Polarisation is a wave property associated with transverse waves only.
Imagine you fixed one end of a rope to a post. Grab the other end of the rope and pull it tight so that it is stretched out horizontally. Move the rope repeatedly vertically up and down. This will produce a transverse wave on the rope. The vibrations of the rope are in just one plane – the vertical plane. The vibrations are described as plane polarised in the vertical plane. You can produce plane polarised vibrations in the horizontal plane by moving the rope repeatedly from side to side. It would also be fun to keep changing the direction of vibration of the rope – in this case, you will produce an unpolarised wave where the vibrations are in more than one plane.
A plane polarised wave incident at a vertical slit will pass through this slit. When the slit is turned through ${90^ \circ }$, the plane polarised wave will be blocked. When an unpolarised wave is incident at a vertical slit, then all vibrations, other than those in the vertical plane, will be blocked (see Figure 12.17). The wave passing through the slit will be a plane polarised wave in the vertical plane.

Only transverse waves can be plane polarised. So, it should be possible to produce plane polarised light waves. In fact, all types of electromagnetic waves can be plane polarised.
Longitudinal waves vibrate along the direction of wave travel, so no matter what the orientation of the slit, the waves will be able to get through. In short, longitudinal waves, such as sound, cannot be polarised. 

Polarised light

Light is a transverse wave. Its transverse nature can be demonstrated by polarising light. As mentioned previously, light consists of oscillating electric and magnetic fields. Light from the Sun, or a filament lamp, is unpolarised. This means it has oscillating electric fields in all planes at right angles to the direction in which it travels. What can we use to plane polarise such light?
We can use transparent polymer material, such as a Polaroid, a type of polarising filter. The Polaroid has long chains of molecules all aligned in one particular direction. Any electric field vibrations along these chains of molecules are absorbed. The energy absorbed is transferred to thermal energy in the Polaroid.
Electric field vibrations at right angles to the chains of molecules are transmitted with negligible absorption. Figure 12.18 shows the unpolarised light incident at a Polaroid–the transmitted light is plane polarised.

What would happen when you view unpolarised light using two Polaroids? Figure 12.19 shows plane polarised light produced by the first Polaroid. This plane polarised light is incident at the second Polaroid, whose transmission axis is initially vertical. The second Polaroid is often known as the analyser. The incident light passes straight through. Now rotate the analyser through ${90^ \circ }$, so its transmission axis is horizontal. This time, the analyser will absorb all the light. The analyser will appear black. Turning the analyser through a further ${90^ \circ }$ will let the light through the analyser again. What happens at angles other than ${0^ \circ }$ and ${90^ \circ }$ is discussed later.
Here are a few things you can try with a single Polaroid.
- Light reflected from the surface of water, or glass, is partially polarised in a plane parallel to the reflecting surface. Holding a Polaroid with its transmission axis vertical, will reduce the glare of reflected light. This is how your Polaroid sunglasses work. Polarising filters help in photography (Figure 12.20).
- Light from your laptop screen is plane polarised. You can completely cut out the display by viewing the screen through a Polaroid. You can observe the same effect with your LCD calculator display.
Twist the Polaroid, and see the display vanish.

Malus’s law

Figure 12.21 shows plane polarised light incident at a Polaroid. The transmission axis of this Polaroid is at an angle $\theta $ to the plane of the incident light. Now you already know that when $\theta  = 0$, then the light will go through the Polaroid, and when $\theta  = {90^ \circ }$, there is no transmitted light. The intensity of the transmitted light depends on the angle $\theta $.

Consider the incident plane polarised light of amplitude ${A_o}$. The component of the amplitude transmitted through the Polaroid along its transmission axis is ${A_o}\,\cos \theta $. You know that the intensity of light is directly proportional to the amplitude squared. So, the intensity of light transmitted will be given by the expression:

$I = {I_o}\,{\cos ^2}\theta $

where I0 is the intensity of the incident and I is the transmitted intensity at an angle $\theta $ between the transmission axis of the Polaroid and the plane of the incident polarised wave.
The relationship is known as Malus’s law.

Note that the fraction of the light intensity transmitted is equal to ${\cos ^2}\theta $. This means that a graph of I against $\theta $ is a cosine squared graph, see Figure 12.22.

EXAM-STYLE QUESTIONS

 

What is the correct unit for intensity? [1]
A: $J\,{m^2}$
B: $J\,{s^{ - 1}}$
C: $W\,{m^2}$
D: $W\,{s^{ - 1}}$

2) This image shows the screen of an oscilloscope. The time-base of the oscilloscope is set at $500\,\mu s\,di{v^{ - 1}}$.

Calculate the time period of the signal and hence its frequency. [3]

3) a: State two main properties of electromagnetic waves. [2]
b: State one major difference between microwaves and radio waves. [1]
c: i- Estimate the wavelength in metres of X-rays. [1]
ii- Use your answer to i to determine the frequency of the X-rays. [1]
[Total: 5]
A student is sitting on the beach, observing a power boat moving at speed on the sea. The boat has a siren emitting a constant sound of frequency $420 Hz$.
The boat moves around in a circular path with a speed of $25\,m\,{s^{ - 1}}$. The student notices that the pitch of the siren changes with a regular pattern.
a: Explain why the pitch of the siren changes, as observed by the student. [1]
b: Determine the maximum and minimum frequencies that the student will hear. [4]
c: At which point in the boat’s motion will the student hear the most highpitched note? [1]
(Speed of sound in air $ = 330\,m\,{s^{ - 1}}$.)
[Total: 6]

5) This diagram shows some air particles as a sound wave passes.

a: On a copy of the diagram, mark:
i- a region of the wave that shows a compression–label it C [1]
ii- a region of the wave that shows a rarefaction–label it R. [1]
b: Describe how the particle labelled P moves as the wave passes. [2]
c: The sound wave has a frequency of $240 Hz$. Explain, in terms of the movement of an individual particle, what this means. [2]
d: The wave speed of the sound is $320\,m\,{s^{ - 1}}$. Calculate the wavelength of the wave.
[2]
[Total: 8]

6) In an experiment, a student is determining the speed of sound using the equation v = fλ. The values of frequency f and wavelength λ are shown below:
$f = 1000 \pm 10Hz$
$\lambda  = 33 \pm 2cm$
Determine the speed v including the absolute uncertainty. [5]

7) This diagram shows a loudspeaker producing a sound and a microphone connected to a cathode-ray oscilloscope (CRO).

a: Sound is described as a longitudinal wave. Describe sound waves in terms of the movements of the air particles. [1]
b: The time-base on the oscilloscope is set at $5\,ms\,di{v^{ - 1}}$. Calculate the frequency of the CRO trace. [2]
c: The wavelength of the sound is found to be $1.98 m$.
Calculate the speed of sound. [2]
[Total: 5]

8) The Doppler effect can be used to measure the speed of blood. Ultrasound, which is sound of high frequency, is passed from a transmitter into the body, where it reflects off particles in the blood. The shift in frequency is measured by a stationary detector, placed outside the body and close to the transmitter.
In one patient, particles in the blood are moving at a speed of $30\,cm\,{s^{ - 1}}$ in a direction directly away from the transmitter. The speed of ultrasound in the body is $1500\,cm\,{s^{ - 1}}$.
This situation is partly modelled by considering the particles to be emitting sound of frequency $4.000 MHz$ as they move away from the detector. This sound passes to the detector outside the body and the frequency measured by the detector is not $4.000 MHz$.
a: i- State whether the frequency received by the stationary detector is higher or lower than the frequency emitted by the moving particles. [1]
ii- Explain your answer to part i. [3]
b: Calculate the difference between the frequency emitted by the moving particles and the frequency measured by the detector. [3]
c: Suggest why there is also a frequency difference between the sound received by the particles and the sound emitted by the transmitter. [1]
[Total: 8]

9) a: State what is meant by plane polarised light. [1]
b: Reflected light from the surface of water is partially plane polarised.
Describe briefly how you could demonstrate this. [2]
c: Vertically plane polarised light is incident on three polarising filters. The transmission axis of the first Polaroid is vertical. The transmission axis of the second filter is ${45^ \circ }$ to the vertical and the transmission axis of the last filter is horizontal.
Show that the intensity of light emerging from the final filter is not zero. [4]
[Total: 7]