Superposition and Interference

When two or more waves meet at the same point, the resulting displacement is found by adding the individual displacements. This deceptively simple rule — the principle of superposition — gives rise to the rich phenomena of interference and is the key to understanding Young's double-slit experiment.

The Principle of Superposition

The principle of superposition states that when two or more waves overlap, the resultant displacement at any point is the vector sum of the individual displacements due to each wave at that point.

If two crests arrive at the same point simultaneously, their displacements add to produce a larger crest. If a crest and a trough of equal amplitude arrive together, they cancel each other out, and the resultant displacement is zero.

This principle applies to all types of wave — sound, light, water, and electromagnetic — and it works regardless of how many waves are overlapping.

Constructive and Destructive Interference

Constructive interference occurs when two waves arrive in phase (crest meets crest, or trough meets trough). Their amplitudes add together, producing a resultant wave with a larger amplitude. If both waves have amplitude A, the resultant has amplitude 2A.

Destructive interference occurs when two waves arrive in antiphase (crest meets trough). Their displacements cancel. If both waves have equal amplitude A, the resultant displacement is zero — complete destructive interference.

Between these extremes, partial constructive or destructive interference can occur, depending on the exact phase relationship.

graph TD
    A["Two waves meet\nat a point"] --> B{"What is the\npath difference?"}
    B -->|"nλ (whole number\nof wavelengths)"| C["Phase difference = 0, 2π, 4π ...\nWaves arrive IN PHASE"]
    B -->|"(n+½)λ (half-integer\nwavelengths)"| D["Phase difference = π, 3π, 5π ...\nWaves arrive IN ANTIPHASE"]
    B -->|"Other values"| E["Partial interference\n(intermediate amplitude)"]
    C --> F["CONSTRUCTIVE\ninterference\nAmplitude = 2A"]
    D --> G["DESTRUCTIVE\ninterference\nAmplitude = 0"]

Path Difference and Phase Difference

Path difference is the difference in the distance travelled by two waves from their sources to the point where they meet. It is measured in metres.

The relationship between path difference and the type of interference is:

Constructive interference occurs when the path difference = nλ (a whole number of wavelengths), where n = 0, 1, 2, 3, ...
Destructive interference occurs when the path difference = (n + ½)λ (a half-integer number of wavelengths), where n = 0, 1, 2, 3, ...

Path difference is related to phase difference by:

$\Delta\phi = \frac{2\pi \cdot \text{path difference}}{\lambda}$

Constructive interference corresponds to Δφ = 0, 2π, 4π, ... (multiples of 2π). Destructive interference corresponds to Δφ = π, 3π, 5π, ... (odd multiples of π).

Condition	Path difference	Phase difference	Result
Constructive	0, λ, 2λ, 3λ ...	0, 2π, 4π, 6π ...	Bright fringe / loud sound
Destructive	λ/2, 3λ/2, 5λ/2 ...	π, 3π, 5π ...	Dark fringe / silence

Coherence

For a stable, observable interference pattern to form, the two wave sources must be coherent. This means they must:

Have the same frequency (and therefore the same wavelength in a given medium)
Maintain a constant phase difference between them

If the sources are incoherent (e.g., two separate light bulbs), the phase difference varies randomly over time, and the interference pattern shifts too rapidly to observe — you just see uniform illumination.

Two slits illuminated by a single light source are coherent because each slit acts as a secondary source of the same wave. Lasers are inherently coherent sources, which is why they are so useful for interference experiments.

Young's Double-Slit Experiment

In 1801, Thomas Young demonstrated the wave nature of light using a beautifully simple experiment. Monochromatic light passes through a single narrow slit (to create a coherent source) and then through two closely spaced parallel slits. On a screen beyond the slits, an interference pattern of alternating bright and dark fringes is observed.

Bright fringes occur where the path difference from the two slits is a whole number of wavelengths (constructive interference).

Dark fringes occur where the path difference is a half-integer number of wavelengths (destructive interference).

The fringe spacing w (the distance between consecutive bright fringes) is given by:

$w = \frac{\lambda D}{a}$

where:

w = fringe spacing (m)
λ = wavelength of the light (m)
D = distance from the slits to the screen (m)
a = slit separation (the distance between the centres of the two slits) (m)

Worked example 1: In a Young's double-slit experiment, the slit separation is 0.50 mm, the screen distance is 2.0 m, and the wavelength of light used is 590 nm (yellow sodium light). Calculate the fringe spacing.

w = λD / a = (590 × 10⁻⁹ × 2.0) / (0.50 × 10⁻³)

w = 1.18 × 10⁻⁶ / 5.0 × 10⁻⁴ = 2.36 × 10⁻³ m = 2.4 mm

Worked example 2: A student measures 8 bright fringes spanning a total distance of 14.4 mm. The slit separation is 0.40 mm and the screen is 1.5 m away. Calculate the wavelength of the light.

The distance for 8 fringes spans 7 fringe spacings (from the 1st bright fringe to the 8th), so:

w = 14.4 mm / 7 = 2.057 mm

Rearranging: λ = wa / D = (2.057 × 10⁻³ × 0.40 × 10⁻³) / 1.5

λ = 8.229 × 10⁻⁷ / 1.5 = 5.49 × 10⁻⁷ m ≈ 549 nm

This is green light.

Common exam mistake: When counting fringes, students often confuse the number of fringes with the number of fringe spacings. If you see 8 bright fringes, the distance between the first and last spans only 7 spacings, not 8. Always subtract 1.

Worked example 3: White light is used in a double-slit experiment instead of monochromatic light. Describe the pattern observed.

The central bright fringe is white (all wavelengths have zero path difference and interfere constructively). Away from the centre, each wavelength produces its own set of fringes with different spacings (since w ∝ λ). The result is that the fringes become coloured — with blue on the inner edge (shorter λ, smaller spacing) and red on the outer edge (longer λ, larger spacing). After a few orders, the fringes from different colours overlap and the pattern becomes an indistinct white blur.

Effect of Changing Variables

Understanding how changes affect the fringe pattern is a common exam topic:

Change	Effect on fringe spacing w
Increase wavelength λ	w increases (w ∝ λ)
Increase screen distance D	w increases (w ∝ D)
Increase slit separation a	w decreases (w ∝ 1/a)
Use white light instead of monochromatic	Coloured fringes; central fringe is white
Increase intensity of source	Fringes are brighter but spacing unchanged
Cover one slit	Interference pattern disappears; single-slit diffraction pattern seen

Interference of Sound Waves

Interference is not limited to light. When two loudspeakers connected to the same signal generator face the same direction, a listener walking across in front of them will hear alternating loud and quiet regions — constructive and destructive interference of sound waves.

Worked example 4: Two loudspeakers are placed 1.5 m apart, both emitting sound of frequency 2000 Hz. A listener stands 4.0 m away. The speed of sound in air is 340 m s⁻¹. Estimate the spacing between consecutive loud points along a line parallel to the speakers.

First find the wavelength: λ = v/f = 340/2000 = 0.17 m

Using the same formula as for double-slit (the geometry is equivalent):

w = λD/a = (0.17 × 4.0) / 1.5 = 0.68/1.5 = 0.45 m

Consecutive loud points are about 0.45 m apart.

Thin Film Interference (Extension)

When light reflects from a thin film (such as an oil film on water or a soap bubble), interference occurs between the ray reflected from the top surface and the ray reflected from the bottom surface. The two reflected rays have travelled different path lengths, and the resulting constructive or destructive interference creates the colourful patterns you see.

The colours vary across the film because the thickness varies, changing the path difference. At any point, the colour you see corresponds to the wavelength(s) that interfere constructively. Where the film is very thin (much less than a wavelength), you often see a dark region because there is an additional π phase change on reflection from a denser medium.

Real-world application: Anti-reflection coatings on camera lenses and spectacles use thin-film interference. A coating of thickness λ/4 (for a chosen wavelength) causes destructive interference of the reflected light, reducing reflections and increasing the light transmitted to the sensor or your eye. The purple-green tint you see on coated lenses is the residual reflection of wavelengths not perfectly cancelled.

A-Level Deep Dive: Superposition and Interference

Spec mapping

Edexcel 9PH0 specification Topic 5 — Waves and the Particle Nature of Light, in the sub-strands covering superposition and two-source interference, requires students to state the principle of superposition; describe coherence in terms of constant phase relationship and equal frequency; relate path difference and phase difference via $\Delta\phi = 2\pi\Delta x/\lambda$ ; apply the constructive condition $\Delta x = n\lambda$ and the destructive condition $\Delta x = (n + \tfrac{1}{2})\lambda$ ; and use Young's double-slit fringe-spacing relation $w = \lambda D/a$ to determine wavelength experimentally (refer to the official Pearson Edexcel specification document for exact wording). These ideas are the load-bearing prerequisite for the diffraction-grating sub-strand $d\sin\theta = n\lambda$ , the standing-wave sub-strand (where superposition of two oppositely-travelling progressive waves produces nodes and antinodes), and the photon-model sub-strand (where the failure of the classical wave picture to account for photoelectric thresholds drives the wave-particle duality discussion). The Edexcel data-and-formulae booklet lists $w = \lambda D/a$ and $d\sin\theta = n\lambda$ , but does not list either the constructive/destructive path-difference conditions or the phase-difference formula — those must be memorised.

Worked example with full mark scheme

Question (8 marks):

A student carries out a Young's double-slit experiment using a helium-neon laser. The two slits are separated by $0.30\ \text{mm}$ and the screen is placed $2.5\ \text{m}$ from the slits. The student measures the distance across $11$ consecutive bright fringes (from the centre of the first to the centre of the eleventh) and obtains a value of $26.4\ \text{mm}$ .

(a) Calculate the fringe spacing. (2)

(b) Calculate the wavelength of the laser light. (3)

(c) The screen is then moved to a new distance and the fringe spacing is found to have doubled. State and explain the new screen distance. State and explain what happens to the fringe spacing if, in addition, the slit separation is also doubled. (3)

Solution with mark scheme:

(a) Step 1 — convert "11 fringes" to fringe spacings.

Eleven bright fringes contain $11 - 1 = 10$ fringe spacings between the first and the eleventh.

$w = \frac{26.4\ \text{mm}}{10} = 2.64\ \text{mm}$

M1 — explicit recognition that $11$ fringes correspond to $10$ spacings (most common slip is dividing by $11$ , giving $w = 2.40\ \text{mm}$ ).

A1 — correct value $w = 2.64\ \text{mm}$ (or $2.6\ \text{mm}$ to 2 sf).

(b) Step 1 — rearrange the fringe-spacing equation.

From $w = \lambda D / a$ :

$\lambda = \frac{wa}{D}$

M1 — algebraically correct rearrangement.

Step 2 — substitute with consistent SI units.

$\lambda = \frac{(2.64 \times 10^{-3})(0.30 \times 10^{-3})}{2.5}$

M1 — converting $w$ to metres ( $\text{mm} \to \text{m}$ ) and $a$ to metres ( $\text{mm} \to \text{m}$ ). Forgetting either conversion gives an answer off by $10^3$ or $10^6$ .

Step 3 — evaluate.

$\lambda = \frac{7.92 \times 10^{-7}}{2.5} = 3.17 \times 10^{-7}\ \text{m}$

A1 — final value with correct unit and an appropriate number of significant figures, carried forward from part (a).

Doubling $w$ at fixed $\lambda$ and $a$ requires doubling $D$ , so the new screen distance is $2 \times 2.5 = 5.0\ \text{m}$ .

M1 — using the proportionality $w \propto D$ explicitly.

A1 — $D' = 5.0\ \text{m}$ .

Step 2 — apply $w \propto 1/a$ to the doubled- $a$ scenario.

Doubling $a$ halves $w$ . Combined with the doubled- $D$ effect (which doubled $w$ ), the net change is $2 \times \tfrac{1}{2} = 1$ — the fringe spacing returns to its original value of $2.64\ \text{mm}$ .

B1 — explicit statement that the two changes cancel and $w$ returns to its original value, with reasoning.

Total: 8 marks (M3 A3 B1, plus the M1 in (b) Step 1 — adjust per individual mark scheme).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): Two coherent loudspeakers are placed $1.20\ \text{m}$ apart and emit sound of frequency $1700\ \text{Hz}$ in phase. A listener walks along a line parallel to the line joining the speakers, at a perpendicular distance of $4.0\ \text{m}$ from the midpoint of the speakers. The speed of sound in air is $340\ \text{m s}^{-1}$ .

(a) Calculate the wavelength of the sound. (1)

(b) Calculate the spacing between consecutive points of maximum loudness along the listener's line. (2)

(c) State and explain how the spacing of loud points would change if either the frequency were halved or the speakers were moved closer together (treat the two cases separately). (3)

Mark scheme decomposition by AO:

(a)

A1 (AO1) — $\lambda = v/f = 340/1700 = 0.20\ \text{m}$ .

(b)

M1 (AO2) — recognising the geometry as equivalent to Young's double-slit, with the speakers acting as the two coherent sources, so $w = \lambda D / a = (0.20)(4.0)/1.20$ .
A1 (AO1) — $w = 0.67\ \text{m}$ (or $2/3\ \text{m}$ ).

(c)

M1 (AO2) — halving frequency doubles the wavelength (from $v = f\lambda$ ); since $w \propto \lambda$ , the spacing of loud points doubles.
M1 (AO2) — moving the speakers closer together decreases $a$ ; since $w \propto 1/a$ , the spacing of loud points increases.
A1 (AO3) — both effects stated quantitatively (or with explicit proportionality reasoning) and linked back to the underlying $w = \lambda D/a$ relationship.

Total: 6 marks split AO1 = 2, AO2 = 3, AO3 = 1. This is a typical Edexcel pattern — substituting once for AO1, then asking the candidate to manipulate proportionalities in (c) to test the relationship rather than the formula.

Synoptic links

Connects to:

Topic 5, sub-strand on diffraction gratings: the grating equation $d\sin\theta = n\lambda$ generalises Young's double-slit constructive condition $\Delta x = n\lambda$ to a regular array of slits, where $d$ is the slit-to-slit spacing. Sharper, brighter maxima arise because contributions from many slits superpose constructively only at very specific angles, with destructive interference filling the gaps. The same path-difference logic underlies both treatments.
Topic 5, sub-strand on polarisation: while polarisation distinguishes transverse from longitudinal waves (only transverse waves can be polarised), it is the combination of polarisation analysis with two-source interference that shows light's transverse, vector character — an interference pattern formed with crossed polarisations vanishes, confirming that interference requires aligned displacement directions.
Topic 5, sub-strand on the photon model and the photoelectric effect: the photoelectric effect — that light below a threshold frequency cannot eject electrons regardless of intensity — directly contradicts the classical wave model, in which arbitrarily intense waves should always supply enough energy. Yet the same light produces clean Young's-fringes interference patterns. This tension is the empirical foundation of wave-particle duality and motivates $E = hf$ .