A-Level · Physics · AQA · Mark scheme decoded
AQA A-Level Physics: Diffraction Patterns and Gratings — mark scheme explained
The short answer
Diffraction patterns are a fascinating aspect of wave behavior, particularly when light passes through narrow slits or gratings.
The question
A plane transmission grating has a grating spacing of d = 1.5 × 10 -6 m . Monochromatic light with a wavelength of λ = 600 nm is incident on the grating at normal incidence. Calculate the angle of diffraction for the first-order maximum ( n = 1 ).
[Paraphrased for study — not reproduced from any exam paper.]
Mark scheme, decoded
How the examiner actually awards the marks on this topic.
Gradora's own decode of the marking approach — not the exam board's published mark scheme.
How marks are awarded
For numerical calculations, ensure that all steps are shown clearly. For descriptive questions, provide detailed explanations and use appropriate scientific terminology. Always check units and conversions.
What the command words demand
- Calculate
- Perform a numerical calculation using given data and appropriate equations.
- Describe
- Provide a detailed account or explanation of a phenomenon or process.
- Explain
- Give reasons for a particular effect or behavior, often linking cause and effect.
- Determine
- Find the value of a quantity by calculation or measurement.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate approximately 5-7 minutes per question to ensure you have enough time to show all working and provide clear, concise answers.
- Write down the equation: dsinθ = nλ1 mark
- Substitute the given values into the equation: (1.5 × 10 -6 )sinθ = (1)(600 × 10 -9 )1 mark
- Simplify the equation: sinθ = (600 × 10 -9 ) / (1.5 × 10 -6 )0 marks
- Calculate the value of sinθ: sinθ = 0.41 mark
- Find the angle θ using the inverse sine function: θ = sin -1 (0.4)0 marks
- θ ≈ 23.6°2 marks
Final answer: 23.6°
Work through every step correctly and you earn all 5 marks.
Another worked example
A single slit of width a = 0.2 mm is illuminated by monochromatic light with a wavelength of λ = 500 nm . Calculate the angular width of the central maximum.
- The angular width of the central maximum is given by: θ ≈ (2λ) / a1 mark
- Substitute the given values into the equation: θ ≈ (2 × 500 × 10 -9 ) / (0.2 × 10 -3 )1 mark
- Simplify the equation: θ ≈ (1000 × 10 -9 ) / (0.2 × 10 -3 )0 marks
- Calculate the value of θ: θ ≈ 5 × 10 -3 radians1 mark
- Convert radians to degrees if necessary: θ ≈ (5 × 10 -3 ) × (180 / π) ≈ 0.29°1 mark
Final answer: 0.29°
Work through every step correctly and you earn all 4 marks.
Common mistakes
Confusing the width of the central maximum with the distance between slits in a diffraction grating.
Why it happens: Students may mix up the concepts of slit width and grating spacing, leading to incorrect calculations or descriptions.
Fix: Clearly distinguish between the slit width (a) in single-slit diffraction and the grating spacing (d) in diffraction gratings. Use the correct equations for each scenario.
Forgetting to convert units when using the diffraction grating equation.
Why it happens: Students often forget to ensure that all units are consistent, leading to incorrect results.
Fix: Always check and convert units before substituting values into equations. Ensure that wavelengths (λ) are in meters and distances (d) are in meters.
Using the wrong order of maximum (n) in calculations.
Why it happens: Students may use n = 0 instead of n = 1 for the first-order maximum, leading to incorrect results.
Fix: Always specify the correct order of the maximum when using the diffraction grating equation. For the first-order maximum, use n = 1.
Misinterpreting the appearance of the diffraction pattern for white light.
Why it happens: Students may describe the pattern as a single bright central band without mentioning the spectrum of colors in the secondary maxima.
Fix: Clearly state that the central maximum is white (all wavelengths overlap there); the side maxima are fringed with colour, with violet/blue nearest the centre and red furthest out (red diffracts most), and the spectra increasingly overlap outward.
Using the wrong formula for the angular width of the central maximum in single-slit diffraction.
Why it happens: Students may use the formula for the first minimum instead of the correct formula for the angular width of the central maximum.
Fix: Use the correct formula: θ ≈ (2λ) / a . Ensure that you are calculating the angular width, not the position of the first minimum.
Forgetting to use the inverse sine function when solving for the angle of diffraction.
Why it happens: Students may solve for sinθ and then forget to find the angle using the inverse sine function, leading to incorrect results.
Fix: Always use the inverse sine function ( sin -1 ) to find the angle of diffraction after solving for sinθ.
Where the marks go
The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.
| Question type | What you’re asked to do | Marks |
|---|---|---|
| Diffraction Grating Angle | Use the grating equation to find the first-order diffraction angle for given light. | 5 |
| Single-Slit Diffraction | Calculate the angular width of the central diffraction maximum from slit width and wavelength. | 4 |
| Diffraction Grating Angle | Use the diffraction grating equation to find the angle of a given order maximum | 5 |
| White Light Diffraction | Describe the appearance of the single-slit diffraction pattern produced by white light. | 3 |
| Total across these question types | 17 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.