On a hot afternoon you are served a tall glass of nimbu-pani with a straight plastic straw in it. Look at the glass from the side, a little from above. The straw seems to break at the surface of the drink. The part under the drink looks bent upward and shorter than it should be. Pull the straw out, and it is perfectly straight.
Light changes speed
In vacuum, light travels at \(c = 3 \times 10^8\,\text{m/s}\). In air it is slower by only 3 parts in 10 000, so for this chapter air counts as vacuum. In water and in glass, light is much slower. This change of speed at a boundary is the cause of everything in this chapter.
The refractive index of a medium, \(\mu\), tells how many times slower light is in it: \[ \mu = \dfrac{c}{v}, \] where \(v\) is the speed of light in the medium. Glass has \(\mu = 1.5\), so light in glass travels at \(3 \times 10^8 / 1.5 = 2 \times 10^8\,\text{m/s}\). Water has \(\mu = \tfrac43\), and diamond \(\mu = 2.4\). Other books write the refractive index as \(n\); this book writes \(\mu\).
The refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in the medium, \(\mu = c/v\). It is a pure number, never less than 1. A medium with the larger \(\mu\) is called optically denser.
The frequency stays; the wavelength shrinks
Light is a wave. The chapter on waves on a string showed that a wave keeps its frequency when it crosses into a new medium. The reason is simple. Every crest that arrives at the boundary must go on into the glass. So crests leave the boundary at exactly the rate they arrive, and the frequency cannot change.
The speed does change. From \(v = f\lambda\), a smaller speed at the same frequency means a shorter wavelength. Light of wavelength \(600\,\text{nm}\) in air has frequency \(f = 3 \times 10^8 / (600 \times 10^{-9}) = 5 \times 10^{14}\,\text{Hz}\). In glass it keeps that frequency, and its wavelength becomes \(600 / 1.5 = 400\,\text{nm}\).
Watch the boundary in Fig 1.1 while the wave plays. The number of crests that cross it each second is the same on both sides. In diamond the crests crowd to \(250\,\text{nm}\), because light there is 2.4 times slower. The colour we see is set by the frequency, so a red light stays red under water.
The frequency is fixed by the source and never changes at a boundary. Only the speed and the wavelength change, both by the same factor \(\mu\).
Optical path
Light takes longer to cross a slab of glass than the same thickness of air. It is often useful to ask how far light would have gone in air in that time. For a thickness \(d\) of a medium, that distance is the optical path \(\mu d\).
Light crosses \(3\,\text{cm}\) of glass of \(\mu = 1.5\). Find the time it takes, and the optical path.
In glass \(v = c/\mu = 2 \times 10^8\,\text{m/s}\). The time is \(0.03 / (2 \times 10^8) = 1.5 \times 10^{-10}\,\text{s}\).
The optical path is \(\mu d = 1.5 \times 3 = 4.5\,\text{cm}\).
Check: in \(1.5 \times 10^{-10}\,\text{s}\), light in air covers \(3 \times 10^8 \times 1.5 \times 10^{-10} = 0.045\,\text{m}\), which is \(4.5\,\text{cm}\).
Light of wavelength \(600\,\text{nm}\) in air enters water of \(\mu = \tfrac43\). What is its wavelength in the water, in nm?
Yellow light passes from air into glass. Which statement is correct?
A broadband signal travels from Mumbai to Delhi along about \(1400\,\text{km}\) of glass fibre, with \(\mu = 1.5\). Roughly how many milliseconds does the light take?
The ray bends at the boundary
A ray that meets a boundary head on goes straight through, only more slowly. A ray that meets the boundary at a slant changes its direction. This bending of light as it passes from one medium into another is refraction. Draw the normal, the line at right angles to the surface at the point where the ray arrives.
The angle between the incident ray and the normal is the angle of incidence \(i\). The angle between the refracted ray and the normal is the angle of refraction \(r\). A small part of the light is also reflected at the surface, at the angle \(i\), as in the chapter on reflection.
When light passes from a medium of refractive index \(\mu_1\) into a medium of refractive index \(\mu_2\):
From air into glass, \(\sin i = 1.5\sin r\), so \(\sin r\) is smaller than \(\sin i\), and \(r < i\). The ray bends towards the normal. From glass into air the same law runs the other way, and the ray bends away from the normal. So light entering a denser medium bends towards the normal, and light entering a rarer medium bends away from it.
The readout of the figure above is the whole law in one number. For water, \(\sin i/\sin r\) stays at \(1.33\) at every angle; for diamond it stays at \(2.40\). The angle of refraction grows more slowly than \(i\). Even at \(i = 89°\), water refracts the ray to only \(48.6°\).
The deviation of a refracted ray is the angle through which it turns, \(|i - r|\). A ray from air meets glass of \(\mu = \sqrt2\) at \(45°\). Then \(\sin r = \sin 45° / \sqrt2 = \tfrac12\), so \(r = 30°\). The ray turns towards the normal by \(45° - 30° = 15°\).
A ray in air falls on a glass surface at \(60°\). The reflected ray and the refracted ray are found to be at right angles to each other. Find the angle of refraction and the refractive index of the glass.
The reflected ray leaves at \(60°\) to the normal, on the air side. The refracted ray makes an angle \(r\) with the normal, on the glass side.
Along the surface, the three angles at the point make a straight angle: \(60° + 90° + r = 180°\).
\(r = 30°\). Then \(\mu = \sin 60° / \sin 30° = (\sqrt3/2)/(1/2) = \sqrt3 \approx 1.73\).
The angle of refraction is \(30°\), and the glass has \(\mu = \sqrt3\).
A ray in air meets a flat surface of glass of \(\mu = \sqrt2\) at an angle of incidence of \(45°\). What is the angle of refraction, in degrees?
Which of these rays bend towards the normal as they cross the boundary?
Around 984, the mathematician Ibn Sahl in Baghdad wrote a book on burning mirrors and lenses. In it he used the correct law of refraction to shape a lens that focuses light. His work was forgotten for centuries. In 1621 the Dutch astronomer Willebrord Snellius found the law again from careful measurements, but never published it. In 1637 René Descartes printed it in his book on optics. English books name it after Snellius; French books still call it the law of Descartes.
Comparing two media
Snell's law can be written as \(\sin i / \sin r = \mu_2 / \mu_1\). The ratio on the right is the relative refractive index of medium 2 with respect to medium 1, written \({}_1\mu_2\). Because \(\mu = c/v\), it also equals \(v_1/v_2\), the ratio of the two speeds.
Glass (\(1.5\)) with respect to water (\(\tfrac43\)) is \({}_w\mu_g = 1.5 \div \tfrac43 = \tfrac98\). Water with respect to glass is the reciprocal, \(\tfrac89\). A relative index below 1 simply means the second medium is the rarer one.
Why the reciprocal? Reverse any ray and it retraces its path exactly. This is the principle of reversibility. A ray going from water into glass at \(i\) and \(r\) comes back from glass into water at \(r\) and \(i\). So the ratio of sines for the return trip is the inverse of the ratio for the outward trip.
A ray in glass (\(\mu = 1.5\)) meets a boundary with water (\(\mu = \tfrac43\)) with \(\sin i = 0.8\). Find \(\sin r\) in the water. Does the ray bend towards or away from the normal?
Snell's law: \(1.5 \times 0.8 = \tfrac43\sin r\), so \(\sin r = 1.2 \times \tfrac34 = 0.9\).
So \(r \approx 64.2°\), while \(i \approx 53.1°\). The ray bends away from the normal, because water is rarer than glass.
A ray passes from medium 1 into medium 2. It meets the boundary at \(60°\) to the normal and leaves at \(30°\). What is the ratio of the speed of light in medium 1 to that in medium 2?
A straw stands at a slant in a glass of nimbu-pani. Explain why it seems to bend exactly at the surface of the drink and not anywhere else.
Light from the part of the straw under the drink reaches the eye only after crossing the surface from the drink into air. At the surface, the drink being denser, each ray bends away from the normal. The eye traces the ray straight back and sees that part of the straw in a different direction, higher than it really is. Light from the part above the surface does not cross any boundary, so it is seen where it is. The change happens only at the surface, so the straw seems to break there.
Now return to the glass of nimbu-pani. Light from the part of the straw under the drink leaves through the surface and bends away from the normal. The eye traces each ray back in a straight line and sees that part of the straw higher than it is. Light from the part above the drink crosses no boundary. So the straw looks broken exactly at the surface, the only place where the light bends. The next lesson finds how far up the hidden part appears.
- Thinking that the frequency, and so the colour, changes when light enters glass.The frequency is fixed by the source. Only the speed and the wavelength change, both by the factor \(\mu\).
- Measuring the angles of incidence and refraction from the surface.Both angles are measured from the normal. A ray along the normal has \(i = 0\).
- Dividing the angles instead of their sines.Snell's law is \(\mu_1\sin i = \mu_2\sin r\). The ratio of the angles is not constant.
- Writing the relative index upside down.\({}_1\mu_2 = \mu_2/\mu_1\), the medium the light enters on top. It is greater than 1 when the light enters the denser medium.