths of the sodium doublet (589.0 nm and 589.6 nm) were first measured.

Explore: drag the slit separation

Below is an interactive plot of fringe width \beta against slit separation d for fixed wavelength \lambda = 600 nm (red laser) and screen distance D = 1.5 m. Drag the red point along the horizontal axis to change d from 0.1 mm to 2 mm. The curve plots \beta(d) from the formula; the readout shows numerical values. Notice that \beta diverges as d \to 0 (a single slit's fringes fill the whole screen) and shrinks to near-invisibility as d grows large (fringes too close to see).

Fringe width $\beta$ as a function of slit separation $d$ for $\lambda = 600$ nm (red) and $D = 1.5$ m. The relation is a rectangular hyperbola: halving $d$ doubles $\beta$. Drag the red point to see how a typical lab-bench setup tunes fringe visibility — too small a $d$ gives fewer, wider fringes; too large a $d$ gives a tightly packed pattern that blurs beyond the resolving power of the eye.

Intensity profile on the screen — the full answer

The bright/dark classification is a snapshot. The actual intensity at height y varies continuously between zero (fully destructive) and 4I_0 (fully constructive). Write the phase difference between the two waves at P:

\phi \;=\; k\Delta \;=\; \frac{2\pi \Delta}{\lambda} \;=\; \frac{2\pi y d}{\lambda D}

The total amplitude at P is the sum of two sinusoids of the same amplitude E_0 with phase difference \phi. By the cosine sum formula,

E_1 + E_2 \;=\; 2E_0\cos(\phi/2)\cos(\omega t - k(r_1 + r_2)/2)

The intensity is proportional to the time-average of the square:

I \;=\; 4I_0 \cos^2(\phi/2) \;=\; 4 I_0 \cos^2\!\left(\frac{\pi y d}{\lambda D}\right)

Why: the sum of two sinusoids at phase difference \phi is itself a sinusoid of amplitude 2E_0\cos(\phi/2). Squaring and time-averaging kills the fast-oscillating \cos^2(\omega t - \ldots) (which averages to 1/2) and leaves the envelope 4I_0\cos^2(\phi/2). This is the interference intensity profile — a smooth \cos^2 pattern on the screen.

Check the limits:

At y = 0: \phi = 0, \cos^2(0) = 1, I = 4I_0. Central bright fringe.
At y = \beta/2: \phi = \pi, \cos^2(\pi/2) = 0, I = 0. First dark fringe.
At y = \beta: \phi = 2\pi, \cos^2(\pi) = 1, I = 4I_0. First-order bright fringe.

The pattern is 4I_0\cos^2 — all maxima of equal height, all minima going to exactly zero. (Real patterns dim towards the edges because of single-slit diffraction, treated in Diffraction at a Single Slit; but inside the "diffraction envelope," the fringe intensities are equal in the idealised Young experiment.)

Worked examples

Example 1: The classic Class 12 lab with a sodium lamp

A physics lab uses a sodium lamp (\lambda = 589 nm) and two slits separated by d = 0.5 mm. The screen is placed D = 1.2 m beyond the slits. Compute the fringe width. If a student counts 20 bright fringes across a span on the screen, how wide is that span?

Standard Class 12 sodium-lamp setup. Yellow fringes spread across the screen with well-defined spacing.

Step 1. Convert all lengths to SI (metres).

\lambda = 589 \times 10^{-9} m, d = 0.5 \times 10^{-3} m, D = 1.2 m.

Why: working in mixed units is the single largest source of errors on wave optics problems. Convert everything to SI first, compute, then convert back at the end for presentation.

Step 2. Apply the fringe-width formula.

\beta \;=\; \frac{\lambda D}{d} \;=\; \frac{(589 \times 10^{-9})(1.2)}{0.5 \times 10^{-3}}

\beta \;=\; \frac{706.8 \times 10^{-9}}{0.5 \times 10^{-3}} \;=\; 1.414 \times 10^{-3} \text{ m} \;=\; 1.41 \text{ mm}

Why: direct substitution. The answer should be in millimetres for a typical lab setup — much smaller than a centimetre, much larger than a micrometre. 1.4 mm is exactly what a travelling microscope can measure to three significant figures.

Step 3. Twenty bright fringes span 19 fringe widths (from the 0th fringe to the 19th, there are 19 gaps).

\text{Span} \;=\; 19 \beta \;=\; 19 \times 1.41 \text{ mm} \;\approx\; 26.9 \text{ mm}

Why: counting fringes, start with the first at position 0 and end with the twentieth at position 19\beta. A common error is to use 20 instead of 19 — off by one fringe.

Result: Fringe width \beta = 1.41 mm. Twenty bright fringes span about 27 mm. Well-resolved with a travelling microscope.

What this shows: The numbers \lambda, D, and d each contribute linearly to \beta. The answer in millimetres, as measured on a lab bench, emerges naturally from SI-unit inputs. Young's experiment is self-calibrating: knowing three of these four quantities gives you the fourth, and that is how the sodium doublet wavelength was measured in the first place.

Example 2: Distinguishing red and green laser pointers

A green laser pointer (\lambda_g = 532 nm) and a red laser pointer (\lambda_r = 650 nm) are shone separately through the same double-slit apparatus (d = 0.1 mm, D = 2 m). Find the fringe widths. If the two beams were shone simultaneously (two separate experiments overlaid), at what position would a green bright fringe fall on top of a red bright fringe?

Red laser produces fringes at 13 mm spacing; green laser at 10.6 mm spacing. Overlaid patterns coincide at the 0th order (both bright on-axis) and at higher orders where $n_r \lambda_r = n_g \lambda_g$.

Step 1. Compute each fringe width.

\beta_r \;=\; \frac{\lambda_r D}{d} \;=\; \frac{(650\times 10^{-9})(2)}{0.1 \times 10^{-3}} \;=\; 13.0 \times 10^{-3} \text{ m} \;=\; 13.0 \text{ mm}

\beta_g \;=\; \frac{\lambda_g D}{d} \;=\; \frac{(532 \times 10^{-9})(2)}{0.1 \times 10^{-3}} \;=\; 10.64 \times 10^{-3} \text{ m} \;\approx\; 10.6 \text{ mm}

Why: shorter wavelength → tighter fringes. Green fringes are 18% narrower than red. This is the \beta \propto \lambda scaling.

Step 2. Fringes coincide where a red bright fringe (at y = n_r \beta_r) sits exactly on a green bright fringe (at y = n_g \beta_g):

n_r \beta_r \;=\; n_g \beta_g \;\Rightarrow\; \frac{n_r}{n_g} \;=\; \frac{\beta_g}{\beta_r} \;=\; \frac{\lambda_g}{\lambda_r} \;=\; \frac{532}{650}

Why: the fringe-width ratio is \lambda ratio — the D/d factor cancels. This is a sensitive test of the wave theory: only a wave gives fringe spacings in exact proportion to wavelength.

Simplify 532/650 = 266/325. So the first coincidence after the 0th order is at n_g = 325, n_r = 266. Position:

y \;=\; 266 \times 13 \text{ mm} \;=\; 3458 \text{ mm} \;\approx\; 3.46 \text{ m}

That is far off to the side, well outside the visible pattern. In practice, the only coincidence visible on a lab screen is at y = 0 — the central bright fringe of both colours overlap exactly at the centre. Everywhere else, the two patterns slide past each other.

Step 3. Alternative — look for approximate coincidence.

At small n, the fringes of the two colours are barely offset. At n = 5 red (y = 65 mm), the corresponding green fringe number is y/\beta_g = 65/10.64 \approx 6.11 — so the 6th green fringe is at 63.8 mm, about 1.2 mm away from the 5th red. They are close but not coincident.

Result: \beta_r = 13 mm, \beta_g \approx 10.6 mm. Perfect coincidence only at the central fringe within any reasonable screen area.

What this shows: Two independent coherent sources produce two independent fringe patterns with spacing proportional to wavelength. They coincide only where n_r/n_g = \lambda_g/\lambda_r is a rational number with small numerator and denominator — a condition rarely satisfied for arbitrary laser pointers. This is also why white-light fringes (a continuous spectrum of \lambda) only produce a clean central maximum and then wash into a smear of colours: each \lambda has its own fringe pattern, and only n = 0 is \lambda-independent.

Example 3: Designing an experiment to measure an unknown wavelength

An Indian college lab has a laser of unknown wavelength (some kind of diode, from a surplus electronics bin in Lamington Road, Mumbai). A student sets up a double-slit apparatus with d = 0.25 mm and a screen at D = 1.5 m. They measure the distance between the first and twelfth bright fringes as 33.0 \pm 0.3 mm. Find the laser's wavelength and its uncertainty.

Measure 11 fringe widths, divide by 11 to get $\beta$, then invert $\beta = \lambda D/d$ to solve for $\lambda$.

Step 1. The distance from the 1st to the 12th bright fringe spans 11 fringe widths.

\beta \;=\; \frac{33.0 \text{ mm}}{11} \;=\; 3.00 \text{ mm}

Why: between the n=1 fringe and the n=12 fringe there are 12 - 1 = 11 intervals, each of width \beta. Dividing the measured span by the number of intervals (not the number of fringes) is the standard trick for reducing measurement uncertainty.

Step 2. Invert the fringe-width formula.

\lambda \;=\; \frac{\beta d}{D} \;=\; \frac{(3.00 \times 10^{-3})(0.25 \times 10^{-3})}{1.5}

\lambda \;=\; \frac{7.5 \times 10^{-7}}{1.5} \;=\; 5.0 \times 10^{-7} \text{ m} \;=\; 500 \text{ nm}

Why: a 500 nm laser is green — a typical surplus-electronics green diode. Within the visible range, between yellow (589 nm, sodium) and blue (450 nm).

Step 3. Propagate the uncertainty. The dominant error is in the measured span (\pm 0.3 mm out of 33.0 mm), so the fractional uncertainty is

\frac{\delta\beta}{\beta} \;=\; \frac{0.3}{33.0} \;\approx\; 0.91\%

d and D are usually known better than this (slit cards are manufactured to \pm 1\% or tighter, and a metre stick reads D to \pm 1 mm in 1500 mm = 0.07\%). So the wavelength's uncertainty is dominated by the fringe measurement:

\frac{\delta\lambda}{\lambda} \;\approx\; 1\% \;\Rightarrow\; \delta\lambda \;\approx\; 5 \text{ nm}

Result: \lambda = 500 \pm 5 nm. A green diode laser, consistent with the "515 nm" type commonly sold as a laser pointer.

What this shows: Young's experiment is a precision tool even in an undergraduate lab — a millimetre-level measurement of fringe spacing converts directly to nanometre-level knowledge of wavelength. The sensitivity comes from the lever arm D/d (here, 1500/0.25 = 6000), which magnifies \lambda-scale path differences to millimetre-scale fringe separations.

Common confusions

"The slits must be identical for interference to work." Close, but what really matters is that the two slits are coherent sources — their waves have a fixed phase relationship. If both slits are illuminated by the same plane wave, they are coherent by construction. Their intensities (and hence the fringe contrast) are best when the slits are identical, but some fringes form even with slightly unequal slits, just with lower contrast.
"Light from a bulb doesn't interfere because light isn't a wave from a bulb." Light is always a wave. A bulb is incoherent — each atom emits a short wavetrain independently, and the phases of different atoms are random. The fringe patterns still form from each individual atomic wavetrain, but they form at slightly different positions for each wavetrain, so they wash out. This is why Young originally used a small pinhole to select a spatially coherent patch of sunlight. Modern students use lasers, which are coherent by design.
"Energy is destroyed at a dark fringe." No. Energy cancels at one spot because it piles up at another. Integrated across the whole screen, the total light energy equals what would arrive with no interference — just 2I_0 times the total area. The interference redistributes energy, and the redistribution conserves it.
"The fringes get smaller because the light is weaker far from the axis." They do not. In the idealised Young experiment, every fringe within the central diffraction envelope has intensity 4I_0 — all maxima equal. The dimming you see on a real screen is from single-slit diffraction superimposed on double-slit interference, covered separately. Pure interference gives all-equal fringes; diffraction envelops them in a sinc² profile.
"If you close one slit, you still get the same pattern just dimmer." Wrong — you lose the interference entirely. What remains is a single-slit diffraction pattern, which is a single broad central peak with much weaker side lobes, very different from the \cos^2 fringe pattern. The dramatic disappearance of the fringes when you cover one slit is the experiment's most striking qualitative result.
"The distance D must be huge for interference to work." D just needs to be much larger than d for the small-angle approximations \sin\theta \approx \tan\theta \approx y/D to hold. For d = 0.5 mm, D = 10 cm is already 200\times d — perfectly fine. In a classroom setup, D = 1{-}2 m is typical, making the fringes big enough to see with the naked eye.

If you came here to understand why fringes form, compute their spacing, and analyse standard problems, you have what you need. What follows is for readers heading into JEE Advanced and beyond — the effect of a thin plate, the Fresnel biprism variant, and the astounding quantum single-photon version.

Introducing a thin transparent plate in front of one slit

Place a thin glass plate (refractive index \mu, thickness t) in front of slit S_1. The wave from S_1 now travels through glass over distance t and through air over distance r_1 - t. Its optical path (which is what controls phase) is \mu t + (r_1 - t) = r_1 + (\mu - 1)t. The effective path difference changes:

\Delta_{\text{new}} \;=\; \Delta_{\text{old}} - (\mu - 1)t \;=\; \frac{yd}{D} - (\mu - 1)t

The central bright fringe moves to

y_0^{\text{new}} \;=\; \frac{(\mu - 1)t D}{d}

— the whole pattern shifts sideways by this amount. Fringe width is unchanged (\beta still equals \lambda D/d); only the centre has moved. This is the standard plate-shift problem: measure the shift, deduce (\mu - 1)t. Every serious JEE problem on double-slit interference has a plate in it somewhere.

Numerical example: plate thickness t = 10 μm, refractive index \mu = 1.5, with D = 1 m, d = 0.5 mm. Shift = (0.5)(10^{-5})(1)/(0.5 \times 10^{-3}) = 10^{-2} m = 10 mm. A small glass plate shifts the central fringe by 10 mm — easily visible.

The Fresnel biprism — same physics, different geometry

Lacking a lab setup with two slits, you can use a Fresnel biprism — a single prism ground so that its two faces make equal small angles with the base. A single point source illuminating the biprism produces two virtual sources (one refracted by the left face, one by the right face) separated by a small distance d. The rest of the experiment is identical to Young's: light from the two virtual sources interferes on a screen beyond the biprism.

The practical advantage is that with a biprism you lose no light to edge diffraction (slits necessarily block most of the incident light), so the fringes are much brighter. This is why the biprism variant is preferred in some Indian college labs — brighter fringes are easier to measure. The analysis is identical: fringe width \beta = \lambda D/d, with d now the separation of the two virtual sources.

Quantum single-photon interference — the deep mystery

Attenuate the laser until single photons arrive one at a time — seconds between photons. Each photon arrives as a single dot on a CCD: a particle landing at a definite place. Over hours, as thousands of dots accumulate, they build up the double-slit interference pattern.

A single photon, when passed through the apparatus, seems to interfere with itself. There is no other photon present to interfere with. The only way for the pattern to emerge is if each photon's probability amplitude at the screen is the sum of amplitudes for the two paths (S_1-to-P and S_2-to-P), and the probability is the square of that sum. This is the content of the Born rule.

Even more remarkably, this pattern builds up for electrons, neutrons, and even molecules of C_{60} (buckminsterfullerene — demonstrated experimentally). Matter is also a wave. The same fringe-spacing formula \beta = \lambda D/d applies, with \lambda now the de Broglie wavelength h/p of the particle.

The most unsettling version: perform the double-slit experiment with single photons, and place a "which-path" detector at one of the slits. As soon as the detector determines which slit the photon went through, the interference pattern vanishes — you get a sum of two single-slit patterns instead. Measuring the path destroys the interference. This is the content of complementarity: wave and particle behaviours are mutually exclusive features of the same underlying quantum object. Young's 1801 classical experiment, 200+ years on, still defines the frontier of how to think about reality.

Why the wave theory needed this experiment

Before Young, the debate between Newton's corpuscular theory of light (light is a stream of tiny particles) and Huygens' wave theory (light is a wave in an ether) had been stalled for nearly a century. Newton's reputation gave his theory such weight that most physicists defaulted to it. Young's experiment showed cancellation — two beams of light producing darkness where one alone would produce brightness. No particle theory can explain cancellation by addition. The experiment was so decisive that within two decades, Fresnel developed the full mathematical wave theory, Poisson sneered that it predicted a bright spot in the centre of the shadow of a disc, and Arago promptly observed the bright spot — vindicating Fresnel and closing the debate. Light was a wave. (Einstein's 1905 photoelectric effect paper revived the particle side, and the quantum mechanics of 1925 fused the two into the probability-amplitude picture above. But that is another chapter.)

Where this leads next

Coherence and Conditions for Interference — why two lamps shining on the same screen do not produce fringes, and the formal requirements (same frequency, stable phase relation, comparable amplitudes) for interference to be observable.
Huygens' Principle — the wave-optics foundation that each slit acts as a new source of secondary wavelets. The double-slit experiment is Huygens' principle made visible.
Diffraction at a Single Slit — the envelope that modulates the ideal \cos^2 fringe pattern when each slit has finite width. The place where the interference pattern fades into the dark.
Polarisation of Light — another wave property that light exhibits (particles cannot), complementing the interference argument.
Thin Lenses — in every real interference setup, lenses are used to collimate the source and focus the screen image. The wave-optics and geometrical-optics worlds interlock here.