WJEC A-Level Physics 1420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Solids Under Stress
Using Radiation to Investigate Stars
Particles and Nuclear Structure
Conduction of Electricity
Resistance
D.C. Circuits
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Nuclear Decay
Nuclear Energy
Capacitance
Electrostatic and Gravitational Fields of Force
Orbits and the Wider Universe
Magnetic Fields
Electromagnetic Induction
Option A: Alternating CurrentsOption A
Option B: Medical PhysicsOption B
Option C: The Physics of SportsOption C
Option D: Energy and the EnvironmentOption D
Mathematical Requirements (Appendix C)
Refraction of Light
Optical fibres: core, cladding and pulse broadening
- 3 things to know
- AS Unit 2
Overview
Recall questions on core, cladding broadening and pulse broadening.
What you need to know
Answer each one before you reveal it.
What is material dispersion in an optical fibre?
Why is a very nearly monochromatic source used to send signals along a long optical fibre?
Questions from this subtopic
Each one with the model answer underneath.
- 1
Why is the core of an optical fibre given a higher refractive index than the cladding?
Model answer
Total internal reflection can only occur when light travels from the optically denser medium towards the less dense one. Making the core the denser of the two traps the light inside it.
- 2
Give two functions of the cladding on an optical fibre other than providing the refractive index step.
Model answer
It protects the surface of the core from scratches and dirt that would otherwise let light escape at the point of damage. It also keeps the cores of neighbouring fibres apart so that light cannot cross from one fibre into another.
- 3
What is meant by crossover in a bundle of optical fibres?
Model answer
Crossover is light escaping from one fibre and entering an adjacent one. The signal is corrupted, or the image the bundle is carrying is blurred.
- 4
An optical fibre has a core of refractive index 1.50 and cladding of refractive index 1.46. Calculate the critical angle at the core–cladding boundary.
Model answer
Using n₁ sin C = n₂ gives sin C = 1.46 ÷ 1.50 = 0.9733, so C = 76.7°. Substituting back, 1.50 × sin 76.7° = 1.50 × 0.973 = 1.46, the refractive index of the cladding.
- 5
Why is the critical angle inside an optical fibre so much larger than that at a glass–air boundary?
Model answer
The refractive indices of the core and the cladding are very close to one another, so their ratio is only just below one and the critical angle is only just below 90°.
- 6
In a fibre whose critical angle is 76.7°, what is the greatest angle to the fibre axis at which a ray can travel and still be trapped?
Model answer
A ray at an angle α to the axis meets the wall at (90° − α) to the normal, so the largest permitted value of α is 90° − 76.7° = 13.3°.
Other ways to work this
The same content, four other ways.
What is inside Levo
Six ways to work the same material.
Every question in Refraction of Light, on a schedule.
Attempt a question from memory, write the correct version, then attempt it again. Levo records the jump between your first attempt and your last, and brings each question back on the day you are due to forget it.
Or £46 a year, about £3.83 a month. Cancel anytime.