Learning outcomes
- Define critical angle and total internal reflection.
- Recall and use n = 1/sin c.
- State the two conditions for total internal reflection.
- Explain optical-fibre transmission and advantages.
8.1 From refraction to the critical angle
Consider light travelling from glass to air. As the angle of incidence in the glass increases, the refracted ray bends farther away from the normal. At one particular incidence angle, the refracted ray is at 90° and travels along the boundary. This incidence angle is the critical angle c.
For incidence angles greater than c, no refracted ray emerges. The light is completely reflected back inside the denser medium. This is total internal reflection, often abbreviated TIR.
8.2 Conditions for total internal reflection
Two conditions are essential. First, light must travel from a medium of higher refractive index to a medium of lower refractive index. Second, the incidence angle in the higher-index medium must be greater than the critical angle.
If light travels from air into glass, TIR cannot occur at that boundary. If it travels from glass toward air but i is less than c, some light refracts out and some is reflected; the reflection is not total.
8.3 Critical-angle equation
For a material surrounded by air, n = 1/sin c. Equivalently, sin c = 1/n. A material with larger refractive index has a smaller critical angle. This means TIR can occur over a wider range of incidence angles.
The equation assumes the outside medium is air or vacuum. In more advanced work, the relationship depends on both refractive indices, but the O Level form should be used when the question gives one n and air outside.
8.4 Demonstrating TIR
A semicircular glass or acrylic block is useful. Direct a ray through the curved surface toward the centre, so it enters without bending. At the flat face, increase the incidence angle. Observe refraction for i < c, a grazing refracted ray at i = c, and only reflection for i > c.
Use a narrow beam and mark the centre accurately. The boundary may produce weak reflections even below c, so identify TIR by the disappearance of the transmitted ray and the strong reflected ray.
8.5 Optical fibres
An optical fibre contains a high-index core surrounded by lower-index cladding. Light travelling within the core strikes the boundary above the critical angle and undergoes repeated TIR, following the curved fibre. Cladding protects the core and ensures the refractive-index difference needed for reliable guiding.
In telecommunications, fibres carry rapid light pulses with high bandwidth, low attenuation and little electromagnetic interference. They are lightweight and difficult to tap without detection. In medical endoscopes, bundles of fibres carry illumination and images through narrow flexible paths. The syllabus particularly expects the telecommunications advantages to be stated in context.
8.6 Limitations and signal quality
Fibres are not perfect. Absorption and scattering reduce signal intensity, so repeaters may be needed over long distances. Very sharp bends can make incidence angles fall below the critical angle, allowing light to escape. Pulse spreading can limit the maximum data rate.
Nevertheless, compared with copper cables, fibre systems can carry more information over long distances with less signal loss and no electrical sparks, which is useful in many environments.
Worked examples
Critical angle from n
For glass with n = 1.50, sin c = 1/1.50 = 0.6667, so c = 41.8°.
Checking TIR
Light inside glass of critical angle 42° strikes the glass–air boundary at 50°. It travels from higher to lower n and i > c, so TIR occurs.
Practical focus
Investigation
Use a semicircular block to measure c. Rotate the block until the refracted ray just travels along the flat face. Repeat several times from slightly smaller and larger angles and average the transition value.
Examination guidance
- State both TIR conditions.
- At i = c the refracted angle is 90°; TIR requires i > c.
- Do not say light “bounces because the fibre is shiny”; explain the refractive indices and incidence angle.
Check your understanding
- Find c for n = 1.33.
- Why is cladding needed?
- Give two advantages of optical fibre in communications.
Answers
- c = sin⁻¹(1/1.33) ≈ 48.8°.
- It has lower refractive index than the core, enabling TIR and protecting the core.
- Any two: high bandwidth, low attenuation, immunity to electromagnetic interference, low mass, improved security.
