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
Total internal reflection and optical fibres
- 3 things to know
- AS Unit 2
Overview
Recall questions on Total internal reflection and optical fibres.
What you need to know
Answer each one before you reveal it.
Define the critical angle.
Calculate the critical angle for a glass–air boundary where the glass has refractive index 1.50.
Questions from this subtopic
Each one with the model answer underneath.
- 1
State the two conditions for total internal reflection to occur.
Model answer
The light must be travelling from a medium of higher refractive index to one of lower refractive index, and the angle of incidence must exceed the critical angle.
- 2
Define the critical angle.
Model answer
The angle of incidence in the denser medium for which the angle of refraction is 90°.
- 3
Derive the equation for the critical angle between two media.
Model answer
Applying n₁ sin θ₁ = n₂ sin θ₂ with θ₂ = 90° gives n₁ sin C = n₂, so sin C = n₂/n₁.
- 4
Calculate the critical angle for a glass–air boundary where the glass has refractive index 1.50.
Model answer
sin C = 1.00 ÷ 1.50 = 0.667, so C = 41.8°.
- 5
What happens to light striking a boundary at exactly the critical angle?
Model answer
It is refracted along the boundary, at 90° to the normal.
- 6
How does an optical fibre use total internal reflection?
Model answer
Light entering the core meets the core–cladding boundary at more than the critical angle, so it is totally internally reflected and stays within the core.
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