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Chapter 24 · Physics module 15

Waves

One set of ideas. A world of wave phenomena.

Build from oscillations to interference, light and sound. Explore ten focused lessons with interactive models, worked examples and feedback.

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OSCILLATE · SUPERPOSE · EXPLOREv = fλ

Change an input. Discover a relationship.

Chapter Overview

Simple harmonic motion, the wave equation, interference, double slits, gratings, standing waves, and Doppler.

Each lesson brings the key relationships, a model you can control, assumptions, worked examples and a five-question quiz together. Lesson quiz answers remain in the open page only.

Learning Objectives

  • Connect SHM with sinusoidal waves
  • Describe wavelength, frequency and phase
  • Explain interference and diffraction
  • Apply stationary-wave boundary conditions
  • Distinguish sound and light Doppler models

Explore the Ten Lessons

Follow the sequence or select the idea you need.

Try It: The Oscillation Behind a Wave

Move the time probe. Where are speed and acceleration greatest?

Watch the acceleration mirror the displacement with opposite sign — that is the SHM condition a = −ω²x.

x0T = 2 s2T = 4 s

Displacement x0 cm

Velocity-15.7 cm/s

Acceleration0 cm/s²

ω = 2π/T3.14 rad/s

Ideal, undamped, sinusoidal motion. The two-period time window rescales when T changes. Pendulum SHM is a small-angle approximation.

Open the full SHM lesson →

Key Relationships

Wave speed
v = fλ
Angular frequency
ω = 2πf
SHM acceleration
a = −ω²x
Polarisation
I = Iₚ cos²θ
Double-slit spacing
x = λD/a
Grating orders
d sin θ = nλ

Check the assumptions and units in each lesson before applying a formula.

Exam diagrams for this topic1 figure to inspect and practiseQuestions, hints and marking points in one compact subsection.

See it. Read it. Work it.

These figures come from GioPhysics practice papers. Open one, decode the drawing, work the guided questions, then follow its link to the full paper question.

  1. 01

    InspectRead the figure comment.

  2. 02

    TraceFollow labels, arrows and axes.

  3. 03

    AnswerWork one part at a time.

  4. 04

    CheckReveal hints and marking points.

A Level

01Fig. 9.1Waves · SuperpositionA Level
Light diffracted by a grating into orders either side of the straight-through directionλ = 590 nmdiffraction grating500 lines per mmn = 0n = 1n = 1n = 2n = 2θnot to scale

Figure comment

Fig. 9.1A parallel beam of light of wavelength 590 nm travels from the left and meets a diffraction grating at normal incidence; the grating is drawn edge-on as a narrow ruled strip and labelled 500 lines per mm. On the far side five beams spread from the grating: one continues straight through and is labelled n = 0, and above and below it lie beams labelled n = 1 and, at larger angles, n = 2. An arc marks the angle θ between the straight-through direction and the upper second-order beam. The angles drawn are not to scale.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

Guided questions 5 parts

Reading cue. θ is measured from the straight-through n = 0 beam, not from the grating face, and 500 lines per mm must first be inverted to give the line spacing d.

  1. aCalculate Calculate the distance between the centres of adjacent lines on the grating labelled in Fig. 9.1.

    recall2 marks

    Check answer 2 marks
    1. d = 1 mm / 500
    2. d = 2.0 × 10⁻⁶ m
  2. bDetermine Determine the angle between the two first-order beams drawn either side of the straight-through direction.

    routine3 marks

    Check answer 3 marks
    1. sin θ = λ/d = 590 × 10⁻⁹ / 2.0 × 10⁻⁶ = 0.295
    2. θ = 17.2°
    3. the two n = 1 beams are symmetrical about n = 0, so the angle between them is 34.3°
  3. cDetermine The source is replaced by one giving white light of wavelengths from 400 nm to 700 nm. Determine the angular width of the first-order spectrum.

    demanding3 marks

    Check answer 3 marks
    1. sin θ = 400 × 10⁻⁹ / 2.0 × 10⁻⁶ = 0.200, giving θ = 11.5°
    2. sin θ = 700 × 10⁻⁹ / 2.0 × 10⁻⁶ = 0.350, giving θ = 20.5°
    3. angular width = 20.5 − 11.5 = 9.0°
  4. dDeduce With the white-light source still in place, deduce whether the second-order and third-order spectra overlap.

    top of the paper4 marks

    Check answer 4 marks
    1. second order at 700 nm: sin θ = 2 × 700 × 10⁻⁹ / 2.0 × 10⁻⁶ = 0.700, θ = 44.4°
    2. third order at 400 nm: sin θ = 3 × 400 × 10⁻⁹ / 2.0 × 10⁻⁶ = 0.600, θ = 36.9°
    3. the third-order violet leaves at a smaller angle than the second-order red, so the two spectra overlap
    4. the third order is incomplete: sin θ would exceed 1 beyond about 670 nm, so its red end is missing

Transfer challenge

Light of wavelength 590 nm falls on a double slit whose slits are 0.45 mm apart, and fringes are formed on a screen 2.4 m away. Calculate the separation of adjacent bright fringes.

Check answer 3 marks
  1. x = λD/a
  2. x = 590 × 10⁻⁹ × 2.4 / 0.45 × 10⁻³
  3. x = 3.1 × 10⁻³ m

Practise & Teach