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.
Diffraction-grating maxima and wavelength 11(i)–(j)
Question 1
A grating has 500 lines mm⁻¹. Find its spacing and the angle of the second-order maximum for 600 nm light.
Check the model response
The line density is 5.00 × 10⁵ m⁻¹, so a = 1/N = 2.00 × 10⁻⁶ m. From a sinθ = nλ, sinθ = 2(600 × 10⁻⁹)/(2.00 × 10⁻⁶) = 0.600, giving θ = 36.9°.
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.
Diffraction-grating maxima and wavelength 11(i)–(j)
Check this idea
Misconception: Lines per metre can be inserted directly as a in a sinθ = nλ.
Repair: Convert line density N to adjacent-line spacing a = 1/N before applying the grating equation.
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.
Diffraction-grating maxima and wavelength 11(i)–(j)
Model 1
A 600 lines mm⁻¹ grating receives 500 nm light normally. Find first-order angle and highest possible order, and state how wavelength is determined experimentally.
Check the model response
a = 1/(6.00 × 10⁵) = 1.67 × 10⁻⁶ m. First order: sinθ = λ/a = 0.300, so θ = 17.5°. Since nλ ≤ a, nmax = 3. Measure a principal-maximum angle for known a and n, then calculate λ = a sinθ/n; spectrometer structure is not required.
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.
Diffraction-grating maxima and wavelength 11(i)–(j)
Question 1
A first-order maximum for 450 nm light is at 15.0°. Find the grating spacing and line density in lines mm⁻¹.
Hint: Find a from a sinθ = λ, then use N = 1/a and convert metres to millimetres.
Check the model response
a = 450 × 10⁻⁹/sin15.0° = 1.74 × 10⁻⁶ m. N = 5.75 × 10⁵ m⁻¹ = 575 lines mm⁻¹.
independent practice
5. Independent practice
Solve without repair notes and state the graph, source, boundary, coherence and aperture assumptions used.
Diffraction-grating maxima and wavelength 11(i)–(j)
Question 1
A grating spacing is 2.50 μm. For 625 nm light, find the first two principal-maximum angles and state why no fifth order exists.
Check the model response
For n = 1, sinθ = 0.250, so θ = 14.5°. For n = 2, sinθ = 0.500, so θ = 30.0°. A fifth order would require sinθ = 5λ/a = 1.25, which is impossible.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Diffraction-grating maxima and wavelength 11(i)–(j)
Question 1
A 400 lines mm⁻¹ grating gives a second-order maximum at 30.0°. Calculate wavelength and describe the measurement needed, respecting the syllabus exclusion.
Check the model response
a = 1/(4.00 × 10⁵) = 2.50 × 10⁻⁶ m. λ = a sinθ/n = (2.50 × 10⁻⁶)(0.500)/2 = 625 nm. Measure a principal-maximum angle for known grating spacing and order; spectrometer structure and use are not required.
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.
Diffraction-grating maxima and wavelength 11(i)–(j)
Question 1
Light of unknown wavelength gives a third-order maximum at 48.6° with a 300 lines mm⁻¹ grating. Find the wavelength.
Check the model response
a = 1/(3.00 × 10⁵) = 3.33 × 10⁻⁶ m. λ = a sin48.6°/3 = 8.33 × 10⁻⁷ m = 833 nm.