AQA AS Level Physics Paper 1, November 2021: Question 7
9 marks · Medium difficulty · Short Answer
Analyze modal dispersion, transit time, speed of light, and refractive index effects in optical fibres.
Practise this questionQuestion
Question text
07 Optical fibres are used to carry pulses of light.
07.1 Explain what is meant by modal dispersion in an optical fibre.
[2 marks]
Figure 15 shows a ray of light incident on the central axis of an optical fibre at an
angle of incidence of 30°. The optical fibre is straight and horizontal and has a length
of 10.0 km.
Figure 15
For light incident on the core at a given angle of incidence, the angle of refraction θR
varies with the frequency f of the light.
Figure 16 shows how sin θR varies with f when the angle of incidence is 30°.
Figure 16
The transit time is the time between a pulse of light entering and leaving the optical
fibre.
A single pulse of blue light is incident on the air−core boundary at an angle of
incidence of 30°.
The transit time of this pulse along the 10 km length of the optical fibre
is 5.225 × 10−5 s.
07.2 Show that the horizontal component of the velocity of the pulse is
approximately 1.9 × 108 m s−1.
[1 mark]
07.3 The frequency of the blue light in the pulse is 720 THz.
Calculate the speed of the blue light in the core of the optical fibre.
[3 marks]
26 speed = m s−1
07.4 Two pulses of monochromatic light are incident normally on the air−core boundary.
They then travel along the central axis of the core.
One pulse consists of blue light; the other consists of red light.
Explain, with reference to refractive index, why the pulse of red light has a shorter
transit time than the pulse of blue light.
[2 marks]
07.5 Another two pulses, identical to the pulses in Question 07.4, are incident on the
central axis of the optical fibre and travel along its length.
However, the pulse of red light and pulse of blue light are now incident on the air−core
boundary at an angle of incidence of 30°.
Suggest one reason why the difference in their transit times may not be the same as
in Question 07.4.
[1 mark]
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidance Mark AO
Do not credit material dispersion.
07.1 Spreading of pulse / parts of a pulse take different times to 2 AO1.1a
owtte
travel through the fibre / pulse broadening AO1.1a
Due to different paths through the optical fibre / due to
Accept a diagram showing different paths.
entering the optical fibre at different angles
.
07.2 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑒𝑒 10 × 103 1 AO1.1a
speed (= )= (= 1.91 x 108)
𝑑𝑑𝑑𝑑𝑡𝑡𝑒𝑒 5.225 × 10−5
07.3 Reads off Sin θR = 0.3391 3 AO1.1a
or AO2.1b
AO2.1b
use of n1 sinθ1 = n2 sinθ2 With their Sin θR
(Refractive index of core = 1.47)
c
Use of n = seen Allow use of their refractive index where cs
cs
is the subject of the formula
c = 2.03 × 108 25
s
Alternative:
Reads off Sin θR = 0.3391
or
θ = 19.8° Allow finding θR for their read off
c cos 19.8 = 1.9 × 108 Allow use of their θ
s R
c = 2.03 × 108
s
07.4 The refractive index of core for blue light is greater than the Max 1 mark for stating that the refractive 2 AO1.1b
26 refractive index for red / The refractive index of core for red indices are different because their speeds are
light is less than the refractive index for blue different
AO2.1a
MP1 can come from graph or prior knowledge
The speed of the blue light is less than the speed of the red
light and travel the same distance / The speed of the red light
is greater than the speed of the blue light and travel the same
distance
07.5 the blue now travels a shorter distance than the red light 1 AO2.1a
(compared to 07.4)
or
the red light now travels a greater distance than the blue light
(compared to 07.4)
or Allow: now travel different distances whereas
the difference between the blue’s velocity parallel to the previously travelled the same distance.
central axis and the red’s velocity (parallel to the central axis)
has decreased (compared to 07.4).
or
the difference between the horizontal velocity of the red light
and the horizontal velocity of the blue light has decreased
(compared to 07.4).
Total 9
How to answer it
AQA AS Physics Study Guide: Optical Fibres & Dispersion
What this question tests
This question assesses your understanding of optical communication systems, specifically focusing on modal dispersion, refraction at boundaries (Snell's Law), the relationship between refractive index and frequency (material dispersion), and resolving vector components of velocity for pulses travelling through optical fibre cores.
Explaining Modal Dispersion
✅ Correct Answer
- Mark 1: Spreading of a pulse / different parts of a pulse take different times to travel through the fibre (pulse broadening).
- Mark 2: Due to different paths taken through the optical fibre / light entering the fibre at different angles.
💡 Key Knowledge
Modal dispersion happens because light rays enter the core at different angles, creating paths of varying physical lengths. Rays undergoing more total internal reflections take longer to traverse the fibre than axial rays.
❌ Common Errors
Students frequently lose marks by confusing modal dispersion with material dispersion. Mentioning different wavelengths or frequencies here will not gain credit.
Calculating Horizontal Velocity Component
📐 Step-by-Step Calculation
- Identify formula: velocity = distance / time
- Substitute values: distance = 10.0 km = 10 × 10³ m , time = 5.225 × 10⁻⁵ s
- Compute: v = (10 × 10³) / (5.225 × 10⁻⁵) = 1.9138... × 10⁸ m s⁻¹
- State to required sig figs: 1.9 × 10⁸ m s⁻¹
🧠 Exam Technique
This is a "show that" question. Your substitution must be crystal clear and your unrounded answer (or direct verification) must support the printed 1.9 × 10⁸ m s⁻¹ value.
Calculating the Speed of Blue Light in the Core
📐 Step-by-Step Calculation
- Read Graph: At f = 720 THz , find sin θ_R = 0.3391 from Figure 16.
- Apply Snell's Law: 1.00 × sin(30°) = n_core × sin θ_R
- Find Refractive Index: n_core = 0.5 / 0.3391 = 1.4745...
- Calculate Speed (c_s): Using n = c / c_s , we get c_s = (3.00 × 10⁸) / 1.4745 = 2.03 × 10⁸ m s⁻¹ .
❌ Common Errors
Misreading scale divisions on Figure 16 is the most common pitfall. Always use a ruler to align grid lines carefully when extracting data points from exam insert graphs.
Transit Time Comparison (Red vs. Blue Light)
✅ Correct Answer
- Point 1: The refractive index of the core for blue light is greater than that for red light (or vice versa).
- Point 2: Therefore, the speed of blue light is less than the speed of red light. Since both travel the same distance along the central axis, red light has a shorter transit time.
🧠 Exam Technique
To secure both marks, link optical properties sequentially: Frequency difference → Refractive index difference → Speed difference → Time difference.
Varying Angle of Incidence Effects
✅ Correct Answer
Accept any valid reasoning showing a change in conditions, such as: "The blue light now travels a shorter distance than the red light (due to different refraction angles at the boundary)" or "The difference in their velocities parallel to the central axis has changed."
💡 Key Knowledge
When incidence angles change from normal to 30° , refraction occurs at entry. Because blue and red light have different refractive indices, their angles of refraction inside the core will differ, altering path geometries compared to Question 07.4.
Topics
Physics · 3.3 Waves
Question and mark scheme from the AQA AS Level Physics examination, Paper 1, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.