Describe the passage of light through a transparent material (limited to the boundaries between two mediums only)

3.2.2 Refraction of Light

Objective

Describe the passage of light through a transparent material, focusing on the boundary between two different media.

Key Concepts

• Refraction – the change in direction of a light ray when it passes from one transparent medium to another.
• Normal – an imaginary line perpendicular to the surface at the point of incidence.
• Angle of incidence (\$i\$) – the angle between the incident ray and the normal.
• Angle of refraction (\$r\$) – the angle between the refracted ray and the normal.
• Refractive index (\$n\$) – a dimensionless number that describes how much light slows down in a material.

Why Light Bends

When light enters a second medium, its speed changes. The change in speed causes the wavefronts to pivot around the point where the ray meets the surface, resulting in a change of direction.

Snell’s Law (Law of Refraction)

The quantitative relationship between the angles and the refractive indices of the two media is given by Snell’s law:

\$n1 \sin i = n2 \sin r\$

where:

• \$n_1\$ – refractive index of the first medium (where the light originates).
• \$n_2\$ – refractive index of the second medium.
• \$i\$ – angle of incidence.
• \$r\$ – angle of refraction.

Refractive Index

The refractive index of a medium is defined as the ratio of the speed of light in vacuum (\$c\$) to the speed of light in the medium (\$v\$):

\$n = \frac{c}{v}\$

Because \$c\$ is constant, a higher \$n\$ means a lower speed \$v\$ in that medium.

Behaviour at the Boundary

1. If \$n2 > n1\$ (light entering a denser medium), the ray bends towards the normal (\$r < i\$).
2. If \$n2 < n1\$ (light entering a rarer medium), the ray bends away from the normal (\$r > i\$).
3. When the incident ray is perpendicular to the surface (\$i = 0^\circ\$), no bending occurs; the ray continues straight.

Typical Refractive Indices

Worked Example

Calculate the angle of refraction when a ray of light in air (\$n1 = 1.00\$) strikes a glass surface (\$n2 = 1.50\$) at an incidence angle of \$30^\circ\$.

Using Snell’s law:

\$1.00 \sin 30^\circ = 1.50 \sin r\$

\$\sin r = \frac{\sin 30^\circ}{1.50} = \frac{0.5}{1.50} = 0.333\$

\$r = \sin^{-1}(0.333) \approx 19.5^\circ\$

The ray bends towards the normal, reducing the angle from \$30^\circ\$ to about \$19.5^\circ\$.

Important Points for IGCSE Exams

• State the definition of refraction and the normal.
• Write Snell’s law correctly and identify each term.
• Explain why the ray bends towards the normal when entering a medium of higher refractive index.
• Be able to rearrange Snell’s law to solve for an unknown angle or refractive index.
• Remember that the refractive index is a ratio of speeds, not a speed itself.