AQA A-Level Physics Paper 1, June 2025: Question 20

Total Maxima in Diffraction Gratings

📌 What this question tests

This question assesses your ability to apply the diffraction grating equation to calculate the theoretical limit on the number of observable interference maxima. Core competencies tested include:

Handling SI unit prefixes: nanometres ( nm = 10⁻⁹ m ) and micrometres ( μm = 10⁻⁶ m ).
Applying the boundary condition for diffraction angles: sin θ ≤ 1 (since θ ≤ 90° ).
Truncating to an integer order (always round down, never round up).
Accounting for the full diffraction pattern: maxima on both sides plus the central zero-order maximum ( Total = 2n + 1 ).

Question 20 Analysis

Monochromatic light on a transmission diffraction grating

✅ Correct Answer

D (17)

Awarded 1 mark under AO2 (Application of knowledge in a practical context).

📐 Step-by-Step Calculation

Convert all quantities to standard SI units (metres):
Wavelength, λ = 420 nm = 420 × 10⁻⁹ m
Slit spacing, d = 3.6 μm = 3.6 × 10⁻⁶ m
State the grating formula and physical limit:
d sin θ = nλ
The maximum possible angle for light transmitted through the grating is θ = 90° , which gives sin θ = 1 .
Calculate theoretical maximum order, n:
n ≤ d / λ
n ≤ (3.6 × 10⁻⁶) / (420 × 10⁻⁹) = 8.57
Determine the highest integer order:
Because an order must be a complete integer, we round down: n_max = 8 .
Count the total number of maxima:
The pattern contains:
- 8 maxima on the positive side ( n = +1 to +8 )
- 8 maxima on the negative side ( n = -1 to -8 )
- 1 central maximum ( n = 0 )
Total = 2n_max + 1 = 2(8) + 1 = 17

💡 Key Knowledge

Diffraction Formula: d sin θ = nλ , where d is slit separation, θ is diffraction angle, n is order number, and λ is wavelength.
Line density relation: If lines per mm ( N ) is given instead of slit spacing, d = 1 / N (in appropriate units). Here, d is given directly.
Central maximum: The straight-through beam corresponds to n = 0 (zero path difference). It is always present and is usually the brightest maximum.

🧠 Exam Technique & Speed Tip

Memorise the master formula: Whenever an exam question asks for total number of maxima/fringes/beams, use:
Total = 2 × int(d / λ) + 1
Check your units first: Notice that 3.6 μm = 3600 nm . Doing 3600 / 420 = 8.57 in your head or calculator takes under 5 seconds!

❌ Common Traps & Wrong Options

Choosing A (8): Found n = 8 correctly, but forgot that maxima appear on both sides of the centre, plus the centre itself.
Choosing B (9): Added the central maximum to one side only ( 8 + 1 = 9 ), or rounded 8.57 up to 9.
Choosing C (16): Doubled the order for both sides ( 2 × 8 = 16 ) but forgot the central zero-order maximum ( n = 0 ). This is the single most common student error on this paper.
Rounding Up: Never round 8.57 up to 9. The 9th order would require sin θ > 1 , which is physically impossible.

AQA A-Level Physics Paper 1, June 2025: Question 20

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