Superposition and Coherence

Spec mapping (OCR H556 Module 4.5): This lesson develops the principle of superposition and the conditions required for stable two-source interference — coherence (constant phase relation plus equal frequency) and path-difference rules for constructive ($n\lambda$nλ) and destructive ($(n+\tfrac{1}{2})\lambda$(n+21​)λ) interference. It is the foundation for Young's fringes, the diffraction-grating equation, single-slit envelopes and stationary waves later in the module.

So far we have treated waves singly. The real world is full of situations where two or more waves overlap in space and time. When this happens, the principle of superposition tells us that the resultant disturbance at any point is found by adding the individual wave displacements. This simple rule has profound consequences: it produces interference, diffraction patterns, beats, stationary waves and most of the beautiful phenomena of wave physics.

This lesson establishes the principle of superposition, defines the closely-related idea of coherence, and introduces the concepts of path difference, phase difference, constructive interference and destructive interference. These ideas are the foundation of the next five lessons.

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The Principle of Superposition

The principle of superposition states:

When two or more waves meet at a point, the resultant displacement is the vector sum of the individual displacements that each wave would produce at that point if the others were absent.

In symbols, for two waves whose displacements at point $P$P at time $t$t are $y_1(P,t)$y1​(P,t) and $y_2(P,t)$y2​(P,t):

$y(P,t) = y_1(P,t) + y_2(P,t)$y(P,t)=y1​(P,t)+y2​(P,t)

This applies to every type of wave — mechanical, sound, electromagnetic. It is an instantaneous rule: at every instant you add the displacements that each wave alone would produce, and the sum gives the actual disturbance of the medium. The rule extends naturally to any number of waves.

What It Does and Does Not Say

• It does say that waves pass through each other without destroying each other. After the overlap, each wave continues on its way unaltered — superposition is reversible.
• It does say that where two crests meet you get a bigger crest, and where a crest meets a trough you get partial (or total) cancellation at that instant.
• It does not say anything about energy being destroyed. In regions of destructive interference no energy arrives, but this is compensated by constructive interference elsewhere. Total energy is conserved.
• It does not apply at very large amplitudes where the medium behaves non-linearly (e.g. shock waves in air, or extreme optical intensities where the index of refraction itself depends on intensity). For all A-Level situations the linear superposition rule is exact.

Why Superposition Is True

Superposition is a direct consequence of the linearity of the wave equation:

$\frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2}$∂t2∂2y​=v2∂x2∂2y​

If $y_1$y1​ and $y_2$y2​ each satisfy this equation, then so does their sum $y_1 + y_2$y1​+y2​ — both sides of the equation add. So any linear combination of valid solutions is also a valid solution. This is exactly what superposition asserts.

You do not need to derive this at A-Level, but the message matters: superposition is not an extra rule of nature — it is built into the wave equation itself.

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Constructive and Destructive Interference

When two waves of the same frequency and amplitude superpose, the result depends on their phase difference at the meeting point.

Constructive Interference

If the two waves arrive in phase (phase difference $0$0, equivalently a whole number of wavelengths of path difference), crests line up with crests and troughs with troughs. The result is a wave of double amplitude — constructive interference.

Constructive interference: path difference $\Delta = n\lambda$Δ=nλ, where $n = 0, 1, 2, 3, \ldots$n=0,1,2,3,… (a whole number of wavelengths).

Destructive Interference

If the two waves arrive in antiphase (phase difference $\pi$π rad $= 180°$=180°, equivalently a path difference of an odd number of half-wavelengths), crests line up with troughs. If the amplitudes are equal the result is zero — total destructive interference.

Destructive interference: path difference $\Delta = (n + \tfrac{1}{2})\lambda$Δ=(n+21​)λ, where $n = 0, 1, 2, 3, \ldots$n=0,1,2,3,… (an odd number of half-wavelengths).

Partial Interference

For phase differences between $0$0 and $\pi$π, the two waves partially reinforce. For phase differences between $\pi$π and $2\pi$2π, they partially cancel. The resultant amplitude varies continuously between $2A$2A (fully constructive) and $0$0 (fully destructive). The general result for two coherent waves of equal amplitude $A$A with phase difference $\varphi$φ is:

$A_{\text{resultant}} = 2A \cos\!\left(\frac{\varphi}{2}\right)$Aresultant​=2Acos(2φ​)

You will not need to derive this at A-Level, but it explains the sinusoidal intensity envelope you see across a fringe pattern: at $\varphi = 0$φ=0 amplitude is $2A$2A, at $\varphi = \pi$φ=π it is $0$0, and in between it sweeps smoothly through intermediate values.

flowchart TB
    Start[Two sources reach point P] --> Q[Compute path difference delta]
    Q --> Phi[Phase difference phi equals two pi delta over lambda]
    Phi --> C{Is phi a multiple of two pi?}
    C -- Yes --> CON[Constructive: amplitude 2A, intensity 4I]
    C -- No --> D{Is phi an odd multiple of pi?}
    D -- Yes --> DES[Destructive: amplitude 0, intensity 0]
    D -- No --> P[Partial: amplitude 2A cos half phi]

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Path Difference and Phase Difference

When two waves reach a point $P$P from two different sources $S_1$S1​ and $S_2$S2​, the distances travelled are $S_1P$S1​P and $S_2P$S2​P. The path difference is the difference of these two distances:

$\Delta = |S_1P - S_2P|$Δ=∣S1​P−S2​P∣

The corresponding phase difference $\varphi$φ (in radians) is:

$\varphi = \frac{2\pi}{\lambda}\,\Delta$φ=λ2π​Δ

In words: each wavelength of path difference corresponds to a full $2\pi$2π of phase difference. Half a wavelength of path difference $\Rightarrow \pi$⇒π rad of phase shift $\Rightarrow$⇒ antiphase.

Path difference $\Delta$Δ	Phase difference $\varphi$φ	State at $P$P
$0$0	$0$0	In phase — constructive
$\lambda/4$λ/4	$\pi/2$π/2	Quadrature — partial
$\lambda/2$λ/2	$\pi$π	Antiphase — destructive
$3\lambda/4$3λ/4	$3\pi/2$3π/2	Quadrature — partial
$\lambda$λ	$2\pi$2π	In phase — constructive
$3\lambda/2$3λ/2	$3\pi$3π	Antiphase — destructive
$2\lambda$2λ	$4\pi$4π	In phase — constructive

You must be fluent in converting between path difference and phase difference in both directions: this is the single most common arithmetic step in any interference problem.

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Worked Example 1 — Path and Phase Difference

Q. Two loudspeakers emit sound of wavelength $\lambda = 0.40\text{ m}$λ=0.40 m. At a point $P$P, the distances from the two speakers are $3.60\text{ m}$3.60 m and $4.20\text{ m}$4.20 m respectively. State whether $P$P is a point of constructive or destructive interference.

A. Path difference:

$\Delta = 4.20 - 3.60 = 0.60\text{ m}$Δ=4.20−3.60=0.60 m

As a multiple of $\lambda$λ:

$\frac{\Delta}{\lambda} = \frac{0.60}{0.40} = 1.5$λΔ​=0.400.60​=1.5

This is an odd half-wavelength multiple ($1.5 = 3/2$1.5=3/2). The two waves arrive in antiphase, so they interfere destructively — $P$P is a quiet point. If the speakers are equally loud and matched in waveform, $P$P would be (in principle) a silent null.

The corresponding phase difference is

$\varphi = \frac{2\pi}{0.40}\times 0.60 = 3\pi\text{ rad}$φ=0.402π​×0.60=3π rad

which is indeed an odd multiple of $\pi$π — antiphase — consistent with the path-difference analysis.

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Worked Example 2 — Locating a Bright Fringe

Q. Two coherent radio transmitters emit waves of wavelength $2.0\text{ m}$2.0 m. A receiver is positioned so that the path difference between the two transmitters is exactly $6.0\text{ m}$6.0 m. Is the signal strong or weak? Explain.

A. $6.0\text{ m} / 2.0\text{ m} = 3$6.0 m/2.0 m=3, a whole number of wavelengths. The waves arrive in phase and interfere constructively. The signal is strong (in fact four times the intensity each transmitter would deliver on its own; see the section on intensity below).

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Coherence

For a stable, observable interference pattern, the two sources must be coherent. This is one of the most-tested concepts in the OCR module.

Definition of Coherence

Two sources are coherent if they have:

1. The same frequency (and hence the same wavelength in a given medium), and
2. A constant phase difference between them.

If either condition fails, the interference pattern changes so rapidly with time that it averages out and becomes invisible to any normal detector.

Common slip. Students often write "the same phase" — that is wrong. Coherence requires a constant phase difference, which is not necessarily zero. A constant non-zero offset still produces stable fringes; a random, fluctuating offset destroys them.

Why Coherence Is Necessary

Imagine two independent sources — say, two filament bulbs — each producing random bursts of waves lasting a few nanoseconds. The phase difference between them changes randomly hundreds of millions of times a second. At a given point in space the waves might momentarily be in phase, then a nanosecond later in antiphase, then somewhere in between. Any interference pattern averages to zero over any observable timescale. You cannot see interference between two independent bulbs — or even between two independent lasers, except over very short timescales with specialised equipment.

But if the two sources have a fixed phase relationship — if they are coherent — the interference pattern at each point is stable in time and can be observed directly with the eye or a camera.

How Coherence Is Achieved in Practice

The simplest way is to use a single source split into two paths:

• Young's double-slit experiment (Lesson 8): a single light source illuminates two narrow slits. Waves emerging from the two slits are coherent because they come from the same initial wavefront — any random phase wobble of the source affects both slits in step.
• Diffraction gratings (Lesson 9): the same principle extended to many slits.
• Beam-splitters and fibre couplers: split a single laser beam into two paths in a Mach–Zehnder or Michelson interferometer.

Lasers are highly coherent intrinsically — they emit a nearly-pure single frequency with a long phase-coherence time (microseconds to seconds, depending on laser quality). Ordinary light bulbs are incoherent and must be split into two paths from a single source to produce usable interference.

Coherent vs Incoherent Sources — A Decision Tree

flowchart TD
    Q1[Two sources of the same frequency?]
    Q1 -- No --> No1[No interference]
    Q1 -- Yes --> Q2[Constant phase relation?]
    Q2 -- No --> No2[No stable pattern: time-averages to uniform intensity]
    Q2 -- Yes --> Yes[Coherent: stable interference observable]
    Yes --> A1[Use single-source split: Young, grating, beam-splitter]
    Yes --> A2[Or use a single laser]

OCR questions often ask "Why must the two sources be coherent?" The complete answer states both conditions (same frequency, constant phase difference) and explains that otherwise the pattern of maxima and minima would shift rapidly with time and be unobservable.

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Monochromatic Light

A closely related term is monochromatic, meaning "single-coloured". A monochromatic source emits light of a single frequency (or a very narrow band of frequencies around a single value). Lasers are nearly perfectly monochromatic; a sodium lamp is roughly monochromatic (two yellow lines very close together at $589.0$589.0 nm and $589.6$589.6 nm); a filament lamp is very far from monochromatic (it emits light of every visible frequency, plus a great deal of infrared).

Monochromaticity is a prerequisite for clear interference patterns because different wavelengths produce differently-spaced fringes; a composite source gives overlapping patterns. This is why Young's fringes are most clearly seen with a laser or filtered lamp, and why white-light fringes in a Young's experiment are visible only very close to the central maximum (where all wavelengths still nearly coincide).

Note that monochromatic and coherent are different conditions. A sodium lamp is roughly monochromatic but not coherent — the atoms emit independently. A laser is both. Two separated lasers tuned to the same frequency are nearly monochromatic but typically not phase-coherent with each other.

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Amplitude vs Intensity in Superposition

Care is needed about amplitude addition versus intensity addition.

• Amplitudes add algebraically (including signs), because the wave equation is linear.
• Intensities do not add directly, because intensity is proportional to the square of amplitude: $I \propto A^2$I∝A2.

For two coherent waves of equal amplitude $A$A meeting in phase:

$A_{\text{total}} = A + A = 2A$Atotal​=A+A=2A $I_{\text{total}} \propto (2A)^2 = 4A^2 = 4I_1$Itotal​∝(2A)2=4A2=4I1​

That is, two coherent in-phase waves of intensity $I_1$I1​ each superpose to give an intensity of $4I_1$4I1​, not $2I_1$2I1​. Twice the amplitude means four times the intensity.

For two coherent waves meeting in antiphase ($\varphi = \pi$φ=π):

$A_{\text{total}} = A - A = 0,\qquad I_{\text{total}} = 0$Atotal​=A−A=0,Itotal​=0

For two incoherent in-phase waves (no fixed phase relation, so the cross-term averages to zero):

$\langle I_{\text{total}} \rangle = 2I_1$⟨Itotal​⟩=2I1​