E-StorePhysics of Polarisation
Introduction |    |
One of the properties of a transverse wave is that it can be polarised. This means that all the oscillations of the wave are in the same plane. In this Topic we will investigate the production and properties of polarised waves. Most of this Topic will deal with light waves, and some of the applications of polarised light will be described at the end of the Topic. Light waves are electromagnetic waves, made up of orthogonal (perpendicular) oscillating electric and magnetic fields. When we talk about the oscillations of a light wave, we will be describing the oscillating electric field. For clarity, the magnetic fields will not be shown on any of the diagrams in this chapter - this is the normal practice when describing electromagnetic waves. |
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Polarised and unpolarised waves | |
Let us consider a transverse wave travelling in the x-direction. Although we will be concentrating on light waves in this Topic, it is useful to picture transverse waves travelling along a rope. Figure 1 shows transverse waves oscillating in the y-direction.
The oscillations are not constrained to the y-direction (the vertical plane). The wave can make horizontal oscillations in the z-direction, or at any angle
When all the oscillations occur in one plane, as shown in Figure 1 and Figure 2, the wave is said to be polarised. If oscillations are occurring in many or random directions, the wave is unpolarised. The difference between polarised and unpolarised waves is shown in Figure 3.
Light waves produced by a filament bulb or strip light are unpolarised. In the next two Sections of this Topic different methods of producing polarised light will be described. You should note that longitudinal waves cannot be polarised since the oscillations occur in the direction in which the wave is travelling. This means that sound waves, for example, cannot be polarised. |
in the y-z plane, so long as the oscillations are at right angles to the direction in which the wave is travelling (see 
in the y-z plane.
Before we look at how to polarise light waves, let us think again about a (polarised) transverse wave travelling along a rope in the x-direction with its transverse oscillations in the y-direction. In Figure 4 the rope passes through a board with a slit cut into it. Figure 4 (a) shows what happens if the slit is aligned parallel to the y-axis. The waves pass through, since the oscillations of the rope are parallel to the slit. In Figure 4 (b), the slit is aligned along the z-axis, perpendicular to the oscillations. As a result, the waves cannot be transmitted through the slit.
What would happen if the wave incident on the slit was oscillating in the y-z plane, making an angle of 45° with the y-axis, for instance? Under those conditions, the amplitude of the incident wave would need to be resolved into components parallel and perpendicular to the slit. The component parallel to the slit is transmitted, whilst the component perpendicular to the slit is blocked. The transmitted wave emerges polarised parallel to the slit. Later in this Section we will derive the equation used for calculating the intensity of light transmitted through a polariser whose transmission axis is not parallel to the plane of polarisation of the light waves. |
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Although the mechanical analogy is helpful it cannot be carried over directly to the comparable situation involving light waves. Consider a sheet of Polaroid®, a material consisting of long, thin polymer molecules (doped with iodine) that are aligned with each other. Because of the way a polarised light wave interacts with the molecules, the sheet of Polaroid® only transmits the components of the light with the electric field vector perpendicular to the molecular alignment. The direction which passes the polarised light waves is called the transmission axis. The Polaroid® sheet blocks the electric field component that is parallel to the molecular alignment.
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In Figure 5, light from a filament bulb is unpolarised. This light is incident on a sheet of Polaroid whose transmission axis is vertical. The beam that emerges on the right of the diagram is polarised in the same direction as the transmission axis of the Polaroid. Remember, this means that the electric field vector of the electromagnetic wave is oscillating in the direction shown.
On-screen animation.
This animation shows an electromagnetic wave, namely a plane polarized wave, which propagates in positive x direction. The vectors of the electric field (red) are parallel to the y axis, the vectors of the magnetic field (blue) are parallel to the z axis.
Earlier in this Section, the problem of a polariser acting on a polarised beam of light was introduced. We will now tackle this problem and calculate how much light is transmitted when the transmission axis of a Polaroid is at an angle to the plane of polarisation. The two cases illustrated in Figure 4 show what would happen if the transmission axis is parallel or perpendicular to the polarisation direction of the beam. In the former case all of the light is transmitted, in the latter case none of it is. Figure 6 shows what happens when the transmission axis of a sheet of Polaroid makes an angle 
with the plane of polarisation of an incident beam.
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In Figure 6 an unpolarised beam of light is polarised by passing it through a sheet of Polaroid. The polarised beam is then passed through a second Polaroid sheet, often called the analyser. The transmission axis of the analyser makes an angle 
with the plane of polarisation of the incident beam. The beam that emerges from the analyser is polarised in the same direction as the transmission axis of the analyser.
The following animation shows how the intensity of the transmitted beam depends on the angle between the transmission axes of the polariser and the analyser.
A 'head-on' view of the analyser will help us to find the intensity of the transmitted beam (see Figure 7).
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The incident beam has amplitude A 0. From Figure 7, the component of A 0 parallel to the transmission axis of the analyser is A 0cos
. So the beam transmitted through the analyser has amplitude A, where
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The intensity of a beam, measured in W m-2, is proportional to the square of the amplitude. Thus the intensity I 0 of the incident beam is proportional to A 0 2 and the intensity I of the transmitted beam is proportional to A 2 ( = (A 0cos
)2). From Equation 1
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Equation 2 is known as Malus' law, and gives the intensity of the beam transmitted through the analyser.
Example 1
A sheet of Polaroid is being used to reduce the intensity of a beam of polarised light. What angle should the transmission axis of the Polaroid make with the plane of polarisation of the beam in order to reduce the intensity of the beam by 50%?
We will use Malus' law to solve this problem, with I 0 as the intensity of the incident beam and I 0/2 as the intensity of the transmitted beam. Equation 2 then becomes
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