AQA AS Level Physics Paper 2, June 2024: Question 7

1 mark · Easy difficulty · Multiple Choice

Calculate the wavelength of light that produces a fourth-order maximum at the same angle as the third-order maximum for 520 nm light incident on a diffraction grating.

Practise this question

Question

Multiple choice question 07 asking to find wavelength lambda given that the third-order maximum of 520 nm light and fourth-order maximum of wavelength lambda occur at the same diffraction angle theta on a diffraction grating. Four options are provided: A 260 nm, B 390 nm, C 690 nm, and D 780 nm.
Question text

07 Monochromatic light of wavelength 520 nm is incident normally on a diffraction grating.

The third-order maximum occurs at a diffraction angle θ.

Light of wavelength λ is incident normally on the same grating.

The fourth-order maximum also occurs at angle θ.

What is λ?

[1 mark]

A 260 nm

B 390 nm

C 690 nm

D 780 nm

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer is B, corresponding to 390 nm.

07 B 390 nm

How to answer it

Diffraction Grating Wavelength Comparison

What this question tests: Your understanding of the diffraction grating equation ( d sin(θ) = nλ ), how grating spacing ( d ) and angle ( θ ) relate across different orders, and proportional reasoning with wavelengths.
Question 07

Exam Breakdown

✅ Correct Answer

B (390 nm)

Marks: [1 mark]

💡 Key Knowledge

  • The diffraction grating formula is d sin(θ) = nλ .
  • d is the grating spacing (constant for the same grating).
  • θ is the angle of diffraction.
  • n is the order of maximum.

🧠 Exam Technique

Since the same grating and the same angle θ are used in both scenarios, set up an equation equating the constant d sin(θ) for both wavelengths to save calculation time.

❌ Common Errors

  • Inverting the orders (e.g., multiplying 520 nm by 4 instead of 3).
  • Unnecessarily converting nanometres to metres, wasting valuable exam time since all options share the nm unit.

📐 Step-by-Step Calculation

  1. Write down the diffraction grating equation:
    d sin(θ) = nλ
  2. Apply the formula to the first scenario (3rd order, 520 nm):
    d sin(θ) = 3 × 520
  3. Apply the formula to the second scenario (4th order, wavelength λ):
    d sin(θ) = 4 × λ
  4. Equate the two expressions because d sin(θ) is identical:
    4 × λ = 3 × 520
  5. Rearranging and solving for λ:
    λ = (3 × 520) / 4
    λ = 3 × 130 = 390 nm

Topics

Physics · 3.3 Waves

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.