Sit on the sea wall at Marine Drive in Mumbai on a clear evening. The Sun sinks towards the Arabian Sea. High in the sky it was a dazzling white; near the water it turns orange, then deep red. The sky overhead is still blue. A long, bright path of glare lies across the water. A friend in polaroid sunglasses says the glare is much weaker for her.
Reflection can polarise light
Malus found polarisation in light reflected from a window. Reflection from glass or water favours one direction of \(\vec E\): the direction parallel to the surface, at right angles to the plane of incidence. The plane of incidence is the plane that holds the incident ray and the normal, the plane of the page in a ray diagram. At one angle of incidence the favouring is complete.
In the figure above, the line on the reflected ray shrinks as \(i\) grows. At \(i = 56.3°\) it vanishes. The reflected light then carries only the field at right angles to the page, so it is plane polarised. At the same angle, the readout shows the reflected and refracted rays at exactly \(90°\).
That angle is the polarising angle \(i_p\), also called Brewster's angle. David Brewster found by experiment, in 1815, that it obeys a simple rule.
Let unpolarised light fall on the surface of a transparent medium of refractive index \(\mu\). The reflected light is completely plane polarised when the angle of incidence \(i_p\) satisfies \(\tan i_p = \mu\). The electric field of the reflected light is then at right angles to the plane of incidence.
For glass of \(\mu = 1.5\), \(i_p = \tan^{-1} 1.5 = 56.3°\). For water, \(\mu = \tfrac43\) and \(i_p = 53.1°\). A glass of \(\mu = \sqrt3\) gives \(i_p = 60°\). The chapter on refraction met the last case in a worked example, where the two rays came out at right angles.
Unpolarised light falls on a flat glass plate of \(\mu = 1.5\). At what angle of incidence is the reflected light completely plane polarised, in degrees?
The two rays at right angles
Brewster's law and Snell's law together fix the angle of refraction. Snell's law gives \(\sin i_p = \mu\sin r\). Brewster's law gives \(\mu = \sin i_p/\cos i_p\). Put the second into the first, and \(\sin r = \cos i_p\).
\[ \sin i_p = \mu \sin r \]
Why. Snell's law at the surface, from air into the medium.
\[ \mu = \tan i_p = \dfrac{\sin i_p}{\cos i_p} \]
Why. Brewster's law at the polarising angle.
\[ \sin i_p = \dfrac{\sin i_p}{\cos i_p}\,\sin r \;\Rightarrow\; \sin r = \cos i_p \]
Why. substitute step 2 into step 1 and cancel \(\sin i_p\).
\[ r = 90^\circ - i_p \]
Why. \(\cos i_p = \sin(90° - i_p)\), and both angles lie between \(0°\) and \(90°\).
\[ \text{angle between the rays} = 180^\circ - i_p - r = 90^\circ \]
Read it. the reflected ray leaves at \(i_p\) on one side of the normal, and the refracted ray at \(r\) on the other side, below the surface.
The right angle is also the reason for the polarisation. The reflected light is sent out by the electrons of the glass, set vibrating by the refracted wave. A vibrating charge sends out no light along its own line of vibration. At Brewster's angle, the in-page vibration of the refracted wave points straight along the reflected ray. So that component cannot be reflected, and only the component at right angles to the page is left.
At Brewster's angle the reflected light is completely polarised but weak. For glass it is only about \(15\,\%\) of the field component at right angles to the page, some \(7\,\%\) of the whole beam. The refracted light carries the rest. It has all the in-page component and most of the other, so it is only partially polarised.
(a) Light reflected from a liquid is completely plane polarised at an angle of incidence of \(53.1°\). Find the refractive index of the liquid and the angle of refraction. (b) Light inside glass of \(\mu = \sqrt3\) meets the surface with air. Find its polarising angle.
(a) Use \(\tan i_p = \mu\) and \(r = 90° - i_p\). (b) From glass into air the relative index is \(1/\sqrt3\).
(a) \(\mu = \tan 53.1° = \tfrac43\), and \(r = 90° - 53.1°\). (b) \(\tan i_p' = 1/\sqrt3\).
(a) \(\mu = 1.33\) and \(r = 36.9°\); check: \(\sin 53.1° = 0.80 = 1.33 \times \sin 36.9°\). (b) \(i_p' = 30°\).
(a) The liquid has \(\mu = \tfrac43\), like water, and \(r = 36.9°\). (b) Inside the glass \(i_p' = 30°\); from air into the glass \(i_p = 60°\). The two add to \(90°\): each is the angle of refraction of the other.
The critical angle of the chapter on refraction is linked to Brewster's angle through \(\mu\). Since \(\sin C = 1/\mu\), we have \(\tan i_p = \mu = 1/\sin C\). A material with a critical angle of \(45°\) has \(\mu = \sqrt2\), so \(i_p = \tan^{-1}\sqrt2 = 54.7°\).
A transparent material has a critical angle of \(45°\) for light going into air. What is its polarising angle for light arriving from air, in degrees?
Light falls on water at its polarising angle, \(53.1°\). What is the angle between the reflected ray and the refracted ray?
Glare and polaroid sunglasses
Sunlight reflected from a wet road, a lake or the sea arrives at a range of angles near Brewster's angle. For water \(i_p = 53.1°\), so the strongest polarisation is in light that leaves the water \(36.9°\) above the surface. Near that angle, the reflected glare is mostly polarised with \(\vec E\) parallel to the surface, which is horizontal.
Polaroid sunglasses are made with their transmission axis vertical. By Malus's law they pass little of the horizontal field of the glare. They still pass half of the unpolarised light from everything else. So the glare is cut far more than the scene. Tilt your head by \(90°\), and the glare comes back.
Polaroid sunglasses cut the glare from the surface of a lake. In which direction is their transmission axis set?
Why the sky is blue
Sunlight travelling through the air meets molecules far smaller than its wavelength. The electric field of the light shakes the electrons of each molecule, and the shaken electrons send light out in all directions. This is scattering. For particles much smaller than the wavelength, the scattered intensity goes as \(1/\lambda^4\).
The power four makes a large difference. Blue light of \(450\,\text{nm}\) is scattered \((700/450)^4 \approx 5.9\) times as strongly as red light of \(700\,\text{nm}\). So the light scattered towards us from every part of the sky is rich in blue. High over the Himalaya the air above is thinner, and the sky there is a deeper, darker blue. Clouds are white because their water drops are larger than the wavelength and scatter every colour alike.
How many times more strongly do air molecules scatter light of \(400\,\text{nm}\) than light of \(800\,\text{nm}\)?
Rank these colours by how strongly air molecules scatter them, strongest first.
Skylight is polarised
The scattered light is also polarised. Let the Sun be on the horizon, and look at the sky straight overhead, as in Fig 5.1. The sunlight reaching a molecule above you travels horizontally. Its field has two parts across that direction: one up and down, and one horizontal, at right angles to the Sun's direction.
Each part shakes the electrons of the molecule along its own line. Light scattered straight down to you would need the up-down vibration to send light along its own line. A vibrating charge cannot do that. So only the horizontal part reaches you. Skylight scattered through \(90°\) is plane polarised, with \(\vec E\) horizontal and at right angles to the Sun's direction. At other angles it is partly polarised.
The Sun is setting on the western horizon. You look at the sky straight overhead through a polaroid. In which direction does the field of that skylight vibrate?
Now return to Marine Drive. Near the horizon, sunlight crosses nearly forty times as much air as when the Sun is overhead. Along that long path, the blue is scattered out to the sides, lighting other people's sky. The red, scattered least, comes through, and the setting Sun looks red. Overhead, you see light scattered from sunlight passing above: rich in blue. The glare on the Arabian Sea is light reflected near Brewster's angle, mostly with \(\vec E\) horizontal, and polaroid sunglasses block it.
In 1921 C. V. Raman sailed to England on the ship Narkunda. Crossing the Mediterranean, he wondered why the sea was so deeply blue. Most people said it simply reflected the blue sky. Raman looked at the water through a Nicol prism, a polariser, held at Brewster's angle. It removed the reflected skylight, yet the sea looked bluer than ever. The blue came from light scattered by the water itself. His study of scattering led, in 1928, to the Raman effect and a Nobel Prize in 1930.
- Writing \(\sin i_p = \mu\) or \(\cos i_p = \mu\).Brewster's law is \(\tan i_p = \mu\). A quick check: it is the only one of the three that can exceed 1.
- Thinking the refracted light is plane polarised at Brewster's angle.Only the reflected light is completely polarised. The refracted light keeps the whole in-page component and most of the other, so it is partially polarised.
- Setting sunglasses with a horizontal axis to stop glare.Glare has its field horizontal, parallel to the water, so the axis must be vertical.
- Explaining the red sunset by the Sun itself turning red.The Sun's light is the same; near the horizon it crosses about forty times more air, which scatters the blue out and leaves the red.