Dive to the bottom of a swimming pool and look up. Straight above you, the surface is a bright round window: you can see the sky, the lights, even the people standing at the edge. Farther out, the surface is not transparent at all. It shines like silver, and in it you see the floor and the walls of the pool, upside down.
When the refracted ray runs out
So far light has mostly gone from air into water or glass. Now let it go the other way, from glass into air. The ray bends away from the normal, so \(r\) is larger than \(i\). As \(i\) grows, \(r\) grows faster, and it reaches \(90°\) first. At that moment the refracted ray grazes the surface.
Follow the glass in Fig 3.1. At \(i = 30°\), Snell's law gives \(\sin r = 1.5\sin 30° = 0.75\), so \(r = 48.6°\). At \(i = 41.8°\), \(\sin r = 1.5 \times \tfrac23 = 1\), and the ray grazes the surface. For any larger \(i\), Snell's law would need \(\sin r > 1\), which no angle has. There is no refracted ray at all, and all the light is reflected back into the glass.
The angle of incidence at which the refracted ray grazes the surface is the critical angle \(C\). Above it, the light undergoes total internal reflection. Below it, some light is always reflected too. That weak reflection is the dashed ray in Fig 3.1, which you may have seen as a faint image in a window pane.
For light going from a denser medium towards a rarer one, the critical angle \(C\) is the angle of incidence that gives refraction at \(90°\). Total internal reflection happens only when both conditions hold:
Snell's law at the critical angle gives \(\mu_d\sin C = \mu_r\sin 90° = \mu_r\). So \(\sin C = \mu_r/\mu_d\). With air as the rarer medium, \(\sin C = 1/\mu\).
Glass (\(1.5\)) in air has \(\sin C = \tfrac23\), so \(C = 41.8°\). Water has \(C = 48.6°\), and diamond (\(2.4\)) only \(24.6°\). Problems often use round values: \(\mu = \sqrt2\) gives \(C = 45°\), \(\mu = 2\) gives \(30°\), and \(\mu = \tfrac54\) gives \(\sin C = \tfrac45\), \(C = 53.1°\). The denser the medium, the smaller its critical angle.
Total internal reflection never happens when light goes from a rarer into a denser medium. A ray from air into glass bends towards the normal and always finds an angle of refraction, even at grazing incidence.
A ray from air that grazes a glass surface, at \(i = 90°\), refracts at \(\sin r = 1/\mu = \sin C\). So it enters at exactly the critical angle. Every other ray from air enters at a smaller angle. By reversibility, a ray inside the glass at more than \(C\) could never have come from the air. It cannot get out into the air either.
A ray inside glass of \(\mu = \sqrt2\) meets the surface with air. Find the critical angle. Does the ray leave the glass if it meets the surface at \(60°\)? At \(30°\)?
\(\sin C = 1/\sqrt2\), so \(C = 45°\).
At \(60°\), the angle of incidence is larger than \(C\). The ray is totally reflected and does not leave.
At \(30°\), \(\sqrt2\sin 30° = \sin r\) gives \(\sin r = 1/\sqrt2\), so \(r = 45°\). The ray leaves, bent away from the normal. At \(i = 45°\) exactly, it would graze the surface.
What is the critical angle, in degrees, for a material of \(\mu = 2\) in air?
Glass (\(\mu = 1.5\)) is in contact with water (\(\mu = \tfrac43\)). How does the critical angle at the glass-water surface compare with that at a glass-air surface?
In which of these cases is the ray totally reflected?
The circle of light
Put a small lamp at a depth \(h\) in a pool. Its rays spread in every direction. Those that meet the surface at less than \(C\) leave the water. Those that meet it at more than \(C\) are totally reflected. So the light escapes only through a bright circle straight above the lamp, the circle of illuminance.
\[ \tan i = \dfrac{x}{h} \]
Why. a ray from the lamp reaching the surface at a distance \(x\) from the point straight above it meets the surface at the angle \(i\).
\[ r = h\tan C \]
Why. the edge of the circle is the ray that meets the surface at exactly \(C\).
\[ \sin C = \dfrac{1}{\mu} \;\Rightarrow\; \cos C = \dfrac{\sqrt{\mu^2 - 1}}{\mu}, \quad \tan C = \dfrac{1}{\sqrt{\mu^2 - 1}} \]
Why. draw a right triangle with hypotenuse \(\mu\) and opposite side 1; the adjacent side is \(\sqrt{\mu^2 - 1}\).
\[ r = \dfrac{h}{\sqrt{\mu^2 - 1}} \]
Read it. substitute \(\tan C\) into step 2.
For the lamp in Fig 3.2, \(\mu^2 - 1 = \tfrac{16}{9} - 1 = \tfrac79\), so \(\sqrt{\mu^2 - 1} = \sqrt7/3\), and \(r = \sqrt7 \times 3/\sqrt7 = 3\,\text{m}\). The deeper the lamp, the wider its circle, in proportion to the depth.
The fish's window
Now press the second chip of Fig 3.2. Every ray is reversed, and the lamp becomes the eye of a fish. Light from the whole sky, from the zenith down to the horizon, reaches the fish through that same circle. The rays from the horizon graze the water and enter at \(C\). So the fish sees everything above the water squeezed into a cone of angle \(2C = 97°\).
Outside that cone, the fish looks at the surface at more than \(C\). No light from above can arrive along those directions. What does arrive is light from the bottom of the pool, totally reflected by the surface. So outside the window, the surface is a perfect mirror. This is the silver ceiling you see when you look up from the floor of a swimming pool.
A coin lies \(12\,\text{cm}\) below the surface of a liquid of \(\mu = \tfrac54\). A flat opaque disc floats on the surface, centred above the coin. What is the smallest diameter of the disc that hides the coin from every viewer above the surface?
Light from the coin leaves the liquid only inside the circle of illuminance above it. The disc must cover that whole circle.
\(\sin C = \tfrac45\), so \(\cos C = \tfrac35\) and \(\tan C = \tfrac43\). Then \(r = h\tan C = 12 \times \tfrac43\).
\(r = 16\,\text{cm}\), so the diameter is \(32\,\text{cm}\).
A disc \(32\,\text{cm}\) across hides the coin completely. A lamp \(3\,\text{m}\) deep in the same liquid would light a circle of radius \(4\,\text{m}\) and area \(16\pi \approx 50\,\text{m}^2\).
A small bulb is \(1.2\,\text{m}\) below the surface of a liquid of \(\mu = \tfrac54\). What is the radius of the circle through which its light leaves the surface, in metres?
Light in a glass thread
An optical fibre is a glass thread thinner than a hair. Its core has a higher refractive index than the cladding around it. A ray inside the core meets the core-cladding boundary at a large angle, more than the critical angle. It is totally reflected, crosses the core, and is totally reflected again, thousands of times a metre. No light leaks out, even when the fibre bends gently.
An optical fibre has a core of \(\mu = 1.5\) and a cladding of \(\mu = 1.2\). Find the critical angle at the core-cladding boundary, and the largest angle a ray in the core may make with the axis. Then find the largest angle of incidence in air, at the flat end, for which the ray is guided.
Inside the core the ray makes an angle \(a\) with the axis. The wall is parallel to the axis, so the ray meets the wall at \(90° - a\).
\(\sin C = 1.2/1.5 = 0.8\), so \(C = 53.1°\). Guiding needs \(90° - a \ge C\), that is \(a \le 36.9°\). At the end face \(\sin i = 1.5\sin a\).
\(\sin i_{\max} = 1.5\sin 36.9° = 1.5 \times 0.6 = 0.9\), so \(i_{\max} = 64.2°\).
Every ray that meets the flat end within \(64.2°\) of the axis is guided. The figure above shows the change between \(64°\) and \(65°\).
Prisms that reflect
A right-angled prism with two \(45°\) corners makes a better mirror than silvered glass. A ray enters one short face at right angles and meets the long face at \(45°\). If \(C < 45°\), the ray is totally reflected: all the light, with no silver to tarnish. So the glass needs \(\sin 45° \ge 1/\mu\), that is \(\mu \ge \sqrt2 \approx 1.41\). Ordinary glass, with \(\mu = 1.5\) and \(C = 41.8°\), is good enough.
Diamond has \(\mu = 2.4\), so \(\sin C = 1/2.4 = \tfrac{5}{12}\) and \(C = 24.6°\). A cut diamond is shaped so that light which enters its top meets the inner faces at more than this tiny critical angle. The light is totally reflected again and again inside the stone. It escapes only through a few faces at the top, in bright flashes. That is the sparkle of a diamond. Glass cut in the same shape, with \(C = 41.8°\), lets most of the light leak out of the back and looks dull.
A prism has angles of \(45°\), \(45°\) and \(90°\). It is to turn a ray through \(90°\) by total reflection at its long face, as in Fig 3.3. What is the least refractive index it needs?
Explain in two or three sentences why a cut diamond sparkles far more than a piece of glass of the same shape.
Diamond has a very high refractive index, 2.4, so its critical angle is only about 24.6°, against 41.8° for glass. Light that enters a cut diamond meets most inner faces at more than 24.6° and is totally reflected many times. It comes out only through a few faces at the top, in bright flashes; in glass much of the light leaks out through the back.
Narinder Singh Kapany was born in 1926 in Moga, Punjab. As a student in India he was told that light always travels in straight lines. He wondered whether it could be made to go round a corner. In London, in 1954, he and Harold Hopkins reported that a bundle of fine glass fibres could carry an image along a bent path. In 1960 he wrote an article that gave the new field its name, fibre optics. Every broadband cable today carries light by the total internal reflection he studied.
A fibre-to-the-home connection brings the internet as flashes of infrared light in a glass fibre. The flashes are guided by total internal reflection for many kilometres from the exchange to the router. One fibre carries far more data than a copper wire of the same thickness, and electrical noise from motors or lightning cannot disturb it. The endoscope a doctor passes down a patient's throat uses the same idea, with thousands of fibres carrying a picture.
- Expecting total internal reflection when light goes from air into glass.Only a ray going towards the rarer medium can be totally reflected. From the rarer side there is always a refracted ray.
- Writing \(\sin C = \mu\), or taking the critical angle as the angle of refraction.\(\sin C = \mu_r/\mu_d\), always less than 1. \(C\) is an angle of incidence in the denser medium.
- Using \(\sin C = 1/\mu\) at a boundary between two liquids or glasses.\(1/\mu\) holds only with air or vacuum outside. In general use the ratio \(\mu_r/\mu_d\).
- Measuring the fibre's angle at the wall from the axis.The normal to the wall is perpendicular to the axis. A ray at angle \(a\) to the axis meets the wall at \(90° - a\).