A Level Physics 9702 · AS Level · Topic 8 · ~9 min read
Superposition
The principle of superposition, how stationary waves form and how to find λ from nodes, experiments with strings, microwaves and air columns, diffraction, interference and coherence, the double slit λ = ax/D and the diffraction grating d sin θ = nλ.
SuperpositionStationary wavesNodes & antinodesMicrowaves, strings, air columnsDiffractionInterference & coherenceDouble slit λ = ax/DDiffraction grating
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In 30 seconds
- Superposition: where two waves meet, the resultant displacement is the vector sum of the individual displacements.
- A stationary wave forms when two waves of the same frequency (and similar amplitude) travel in opposite directions, usually a wave and its reflection. Node to node is λ/2.
- Diffraction is the spreading of a wave through a gap or round an edge. It is most noticeable when the gap is about the same size as the wavelength.
- Interference fringes need coherent sources: a constant phase difference (so the same frequency).
- Double slit: λ = ax/D. Grating: d sin θ = nλ, with d = 1/(lines per metre).
By the end you can
- State and apply the principle of superposition.
- Explain how stationary waves form and find λ from node or antinode positions.
- Describe stationary-wave experiments with microwaves, strings and air columns.
- Describe diffraction and how it depends on the gap width.
- Explain interference and coherence and the conditions for observable fringes.
- Use λ = ax/D and d sin θ = nλ, and describe how a grating measures wavelength.
1Superposition and stationary waves
Definitions
Principle of superposition — when two or more waves meet at a point, the resultant displacement is the (vector) sum of the displacements of the individual waves.
Node — a point on a stationary wave with zero amplitude (always at rest).
Antinode — a point on a stationary wave with maximum amplitude.
How a stationary wave forms
- A wave travels to a boundary and is reflected. The incident and reflected waves have the same frequency, speed and (nearly) the same amplitude, and travel in opposite directions.
- They superpose. At some points they always cancel (nodes); at others they reinforce to give maximum amplitude (antinodes).
- Write all three ideas: reflection, superposition (or interference), and same frequency/speed in opposite directions. In 2024 answers were too generic, and some wrongly mentioned "coherence".
| Progressive wave | Stationary wave | |
|---|---|---|
| Energy | transferred along the wave | not transferred (stored) |
| Amplitude | the same for every point | varies from zero (node) to maximum (antinode) |
| Phase | changes continuously along the wave | all points between adjacent nodes are in phase; across a node, antiphase |
| Wavelength | distance between adjacent points in phase | twice the node-to-node distance |
Three experiments
- Stretched string: a vibration generator drives a string over a pulley with a hanging mass. Adjust the frequency to see 1, 2, 3… loops. Each loop is λ/2; v = fλ.
- Microwaves: a transmitter faces a metal plate. Move a detector along the line between them: the reading rises and falls. Successive minima (nodes) are λ/2 apart.
- Air column: a tube closed at one end (or a tube in water) resonates with a tuning fork or loudspeaker. There is a node at the closed end and an antinode near the open end. Two successive resonance lengths differ by λ/2, which avoids the end correction.
Worked example
A string 1.2 m long, fixed at both ends, vibrates in three loops at 150 Hz. Find the wave speed.
- 3 × λ/2 = 1.2 ⇒ λ = 0.80 m C1
- v = fλ = 150 × 0.80 = 120 m s⁻¹ A1
Worked example
Microwaves reflect from a metal plate. A detector finds successive minima 1.4 cm apart. Find the frequency.
- λ/2 = 1.4 cm ⇒ λ = 2.8 cm C1
- f = c/λ = 3.0 × 10⁸/0.028 = 1.1 × 10¹⁰ Hz A1
Worked example
A tube closed at one end resonates at 500 Hz at lengths 0.16 m and 0.50 m. Find the speed of sound.
- successive resonances differ by λ/2: λ = 2 × 0.34 = 0.68 m C1
- v = 500 × 0.68 = 340 m s⁻¹ A1
Common mistakeTaking the distance between two points as λ instead of λ/2. Common in 2024 Paper 1. Node to node (or antinode to antinode) is half a wavelength.
2Diffraction
Spreading through a gap
Diffraction — the spreading of a wave as it passes through a gap or around the edge of an obstacle.
- Ripple tank demonstration: plane waves pass through a gap. With the gap about one wavelength wide, the emerging wavefronts are nearly circular; with a wide gap, they stay mostly straight with a little curving at the edges. Wavelength and speed are unchanged by diffraction.
3Interference
Coherence and conditions
Interference — the superposition of waves from two (or more) coherent sources, giving a stable pattern of maxima and minima.
Coherent — having a constant phase difference (which requires the same frequency).
- Conditions for observable fringes: coherent sources; similar amplitudes (so minima are dark); the waves overlap (diffraction at the slits makes this happen); slit separation not too large compared with λ so fringes are resolved.
- Path difference (sources in phase): constructive (maximum) when the path difference is nλ; destructive (minimum) when it is (n + ½)λ.
- Demonstrations: two dippers in a ripple tank (water); two loudspeakers driven by the same signal generator (sound); a laser through a double slit (light); microwaves through two gaps in a metal sheet.
The double slit
λ = ax/D LEARN IT
a = slit separation, x = fringe separation, D = slit–screen distance (D ≫ a)- Measure x across many fringes and divide (e.g. 10 fringe spacings) to reduce the percentage uncertainty.
Worked example
Light passes through slits 0.50 mm apart. On a screen 2.4 m away, 10 fringe spacings measure 30 mm. Find λ.
- x = 30/10 = 3.0 mm C1
- λ = ax/D = 0.50 × 10⁻³ × 3.0 × 10⁻³/2.4 = 6.3 × 10⁻⁷ m A1
4The diffraction grating
d sin θ = nλ
d sin θ = nλ LEARN IT
d = slit spacing = 1/N (N = lines per metre); n = order- Using a grating to measure λ: shine monochromatic light normally onto the grating; measure the angle θ of a known order (measure both sides and halve 2θ to reduce error); λ = d sin θ/n. Higher orders give larger angles and smaller percentage uncertainty.
- Highest order: sin θ ≤ 1, so n ≤ d/λ. Round down. Total number of maxima = 2n_max + 1 (including the zero order).
Worked example
A grating has 600 lines per mm and is illuminated normally with light of wavelength 550 nm. Find the angle between the 2nd and 3rd order maxima, and the highest order.
- d = 1/(600 × 10³) = 1.67 × 10⁻⁶ m C1
- θ₂ = sin⁻¹(2 × 550 × 10⁻⁹/1.67 × 10⁻⁶) = 41.3°; θ₃ = sin⁻¹(0.990) = 81.9° C1
- angle between = 40.6° A1
- n ≤ d/λ = 3.03 ⇒ highest order 3 A1
Common mistakeMixing up N and d, giving the angle of one order when the angle between two was asked, or using sin 180° for the limit instead of sin 90°. All three are in the 2024 reports.
Linking questionSuperposition needs the wave terms from Waves; electron diffraction is evidence for wave behaviour of particles in Quantum physics; and the grating's line spectra link to energy levels in the same note.
Test yourself — cover the answers first
Quick check
- State the principle of superposition.
- What is the distance between adjacent nodes?
- Define coherence.
- What path difference gives a minimum for in-phase sources?
- A grating has 300 lines per mm. What is d?
Exam-style
- Explain how a stationary wave is formed on a stretched string fixed at both ends.
- Red light of wavelength 600 nm gives a second-order maximum at 52° with a grating. Find the number of lines per millimetre.
- In a double-slit experiment, state the effect on the fringe separation of (a) doubling D, (b) using blue light instead of red, (c) halving the slit separation.
Show answers
- resultant displacement = vector sum of individual displacements
- λ/2
- a constant phase difference (same frequency)
- (n + ½)λ
- 3.3 × 10⁻⁶ m
- the wave travels to the end and is reflected; incident and reflected waves (same frequency, speed and similar amplitude) travel in opposite directions and superpose; nodes where they always cancel (including the fixed ends), antinodes between
- d = 2 × 600 × 10⁻⁹/sin 52° = 1.52 × 10⁻⁶ m ⇒ 660 lines per mm
- (a) doubles (b) decreases (shorter λ) (c) doubles
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