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.
Polarisation and Malus' law 10(i)–(j)
Question 1
Plane-polarised light of intensity 80 W m⁻² reaches an analyser at 30° to its polarisation direction. Find transmitted intensity and amplitude factor, and state what polarisation shows about the wave.
Check the model response
I = I₀cos²30° = 80(0.75) = 60 W m⁻². Electric-field amplitude is multiplied by cos30° = 0.866. Polarisation is associated with transverse waves, so the electromagnetic oscillation is transverse.
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.
Polarisation and Malus' law 10(i)–(j)
Check this idea
Misconception: A first ideal polariser leaves unpolarised intensity unchanged.
Repair: It transmits half the intensity of unpolarised light on average. Malus' law then applies between a known plane-polarisation direction and another axis.
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.
Polarisation and Malus' law 10(i)–(j)
Model 1
Unpolarised light of intensity 120 W m⁻² passes through an ideal polariser and then an analyser at 60°. Find final intensity and the analyser's amplitude factor.
Check the model response
The first polariser transmits 60 W m⁻². The analyser gives I = 60cos²60° = 15 W m⁻². Relative to the already polarised wave, field amplitude is multiplied by cos60° = 0.50.
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.
Polarisation and Malus' law 10(i)–(j)
Question 1
Plane-polarised intensity 50 W m⁻² passes through analysers at 20° and then 70° to the original direction. Find the final intensity.
Hint: Apply Malus' law successively; the second relative angle is 70° − 20°.
Check the model response
After the first analyser I₁ = 50cos²20°. The second is 50° from the new polarisation axis, so I₂ = I₁cos²50° = 50cos²20°cos²50° = 18.2 W m⁻².
independent practice
5. Independent practice
Solve without repair notes and state the graph, source, boundary, coherence and aperture assumptions used.
Polarisation and Malus' law 10(i)–(j)
Question 1
Unpolarised light of intensity I enters a polariser followed by an analyser at 45°. Find final intensity and field-amplitude factor relative to the light leaving the first polariser.
Check the model response
The first polariser transmits I/2. The analyser transmits cos²45° = 1/2 of that, so final intensity is I/4. Field amplitude after the analyser is cos45° = 1/√2 of the amplitude leaving the first polariser.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Polarisation and Malus' law 10(i)–(j)
Question 1
Explain why polarisation distinguishes transverse from longitudinal waves. Unpolarised light of intensity 200 W m⁻² passes through a polariser and an analyser at 40°; find final intensity and analyser amplitude factor.
Check the model response
Only an oscillation perpendicular to propagation can be restricted to a chosen transverse direction; a purely longitudinal wave cannot be plane-polarised. The first polariser transmits 100 W m⁻², then I = 100cos²40° = 58.7 W m⁻². The analyser amplitude factor is cos40° = 0.766.
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.
Polarisation and Malus' law 10(i)–(j)
Question 1
Plane-polarised light has intensity 90 W m⁻². An analyser transmits 22.5 W m⁻². Find the acute angle between axes and the field-amplitude factor.
Check the model response
I/I₀ = 0.25 = cos²θ, so cosθ = 0.50 and θ = 60°. The field-amplitude factor is 0.50.