AQA AS Level Physics Paper 2, June 2022: Question 19

Calculating Total Observable Maxima in a Diffraction Grating

AQA AS Level Physics • Waves • Multiple Choice

What this question tests

This question assesses your ability to apply the diffraction grating formula ( d sin θ = nλ ), manipulate grating line density to find slit spacing ( d = 1 / N ), determine the wavelength of incident light, and logically calculate the maximum possible order ( n_max ) before converting that order count into the total number of observable bright fringes.

Question 19: Full Breakdown

Determining total visible maxima

✅ Correct Answer: D (7)

The total number of observable maxima is 7 (made up of the 3rd order on each side plus the central zero-order maximum).

💡 Key Knowledge

Grating equation: d sin θ = nλ
Grating spacing d is the reciprocal of the line density N : d = 1 / 4.8×10⁵ m
The maximum possible order n_max is found by setting sin θ = 1 (since max value of sine is 1).

🧠 Exam Technique

For "total observable maxima" questions, remember the golden rule: Calculate n_max (integer floor), multiply by 2 (for both sides), and add 1 (for the central n = 0 maximum).

❌ Common Errors

Forgetting to add the central maximum ( n = 0 ), leading to an answer of 6.
Rounding n up instead of down (e.g., rounding 3.12 to 4). Orders must be whole integers that actually fit within 90 degrees ( sin θ ≤ 1 ).

📐 Step-by-Step Calculation

Find grating spacing ( d ):
d = 1 / (4.8 × 10⁵) = 2.083 × 10⁻⁶ m
Find wavelength ( λ ) using first-order data ( n = 1 , θ = 16° ):
λ = (d sin θ) / n
λ = (2.083 × 10⁻⁶ × sin(16°)) / 1 = 5.744 × 10⁻⁷ m
Find maximum possible order ( n_max ) by setting sin θ = 1 :
n_max = d / λ = (2.083 × 10⁻⁶) / (5.744 × 10⁻⁷) = 3.62
Since order must be an integer, truncate (round down) to n = 3 .
Calculate total number of maxima:
Total maxima = (2 × n_max) + 1 = (2 × 3) + 1 = 7 .

Examiner Note (AO2): Top-level responses quickly established the wavelength from the first-order data and directly compared d / λ to find the limiting order without getting bogged down in intermediate rounding errors. Option D was successfully selected by candidates who remembered to account for both positive/negative sides and the central fringe.