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.
Single-aperture diffraction and Rayleigh resolution 11(k)–(m)
Question 1
For 500 nm light through a 0.200 mm slit, find the first-minimum angle. State how a narrower gap changes water-wave diffraction and give the Rayleigh limiting angle for a 0.200 mm aperture.
Check the model response
b sinθ = λ gives sinθ = 500 × 10⁻⁹/(0.200 × 10⁻³) = 2.50 × 10⁻³, so θ ≈ 2.50 mrad. A gap closer to the wavelength gives more pronounced spreading. Rayleigh gives the same small-angle scale θ ≈ λ/b = 2.50 mrad for just resolving two sources.
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.
Single-aperture diffraction and Rayleigh resolution 11(k)–(m)
Check this idea
Misconception: Diffraction begins only when an aperture is narrower than the wavelength.
Repair: Diffraction always occurs; its spreading becomes pronounced when aperture size is comparable to wavelength.
Check this idea
Misconception: The grating spacing and single-slit width are interchangeable.
Repair: A grating uses adjacent-slit separation in a sinθ = nλ. A single slit uses its width in b sinθ = λ for the first minimum.
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.
Single-aperture diffraction and Rayleigh resolution 11(k)–(m)
Model 1
Light of wavelength 600 nm passes through a 0.250 mm slit onto a screen 2.00 m away. Estimate central-maximum width. Also find the Rayleigh limit for a 50.0 mm aperture at 550 nm.
Check the model response
First minima have θ ≈ λ/b = 2.40 × 10⁻³ rad. Central width ≈ 2Dθ = 2(2.00)(2.40 × 10⁻³) = 9.60 mm. Rayleigh limit = 550 × 10⁻⁹/0.0500 = 1.10 × 10⁻⁵ rad.
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.
Single-aperture diffraction and Rayleigh resolution 11(k)–(m)
Question 1
A 0.100 mm slit produces first minima at ±0.0050 rad. Estimate wavelength, then state how doubling slit width changes the first-minimum angle.
Hint: For small angles, bθ ≈ λ.
Check the model response
λ ≈ bθ = (0.100 × 10⁻³)(0.0050) = 5.0 × 10⁻⁷ m. Since θ ≈ λ/b, doubling b halves the first-minimum angle and narrows the central maximum.
independent practice
5. Independent practice
Solve without repair notes and state the graph, source, boundary, coherence and aperture assumptions used.
Single-aperture diffraction and Rayleigh resolution 11(k)–(m)
Question 1
Explain why sound is heard around a doorway more readily than light is seen around it, then compare first-minimum and Rayleigh equations.
Check the model response
Sound wavelengths are often comparable to doorway dimensions, giving pronounced diffraction; visible wavelengths are far smaller. For a single slit, b sinθ = λ locates the first intensity minimum. The Rayleigh estimate θ ≈ λ/b gives the angular separation at which one diffraction maximum falls at the other's first minimum.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Single-aperture diffraction and Rayleigh resolution 11(k)–(m)
Question 1
For 650 nm light through a 0.300 mm slit, find the first-minimum angle. Explain one everyday diffraction phenomenon and find the Rayleigh limit for a 6.0 mm aperture at the same wavelength.
Check the model response
θ ≈ λ/b = 650 × 10⁻⁹/(0.300 × 10⁻³) = 2.17 mrad. Sound around corners or water-wave spreading at a narrow gap demonstrates diffraction. The Rayleigh limit is 650 × 10⁻⁹/0.0060 = 1.08 × 10⁻⁴ rad.
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.
Single-aperture diffraction and Rayleigh resolution 11(k)–(m)
Question 1
A telescope aperture is doubled while wavelength is unchanged. State the Rayleigh-limit factor and relate it to the single-slit first-minimum angle.
Check the model response
Since θR ≈ λ/b, doubling aperture halves the minimum resolvable angle, improving resolution. The same inverse-aperture scale sets the first-minimum angle b sinθ = λ; a wider aperture gives a narrower diffraction pattern.