AQA A-Level Physics Paper 1, June 2025: Question 19

1 mark · Medium difficulty · Multiple Choice

Calculate the speed of light in a second medium given a critical angle of 53° at the boundary and the speed of light in one medium as 2.6 × 10⁸ m s⁻¹.

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Question

Question 19 asks: 'The critical angle at a boundary between two media is 53°. The speed of light in one medium is 2.6 × 10⁸ m s⁻¹. What is the speed of light in the other medium?' Four multiple-choice options are given: A is 1.6 × 10⁸ m s⁻¹, B is 2.1 × 10⁸ m s⁻¹, C is 3.0 × 10⁸ m s⁻¹, and D is 3.2 × 10⁸ m s⁻¹.
Question text

19 The critical angle at a boundary between two media is 53°.

The speed of light in one medium is 2.6 × 108 m s−1.

What is the speed of light in the other medium?

[1 mark]

A 1.6 × 108 m s−1

B 2.1 × 108 m s−1

C 3.0 × 108 m s−1

D 3.2 × 108 m s−1

Mark scheme

Show the mark scheme Mark scheme table showing question number 19, key answer B, value 2.1 × 10⁸ m s⁻¹, and assessment objective AO2.

19 B 2.1 × 108 m s−1 AO2

How to answer it

Optics: Critical Angle & Wave Speed

📌 What this question tests

This question assesses your understanding of total internal reflection, Snell's law, and the fundamental link between refractive index and wave speed in different media:

  • Relating refractive index to wave speed ( n = c / v ).
  • Applying the critical angle formula across a boundary of two generic media ( sin θc = n₂ / n₁ = v₁ / v₂ ).
  • Evaluating physical boundary conditions (recognising that the speed of light in any medium cannot exceed c = 3.0 × 10⁸ m s⁻¹ ).

Question 19 Walkthrough

Multiple Choice Question (1 Mark)

✅ Correct Answer

Option B: 2.1 × 10⁸ m s⁻¹

Awarded 1 mark for selecting option B (AO2 application).

🧠 Quick Deduction Strategy

You can instantly rule out D ( 3.2 × 10⁸ m s⁻¹ ) because light cannot travel faster than its speed in a vacuum ( c = 3.0 × 10⁸ m s⁻¹ ).

Since total internal reflection requires light to travel from a slower (optically denser) medium into a faster (less dense) medium, setting v₁ = 2.6 × 10⁸ m s⁻¹ would force v₂ to exceed c . Therefore, 2.6 × 10⁸ m s⁻¹ must be the faster medium!

📐 Step-by-Step Calculation

  1. Recall Snell's Law at the critical angle:
    Total internal reflection occurs when the refracted angle reaches 90°:
    n₁ sin(θc) = n₂ sin(90°) = n₂
    Therefore: sin(θc) = n₂ / n₁ (where n₁ > n₂ ).
  2. Express refractive index in terms of speed:
    Since n = c / v :
    n₂ / n₁ = (c / v₂) / (c / v₁) = v₁ / v₂
    So, sin(θc) = v₁ / v₂ (where v₁ < v₂ ).
  3. Determine which speed is given:
    If v₁ = 2.6 × 10⁸ m s⁻¹ :
    v₂ = v₁ / sin(53°) = (2.6 × 10⁸) / 0.7986 = 3.26 × 10⁸ m s⁻¹
    This is greater than c (speed of light in a vacuum), which is physically impossible.
    Therefore, the given speed must be the faster medium: v₂ = 2.6 × 10⁸ m s⁻¹ .
  4. Calculate the unknown speed (v₁):
    v₁ = v₂ × sin(53°)
    v₁ = (2.6 × 10⁸) × sin(53°)
    v₁ = (2.6 × 10⁸) × 0.798635... ≈ 2.08 × 10⁸ m s⁻¹
    To 2 significant figures, this gives 2.1 × 10⁸ m s⁻¹.

💡 Key Knowledge Recap

  • Condition for TIR: Ray must travel from higher refractive index to lower refractive index ( n₁ > n₂ ), which means moving from a slower to a faster medium ( v₁ < v₂ ).
  • Sine values: Since sin(θ) ≤ 1 , the ratio must always be vslow / vfast or nlow / nhigh .
  • Universal Speed Limit: In all A-Level physics problems, v ≤ 3.00 × 10⁸ m s⁻¹ .

❌ Common Errors & Traps

  • Inverting the ratio: Confusing n and v . Because n ∝ 1/v , the ratio n₂/n₁ equals v₁/v₂ , not v₂/v₁ .
  • Choosing D (3.2 × 10⁸ m s⁻¹): Dividing by sin(53°) without noticing that the resulting speed exceeds the speed of light in a vacuum.
  • Calculator in radians: Computing sin(53 rad) ≈ -0.396 will produce errors or nonsense results. Ensure your calculator is set to DEG.

Topics

Physics · 3.3 Waves

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.