AQA A-Level Physics Paper 1, June 2022: Question 21

1 mark · Medium difficulty · Multiple Choice

Calculate the total number of maxima produced by a plane transmission diffraction grating given the wavelength of incident monochromatic light and the slit separation.

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Question

Multiple-choice question 21 asks for the number of maxima produced by a diffraction grating with slit separation 2.5 x 10^-6 m when illuminated normally by monochromatic light of wavelength 5.8 x 10^-7 m. Four options are given: A 4, B 5, C 8, D 9.
Question text

21 Monochromatic light of wavelength 5.8 × 10−7 m is incident normally on a plane

transmission diffraction grating that has a slit separation of 2.5 × 10−6 m.

How many maxima are produced by the grating?

[1 mark]

A 4

B 5

C 8

D 9

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 21 is D (AO2), corresponding to 9 maxima.

21 D (AO2) 9

How to answer it

Calculating Total Maxima for a Diffraction Grating

What this question tests

This question assesses your understanding of wave optics, specifically diffraction gratings. You must recall the grating equation ( d sin(theta) = n lambda ), understand the physical limits of angle ( sin(theta) <= 1 ), correctly calculate the maximum possible order n , and use symmetry to find the total number of observable maxima across both sides of the central zero-order beam.

Question 21

Exam Breakdown & Solution

✅ Correct Answer

D (9)

The total number of maxima produced by the grating is 9 (orders from n = -4 to n = +4 inclusive).

💡 Key Knowledge

  • The Grating Equation: d sin(theta) = n lambda , where d is slit separation, theta is the angle of diffraction, n is the order, and lambda is wavelength.
  • The Sine Limit: The maximum possible value for sin(theta) is 1 . Therefore, n lambda / d ≤ 1 .
  • Symmetry: Maxima occur in symmetrical pairs on either side of the central maximum ( n = 0 ).

📐 Step-by-Step Calculation

  1. Rearrange the grating equation for order n :
    n = (d sin(theta)) / lambda
  2. Substitute sin(theta) = 1 to find the maximum theoretical order:
    n = (2.5 × 10⁻⁶ m × 1) / (5.8 × 10⁻⁷ m)
  3. Evaluate the calculation:
    n = 4.31
  4. Truncate to find the integer order:
    Since order must be an integer, the highest whole order visible is n = 4 .
  5. Calculate total maxima:
    The visible orders range from -4 through 0 to +4 .
    Total maxima = (2 × 4) + 1 (for n = 0) = 9 .

🧠 Exam Technique & Examiner Insight

  • Rounding trap: Students frequently see 4.31 and incorrectly round up to 5 . Remember that diffraction orders cannot exceed the geometric limit; rounding up would imply an angle where sin(theta) > 1 , which is physically impossible.
  • Forgetting the zero order: A common oversight is calculating 2 × 4 = 8 and selecting option C. Always remember to add 1 for the central un-diffracted maximum ( n = 0 ).

❌ Common Student Errors

  • Distractor A (4): Found by only counting positive orders or forgetting to multiply by 2 for both sides.
  • Distractor B (5): Caused by incorrectly rounding 4.31 up to 5 .
  • Distractor C (8): Calculated as 2 × 4 , omitting the crucial central n = 0 maximum.
AQA Mark Scheme Note: AO2 (Application of knowledge) — 1 mark awarded for selecting D through correct calculation of maximum integer order and symmetry.

Topics

Physics · Required Practicals · 3.3 Waves · AS practicals (1–6)

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.