Refraction and Total Internal Reflection

Snell's law n₁ sin θ₁ = n₂ sin θ₂; critical angle; optical fibres

# Refraction and Total Internal Reflection — A-Level Physics

Refraction — the bending of light as it passes between media — is explained by the change in wave speed. Snell's Law provides the quantitative relationship, and total internal reflection has crucial applications in optical fibres and communications.


1. Refractive Index

The refractive index (nn) of a material describes how much light slows down in that medium:

n=cv\boxed{n = \frac{c}{v}}

Where:

  • nn = refractive index (dimensionless)
  • cc = speed of light in vacuum (3×1083 \times 10^8 m/s)
  • vv = speed of light in the medium

n1n \geq 1 for all materials. Vacuum: n=1n = 1. Air: n1.00n \approx 1.00.

Material nn
Air 1.00
Water 1.33
Glass 1.50 (typical)
Diamond 2.42

2. Snell's Law

n1sinθ1=n2sinθ2\boxed{n_1 \sin\theta_1 = n_2 \sin\theta_2}

Where θ1\theta_1 and θ2\theta_2 are measured from the normal.

Alternatively: sinθ1sinθ2=n2n1=v1v2\frac{\sin\theta_1}{\sin\theta_2} = \frac{n_2}{n_1} = \frac{v_1}{v_2}

When Light Enters a Denser Medium (n2>n1n_2 > n_1)

  • Speed decreases
  • Wavelength decreases
  • Light bends towards the normal (θ2<θ1\theta_2 < \theta_1)

When Light Enters a Less Dense Medium (n2<n1n_2 < n_1)

  • Speed increases
  • Wavelength increases
  • Light bends away from the normal (θ2>θ1\theta_2 > \theta_1)

Frequency never changes during refraction.


3. Total Internal Reflection (TIR)

Occurs when light travels from a denser to a less dense medium and the angle of incidence exceeds the critical angle.

Critical Angle

sinθc=n2n1\boxed{\sin\theta_c = \frac{n_2}{n_1}}

For light going from a medium to air (n2=1n_2 = 1): sinθc=1n\sin\theta_c = \frac{1}{n}

Conditions for TIR

  1. Light must travel from denser to less dense medium
  2. Angle of incidence must be greater than the critical angle

4. Optical Fibres

Structure

  • Core: High refractive index glass/plastic — light travels here
  • Cladding: Lower refractive index — ensures TIR keeps light in the core
  • Protective sheath: Physical protection

Why Cladding?

  • Prevents light leaking between adjacent fibres (cross-talk)
  • Protects the core surface from scratches (which would allow light to escape)
  • Provides the interface for TIR

Signal Degradation

Absorption: Some light energy is absorbed by the glass, converting to heat. Reduces signal strength.

Dispersion: Two types:

  • Modal dispersion: Different rays take different paths (different angles), arriving at different times → pulse broadening. Reduced by using narrow (single-mode) fibres.
  • Material dispersion: Different wavelengths travel at different speeds → pulse broadening. Reduced by using monochromatic light (lasers).

Solution: Use repeaters (regenerate signal) at intervals.


Worked Example: Snell's Law

Problem

Light passes from air into glass (n=1.52n = 1.52) at an angle of incidence of 40°. Find the angle of refraction.

n1sinθ1=n2sinθ2n_1\sin\theta_1 = n_2\sin\theta_2 1.00×sin40°=1.52×sinθ21.00 \times \sin40° = 1.52 \times \sin\theta_2 sinθ2=0.643/1.52=0.423\sin\theta_2 = 0.643/1.52 = 0.423 θ2=25.0°\theta_2 = 25.0°

Solution

Worked Example: Critical Angle

Problem

Calculate the critical angle for glass (n=1.50n = 1.50) to air.

sinθc=1/1.50=0.667\sin\theta_c = 1/1.50 = 0.667 θc=41.8°\theta_c = 41.8°

Solution

Worked Example: Speed in Medium

Problem

The speed of light in a material is 2.0×1082.0 \times 10^8 m/s. Find the refractive index.

n=c/v=3×108/2×108=1.5n = c/v = 3 \times 10^8/2 \times 10^8 = 1.5

Solution

Worked Example: Fibre Optic Critical Angle

Problem

An optical fibre has a core of n=1.62n = 1.62 and cladding of n=1.52n = 1.52. Find the critical angle at the core-cladding boundary.

sinθc=n2/n1=1.52/1.62=0.938\sin\theta_c = n_2/n_1 = 1.52/1.62 = 0.938 θc=69.6°\theta_c = 69.6°


Solution

6. Practice Questions

    1. Light passes from water (n=1.33n = 1.33) into glass (n=1.50n = 1.50) at 30°. Calculate the angle of refraction. (2 marks)
    1. Calculate the critical angle for diamond (n=2.42n = 2.42) to air. Explain why diamonds sparkle. (3 marks)
    1. An optical fibre has core index 1.55 and cladding index 1.45. Calculate the critical angle. If light enters at 80° to the core-cladding boundary, will TIR occur? (3 marks)
    1. Explain modal and material dispersion in optical fibres and how each can be reduced. (4 marks)

    Answers

    1. 1.33sin30°=1.50sinθ21.33\sin30° = 1.50\sin\theta_2sinθ2=0.665/1.50=0.443\sin\theta_2 = 0.665/1.50 = 0.443θ2=26.3°\theta_2 = 26.3°.

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Summary

  • n=c/vn = c/v; higher nn = slower light
  • Snell's Law: n1sinθ1=n2sinθ2n_1\sin\theta_1 = n_2\sin\theta_2
  • Critical angle: sinθc=n2/n1\sin\theta_c = n_2/n_1
  • TIR: light from dense → less dense medium, angle > critical angle
  • Optical fibres: core (high nn) + cladding (low nn) → TIR
  • Signal loss: absorption, modal dispersion, material dispersion

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