AQA AS Level Physics Paper 1, June 2025: Question 2

8 marks · Medium difficulty · Short Answer

Calculate the refractive index of optical fibre cladding, the maximum angle of incidence for total internal reflection, and the transit time of light through the fibre.

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Question

Question 02 consists of four parts based on an optical fibre with a glass core (refractive index 1.51) and cladding. Part 02.1 asks what happens to the frequency of light upon entering the core from air (tick box: decreases, unchanged, increases). Part 02.2 gives a critical angle of 80.7 degrees and asks to calculate the refractive index of the cladding. Figure 1 shows a ray incident on the core at angle theta_A to the normal, refracting at angle theta_B to the central axis; part 02.3 asks to calculate the maximum value of theta_A if theta_B cannot exceed 9.3 degrees. Figure 2 shows the zigzag path of a ray inside a 2.5 mm diameter core with theta_B = 9.0 degrees; part 02.4 asks to calculate the transit time for 3.1 x 10^4 total internal reflections.
Question text

02 An optical fibre consists of a glass core surrounded by cladding.

Monochromatic light enters the glass core of the optical fibre from air.

02.1 What happens to the frequency of the light as it enters the glass core?

Tick ( ) one box.

[1 mark]

It decreases.

It is unchanged.

It increases.

The refractive index of the glass core is 1.51

02.2 The critical angle at the core–cladding boundary is 80.7°

Calculate the refractive index of the cladding.

[2 marks]

refractive index of cladding =

Figure 1 shows a ray of light that is incident on the centre of the core.

At the air–core boundary, the angle of incidence of this ray is θA and the angle of

refraction is θB.

Figure 1

When θB is greater than 9.3°, the ray will not be transmitted through the optical fibre.

02.3 Calculate the maximum value of θA for the ray to be transmitted through the

optical fibre.

[2 marks]

maximum value of6 θA = °

02.4 Figure 2 shows the path of a ray of light that enters at the centre of the straight optical

fibre at a smaller value of θA. Only the first two total internal reflections are shown.

Figure 2

The core has a diameter of 2.5 mm.

θB is now 9.0°

3.1 × 104 total internal reflections occur before the light leaves the optical fibre.

*05*Calculate the time for the light to travel through the optical fibre.

refractive index of glass core = 1.51

[3 marks]

time = s

Mark scheme

Show the mark scheme Mark scheme for Question 02 gives marking points: 02.1 accepts 'It is unchanged' (1 mark). 02.2 awards 1 mark for using sin(theta_c) = n2/n1 or n1 sin(theta1) = n2 sin(90) with n1 = 1.51, and 1 mark for cladding refractive index 1.49 (or 1.5 with working). 02.3 awards 1 mark for Snell's law (sin theta_A = 1.51 sin 9.3) and 1 mark for theta_A = 14 degrees. 02.4 awards up to 2 marks from a list including calculating speed of light in core (1.99 x 10^8 m/s), component velocity or geometric path distance per reflection (e.g. horizontal distance 15.8 mm or path distance 15.98 mm), and using 31000 reflections, with a final mark for time = 2.5 x 10^-6 s.

Question Answers Additional Comments/Guidance Mark AO

02.1 It is unchanged. Tick in the 2nd box only 1 AO1

02.2 Any one from Condone mix up in n1 and n2 in 1st bullet point 2 1 × AO1

n in otherwise correct substitution 1 × AO2

θ = 2

• use of sin C Do not allow use of n = 1 in first bullet point

n1

OR Expect to see:

use of n1 sin θ1 = n2 sin θ2 where θ2 = 90(°) 𝑛𝑛2

sin 80.7 = 1.51

OR

• n1 = 1.51 and θC = 80.7 (°) 1.51 sin 80.7 = n sin 90

Sin 80.7 = 0.987

(Refractive index of cladding =) 1.49

Calculator value 1.490152132

Condone an answer of (Refractive index of

cladding =) 1.5 where correct working seen to

support and is correct rounding of 1.49

02.3 use of n1 sinθ1 = n2 sinθ2 Use of equation includes: 2 1 × AO1

sin θA = 1.51 x sin 9.3 1 × AO2

Condone confusing n=1 and n=1.51 in

otherwise correct substitution.

Allow sin 9.3 = 1.51 sin θ (as confusing

refractive indices)

Condone use of n2 = 1.49 in an otherwise

correct substitution.

(θ =) 14° Calculator value 14.12403075

A

02.4 Any two from 3 AO3

c Speed of light in core = 1.99 × 108 m s−1

• use of n = (1)

cs

their distance Do not allow their speed as 3 x 108 m s-1 in 2nd

• use of their speed = (2)

time bullet point or bullet point 3

Vertical component = 3.11 × 107 m s−1 11

• calculates a component of their velocity of light in the core

(3) Horizontal component = 1.96 × 108 m s−1

• evidence of correctly calculating an appropriate distance in Horizontal distance between reflections = 15.8

the core (4) mm

Horizontal distance travelled as light travels

from centre to edge = 7.89 mm

Distance travelled between reflections = 15.98

mm

Distance travelled as light travels from centre

• makes use of the 31000 reflections (5) to edge

= 7.99 mm

Total vertical distance travelled by light = 77.5

m

−6 Total distance travelled by light= 495 m

(time =) 2.5 × 10 (s)

Length of fibre = 489 m

How to answer it

Optics & Total Internal Reflection in Fibres

📌 What This Question Tests

This question evaluates foundational wave properties and geometric optics applied to optical communications:

  • Wave behaviour across boundaries (invariance of wave frequency vs speed and wavelength).
  • Calculation of critical angle and refractive index at the core-cladding boundary using Snell's Law.
  • Refraction at the entrance face of an optical fibre to find acceptance limits.
  • Multi-step kinematics and geometry: calculating transmission transit time involving multiple total internal reflections.
Question 02.1 • 1 Mark (AO1)

Frequency Change on Entering Glass Core

Monochromatic light enters the glass core of an optical fibre from air

✅ Correct Answer

Tick the 2nd box: It is unchanged.

Mark allocation: [1 mark] strictly for ticking only the second box.

💡 Key Knowledge

The frequency of a wave is determined exclusively by its source. When light enters an optically denser medium (higher refractive index):

  • Frequency ( f ): Remains constant.
  • Speed ( v ): Decreases ( v = c / n ).
  • Wavelength ( λ ): Decreases proportionately ( λ = v / f ).

❌ Common Errors

Many students conflate wave speed with frequency and assume light "loses energy" or "slows down in frequency", incorrectly ticking "It decreases".

🧠 Exam Technique

Remember that photon energy is given by E = hf . Since no energy is transferred or lost to another frequency mode upon ideal transmission, frequency must stay identical.

Question 02.2 • 2 Marks (1×AO1, 1×AO2)

Refractive Index of Cladding

Core index n₁ = 1.51 , critical angle θc = 80.7°

📐 Step-by-Step Calculation

  1. State the critical angle relationship:
    sin θc = n₂ / n₁  (where n₁ = ncore and n₂ = ncladding )
  2. Rearrange and substitute values:
    n₂ = n₁ × sin θc
    n₂ = 1.51 × sin(80.7°)
  3. Evaluate:
    sin(80.7°) = 0.98686...
    n₂ = 1.51 × 0.98686... = 1.49015...
  4. Final Value: 1.49 (or 1.5 supported by working).

✅ Mark Scheme Breakdown

  • Mark 1: Correct substitution into formula: sin(80.7°) = n₂ / 1.51 or 1.51 sin(80.7°) = n₂ sin(90°) .
  • Mark 2: Final answer of 1.49 (or 1.5 if exact calculation is shown).
Refractive index has no units.

❌ Common Traps

Swapping indices: Thinking sin θc = n₁ / n₂ yields n₂ = 1.53 . Remember: cladding must always have a lower optical density (lower refractive index) than the core for total internal reflection (TIR) to occur!

Question 02.3 • 2 Marks (1×AO1, 1×AO2)

Maximum Acceptance Angle (θA)

Calculate maximum θA when maximum refracted angle θB = 9.3°

📐 Step-by-Step Calculation

  1. Apply Snell's Law at the air-core boundary:
    nair sin(θA) = ncore sin(θB)
  2. Substitute known values ( nair = 1.00 ):
    1.00 × sin(θA) = 1.51 × sin(9.3°)
  3. Compute intermediate value:
    sin(θA) = 1.51 × 0.16160 = 0.24402
  4. Calculate inverse sine:
    θA = sin⁻¹(0.24402) = 14.124°
  5. Round to appropriate sig figs: 14° (or 14.1°).

🧠 Exam Technique & Geometry Insight

Why does θB = 9.3° correspond to transmission failure?

  • The angle of incidence on the cladding wall is 90° − θB .
  • If θB > 9.3° , the wall angle becomes < 80.7° (less than θc ), meaning light refracts out rather than reflecting internally.
  • Always check your calculator is in degree mode!
Mark 1: Use of sin θA = 1.51 × sin(9.3°)
Mark 2: Final answer 14°.
Question 02.4 • 3 Marks (AO3)

Transit Time Through the Optical Fibre

Core diameter = 2.5 mm , θB = 9.0° , reflections = 3.1 × 10⁴ , ncore = 1.51

📐 Method 1: Total Distance Along Ray Path

  1. Find speed of light in the glass core:
    v = c / n = (3.00 × 10⁸) / 1.51 = 1.987 × 10⁸ m s⁻¹
  2. Distance along ray between reflections:
    Across full diameter ( 2.5 mm = 2.5 × 10⁻³ m ):
    dstep = 2.5 × 10⁻³ / sin(9.0°) = 1.598 × 10⁻² m (15.98 mm)
  3. Total path length for 3.1 × 10⁴ reflections:
    Ltotal = 3.1 × 10⁴ × 0.01598 m = 495.4 m
  4. Time taken:
    t = Ltotal / v = 495.4 / (1.987 × 10⁸) = 2.49 × 10⁻⁶ s
  5. Final Answer: 2.5 × 10⁻⁶ s (to 2 s.f.).

📐 Method 2: Vertical Velocity Component

  1. Find vertical speed component ( vy ):
    v = c / 1.51 = 1.987 × 10⁸ m s⁻¹
    vy = v × sin(9.0°) = 3.108 × 10⁷ m s⁻¹
  2. Total vertical distance travelled:
    Each reflection traverses the diameter ( 2.5 mm ):
    ytotal = 3.1 × 10⁴ × 2.5 × 10⁻³ m = 77.5 m
  3. Calculate transit time:
    t = ytotal / vy = 77.5 / (3.108 × 10⁷)
    t = 2.493 × 10⁻⁶ s ≈ 2.5 × 10⁻⁶ s

✅ Mark Scheme Scheme (Any two from points 1–5, plus final answer)

  • Point 1: Speed in core: v = c / n = 1.99 × 10⁸ m s⁻¹ .
  • Point 2: Use of time = distance / speed with their calculated speed.
  • Point 3: Component of velocity in core ( vx = 1.96 × 10⁸ or vy = 3.11 × 10⁷ m s⁻¹ ).
  • Point 4: Correct step distance ( 15.98 mm ray path, 15.8 mm horizontal, or 7.89 mm half-step).
  • Point 5: Correct scaling with 3.1 × 10⁴ reflections ( L = 495 m or fibre length = 489 m ).
  • Final Mark: 2.5 × 10⁻⁶ s (must round to 2 s.f.).

❌ Common Errors That Cost Marks

  • Using speed in vacuum ( 3.00 × 10⁸ m s⁻¹ ): The mark scheme explicitly states: "Do not allow their speed as 3 × 10⁸ m s⁻¹".
  • Unit conversion error: Forgetting that 2.5 mm = 2.5 × 10⁻³ m , leading to powers of ten errors (e.g. 10⁻³ s ).
  • Trigonometry mix-up: Using cos(9.0°) instead of sin(9.0°) for vertical distance across diameter.

Topics

Physics · 3.3 Waves

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