Total Internal Reflection, Critical Angle and Optical Fibres

When Glass Becomes a Perfect Mirror

Look up at the surface of a swimming pool from underwater and, beyond a certain angle, the surface looks like a silvery mirror. Diamonds sparkle far more than glass cut into the same shape. And pulses of light carry internet data under the Atlantic Ocean, trapped inside hair-thin strands of glass for thousands of kilometres. All three are examples of total internal reflection. This tutorial explains when and why total internal reflection happens, how to calculate the critical angle, how it is used in prisms, diamonds and endoscopes, and how optical fibres work, including the acceptance angle, cladding, dispersion and attenuation. It builds on the tutorial on reflection and refraction of light.

Refraction from a Dense to a Less Dense Medium

When light passes from a medium of refractive index n₁ into one of n₂, Snell's law gives n₁ sin θ₁ = n₂ sin θ₂. Going from an optically denser medium (such as glass, n = 1.5) into a less dense one (such as air, n = 1.0), the light bends away from the normal, so the angle of refraction is larger than the angle of incidence.

As the angle of incidence increases, the refracted ray bends further and further from the normal, getting closer to skimming along the surface. At the same time, a weak reflected ray is always present, and it gets brighter as the angle increases.

The Critical Angle

The critical angle, c, is the angle of incidence in the denser medium for which the angle of refraction is exactly 90°: the refracted ray travels along the boundary.

Putting θ₂ = 90° (sin 90° = 1) into Snell's law: n₁ sin c = n₂, so

sin c = n₂ / n₁, and for a boundary with air, sin c = 1/n.

Example 1. Find the critical angle for glass of refractive index 1.50.

sin c = 1/1.50 = 0.667, so c = 41.8°.

For water (n = 1.33), c = 48.8°. For diamond (n = 2.42), c = 24.4°.

Example 2. Find the critical angle for light going from glass (n = 1.50) into water (n = 1.33).

sin c = 1.33/1.50 = 0.887, so c = 62.5°. The critical angle is larger because the two refractive indices are closer together.

Conditions for Total Internal Reflection

If the angle of incidence is greater than the critical angle, there is no refracted ray at all: all the light is reflected back into the denser medium. This is total internal reflection. Two conditions must both be met:

the light must be travelling from an optically denser medium towards a less dense one (higher n to lower n);
the angle of incidence must be greater than the critical angle.

Total internal reflection reflects 100% of the light, which is better than any metal mirror (an ordinary silvered mirror reflects perhaps 90 to 95%). It never happens when light goes from air into glass, however steep the angle.

Uses of Total Internal Reflection

Reflecting prisms. A right-angled isosceles glass prism has angles of 45°. Light entering one short face at right angles meets the long face at 45°, which is greater than glass's critical angle of 41.8°, so it is totally reflected and turned through 90°. Prisms used this way replace mirrors in periscopes. Arranged differently, a prism can turn light through 180° and is used in binoculars to fold the light path and turn the image the right way up.

Diamonds. Diamond's critical angle is only 24.4°, so light entering a well-cut diamond is totally reflected many times inside before it escapes from the top. This, together with diamond's strong dispersion (splitting white light into colours), gives its brilliance and "fire".

Mirages. On a hot day, the air just above a road is hotter and less dense than the air above it, so its refractive index is slightly lower. Light from the sky heading down at a shallow angle is bent upwards and eventually totally reflected, so you see an image of the sky on the road, which looks like a pool of water.

Rain sensors and fingerprint scanners detect when water or skin touches glass, because this changes the critical angle and stops the total internal reflection.

How Optical Fibres Work

An optical fibre is a very thin strand of extremely pure glass. Light entering one end strikes the side walls at angles greater than the critical angle, so it is totally internally reflected again and again, zig-zagging along the fibre even when it bends gently. Very little light escapes.

A practical fibre has two layers of glass:

the core, which carries the light, with a slightly higher refractive index;
the cladding, surrounding the core, with a slightly lower refractive index.

The cladding is essential. Without it, light would leak out wherever the fibre touched another fibre, a support or a speck of dirt, and the surface would be easily scratched. With cladding, total internal reflection happens at the core–cladding boundary, which is protected inside the fibre.

Example 3. A fibre has a core of refractive index 1.48 and cladding of 1.46. Find the critical angle at the core–cladding boundary.

sin c = 1.46/1.48 = 0.9865, so c = 80.6°. Light must hit the boundary at more than 80.6° to the normal, travelling almost parallel to the fibre axis.

The Acceptance Angle

Light entering the end of the fibre at too steep an angle will meet the core–cladding boundary at less than the critical angle and leak out. The largest angle to the fibre axis at which light can enter (from air) and still be trapped is the acceptance angle, θa. Its sine is the numerical aperture:

sin θa = √(n₁² − n₂²), where n₁ is the core index and n₂ the cladding index.

For Example 3: sin θa = √(1.48² − 1.46²) = √0.0588 = 0.2425, so θa = 14.0°. Light must enter within a cone of half-angle 14° about the axis.

Dispersion and Attenuation in Optical Fibres

Modal dispersion. Rays travelling straight down the middle cover a shorter path than rays zig-zagging at steep angles, so a short pulse of light spreads out as it travels. If pulses spread into each other, the data becomes unreadable, which limits how fast data can be sent. Two solutions are used:

Monomode (single-mode) fibres have a very narrow core, about 8 to 10 μm across, so light can travel essentially only straight along the axis. These carry long-distance internet and telephone traffic.
Graded-index fibres have a core whose refractive index decreases gradually from the centre outwards, so rays on longer outer paths travel faster and keep up with the central ray.

Material dispersion. Different wavelengths travel at slightly different speeds in glass, so a source with a range of wavelengths also spreads pulses. Lasers with a very narrow range of wavelengths reduce this.

Attenuation. Some light is lost to absorption and scattering by tiny impurities. Modern fibres lose only about 0.2 dB per kilometre at the infrared wavelength of 1,550 nm used for long-distance links. Over 100 km that is 20 dB, meaning about 1% of the light is left, so amplifiers or repeaters are placed at intervals along undersea cables.

Advantages of Optical Fibres

Enormous bandwidth: one fibre can carry many terabits per second.
Low loss over long distances.
Immune to electrical interference, and no sparks, so safe in fuel depots.
Difficult to tap secretly, and much lighter than copper cables.

Optical fibres also have medical uses. An endoscope uses one bundle of fibres to carry light into the body and a second, coherent bundle (with the fibres in the same arrangement at both ends) to carry the image back, allowing doctors to see inside the stomach or joints without major surgery.

Summary

Total internal reflection needs light going from higher n to lower n at an angle of incidence greater than the critical angle.
sin c = n₂/n₁, or 1/n for a boundary with air; glass c ≈ 42°, diamond c ≈ 24°.
Reflecting prisms, diamonds, mirages and optical fibres all rely on total internal reflection.
Optical fibres have a core and a lower-index cladding; sin(acceptance angle) = √(n₁² − n₂²).
Monomode and graded-index fibres reduce dispersion; low attenuation lets signals travel tens of kilometres.