Waves and superposition: models, interference and diffraction
Key idea: Use signed displacement and phase consistently across time and space, then apply superposition to standing waves, interference and diffraction without confusing their distinct conditions.
Before you start: Oscillations objective chainQuantities & Measurement objective chain
By the end, you can
- Model mechanical and electromagnetic progressive waves, interpret time and position graphs, and use wave quantities, phase and energy transfer.
- Apply intensity–amplitude and inverse-square relationships, and use polarisation and Malus' law for electromagnetic waves.
- Apply superposition and explain standing-wave experiments, boundary conditions, nodes, antinodes and sound-wavelength measurement.
- Analyse coherent two-source interference and Young double-slit fringes using phase and path difference.
- Use diffraction gratings, single-slit first minima and the Rayleigh criterion with their distinct aperture spacings and assumptions.
Starting-point self-check
1. Check your starting point
Attempt all seven groups without notes and mark the first graph axis, phase conversion, boundary condition or aperture spacing you could not justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Intensity, amplitude and inverse-square spreading 10(g)–(h)
Question 1
An isotropic point source radiates 12.0 W without loss. Find intensity at 4.00 m and at 8.00 m, then state the intensity change if wave amplitude halves.
Check the model response
I = P/(4πr²). At 4.00 m, I = 12.0/[4π(4.00)²] = 5.97 × 10⁻² W m⁻². Doubling distance gives 1.49 × 10⁻² W m⁻². Since I ∝ A², halving amplitude quarters intensity.
repair
2. Repair the eleven common breaks
Use only the repair matching an error, then redraw the axes, signed displacement, boundary conditions, rays or aperture before retrying.
Intensity, amplitude and inverse-square spreading 10(g)–(h)
Check this idea
Misconception: Halving amplitude halves intensity.
Repair: For a progressive wave, I ∝ A², so halving amplitude reduces intensity to one quarter.
Check this idea
Misconception: Every measured intensity obeys an inverse-square law.
Repair: The model requires point-source spreading without energy loss. Absorption, directional emission or nearby boundaries break the simple relation.
worked example
3. Follow seven worked models
Follow how each solution fixes the graph type, energy-spreading model, transmission axis, boundary condition, path difference or aperture before calculating.
Intensity, amplitude and inverse-square spreading 10(g)–(h)
Model 1
An isotropic source has intensity 0.318 W m⁻² at 5.00 m. Find its power and the amplitude factor at 10.0 m.
Check the model response
P = I4πr² = 0.318(4π)(5.00²) ≈ 100 W. At double distance, intensity is one quarter. Since amplitude is proportional to √I, amplitude is one half.
guided practice
4. Guided practice
Use each hint only to select the graph interval, spreading surface, polarisation reference, mode shape, interference condition or aperture equation.
Intensity, amplitude and inverse-square spreading 10(g)–(h)
Question 1
A point source gives 6.0 × 10⁻³ W m⁻² at 2.0 m. Find intensity at 5.0 m and the amplitude ratio A₅/A₂.
Hint: Use I₂/I₁ = (r₁/r₂)², then amplitude ∝ √I.
Check the model response
I₅ = 6.0 × 10⁻³(2.0/5.0)² = 9.6 × 10⁻⁴ W m⁻². A₅/A₂ = √(I₅/I₂) = 2.0/5.0 = 0.40.
independent practice
5. Independent practice
Solve without repair notes and state the graph, source, boundary, coherence and aperture assumptions used.
Intensity, amplitude and inverse-square spreading 10(g)–(h)
Question 1
Without absorption, intensity falls from 0.80 to 0.20 W m⁻². State the distance and amplitude factors.
Check the model response
The intensity factor is 1/4. From I ∝ 1/r², distance doubles. From I ∝ A², amplitude halves.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Intensity, amplitude and inverse-square spreading 10(g)–(h)
Question 1
Define wave intensity. A lossless isotropic source radiates 50 W; find intensity at 10 m and the intensity and amplitude factors at 30 m relative to 10 m.
Check the model response
Intensity is power transferred per unit area. I₁₀ = 50/[4π(10)²] = 3.98 × 10⁻² W m⁻². Tripling distance reduces intensity to 1/9 and amplitude to 1/3.
Re-test practice
7. Delayed re-test practice
Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Intensity, amplitude and inverse-square spreading 10(g)–(h)
Question 1
A point-source wave has amplitude A and intensity I at radius r. State both quantities at radius 4r, assuming lossless spreading.
Check the model response
I ∝ 1/r², so intensity becomes I/16. Since intensity is proportional to amplitude squared, amplitude becomes A/4.