Physics 6 Tutorial 8 - Polarisation of Light



Polarising Filters

Two Polaroid Filters
Malus' Law
Uses for Polarisation

Liquid Crystal Displays

Polarisation by Reflection

Brewster Angle



This is quite a long and challenging tutorial.  Work through it slowly and carefully.  If you don't understand it, ask for help from your tutor.


Before you attempt this tutorial, you should review the section Waves Tutorial 2 that covers polarisation.  The important thing to remember is that only transverse waves can be polarised.  Longitudinal waves cannot be polarised as they oscillate in the direction of propagation. We will consider polarisation of electromagnetic waves.   The polarisation of mechanical waves can be used as a simple model for the polarised properties of electromagnetic waves, which consist of an electric component and a magnetic component in phase.


Polarising Filters

These are often called polaroids after the company that developed them.  These are made of long-chain polymers that are oriented in a particular direction like this:




Light from a light bulb is unpolarised, meaning that the photons have random orientations. The molecules transmit the component of the light parallel to their alignment.  Therefore, if the filter is vertically aligned, the light is vertically polarised.  More specifically, it's the electric field component that is vertically polarised.  If the electric field component is transmitted, so also is the perpendicular magnetic field component.


The orientation of the filter is sometimes called the transmission axis.  If the orientation of light is at 90o to the transmission axis, no transmission occurs.  If, however, the light is oriented at an angle to the transmission axis of a vertical polarising filter, the vertical component of the light is transmitted.  The idea is shown below:


The wave at angle q has the same amplitude, A, as the vertically polarised wave.  Since the electric field component is a vector quantity, we can separate it into vertical and horizontal components:

Av = A cos q


Ah = A sin q


Since the horizontal component is absorbed, we can ignore it.


A vertically polarised wave refers to the electric field component.  The magnetic field component is horizontal, but it is transmitted.  The polariser does NOT split the electric and magnetic components from each other, otherwise the whole physical nature of the electromagnetic wave would be altered.


The amplitude represents the intensity of the light, I.  Intensity is the energy per unit area.  As the production of photons is a random process, and their orientation is random, we can say that, on average, 50 % of the light intensity is transmitted through a vertically oriented polarising filter.  50 % of the intensity is absorbed.


Question 1

A ray of light of intensity 3 W m-2 is oriented at 50o to the transmission axis of a polarising filter 10 cm 10 cm. 

(a) What is the intensity of the light is transmitted?

(b) Assuming that the absorbed light is converted to heat, what is the rate of heating of the filter?  Give the correct unit.



Of course, the same argument can be applied to a horizontally oriented filter.



Two Polaroid Filters

We can put a second filter into the light path as shown below:



The second filter is called the analyser.  The analyser is turned at an angle of q to the incident beam of vertically polarised light.  Let's look straight on at the beam that has been twisted through the angle q:


Suppose the amplitude of the incident beam is A0.  From the diagram we can see that the amplitude component parallel to the transmission axis is A, and:


A = A0 cos q


In Waves 1, we saw the relationship between energy and amplitude:

We developed the theme in Waves 4 to give the relationship:


Malus' Law

The point of this is to say that if energy varies as the square of the amplitude, the power must vary as the square of the amplitude.  Hence the intensity (or irradiance), defined as power per unit area, will vary as the square of the amplitude.  We give the intensity the physics code I, and the units are Watts per square metre (W m-2).


So we can write an expression that gives us the intensity:

A = A0 cos q

If we square the relationship we get:

A2 = (A0 cos q)2


In turn this becomes:

A2 = A02 cos2 q


So we can rewrite this for the intensity:

I = I0 cos2 q


This is called Malus' Law.


Question 2

The intensity of a vertically polarised ray of light is reduced to 50 % of its original value.    Show that the analyser has to be turned through 45o to achieve this.




Experiment to Show Malus' Law

This experiment shows the change in intensity with angle.  The apparatus is set up like this:

Source: AQA Specimen practical paper


The light intensity is measured using a solar cell.  The intensity is represented by the output voltage.


In this experiment the intensity of the light is reflected in the output of the solar cell.  As you change the angle of the second polaroid (the analyser), the output of the solar cell changes.  You take readings every 20 degrees.


Data modelling shows what the graph should look like, if Malus' Law is followed:



It is quite challenging to get decent data from this experiment.  At least two repeat readings should be made and an average taken.  Also the model above assumes that the crossed polaroid filters block out all the light.  In reality they do not.


Question 3

In an experiment like the one above, the voltage given out by the sensor is 60 mV when the transmission axis of the analyser is lined up with the transmission axis of the first polariser.  When the analyser is turned through a certain angle, the sensor voltage drops to 20 mV.  What is the angle that has been turned?



Uses for Polarisation

Polarising filters have many applications.  The most obvious one is the polaroid sunglasses that cut down glare off wet surfaces.  Many drivers use them.


This is a polarising filter for a camera:



The polarising filter reduces glare that is reflected off shiny surfaces.  This can be seen on these two pictures:


(1) Note the reflections from the leaves                                (2)  No reflections from the leaves


Polarising filters also can be used to emphasise clouds:



Notice how the clouds stand out more in the right hand picture.  The polarising filter is cutting out the horizontally polarised light.  You can see how the right hand picture is also brighter, with less of a blue tint.  Both pictures were taken at the same time with the same camera.


Geologists use polarising filters in their petrographic microscopes to study minerals.  Here is a picture of a thin slice of a rock:


Image by Matt Affolter - Wikimedia Commons


The top image is taken in plane polarised light, while the bottom image is taken in cross-polarised light.  You can see how the individual crystals stand out more clearly in the bottom image.


Liquid Crystal Displays

This picture shows shows the liquid crystal display of the laptop on which I am preparing these notes.  I am holding a photographic polarising filter to the screen so that the polarisers are crossed.  You can see the result:


A light source is placed behind the screen.  A polariser with a vertical transmission axis produces vertically polarised light.  This is shown in the picture:



When there is zero voltage across the plates, the liquid crystal molecules from into a spiral shape and twist the vertically polarised light to the horizontal.  The horizontally polarised light can pass through the analyser which has a horizontal transmission axis.


When a voltage is applied across the plates, the liquid crystal molecules arrange themselves so that the vertically polarised light does not change its orientation.




Therefore the vertically polarised light does not pass through the analyser.  Therefore the display is dark.



Question 4

Explain how the figures on this calculator are visible.



LCD displays on computers are often referred to as pixels (picture elements).  1 pixel is about 0.26 mm.  The pixels have red, green, and blue filters to make the different colours.


Some liquid crystal displays have the analyser with a vertical transmission axis.


Question 5

How would you know if the display was of this type?



Many solutions twist polarised light.  Some molecules in chemistry have an isomer that twists the light to the left, while the other isomer twists the light to the right.  The molecules have the same formula, but are mirror images of each other.   They are called optical isomers.  Biological molecules are left-hand optical isomers.  Right-hand isomers cannot be processed in many biological systems.


The extent and sense (whether the light is turned to the left or right) to which a material twists polarised light can be measured with a polarimeter.



Polarisation by Reflection

In the picture below, you can see leaves reflecting light.


With a polarising filter, the reflection is much reduced.  (Yes, they are the same.)



We can conclude that the reflected light is (horizontally) polarised.  When light reflects of a non-metallic (or insulating) surface, it is partially polarised.  At a certain angle, it is fully polarised.  Metallic surfaces reflect rays in random orientations, so the reflected light is unpolarised.  So let's look what happens when unpolarised light strikes a non-metallic surface:



The reflected ray is usually not completely polarised.  However there is a case where the light is completely polarised:



Brewster Angle

This was worked out by a Scottish physicist, David Brewster (1781 - 1868).  In the diagram above, we considered the reflected ray, but ignored the transmitted ray.  Let's look at both rays from the side.


The incident ray is unpolarised.  The refracted ray is polarised vertically.  The reflected ray at this angle is fully polarised in the horizontal orientation.  The key point that is critical for this is that the angle between the refracted ray and the reflected ray is 90o.  This is called Brewster's Law, or the Brewster Angle


The angle of incidence is q1, and the angle of refraction is q2.  The angle of reflection is, of course, q1.  We can see that:


q1 + 90o + q2 = 180o


It doesn't take a genius to see that:

q1 + q2 = 90o



q2 = 90o - q1


Since we have a refraction here, we can apply Snell's Law:

n1 sin q1 = n2 sin q2


Substituting for q2:

n1 sin q1 = n2 sin (90 - q1)



sin (90 - q1) = cos q1


We can write:

n1 sin q1 = n2 cos q1




Since sin cos = tan, we can write:


Since we normally observe the horizontal polarisation in air, the usual value for n1 is 1.00.  So we can write the equation as:


tan q1 = n2


This equation describes Brewster's Law, and the angle q1 is called the Brewster Angle.


Note that in the SQA syllabus, the angle q1 is given the code ip (the incident polarising angle).


Worked example

What is the Brewster angle for glass, of which the refractive index is 1.51, when observed in air?



tan q1 = n2


tan q = 1.51

q = tan-1 1.51 = 56.5o


Question 6

The same block is now immersed in water of refractive index is 1.33.  What is the Brewster angle now?





Some crystals have different refractive indices (plural of index), depending on how the light strikes the boundary of the crystal and its orientation.  Calcite (CaCO3) is one such example.



In the diagram, unpolarised light strikes the crystal, the crystal structure polarises the light into both vertical and horizontal direction.  The vertically polarised ray refracts with a refractive index of 1.6854.  This is called the ordinary ray.  The ray that is polarised in the horizontal direction is called the extraordinary ray and has a refractive index of 1.4864.


If you observe the two rays with a vertical polarising filter, you will observe only the vertically polarised ray (the ordinary ray).  Similarly, a horizontal polarising filter will reveal only the horizontally polarised ray (the extraordinary ray).


Other crystals show the same phenomenon, including quartz, water ice, sodium nitrate, and titanium dioxide.  In some crystals, the extraordinary ray has a higher refractive index.  Calcite has the most marked birefringence.


Question 7

A ray of light strikes a crystal of calcite at an incident angle of 35o from the air.  Calculate the angle of refraction of the ordinary ray and the extraordinary ray.  The refractive indices for the ordinary and extraordinary rays are 1.6854 and 1.4864 respectively.





The property of birefringence is seen in many transparent polymers such as Perspex.  The mechanism is similar to the that described above.  The study of the way such materials behave under polarised light is called photoelasticity.   The material can alter the orientation of polarised light. When a sample of such a material is placed between crossed polaroids, the shape can be seen like this:




Question 8

Why can the sample be seen?



When the object is stressed, the orientation of the polarised light is changed further by the stress lines, to give interference patterns.  In monochromatic light this can be seen (in a very stylised way) as:


If the light is white light, the interference patterns are highly coloured as white light is a mixture of all sorts of different wavelengths.


The stress patterns are used by engineers to predict how stress lines will act in real structures.  This will help to avoid weak points on the structure, the results of which could be catstrophic.