AQA AS Level Physics Paper 2, June 2025: Question 11

1 mark · Easy difficulty · Multiple Choice

Determine the phase difference between two coherent sound waves at a detector given their wavelength and the path difference.

Practise this question

Question

Multiple choice question 11 asking: Two coherent sources generate sound waves of wavelength 0.40 m. The waves are in phase when they leave the sources. The path difference between a detector and each of the two sources is 2.0 m. What is the phase difference between the waves arriving at the detector? Options: A zero, B 45 degrees, C 90 degrees, D 180 degrees.

Mark scheme

Show the mark scheme Mark scheme table row showing question 11, correct answer A (zero), and assessment objective AO2.

How to answer it

Path Difference & Phase Difference of Coherent Waves

📋 What this question tests

This multiple-choice question assesses your ability to relate path difference to phase difference for coherent wave sources:

  • Understanding coherence and initial phase alignment (starting in phase).
  • Calculating how many whole or fractional wavelengths fit into a given path difference ( Δx / λ ).
  • Converting an integer number of wavelengths into an equivalent angular phase difference (cycles of 360° or 2π rad).
Question 11 · Multiple Choice

Sound Waves Arriving at a Detector

AQA AS Physics – Waves & Interference (AO2)

✅ Correct Answer

A — zero

Mark Scheme Breakdown:
• 1 mark awarded for selecting A.
• AO2 assessment target: application of wave superposition and phase concepts.

💡 Key Knowledge

  • Path difference (Δx): The difference in distance traveled by two waves from their sources to a given point.
  • Phase difference (Δφ): Related to path difference by:
    Δφ = (Δx / λ) × 360° (or × 2π rad ).
  • Constructive interference: When Δx = nλ (where n is an integer: 0, 1, 2, 3...), the waves arrive completely in phase, meaning the effective phase difference is 0° (zero).

📐 Step-by-Step Calculation

  1. Identify given variables:
    Wavelength, λ = 0.40 m
    Path difference, Δx = 2.0 m
  2. Calculate number of wavelengths:
    Number of cycles = Δx / λ = 2.0 m / 0.40 m = 5.0
  3. Determine angular phase difference:
    Total phase angle = 5 × 360° = 1800°
  4. Express as equivalent phase difference:
    Since 1800° is an exact multiple of 360° ( 5 × 360° + 0° ), the waves arrive exactly at the same point in their cycle.
    Phase difference = zero

🧠 Exam Technique

  • Check whole numbers first: Divide the path difference by the wavelength immediately. If it produces an integer, the answer is immediately zero (or in phase).
  • Half-integers: If the result ends in .5 (e.g. 2.5λ), the waves are in anti-phase (phase difference of 180° / π rad).
  • Quarter-integers: If it ends in .25 or .75 , the phase difference is 90° or 270°.

❌ Common Misconceptions & Examiner Commentary

  • Confusing non-zero integer multiples with a phase difference: Some students calculate 5 cycles and think that because the path difference is not 0 m, the phase difference cannot be zero. Remember: each full cycle resets the phase difference to 0°!
  • Inverting the division: Calculating λ / Δx = 0.40 / 2.0 = 0.2 and incorrectly choosing option B (45°) or C (90°) through confused trigonometry. Always divide path difference by wavelength.
  • Assuming destructive interference by default: Rushing students sometimes assume any interference question must test destructive interference (180°), overlooking the simple 5.0 whole wavelength relationship.

Topics

Physics · 3.3 Waves

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.