Put on a pair of polaroid sunglasses and look at the screen of a mobile phone or an ATM. The screen looks normal. Now tilt your head slowly to one side. The screen grows dimmer, and at a quarter turn it may go almost black. Tilt back, and it brightens again. Ordinary tinted glasses do nothing of the kind.
A rope through a slot
Tie a long rope to a post and pass it through a vertical slot in a wooden fence. Shake your end up and down. The transverse wave on the rope passes through the slot and reaches the post. Now shake the rope from side to side. The slot stops the sideways motion, and the wave dies at the fence.
The chapter Introduction to Waves called a wave transverse when its particles move across the direction of travel. Across the direction of travel there are many directions: up and down, side to side, and every slant between. A wave whose vibrations keep to one of these directions is polarised. The slot lets through only the vibrations along its own length.
A transverse wave is polarised when its vibrations, all at right angles to the direction of travel, are confined to one fixed direction. The act of confining them is polarisation.
A longitudinal wave cannot be polarised. In sound, the air moves back and forth along the direction of travel, and there is no direction across it to choose. A slot of any shape passes sound equally well. So a wave that can be polarised must be transverse. Light can be polarised, and this is one of the proofs that light is a transverse wave.
Why can light be polarised while sound in air cannot?
The electric field decides
The chapter on Young's double slit described light as a transverse electromagnetic wave. Its electric field \(\vec E\) swings to and fro at right angles to the ray, and it is \(\vec E\) that acts on the eye. The unit on Electromagnetism shows how the wave is made. Here we need only the direction of \(\vec E\), which is the direction of polarisation.
A lamp or the Sun contains countless atoms that emit independently, each for a very short time. Each short burst has its \(\vec E\) in some direction across the ray. The directions change at random many millions of times a second. Over any time the eye can notice, every direction is present equally. Such light is unpolarised light.
If \(\vec E\) swings along one fixed line, the light is plane polarised light. Between the two lies partially polarised light, in which every direction is present but one is stronger than the rest. Fig 4.1 shows all three, seen with the beam coming straight at you.
A polaroid
A polaroid is a thin plastic sheet whose long molecules are lined up in one direction and made conducting with iodine. The component of \(\vec E\) along the molecules drives electrons along them and is absorbed. The component at right angles to them passes. That direction is the transmission axis of the polaroid.
Let \(\vec E\), of amplitude \(E_0\), make an angle \(\theta\) with the axis. Only the component \(E_0\cos\theta\) passes. Intensity grows as the square of the amplitude, so the fraction \(\cos^2\theta\) of the intensity passes. In unpolarised light \(\theta\) takes every value equally. The average of \(\cos^2\theta\) over a full turn is \(\tfrac12\). So a polaroid passes exactly half of unpolarised light: \[ I = \frac{I_0}{2}. \]
The light that leaves is no longer unpolarised. Its \(\vec E\) now swings along the transmission axis, so it is plane polarised. Turning the polaroid about the beam changes nothing: unpolarised light has every direction equally, so half passes at every setting. A steady brightness as the polaroid turns is itself a sign that the light was unpolarised.
Unpolarised light of intensity \(12\,\text{W/m}^2\) falls on a polaroid. The polaroid is then turned slowly through a full turn about the beam. What intensity leaves it, in W/m², at every setting?
Polariser and analyser
Pass unpolarised light through one polaroid, the polariser. The light is now plane polarised. Let it fall on a second polaroid, the analyser, whose axis makes an angle \(\theta\) with the first. The analyser passes the component \(E_0\cos\theta\) of the field, and so the fraction \(\cos^2\theta\) of the intensity.
Let plane polarised light of intensity \(I_0\) fall on an analyser. Let its transmission axis make an angle \(\theta\) with the plane of polarisation. Then the transmitted intensity is \(I = I_0\cos^2\theta\).
Unpolarised light passes a fixed polariser and then an analyser that turns steadily through one full turn. How many times in that turn does the light from the analyser go dark?
In the figure above, \(8\,\text{W/m}^2\) of unpolarised light leaves the polariser as \(4\,\text{W/m}^2\). At \(\theta = 60°\) the analyser passes \(4\cos^2 60° = 4 \times \tfrac14 = 1.0\,\text{W/m}^2\). The graph is \(I = 4\cos^2\theta\). It reaches \(4\) at \(0°\) and \(180°\), and zero at \(90°\) and \(270°\): two bright and two dark settings in each turn.
Four values are worth knowing by heart. At \(30°\), \(\cos^2\theta = \tfrac34\). At \(45°\) it is \(\tfrac12\), at \(60°\) it is \(\tfrac14\), and at \(90°\) it is zero. So an analyser at \(60°\) passes half the amplitude but only a quarter of the intensity. Two polaroids with their axes at \(90°\) are crossed polaroids: together they pass nothing.
Unpolarised light of \(10\,\text{W/m}^2\) passes a polariser and then an analyser whose axis is at \(30°\) to the polariser's. What intensity leaves the analyser, in W/m²?
Plane polarised light falls on an analyser. At what angle between the plane of polarisation and the analyser's axis is the transmitted intensity one quarter of the incident intensity, in degrees?
An analyser turns at a steady five turns per second in a beam of plane polarised light. How many times each second does the transmitted light reach its greatest brightness?
Testing a beam with one polaroid
One polaroid, turned slowly in a beam, tells the three kinds of light apart. If the brightness stays the same, the light is unpolarised. If it falls to zero twice in each turn, the light is plane polarised. If it rises and falls but never reaches zero, the light is partially polarised.
Partially polarised light behaves like a mixture of the two. Suppose a beam has \(4\,\text{W/m}^2\) of unpolarised light and \(2\,\text{W/m}^2\) of plane polarised light. The polaroid always passes half the unpolarised part, \(2\,\text{W/m}^2\). It passes \(2\cos^2\theta\) of the polarised part, between \(2\) and zero. So the brightness swings between \(4\) and \(2\,\text{W/m}^2\), and never reaches zero.
A beam is a mixture of \(6\,\text{W/m}^2\) of unpolarised light and \(4\,\text{W/m}^2\) of plane polarised light. A polaroid in the beam is turned slowly. What is the least intensity it transmits, in W/m²?
Three polaroids
Cross two polaroids, and no light passes. Now slide a third polaroid between them, with its axis at \(45°\). Light comes through. Adding a sheet that absorbs has let more light pass. The reason is that each polaroid resets the direction of \(\vec E\) to its own axis.
Unpolarised light of intensity \(I_0\) passes three polaroids. The first and last are crossed, and the middle one makes an angle \(\theta\) with the first. Find the intensity that leaves, and the angle that makes it greatest.
After \(P_1\): \(I_1 = I_0/2\), polarised along \(P_1\).
After \(P_2\): \(I_2 = I_1\cos^2\theta\). \(P_3\) is at \(90° - \theta\) to \(P_2\), so \(I_3 = I_2\cos^2(90° - \theta) = I_2\sin^2\theta\).
\(I_3 = \tfrac{I_0}{2}\cos^2\theta\sin^2\theta = \tfrac{I_0}{8}(2\sin\theta\cos\theta)^2 = \tfrac{I_0}{8}\sin^2 2\theta\).
\(I_3\) is greatest when \(\sin 2\theta = 1\), at \(\theta = 45°\), and then \(I_3 = I_0/8\). With \(P_2\) removed, \(I_3 = 0\).
In the figure above, \(8\,\text{W/m}^2\) becomes \(4\) after \(P_1\). With \(P_2\) at \(45°\) it becomes \(2\), and \(P_3\) passes \(1.0\,\text{W/m}^2\), one eighth. At \(30°\) or \(60°\), \(P_3\) passes \(0.75\,\text{W/m}^2\). Remove \(P_2\), and the crossed pair passes nothing.
Unpolarised light of \(32\,\text{W/m}^2\) passes two crossed polaroids with a third between them, whose axis is at \(30°\) to the first. What intensity leaves the last polaroid, in W/m²?
Now return to the sunglasses and the phone. The screen of a phone or an ATM sends out plane polarised light, because a polaroid sheet is the top layer of the display. Polaroid sunglasses are analysers. Tilting your head turns their axis. At a quarter turn the axes are crossed, and by Malus's law almost nothing passes.
The display of a pocket calculator is a layer of liquid crystal between two crossed polaroids. The molecules of the crystal are arranged in a gentle twist. They turn the plane of polarisation through \(90°\), so light passes the second polaroid and the display looks grey. A small voltage on one segment straightens its molecules. That segment no longer turns the light, the crossed polaroids block it, and it shows black.
In 1808 Étienne-Louis Malus, an engineer in the French army, stood at his window in Paris. He looked through a crystal of calcite at the setting Sun reflected in the windows of the Luxembourg Palace. Calcite normally shows two images. As he turned the crystal, one image faded and returned. Reflected light, he realised, had been polarised. He measured how the brightness varied with the angle and found the law of the square of the cosine.
- Applying Malus's law to unpolarised light.Unpolarised light through one polaroid gives \(I_0/2\) at every setting. Malus's law applies from the second polaroid on, to light that is already plane polarised.
- Using \(\cos\theta\) instead of \(\cos^2\theta\).The amplitude passes as \(\cos\theta\); the intensity, which goes as the square of the amplitude, passes as \(\cos^2\theta\).
- Measuring each angle from the first polaroid when there are three.Apply Malus's law at each sheet with the angle between that sheet and the one just before it.
- Expecting one dark setting per turn.\(\cos^2\theta\) is zero at \(90°\) and \(270°\). An axis is a line, so the pattern repeats every half turn.