Wave Interference

Two dippers tap the water in step, and circular waves run out from each. Where a crest meets a crest the water churns; where a crest meets a trough it stays calm. Drag the sources and the probe, change the wavelength and the phase shift — and see where the “calm” and “rough” bands come from.

× slow motion: drag left for slower ×0,50

lines of reinforcement (maxima) lines of cancellation (nodes) extra path Δr path r₁ from S₁ (beads every λ) path r₂ from S₂ water: lighter — crest, darker — trough, even — still

What happens at point P

Drag the probe P over the water. On the left are two rotating arrows (phasors), on the right is how the water bobs at that point.

What the screen will show

The screen to the right of the tank shows how bright each height y is. Move the screen left and right and watch the fringes change.

The idea in 30 seconds

Each source makes its own wave

Two dippers tap the water in step and send out circular waves. Each wave on its own is just ordinary rings: crest, trough, crest… Run the “One wave” experiment and see.

The waves add up

At every point of the water the displacements from the two waves simply add. Crest + crest = a double crest (reinforcement). Crest + trough = zero (cancellation): the water stands still although both waves have arrived.

The path difference decides everything

If one wave has travelled a whole number of wavelengths farther than the other to a point (Δr = mλ), they arrive in phase — reinforcement. If one and a half, two and a half… (Δr = (m + ½)λ) — they arrive in antiphase: cancellation. That is why the “rough” and the “calm” places lie along hyperbolas.

Try it yourself

Switch on S₁ only, then add S₂: “calm” lines appear. With a single source there are none.
Drag the probe P over the water: on a yellow line the amplitude A doubles, on a blue dashed one it drops to zero.
Increase the wavelength λ: the fringes get wider and fewer. Decrease it — thin and frequent fringes.
Move the sources apart (d): there are more fringes. Make d smaller than λ — one wide maximum is left.
Set Δφ = π: the symmetry axis is now silent, and everything that was a maximum is now a minimum.
Reduce A₂: the nodes stop being perfectly calm — a stronger wave cannot be fully cancelled by a weaker one.

Where you meet it in real life

Noise-cancelling headphones: a microphone listens to the noise and the speaker plays the same sound in antiphase — silence is left in your ear.
Wi-Fi and radio “dead spots”, “deaf” places in a concert hall: there waves that arrived along different paths cancel each other.
Radar and 5G: an array of antennas with a small phase shift turns the beam without rotating the antennas themselves.
The shimmering colours of a soap bubble or an oil film on water: light reflected from the two surfaces of the film interferes, and the colour depends on the thickness.
Young’s double-slit experiment proved the wave nature of light and later of electrons: fringes appear even when particles arrive one at a time.

Mode

One-click experiments

λ wavelength 10.0

d distance between the sources

Δφ S₂ lags behind S₁ 0

A₁ amplitude of S₁ 1.00

A₂ amplitude of S₂ 1.00

Physics

decay of waves with distance, 1/√r

Show

lines of reinforcement (maxima) and their numbers m

lines of cancellation (nodes)

paths S₁→P and S₂→P with a ruler of λ

wave crests (rings)

far-field ray θₘ (d·sinθ = mλ)

10 cm grid

Every number on the screen is computed from exact formulas, not step by step: the picture depends neither on the frame rate nor on the “playback speed”. Waves do not reflect from the walls of the tank (the shores absorb). The sources are sinusoidal, with one frequency and a constant phase difference — that is, coherent.

m order of the maximum m = 1

KEY FORMULAS

Speed, wavelength and frequency

$$v = \lambda f = \frac{\lambda}{T}, \qquad k = \frac{2\pi}{\lambda}, \quad \omega = 2\pi f$$

In one period a wave travels exactly one wavelength — hence v = λf

▶ Detailed Explanation

During one oscillation T a crest moves away from the source by λ, so v = λ/T = λf. In the simulator the wave speed is fixed (v = 25 cm/s), so only one of λ and f is free: as the frequency grows the wavelength shrinks. The numbers on the readout obey exactly this formula: λ = 10 cm and f = 2.5 Hz give v = 25 cm/s. The wave number k = 2π/λ tells how many radians of phase accumulate per centimetre of path, and the angular frequency ω = 2πf how many per second; k and ω are what stands under the sine in the wave formula.

Superposition of two source waves

$$u = A_1\sin(\omega t - k r_1) + A_2\sin(\omega t - k r_2 - \Delta\varphi)$$

The displacement of the water at a point is the sum of two circular waves from coherent sources

▶ Detailed Explanation

A circular wave runs over the water from each source. The phase ωt − kr means: at distance r the water repeats the oscillation of the source with a delay r/v. The term −Δφ says that S₂ lags behind S₁ in phase by Δφ. By the superposition principle the displacements simply add. The wave “switches on” at t = 0 and its front travels at speed v: until the front arrives the water is calm. In a real tank the amplitude of a circular wave falls as 1/√r — the energy spreads over a circle of length 2πr. This correction is switched on by the “decay” tick: then far away a₁ ≠ a₂, and the nodes keep |a₁ − a₂| instead of zero.

Path difference and phase difference

$$\psi = \frac{2\pi}{\lambda}\,\Delta r - \Delta\varphi, \qquad \Delta r = r_1 - r_2$$

Each extra centimetre of path shifts the phase by 2π/λ; the shift of the sources is added on top

▶ Detailed Explanation

The same path difference can be described in centimetres and in radians: a path of Δr = λ is exactly one turn of phase (2π). So at a point the waves from S₁ and S₂ are shifted in phase by ψ = 2πΔr/λ − Δφ: the first term comes from geometry (where the point is), the second from the sources (how much S₂ lags). In the diagram with rotating arrows ψ is the angle between the arrows, and it does not change in time because both arrows rotate at the same frequency. When ψ is a multiple of 2π the arrows point the same way (reinforcement); when ψ = π, 3π… they point opposite ways (cancellation).

Conditions for maximum and minimum

$$\Delta r = \left(m + \frac{\Delta\varphi}{2\pi}\right)\lambda \;\Rightarrow\; \max, \qquad \Delta r = \left(m + \frac12 + \frac{\Delta\varphi}{2\pi}\right)\lambda \;\Rightarrow\; \min$$

For in-phase sources (Δφ = 0): maximum at Δr = mλ, minimum at Δr = (m + ½)λ

▶ Detailed Explanation

A maximum is ψ = 2πm, a minimum is ψ = (2m + 1)π; substituting ψ from the previous formula gives the conditions on the path difference. For Δφ = 0 they are very simple: a whole number of wavelengths — reinforcement, a half-integer number — cancellation. The set of points with the same path difference r₁ − r₂ = const is a hyperbola with the sources as foci, so on the scene the yellow lines (maxima) and the blue dashed lines (minima) are hyperbolas. The central line m = 0 lies on the symmetry axis. There are not infinitely many lines: |Δr| cannot exceed the distance d between the sources, so m is bounded. The shift Δφ moves all the lines at once; at Δφ = π maxima and minima swap places.

Amplitude of the sum and intensity

$$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\psi}, \qquad I \propto A^2, \qquad \langle I\rangle \propto A_1^2 + A_2^2$$

The length of the sum of two arrows; brightness is proportional to the amplitude squared, and the mean brightness equals the sum of brightnesses

▶ Detailed Explanation

Two oscillations of one frequency add into an oscillation of the same frequency, and its amplitude is the length of the sum of two vectors of lengths A₁ and A₂ with the angle ψ between them (the law of cosines). At ψ = 0 the arrows lie in a line, A = A₁ + A₂; at ψ = π they point against each other, A = |A₁ − A₂|, and with equal amplitudes the water stands still. The intensity is proportional to A², so in a maximum it is (A₁ + A₂)²/(A₁² + A₂²) times the sum of the separate intensities (twice for equal amplitudes), and in a minimum it is less. Averaging cos ψ over all phases gives ⟨I⟩ ∝ A₁² + A₂²: interference neither creates nor destroys energy, it only redistributes it. This is what the dashed line I₁ + I₂ on the screen graph shows.

Number of maxima

$$N = 2\left\lfloor \frac{d}{\lambda} \right\rfloor + 1 \qquad (\Delta\varphi = 0)$$

How many directions of reinforcement a pair of in-phase sources at distance d gives

▶ Detailed Explanation

The path difference Δr = r₁ − r₂ at any point of the plane lies between −d and +d. Maxima are possible only for |mλ| ≤ d, that is |m| ≤ d/λ. The integers m from −⌊d/λ⌋ to +⌊d/λ⌋ number 2⌊d/λ⌋ + 1, including the central m = 0. For λ = 10 cm and d = 30 cm: N = 2·3 + 1 = 7 (the outermost m = ±3 run along the line of the sources). If d < λ then N = 1: one wide maximum forward is left, with no side ones. For Δφ ≠ 0 the formula changes: you have to count the integers m with |(m + Δφ/2π)λ| ≤ d — this number is shown on the “Measurements” tab.

Direction of a maximum in the far field

$$d\sin\theta_m = \left(m + \frac{\Delta\varphi}{2\pi}\right)\lambda$$

The angle θ is measured from the symmetry axis towards the lagging source S₂

▶ Detailed Explanation

Far from the sources the rays to a point are almost parallel, and the path difference depends only on the direction: Δr ≈ d·sinθ. Equating it to the condition for a maximum gives d·sinθ = (m + Δφ/2π)λ. For Δφ = 0 the main maximum m = 0 goes straight ahead (θ = 0), and the others go at angles sinθ = mλ/d. For Δφ ≠ 0 the whole fan turns: for m = 0 we get sinθ = Δφ·λ/(2πd). This is how a phased-array antenna works: by changing the phase shift between the antennas, a radar or 5G beam is turned without rotating the antennas themselves. Run the “Beam steering” experiment and tick “far-field ray” to see it on the scene.

Fringe spacing on the screen

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

The farther the screen and the longer the wave, the wider the fringes; the farther apart the sources, the narrower

▶ Detailed Explanation

On a screen at distance L the position of the m-th maximum is y_m = L·tanθ_m. For small angles tanθ ≈ sinθ = mλ/d, so y_m ≈ mλL/d, and the distance between neighbouring maxima is Δy = λL/d. The formula is approximate: it is exact when the screen is far away (L ≫ d). In the simulator L/d is usually about 2, so the “Measurements” tab shows two numbers side by side — the fringe width read from the picture and the one from the formula; the larger L/d (move the screen away, bring the sources together), the better they agree. The same formula describes Young’s double-slit experiment for light, only there λ ≈ 500 nm and Δy is a fraction of a millimetre.

Interference is the addition of waves. At every point of the medium the displacement is the sum of the displacements from all the sources: u = u₁ + u₂. Everything is decided by the path difference Δr = |r₁ − r₂| — how much farther one wave has travelled than the other. If Δr is a whole number of wavelengths (Δr = mλ), a crest arrives together with a crest and the oscillations reinforce each other: the amplitude is A = A₁ + A₂. If Δr is a half-integer number (Δr = (m + ½)λ), a crest meets a trough and the waves cancel: A = |A₁ − A₂|, and with equal amplitudes the water stands still. The sources must be coherent — of one frequency and with a constant phase difference — otherwise the pattern drifts and no fringes are seen. The simulator computes the field not step by step but from an exact formula: u = A₁·sin(ωt − k·r₁) + A₂·sin(ωt − k·r₂ − Δφ), where k = 2π/λ, ω = 2πf and the wave speed is v = λ·f = 25 cm/s. First ring-shaped fronts spread out from the sources (the “New drops” button restarts the waves), then the pattern fills the whole tank. Three views show the same field in different ways: “Water” is a lit surface (lighter — crest, darker — trough, even — node), “Colour” shows the displacement u as colour (warm — up, cold — down), and “Intensity” shows the brightness ⟨u²⟩ averaged over a period — what the screen would show. Yellow lines mark reinforcement (maxima Δr = (m + Δφ/2π)λ), blue dashed lines mark cancellation (nodes); they are hyperbolas with the sources as foci. The probe P shows what the two waves do at its own point: the red piece of the path is the “extra” path Δr, the beads on the rays count wavelengths, and the rotating arrows (phasors) show how the oscillations add: the length of the sum is A = √(A₁² + A₂² + 2A₁A₂·cos ψ), where ψ = 2πΔr/λ − Δφ. The screen on the right shows the brightness I(y): fringes, their numbers m and a dashed line I₁ + I₂ — the brightness without interference. The mean over the screen equals that dashed line: energy is not lost, it is moved from the minima to the maxima. The number of reinforcement directions between the sources at Δφ = 0 is N = 2⌊d/λ⌋ + 1; far away they go at angles d·sinθ = mλ, and the fringe spacing on the screen is Δy = λL/d (for L ≫ d). Increase d — there are more fringes, increase λ — they get wider; if d < λ, a single wide maximum is left. A phase shift Δφ moves the whole pattern: at Δφ = π maxima and minima swap places, and intermediate values turn the beam — this is how phased-array radar antennas work. The “Quiz” and “Lab” modes check understanding: in the lab the wavelength is hidden and has to be found by measuring path differences in the nodes.