3.20-3.22 - Total internal reflection, critical angle and optical fibres

When refraction stops

When a ray of light travels from glass or plastic into air, it is moving from a higher refractive index material into a lower refractive index material. At the boundary, the ray normally partly reflects and partly refracts. The refracted ray bends away from the normal.

As the angle of incidence inside the higher refractive index material increases, the angle of refraction increases too. Eventually the refracted ray is just along the boundary. If the angle of incidence is increased further, there is no refracted ray leaving the material. All the light is reflected back inside: this is total internal reflection.

The two conditions are both needed. Light must be trying to leave the higher refractive index material, and its angle of incidence must be greater than the critical angle. A ray travelling from air into glass cannot undergo total internal reflection at the air-glass boundary: entering glass, it bends towards the normal, so its angle of refraction is always smaller than its angle of incidence and can never reach $90^\circ$.

[DIAGRAM: critical_angle_sequence: Total internal reflection, critical angle and optical fibres - diagram 01; asset_slug: p30_total_internal_reflection_critical_angle_and_optical_fibres__diagram_01; recommended_method: image_gen; description: Minimal NovaLearn three-panel glass-air ray diagram with rounded pale-blue glass regions, dashed normals, incidence below, at and above the critical angle, partial reflections in the first two cases, a 90-degree refracted ray at the critical angle, and total internal reflection in the final case.]

Critical angle equation

The critical angle, $c$, is the angle of incidence in the higher refractive index material that makes the angle of refraction equal to $90^\circ$. In other words, at the critical angle the refracted ray travels along the boundary. In the middle panel of the diagram above ($i = c$), the critical angle is the angle $i$ between the incident ray and the normal inside the glass; the angle in the air, between the normal and the ray along the boundary, is the $90^\circ$ angle of refraction.

Angles in ray diagrams are measured from the normal, not from the surface. This matters because the critical angle is also measured from the normal inside the material.

You need to recall this equation: it is not given in the exam.

Here $n$ is the refractive index of the material compared with air. Because most transparent solids and liquids have $n > 1$, the value of $1/n$ is less than 1, which is possible for $\sin c$.

Worked example: a glass block has refractive index $n = 1.50$. Find its critical angle at a glass-air boundary.

First substitute into the formula:

Then use inverse sine:

So any ray inside this glass that hits the glass-air boundary at an angle greater than about $42^\circ$ to the normal will undergo total internal reflection.

Because $\sin c = 1/n$, a higher refractive index gives a smaller $\sin c$ and so a smaller critical angle. For example, a material with $n = 2.00$ has $\sin c = 1/2.00 = 0.500$, so $c = 30.0^\circ$, smaller than the $41.8^\circ$ of the glass above.