AQA A-Level Physics Paper 1, November 2020: Question 19

1 mark · Medium difficulty · Multiple Choice

Calculate the wavelength of monochromatic light incident on a diffraction grating given the grating spacing and the angle between second-order diffraction maxima.

Practise this question

Question

Multiple choice question 19. Monochromatic light is incident normally on a diffraction grating with 4.50 x 10^5 lines m^-1. The angle between the second-order diffraction maxima is 44 degrees. Options are A: 208 nm, B: 416 nm, C: 772 nm, D: 832 nm.
Question text

19 Monochromatic light is incident normally on a diffraction grating that has

4.50 × 105 lines m−1.

The angle between the second-order diffraction maxima is 44°.

What is the wavelength of the light?

[1 mark]

A 208 nm

B 416 nm

C 772 nm

D 832 nm

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer is B for question 19.

19 B

How to answer it

Diffraction Grating Wavelength Calculation

What this question tests

This multiple-choice question assesses your understanding of wave optics, specifically diffraction gratings. You are tested on your ability to extract grating spacing from line density, interpret angles given symmetrically about the central maximum, correctly apply the diffraction grating formula ( d sin θ = nλ ), and perform unit conversions between metres and nanometres.

Question Part 19

Determining the Wavelength of Monochromatic Light

✅ Correct Answer

B: 416 nm

Awarded for correctly identifying option B as the right choice.

💡 Key Knowledge

  • Grating spacing (d): Calculated using d = 1 / N , where N is the number of lines per metre.
  • Angle θ : The angle given is the total angle between the second-order maxima. The angle from the central maximum ( θ ) is half of this value.
  • Order ( n ): Explicitly stated as the second-order, so n = 2 .

🧠 Exam Technique

  • Always halve angles described as "the angle between the Xth-order maxima" before plugging them into trigonometric functions.
  • Keep intermediate values in your calculator with full precision and only round your final answer to a sensible number of significant figures matching the data.

❌ Common Errors

  • Using the full angle: Using 44° directly instead of halving it to 22° .
  • Inverting grating density incorrectly: Forgetting that d = 1 / (4.50 × 10⁵) and instead multiplying by N .
  • Unit conversion slips: Forgetting to convert the final answer from metres ( m ) into nanometres ( nm ) by multiplying by 10⁹ .

📐 Step-by-Step Calculation

  1. Find the grating spacing ( d ):
    d = 1 / N = 1 / (4.50 × 10⁵ m⁻¹) = 2.222 × 10⁻⁶ m
  2. Determine the angle ( θ ) for n = 2 :
    The angle between the +2nd and -2nd order maxima is 44° .
    Therefore, the diffraction angle from the normal is θ = 44° / 2 = 22° .
  3. Rearrange the diffraction grating equation for wavelength ( λ ):
    d sin θ = nλ → λ = (d sin θ) / n
  4. Substitute values:
    λ = (2.222 × 10⁻⁶ m × sin(22°)) / 2
    λ = (2.222 × 10⁻⁶ × 0.3746) / 2 = 4.160 × 10⁻⁷ m
  5. Convert to nanometres:
    λ = 4.160 × 10⁻⁷ m × 10⁹ = 416 nm (Matches Option B).

Topics

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

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.