The key idea
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.
Understand it · 01
Track a disturbance without confusing it with particles
A progressive wave carries a pattern of oscillation through space. In a transverse wave the oscillations are perpendicular to the direction of propagation; in a longitudinal wave they are parallel. The medium’s particles oscillate about equilibrium positions and do not travel along with the whole wave. A displacement–distance snapshot reveals wavelength, whereas a displacement–time trace at one position reveals period. Wave speed is fλ because one wavelength advances in one period.
Phase describes the stage of an oscillation. For a sinusoidal progressive wave, a separation of one wavelength corresponds to 2π radians of phase difference, while half a wavelength corresponds to π. The sign of the phase relationship depends on which point leads, but many interference questions need only the magnitude or whether points are in phase. Always check whether an axis is position or time before using it to calculate phase.
Understand it · 02
Add displacement, then interpret intensity
The principle of superposition states that overlapping waves produce a resultant displacement equal to the sum of their individual displacements. Constructive interference occurs where they arrive in phase, and destructive interference where they arrive in antiphase. Stable interference requires a constant phase relationship; coherent sources provide this. Equal amplitudes are needed for complete cancellation, but unequal coherent amplitudes can still produce a stable pattern with minima that are not zero.
For two initially in-phase sources, an integer number of wavelengths of path difference produces constructive interference. An odd half-wavelength produces destructive interference. In the familiar double-slit approximation, fringe separation is λD/a when the screen is far from the slits and the angles are small. Increasing wavelength widens the pattern; increasing slit separation narrows it. Intensity is proportional to amplitude squared, so doubling resultant amplitude gives four times the intensity.
Understand it · 03
Use the boundary conditions to find the pattern
A standing wave can be formed by two waves of the same frequency and amplitude travelling in opposite directions. Nodes remain at zero displacement; antinodes have maximum oscillation amplitude. Adjacent nodes are separated by half a wavelength. Points in a loop between neighbouring nodes oscillate in phase, while points in neighbouring loops are in antiphase. Unlike a progressive wave, an ideal standing wave has no net energy transport along the pattern.
For a string fixed at both ends, both ends are displacement nodes. The fundamental fits half a wavelength into the vibrating length and higher harmonics fit additional half-wavelengths. In a practical resonance experiment, change the driving frequency gradually and identify the largest stable amplitude without confusing a transient with resonance. Keep tension and vibrating length controlled. Air-column boundary conditions differ from a fixed string, so do not transfer the same harmonic formula without checking which ends are open or closed.
Key definitions
- 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.
Equations, units & conditions
Double slit
λ = a x ÷ D
a = slit separation, x = fringe spacing, D = distance to screen
Diffraction grating
d sin θ = n λ
Intensity
I ∝ amplitude²
Worked examples, step by step
Example 1
Light of wavelength 600 nm falls on a diffraction grating with 400 lines per mm. Find the angle of the second-order maximum.
Follow the method
- d = 1 ÷ 400,000 = 2.5 × 10⁻⁶ m.
- d sin θ = nλ: sin θ = (2 × 600 × 10⁻⁹) ÷ (2.5 × 10⁻⁶) = 0.48.
- θ = sin⁻¹(0.48).
θ ≈ 28.7°
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Build · 3 suggested marks
1. Double-slit separation
Light of wavelength 520 nm passes through slits separated by 0.25 mm. A screen is 2.0 m away. Calculate the fringe separation using the small-angle approximation.
Show solution and self-check: question 1
- Convert λ = 5.20 × 10⁻⁷ m and a = 2.5 × 10⁻⁴ m.
- w = λD/a = (5.20 × 10⁻⁷ × 2.0)/(2.5 × 10⁻⁴).
4.16 mm, approximately 4.2 mm.
Self-check · one mark per point
- Convert both small lengths to metres.
- Apply w = λD/a.
- Obtain 0.00416 m.
Build · 3 suggested marks
2. Second harmonic on a string
A 0.80 m string is fixed at both ends. Its wave speed is 64 m s⁻¹. Find the wavelength and frequency of its second harmonic.
Show solution and self-check: question 2
- Second harmonic contains two half-wavelengths, so L = λ = 0.80 m.
- f = v/λ = 64/0.80.
Wavelength 0.80 m; frequency 80 Hz.
Self-check · one mark per point
- Apply the fixed-end boundary condition.
- Find wavelength 0.80 m.
- Use v = fλ to obtain 80 Hz.
Common mistakes & how to avoid them
- Interference explanations without path difference: for initially in-phase coherent sources, constructive interference occurs at path difference nλ and destructive interference at (n + ½)λ. The phrase 'path difference' is the mark.
- 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.
- 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.
- Pipe problems with the wrong ends: a closed end is a displacement node and an open end a displacement antinode; the pressure pattern has the opposite boundary conditions. Draw the wave inside the pipe before touching the algebra.
- Doubling errors with node spacing: adjacent nodes are λ/2 apart, not λ.
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
What conditions are needed for observable interference?
A stable interference pattern requires overlapping waves with a constant phase relationship, usually from coherent sources of the same frequency. Similar amplitudes improve contrast, and light waves need compatible polarisation. Equal amplitude is required for complete cancellation, not for interference itself.
If the relative phase changes unpredictably during observation, maxima and minima shift and the measured pattern can average out. Splitting light from one source provides a reliable way to preserve a phase relationship, whereas independent ordinary lamps do not normally form a stable visible pattern. For initially in-phase sources, use path difference to identify constructive and destructive locations. If the sources already have a nonzero phase difference, include that as well; a whole-wavelength path difference alone no longer guarantees a maximum. Amplitude cancellation refers to displacement or field, while intensity depends on its square.
How is a stationary wave different from a progressive wave?
An ideal standing wave has stationary nodes and antinodes with no net energy transport along its pattern. A progressive wave carries a travelling phase pattern and transfers energy; its amplitude can also vary if it spreads or is attenuated.
Between adjacent nodes of a standing wave, points oscillate in phase but with different amplitudes. Crossing a node changes the phase by π. In a progressive sinusoidal wave, phase changes continuously with position, so a separation of half a wavelength gives antiphase. A standing wave can be understood as two equal waves travelling in opposite directions, which explains why energy transport cancels ideally. Real driven systems still require energy input to replace losses, so saying that no energy ever enters a standing-wave apparatus would be too strong.
Why does a diffraction grating give sharper maxima than a double slit?
A grating uses many regularly spaced slits. At a principal maximum their contributions reinforce, but slightly away from that direction many contributions cancel. Increasing the illuminated slit count therefore makes principal maxima narrower and easier to locate.
The principal-maximum directions still satisfy d sin θ = nλ for normal incidence. Here d is the separation between neighbouring slits, not the width of an individual slit or the total grating width. Convert a line density into spacing by taking its reciprocal in consistent units. The finite width of each slit also produces a diffraction envelope that affects the brightness of different orders. A sharper peak helps wavelength measurement, but it does not eliminate angular reading uncertainty, alignment errors or an incorrectly stated line density.
First-order angle
A grating has 300 lines per millimetre and is illuminated normally with 650 nm light. Find the first-order angle.
- d = 1/(300 × 10³) = 3.33 × 10⁻⁶ m.
- sin θ = λ/d = 650 × 10⁻⁹/(3.33 × 10⁻⁶) = 0.195.
- θ = sin⁻¹(0.195).
Approximately 11.2°.
Why are adjacent nodes half a wavelength apart?
Adjacent nodes in a standing wave are half a wavelength apart. A node to its nearest antinode is one quarter wavelength. These distances follow from the cancellation and reinforcement pattern of the two opposing waves.
Draw the complete loop before assigning a wavelength. A single loop between two adjacent nodes spans only half a wavelength, even though it may look like one complete hump. For a string fixed at both ends, the fundamental has one loop and wavelength twice the string length. The second harmonic has two loops and wavelength equal to the length. In an air column, identify whether the diagram represents displacement or pressure because their nodes and antinodes exchange roles. The boundary conditions determine which modes can fit; the same formula does not automatically apply to every pipe.
Wave speed from nodes
Adjacent nodes on a string are 0.18 m apart and its frequency is 150 Hz. Find wavelength and wave speed.
- Node separation = λ/2, so λ = 0.36 m.
- v = fλ = 150 × 0.36.
Wavelength 0.36 m; speed 54 m s⁻¹.
How do I convert a path difference into a phase difference?
For a sinusoidal progressive wave, phase-difference magnitude is 2π times path difference divided by wavelength. Whole wavelengths correspond to full cycles; half a wavelength corresponds to π radians or 180°.
Phase is periodic, so phase differences differing by 2π describe the same relative stage of oscillation. For interference, first decide whether the sources begin in phase; any initial phase difference must be combined with the travel difference. In a time trace, use 2πΔt/T instead of a spatial path difference. The result may describe a phase lead or lag, which depends on propagation direction and which point is used as reference. Do not infer standing-wave phase from position using the progressive-wave rule without accounting for the opposing wave.
What is the difference between diffraction and interference?
Diffraction is the spreading and pattern formation associated with a wave passing an aperture or obstacle. Interference describes the superposition of overlapping contributions. The ideas are connected: diffraction patterns can be analysed as interference between different parts of a wavefront.
The distinction is useful when predicting how a pattern changes. Narrowing one slit generally broadens its diffraction pattern, while reducing the separation of two slits increases their interference-fringe spacing. These are different geometric quantities, even though a double-slit apparatus displays both effects at once. The broad single-slit envelope affects the brightness of the finer double-slit fringes. Strong spreading is particularly apparent when the opening size is comparable with wavelength. Avoid explaining diffraction merely as waves bending around corners without relating the amount of spreading to the relevant size ratio.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- 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
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
Reference this page
Dr Desouky Physics Academy. Edexcel IAL Physics: Waves and superposition. Updated 2026-09-25. https://www.drdesouky.com/revision/a-level-physics/waves-and-superposition
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