Define refractive index, n, as the ratio of the speeds of a wave in two different regions

Published by Patrick Mutisya · 14 days ago

IGCSE Physics 0625 – Refraction of Light

3.2.2 Refraction of Light

Objective

Define the refractive index, \$n\$, as the ratio of the speeds of a wave in two different regions.

Key Concepts

  • When a wave passes from one medium to another, its speed changes.
  • The change in speed causes the wave to change direction – this is refraction.
  • The refractive index quantifies how much the speed changes.

Definition of Refractive Index

The refractive index of medium 2 relative to medium 1 is given by

\$\$

n{12} = \frac{v1}{v_2}

\$\$

where \$v1\$ is the speed of the wave in the first medium and \$v2\$ is the speed in the second medium.

When the first medium is vacuum (or air, approximated as vacuum) the symbol \$n\$ is often used for the absolute refractive index of a medium:

\$\$

n = \frac{c}{v}

\$\$

with \$c = 3.00 \times 10^8\ \text{m s}^{-1}\$ the speed of light in vacuum and \$v\$ the speed of light in the material.

Typical \cdot alues of Refractive Index for Common Materials

MaterialRefractive Index \$n\$ (at 589 nm)
Air (≈ vacuum)1.0003
Water1.33
Glass (crown)1.52
Glass (flint)1.62
Diamond2.42

Using the Refractive Index

  1. Determine the speed of light in a medium using \$v = \dfrac{c}{n}\$.
  2. Apply Snell’s Law to relate the angles of incidence and refraction:

    \$\$

    n1 \sin \theta1 = n2 \sin \theta2

    \$\$

    where \$\theta1\$ and \$\theta2\$ are measured from the normal.

  3. Predict whether the ray will bend towards or away from the normal:

    • If \$n2 > n1\$, the ray bends towards the normal (speed decreases).
    • If \$n2 < n1\$, the ray bends away from the normal (speed increases).

Worked Example

Light travels from air into water. The refractive index of water is \$n_{\text{water}} = 1.33\$.

  1. Calculate the speed of light in water.

Using \$v = \dfrac{c}{n}\$:

\$\$

v_{\text{water}} = \frac{3.00 \times 10^8\ \text{m s}^{-1}}{1.33}

\approx 2.26 \times 10^8\ \text{m s}^{-1}

\$\$

  1. If the angle of incidence in air is \$30^\circ\$, find the angle of refraction in water.

Apply Snell’s Law with \$n_{\text{air}} \approx 1.00\$:

\$\$

1.00 \sin 30^\circ = 1.33 \sin \theta_2 \\

\sin \theta_2 = \frac{0.5}{1.33} \approx 0.376 \\

\theta_2 = \sin^{-1}(0.376) \approx 22^\circ

\$\$

The ray bends towards the normal, as expected.

Common Misconceptions

  • Confusing refractive index with the angle of refraction – they are different quantities.
  • Assuming \$n\$ is always greater than 1; gases at low pressure can have \$n\$ very close to 1.
  • Thinking that light slows down permanently – the speed changes only while the light is in the medium.

Suggested diagram: Ray of light incident on a flat interface between air and water, showing incident angle \$\theta1\$, refracted angle \$\theta2\$, normal line, and labels for \$n{\text{air}}\$ and $n{\text{water}}.

Summary

  • The refractive index \$n\$ is the ratio of the speed of a wave in one medium to its speed in another.
  • Absolute refractive index: \$n = c/v\$.
  • Snell’s Law links refractive indices to the angles of incidence and refraction.
  • Higher \$n\$ means slower light in that medium and greater bending towards the normal.