Edexcel IAL Physics revision · AS — Waves
Waves and superposition
The IGCSE wave model survives intact, then AS adds three ideas central to Paper 2: interference from two sources, the diffraction grating equation, and stationary waves on strings and in pipes.
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What the syllabus demands
- —Use intensity ∝ amplitude²; describe the Doppler effect and use the frequency formula
- —State the principle of superposition
- —Describe two-source interference; use λ = ax/D for double slits
- —Use d sin θ = nλ for diffraction gratings
- —Explain the formation of stationary waves; identify nodes and antinodes; solve string and pipe problems
Definitions that earn marks
Clear definitions to practise — check your course mark scheme
- Principle of superposition
- When two or more waves meet at a point, the resultant displacement is the vector sum of the individual displacements.
- Coherence
- Two sources are coherent when they have a constant phase difference (which requires the same frequency).
- Stationary (standing) wave
- The pattern formed when two progressive waves of equal frequency and amplitude travelling in opposite directions superpose — with nodes of zero amplitude and antinodes of maximum amplitude, storing energy rather than transferring it.
- Node
- A point on a stationary wave where the amplitude is always zero.
The equations
More equations to practise: the Edexcel IAL formula sheet.
Where the marks die
Common mistakes to check
- 01
Interference explanations without path difference: constructive where the path difference is a whole number of wavelengths (nλ), destructive at (n + ½)λ. The phrase 'path difference' is the mark.
- 02
Confusing a (slit separation) with d (grating spacing) — and forgetting d = 1 ÷ lines per metre. If the grating says 500 lines/mm, d = 2 × 10⁻⁶ m.
- 03
Treating a stationary wave like a progressive one. Between adjacent nodes every point oscillates in phase; across a node the phase flips by 180°. Progressive-wave phase logic does not apply.
- 04
Pipe problems with the wrong ends: a closed end is a node, an open end is an antinode. Draw the wave inside the pipe before touching the algebra.
- 05
Doubling errors with node spacing: adjacent nodes are λ/2 apart, not λ.
One worked example, done properly
Question
Light of wavelength 600 nm falls on a diffraction grating with 400 lines per mm. Find the angle of the second-order maximum.
Method
- 1.d = 1 ÷ 400,000 = 2.5 × 10⁻⁶ m.
- 2.d sin θ = nλ: sin θ = (2 × 600 × 10⁻⁹) ÷ (2.5 × 10⁻⁶) = 0.48.
- 3.θ = sin⁻¹(0.48).
θ ≈ 28.7°