9.2 Refraction and Total Internal Reflection

Almost every byte of data that leaves India for the rest of the world — your video calls, the page you are reading — travels as pulses of infrared light inside glass fibres thinner than a hair, laid along the sea floor in cables that land at Mumbai and Chennai. A pulse launched at one end of such a fibre is still inside it thousands of kilometres later, although the fibre curves and twists all the way. Nothing reflects the light back into the fibre except the boundary between two kinds of glass — and that boundary reflects all of it, better than any silvered mirror could.

This section explains how. We state Snell's law of refraction in terms of the refractive index, use it to see why a swimming pool looks shallower than it is, and then follow a ray from a dense medium towards a rarer one. As the angle of incidence grows, the refracted ray swings towards the surface, and at the critical angle it disappears: beyond it, light undergoes total internal reflection. That one idea explains the sparkle of a diamond, the reflecting prisms in binoculars, the mirage over a hot road and the optical fibre.

Refraction and the Refractive Index

When light crosses a boundary between two transparent media at an angle, part of it is reflected and part passes on, bent — refracted. The angle of incidence $i$ and the angle of refraction $r$ are both measured from the normal to the boundary.

Going into a denser medium ($n_2\gt n_1$) the ray bends towards the normal ($r\lt i$); going into a rarer medium it bends away from it. A ray along the normal ($i=0$) passes straight through. The frequency of the light does not change on refraction — the wave on each side of the boundary must keep step with the wave on the other — so its wavelength changes in proportion to its speed: $\lambda=\lambda_0/n$. Refraction is reversible: a ray sent back along the refracted path retraces the incident path.

A rectangular glass slab has parallel faces, so the ray refracted at the first face meets the second at the angle $r$ and leaves at the original angle $i$: the emergent ray is parallel to the incident ray but shifted sideways. If the slab has thickness $t$, the perpendicular distance between the incident and emergent lines — the lateral shift — is $$d=\frac{t\,\sin(i-r)}{\cos r},$$ because the ray travels a distance $t/\cos r$ inside the slab at an angle $i-r$ to its original line.

Apparent depth. Look straight down into a pool and it seems shallower than it is. Rays from a point $O$ on the bottom, at real depth $h$, leave the water bending away from the normal; traced back, they seem to come from a point $I$ at depth $h'$ above $O$. For a ray at small angle $r$ in water emerging at angle $i$ in air, both reach the surface at the same horizontal distance $x$ from the normal through $O$, so $x=h\tan r=h'\tan i$. For near-normal viewing $\tan\approx\sin$, and Snell's law ($n\sin r=\sin i$) gives $$\frac{h}{h'}=\frac{\tan i}{\tan r}\approx\frac{\sin i}{\sin r}=n.$$

Atmospheric refraction. The air thins with height, so its refractive index falls slightly from the ground upwards. Light from the Sun near the horizon bends gently towards the Earth as it comes down through the atmosphere, and we see the Sun about half a degree higher than it really is. As a result the Sun is visible about $2$ minutes before it actually rises above the horizon, and for about $2$ minutes after it has set; the flattened look of the setting Sun has the same cause.

Total Internal Reflection

Now send light from a denser medium into a rarer one — from glass into air, say. Snell's law gives $\sin r=n\sin i$ with $n=n_\text{glass}/n_\text{air}\gt1$, so $r\gt i$: the refracted ray bends away from the normal, and as $i$ increases it swings down towards the surface. At the same time the reflected ray, weak at first, grows brighter. When $\sin i$ reaches $1/n$, $\sin r=1$: the refracted ray grazes along the surface ($r=90^\circ$). For any larger angle of incidence Snell's law would need $\sin r\gt1$, which is impossible — there is no refracted ray, and all the light is reflected back into the denser medium.

Total internal reflection really is total: no energy is lost at the boundary, whereas even a fresh silver mirror absorbs a few per cent of the light. That is why it is used wherever light must be reflected many times.

Total Internal Reflection at Work: Prisms, Diamonds, Mirages and Optical Fibres

Reflecting prisms. A glass prism whose cross-section is a right-angled isosceles triangle ($45^\circ$–$90^\circ$–$45^\circ$) makes an excellent mirror. Light entering normally through one of the short faces meets the long face at $45^\circ$, which is more than the critical angle of glass ($41.8^\circ$), so it is totally reflected and turned through $90^\circ$. Entering through the long face, it is totally reflected twice and turned back through $180^\circ$, with the image inverted. Periscopes use the first arrangement; binoculars use pairs of the second to fold a long light path into a short tube and to turn the image the right way up.

Diamonds. The critical angle of diamond is only $24.4^\circ$. A brilliant-cut diamond is shaped so that light entering the top meets the lower facets at angles above $24.4^\circ$ and is totally reflected several times before it emerges from the top, making the stone sparkle.

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 smaller. Light from the sky heading down at a shallow angle passes into ever-rarer layers and bends away from the normal until, near the road, it is turned upwards (in the last layer it is totally reflected). A driver sees an image of the sky on the road ahead and reads it as a pool of water.

Optical fibres. An optical fibre is a strand of very pure glass with a core of refractive index $n_1$ surrounded by a cladding of slightly smaller index $n_2$. Light entering one end travels along the core by repeated total internal reflection at the core–cladding boundary, even round gentle bends, with almost no loss — modern fibres lose less than $5\%$ of the light per kilometre. The cladding keeps the reflecting surface clean: dirt or a touch on a bare glass rod would spoil the reflection. Bundles of fibres carry images in medical endoscopes; single fibres carry telephone, television and internet signals as on–off pulses of light.

Not every ray that enters a fibre is trapped. A ray entering the flat end from air at angle $\theta_a$ to the axis is refracted to an angle $\theta$ with $\sin\theta_a=n_1\sin\theta$, and then meets the cladding at $90^\circ-\theta$. It is trapped if $90^\circ-\theta\gt C$, i.e. $\sin\theta\lt\cos C$. With $\sin C=n_2/n_1$, $\cos C=\sqrt{1-n_2^2/n_1^2}$, so the largest acceptable $\theta_a$ satisfies $$\sin\theta_{a,\max}=n_1\cos C=\sqrt{n_1^2-n_2^2}.$$ The angle $\theta_{a,\max}$ is the fibre's acceptance angle; rays inside this cone are guided.

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