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When two or more waves meet at a point, they combine according to the principle of superposition. This leads to the phenomena of constructive and destructive interference, which provide some of the most striking evidence for the wave nature of light and sound.
Spec mapping: This lesson sits under AQA 7408 section 3.3.2 and develops the principle of superposition, the conditions for observable interference, the concept of coherence (constant phase difference plus equal frequency), the relationship between path difference and interference type, and quantitative path-difference analysis for two coherent sources. It is the conceptual prerequisite for Young's double slits, diffraction gratings, and stationary waves. (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- Section 3.3.2 (Young's double slits, gratings): the constructive condition Δx = nλ and destructive condition Δx = (n + ½)λ derived here become the foundation of every fringe-spacing calculation in the next two lessons.
- Section 3.3.1 (stationary waves): stationary waves are superposition — two progressive waves of equal amplitude and frequency travelling in opposite directions create the nodal/antinodal pattern via the same principle introduced here.
- Section 3.2.2 (photoelectric effect and de Broglie waves): electron diffraction experiments (Davisson–Germer, 1927) show that the superposition principle applies to matter waves as well as classical waves — a synoptic bridge between waves and quantum phenomena.
The Principle of Superposition: When two or more waves meet at a point, the resultant displacement at that point is the vector sum of the individual displacements due to each wave.
This principle applies to all types of wave. The key word is vector sum — displacements have a sign (positive or negative relative to the equilibrium position), and these must be added algebraically.
After passing through each other, the waves continue unchanged. Superposition does not permanently alter the waves.
Constructive interference occurs when two waves arrive in phase — their crests (and troughs) coincide. The resultant amplitude is the sum of the individual amplitudes.
For two waves of equal amplitude A arriving in phase:
Destructive interference occurs when two waves arrive in antiphase (180° or π radians out of phase) — the crest of one coincides with the trough of the other.
For two waves of equal amplitude A arriving in antiphase:
Common Misconception: Energy is not destroyed in destructive interference. It is redistributed — regions of destructive interference have zero intensity, but regions of constructive interference have enhanced intensity. The total energy is conserved.
For a stable, observable interference pattern, the sources must be coherent:
Two sources are coherent if they have:
If the phase difference between two sources varies randomly over time, the interference pattern shifts so rapidly that no stable pattern is observed — the waves are said to be incoherent.
Two separate light bulbs are incoherent because each atom emits light independently in short, random bursts. A laser produces highly coherent light. To produce coherent light sources without a laser, a single source is split into two (e.g., using a double slit).
Exam Tip: "Coherent" does NOT mean "in phase." Two coherent sources have the same frequency and a constant phase difference. If their phase difference is zero, they are coherent AND in phase.
The path difference is the difference in the distances travelled by two waves from their respective sources to the point where they meet.
For two coherent sources in phase, the interference conditions are:
| Condition | Path Difference | Result |
|---|---|---|
| Constructive interference | nλ (n = 0, 1, 2, 3, ...) | Maximum amplitude |
| Destructive interference | (n + ½)λ (n = 0, 1, 2, ...) | Zero amplitude |
If the two sources have a constant phase difference of π (antiphase), the conditions are reversed:
| Condition | Path Difference | Result |
|---|---|---|
| Constructive interference | (n + ½)λ | Maximum amplitude |
| Destructive interference | nλ | Zero amplitude |
Worked Example 1 — Two coherent loudspeakers, S₁ and S₂, emit sound of frequency 680 Hz in phase. A listener stands at a point P where S₁P = 4.00 m and S₂P = 4.25 m. The speed of sound is 340 m s⁻¹. Determine whether the listener hears a maximum or minimum at P.
Step 1: Calculate the wavelength. λ = v/f = 340/680 = 0.50 m
Step 2: Calculate the path difference. Path difference = S₂P − S₁P = 4.25 − 4.00 = 0.25 m
Step 3: Express as a multiple of λ. Path difference/λ = 0.25/0.50 = 0.5
Path difference = 0.5λ = ½λ
Since the sources are in phase and the path difference is ½λ (a half-integer multiple), there is destructive interference — the listener hears a minimum.
Worked Example 2 — In the same setup, the listener moves along a line parallel to the line joining the speakers. At what path difference will the next maximum be heard?
The current path difference is 0.5λ (a minimum). The next maximum requires a path difference of 1.0λ (the next integer multiple):
Path difference = 1.0λ = 1.0 × 0.50 = 0.50 m
Worked Example 3 — Two coherent microwave sources, S₁ and S₂, emit waves of wavelength 3.0 cm in phase. A detector is placed at point Q where S₁Q = 42.0 cm and S₂Q = 48.0 cm. Is Q a maximum or minimum?
Path difference = 48.0 − 42.0 = 6.0 cm
Number of wavelengths = 6.0/3.0 = 2.0
Path difference = 2λ (an integer multiple)
Since the sources are in phase: constructive interference — Q is a maximum.
Sound interference can be demonstrated by:
This experiment works well because sound wavelengths (typically cm to m) are comparable to the speaker separation, making the maxima and minima well spaced.
Worked Example 4 — Two loudspeakers are 1.5 m apart and face forward in the same direction. They emit coherent sound at 2000 Hz in phase. A student walks along a line 8.0 m in front of the speakers. Calculate the separation between adjacent maxima along this line. (Speed of sound = 340 m s⁻¹.)
λ = v/f = 340/2000 = 0.17 m
This is analogous to a double-slit experiment with slit separation s = 1.5 m and screen distance D = 8.0 m.
Fringe spacing w = λD/s = (0.17 × 8.0)/1.5 = 1.36/1.5 = 0.91 m
Adjacent maxima are approximately 0.91 m apart.
Light can also produce interference patterns, but because the wavelength of visible light is very small (400–700 nm), the effects are only visible under carefully controlled conditions. These are explored in detail in the next lesson (Young's Double Slit).
Key points:
For two waves of equal frequency and amplitude, the resultant amplitude can be visualised using phasors — vectors whose lengths represent amplitudes and whose orientation represents phase. Each wave is drawn as an arrow, and the resultant is the vector sum.
For two equal-amplitude coherent waves with phase difference φ:
I_resultant = 4A² cos²(φ/2)
This formula reproduces the four special cases above and shows that the intensity oscillates sinusoidally with cos²(φ/2) as the phase difference changes. The maxima are at φ = 0, 2π, 4π, ...; the minima are at φ = π, 3π, 5π, ...
For two coherent waves in phase at equal amplitudes, the resultant intensity is 4I₀ (where I₀ is the intensity from one speaker).
I_resultant = 4I₀ — four times the single-speaker intensity, NOT twice.
The factor of 4 arises because intensity ∝ amplitude², and the amplitudes (not intensities) add coherently. This is a perennial trap for candidates who naively double the intensity.
Using I_resultant = 4I₀ cos²(φ/2):
I = 4I₀ cos²(30°) = 4I₀ × (0.866)² = 4I₀ × 0.75 = 3I₀
So the slightly out-of-phase combination still produces three times the single-speaker intensity — but a quarter is "missing" relative to the in-phase case. Where has that energy gone? It has been redistributed to other points in space where the same two waves arrive in antiphase, producing minima — energy is conserved over the whole interference field.
When two coherent point sources S₁ and S₂ in a plane (e.g. ripple-tank wavefronts, or two loudspeakers in a horizontal room) emit waves of wavelength λ and separation s, the locus of constructive interference in the plane is a family of hyperbolae with S₁ and S₂ as their foci. Each hyperbola corresponds to a path difference of nλ for integer n; the n = 0 hyperbola is the perpendicular bisector of the line S₁S₂ (since at any point on that line the path difference is zero).
Between the hyperbolic maxima are hyperbolic minima at path differences of (n + ½)λ. The full pattern is a set of nested hyperbolic curves of maxima and minima, and far from the sources (D ≫ s) the hyperbolae are approximately straight lines fanning out from the midpoint of S₁S₂.
In the limit D ≫ s, a screen placed at distance D from the source plane intersects the hyperbolae at evenly spaced positions (in the small-angle limit), giving the familiar equally spaced bright-and-dark fringe pattern. The next lesson develops this in detail; here it suffices to note that the Young's pattern is the far-field projection of the 2D hyperbolic interference field.
Two coherent dippers in a water tank produce a visible hyperbolic interference field at the water surface. Stroboscopic illumination freezes the pattern, allowing direct measurement of wavelength (from the spacing of crests) and path difference (from the position of nodal lines). This is the classical undergraduate demonstration of two-source interference and is on the AQA spec as a recommended demonstration (though not a formally required practical).
Wave speed: v = fλ = 12 × 0.020 = 0.24 m s⁻¹
The number of nodal lines is determined by the largest n satisfying (n + ½)λ < s. With s = 8.0 cm and λ = 2.0 cm:
(n + ½) × 2.0 < 8.0 → n + ½ < 4.0 → n < 3.5
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