AQA AS Level Physics Paper 2, June 2025: Question 12

1 mark · Easy difficulty · Multiple Choice

Identify the nature of radio waves transmitted from a vertical aerial and the alignment of the receiving aerial required to detect the maximum signal.

Practise this question

Question

Diagram showing a vertical transmitting aerial on the ground and a portable radio with an adjustable receiving aerial some distance away. A multiple-choice table follows with options A to D, combining nature of transmitted radio wave (polarised or unpolarised) and alignment of receiving aerial (vertical or horizontal).

Mark scheme

Show the mark scheme Mark scheme table indicating for question 12 the correct answer is option A (polarised, vertical), with an assessment objective of AO1.

How to answer it

Radio Waves: Polarisation and Aerial Alignment

📋 What this question tests

This question assesses recall and application of wave properties (AQA Physics Section 3.3.1.2 - Longitudinal and Transverse Waves). Specifically:

  • Understanding that electromagnetic (EM) waves are transverse and can be plane-polarised.
  • Knowing that radio waves emitted by a linear rod aerial are polarised in the orientation of the transmitting rod.
  • Applying the principle of wave detection: the receiving aerial must align with the electric field oscillations (plane of polarisation) to detect the maximum signal.
Question 12 [1 Mark]

Radio Wave Transmission and Reception

Target: AO1 (Demonstrate knowledge and understanding of scientific ideas, processes, techniques, and procedures)

✅ Correct Answer

A — polarised | vertical

Mark Scheme Breakdown:
• 1 mark for identifying option A correctly.

💡 Key Knowledge

  • Nature of EM waves: Radio waves are transverse waves consisting of oscillating electric ( E ) and magnetic ( B ) fields perpendicular to each other and to the direction of propagation.
  • Emission mechanism: Electrons accelerate up and down inside the vertical transmitting aerial. This produces an electric field oscillating strictly in the vertical plane. Hence, the emitted wave is plane-polarised.
  • Detection mechanism: The oscillating vertical electric field drives electrons along the receiving aerial to create an alternating potential difference. This signal is maximum when the aerial is parallel to the electric field oscillations (i.e. vertical).

🧠 Exam Technique & Deduction

  1. Step 1: Evaluate column 1 (Nature of wave).
    A single straight aerial forces electrons to oscillate back and forth along one single axis. Waves produced by a single dipole or rod aerial are always polarised (not unpolarised like light from an incandescent bulb). This eliminates C and D immediately.
  2. Step 2: Evaluate column 2 (Alignment of aerial).
    To induce the maximum oscillating voltage in the metal aerial rod, electrons inside the receiving aerial must be pushed along its length. Because the electric field vector is vertical, the receiving aerial must also be aligned vertically. If placed horizontally, it would be perpendicular to the electric field, giving zero (or minimal) signal. This rules out B.

❌ Common Errors & Misconceptions

  • Confusing aerial reception with polarising filters: Some students confuse aerial alignment with crossed Polaroid filters (where perpendicular orientations are thought of). With an aerial, maximum induced emf occurs when the aerial rod is parallel to the E -field, not perpendicular.
  • Assuming all transmitted EM waves are unpolarised: Light from thermal sources (like the Sun or a filament bulb) is unpolarised because oscillations occur in random planes. Radio waves produced by antennas are an engineering source and are naturally polarised along the aerial axis.
  • Thinking horizontal gives maximum signal: If the receiving aerial is horizontal, the vertical electric field cannot induce charge separation along the length of the aerial rod, leading to a minimum/null signal.

📐 Examiner Note on Aerial Diagrams

In longer answer or practical exam questions, you may be asked to sketch or describe this setup:

  • Transmitter: Draw a straight rod with vertical arrows indicating oscillating current / electric field vector.
  • Receiver at 0° (Vertical): Parallel to the transmitter → Maximum reception (Signal = I₀ ).
  • Receiver at 90° (Horizontal): Perpendicular to transmitter → Minimum/Zero reception (Signal ≈ 0).
  • The received intensity follows Malus's Law relationship: I = I₀ cos²θ , where θ is the angle between the receiving aerial and the vertical plane of polarisation.

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.