Controls
Scene
Readings
- I₀ after the first polariser
- 50.0%
- Analyser angle θ
- 30°
- I / I₀
- 0.750
- Final intensity I
- 37.5%
How to use this simulation
- Start in Malus’s law. Unpolarised light leaves the lamp, and the first filter keeps half of it: the meter reads 50 % of the source.
- Drag the analyser angle θ. At θ = 0° the second filter agrees with the first and everything passes. At θ = 90° the filters are crossed and the detector goes dark. At θ = 30° the ratio I / I₀ is 0.75.
- Watch the wave. Between the filters it vibrates in one plane; the little circle under it shows that plane end-on. Past the analyser the vibration turns to the new axis and shrinks.
- Tick the middle filter and put it at 45° while the analyser stays at 90°. Light comes back through two crossed filters. Slide ψ from 0° to 90° and watch the output rise and fall.
- Switch to Brewster angle. Slide the angle of incidence and watch the blue dots (vibration across the page) and amber arrows (vibration in the page) on the reflected ray. At one angle the arrows vanish.
- Tick "Snap to the Brewster angle", then wear the sunglasses. The reflected ray disappears behind the lens, and the reading says almost nothing gets through.
Light vibrates sideways
Shake the end of a long rope up and down and a wave runs along it. Thread the rope through a narrow vertical slit in a fence and the wave passes. Turn the slit horizontal and the wave stops dead. The rope vibrates across the direction it travels, so the slit can choose which direction of vibration gets through.
Light is the same kind of wave, with an electric field vibrating across the direction the light travels. This is called a transverse wave, and polarisation is the proof. Sound in air is a longitudinal wave, its air particles swinging along the direction of travel, so no slit of any orientation can polarise it.
A light bulb or the Sun sends out waves vibrating in every sideways direction at once, changing too fast to follow. That is unpolarised light. In the animation, the first stretch of beam shows six vibration directions together. Plane-polarised light has one direction only.
A polariser, and why it halves the light
A polaroid sheet is plastic whose long molecules have been stretched into parallel lines. It absorbs the part of the vibration along those molecules and passes the part across them. The direction it passes is called the transmission axis; in the animation it is drawn as amber lines on the filter.
Every unpolarised wave can be split into a part along the axis and a part across it. Averaged over all directions, half the energy lands in each. So an ideal polariser passes 50 % of unpolarised light and absorbs the rest. What comes out is polarised along the axis, with the intensity we call I₀. This is a rule to remember: unpolarised in, I₀ = ½ I out.
Where does the other half go? It is absorbed as heat. That is why a second sheet is called an analyser when it is used to find out whether light is polarised and in which direction.
Malus’s law: I = I₀ cos²θ
Now put an analyser behind the polariser, with its axis turned by an angle θ. The polarised light has amplitude A along the first axis. The analyser passes only the component along its own axis, A cos θ. Intensity goes as amplitude squared, so the transmitted intensity is I = I₀ cos²θ, where I₀ is the intensity reaching the analyser. This is Malus’s law.
Take the defaults of the animation. The source gives 100 %, the first filter leaves I₀ = 50 %, and with θ = 30° we get I / I₀ = cos²30° = 0.75, so the final reading is 37.5 % of the source.
The graph below the beam is the shape of cos²: a smooth hump that peaks at 0° and 180° and touches zero at 90°. Notice that it repeats every 180°, not 360°, because turning an axis through half a turn gives the same axis.
I = I₀ cos²θMalus’s law, I₀ the polarised intensity reaching the analyser
I₀ = ½ I(unpolarised)After the first polariser
A = A₀ cos θAmplitude after the analyser
| Angle θ (°) | cos²θ | I / I₀ (%) | I as % of the source |
|---|---|---|---|
| 0 | 1.000 | 100.0 | 50.0 |
| 30 | 0.750 | 75.0 | 37.5 |
| 45 | 0.500 | 50.0 | 25.0 |
| 60 | 0.250 | 25.0 | 12.5 |
| 90 | 0.000 | 0.0 | 0.0 |
The three-filter puzzle
Two filters crossed at 90° block all the light. Now slip a third filter between them, with its axis at 45°. Surprisingly, light comes through. The middle filter is not letting more light by; it is rotating the vibration in two steps, and each step loses less than a single 90° jump would.
Work it out with Malus’s law twice. After the first filter, I₀. After the 45° filter, I₀ cos²45° = ½ I₀. After the analyser, which is 45° from the middle filter, ½ I₀ cos²45° = ¼ I₀. So the detector reads 25 % of I₀, or 12.5 % of the source.
The table varies the middle angle ψ while the analyser stays crossed. The output is largest at 45° and zero at 0° and 90°, because then the middle filter agrees with one of the others and the other pair is still crossed.
I = I₀ cos²ψ cos²(90° − ψ)Crossed outer filters, middle filter at ψ
I = (I₀ / 4) sin²2ψSame result, maximum at ψ = 45°
| Middle angle ψ (°) | I / I₀ (%) | I as % of the source |
|---|---|---|
| 0 | 0.0 | 0.00 |
| 15 | 6.2 | 3.12 |
| 30 | 18.8 | 9.38 |
| 45 | 25.0 | 12.50 |
| 60 | 18.8 | 9.38 |
| 75 | 6.2 | 3.12 |
| 90 | 0.0 | 0.00 |
Polarisation by reflection: the Brewster angle
Light reflecting from glass, water or a road is partly polarised. Light vibrating along the surface (across the plane of incidence) reflects strongly; light vibrating in the plane of incidence reflects weakly. In the animation, the first kind is the blue dots and the second is the amber arrows.
At one special angle of incidence, the in-plane vibration is not reflected at all. The reflected ray is then completely plane-polarised, with its vibration along the surface. This angle is the Brewster angle, and David Brewster’s law gives it as tan θB = n, where n is the refractive index of the surface.
For glass of n = 1.50, θB = 56.31°. Switch on the snap and see the green right-angle mark: at this angle the reflected ray and the refracted ray are exactly 90° apart. The angle of refraction is 33.69°, and 56.31° + 33.69° = 90°. That right angle is the cause. The in-plane vibration of the refracted light would have to drive the glass electrons along the direction of the reflected ray, and a vibrating charge sends no light along its own axis.
Only a fraction of the s-component reflects: 14.8 % at θB for glass. So the reflected beam is fully polarised but weak, 7.4 % of the incident light. The refracted beam is bright but only partly polarised.
tan θB = nBrewster’s law, light going from air into a medium of index n
θB + r = 90°Reflected and refracted rays are perpendicular
R₀ = ((n − 1) / (n + 1))²Reflectance at normal incidence
| Refractive index n | Brewster angle θB (°) | Angle of refraction r (°) | Sun elevation for glare (°) |
|---|---|---|---|
| 1.33 | 53.1 | 36.9 | 36.9 |
| 1.50 | 56.3 | 33.7 | 33.7 |
| 1.60 | 58.0 | 32.0 | 32.0 |
| 2.00 | 63.4 | 26.6 | 26.6 |
How much reflects: Fresnel’s numbers
The table lists the reflectance of each component for glass of n = 1.50. Rs is the part vibrating across the plane of incidence and Rp the part in it. At normal incidence both are 4.0 %, the familiar few percent that makes a shop window show your reflection.
As the angle grows, Rs climbs steadily while Rp first falls to zero at the Brewster angle and then rises again. Between them is the polarisation of the reflected beam: 100 % at θB, falling off on both sides. At grazing incidence, almost everything reflects and the beam is nearly unpolarised again.
| Angle of incidence (°) | Rs (%) | Rp (%) | Reflected beam polarised (%) | Reflected share of the incident (%) |
|---|---|---|---|---|
| 0.0 | 4.0 | 4.0 | 0 | 4.0 |
| 30.0 | 5.8 | 2.5 | 39 | 4.2 |
| 45.0 | 9.2 | 0.8 | 83 | 5.0 |
| 56.3 | 14.8 | 0.0 | 100 | 7.4 |
| 70.0 | 30.0 | 4.2 | 75 | 17.1 |
| 85.0 | 73.2 | 49.3 | 20 | 61.3 |
Polaroid sunglasses
Sunlight bouncing off a lake, a wet road or a car bonnet comes mostly from a flat surface, so its vibration lies mostly along the surface. For a horizontal surface that means horizontal. Polaroid sunglasses are made with the transmission axis vertical, so they absorb that glare and pass the rest of the scene.
For water, n = 1.33, the Brewster angle is 53.1°, which corresponds to the Sun about 36.9° above the horizon. Then the glare is totally horizontal, and the sunglasses remove 100 % of it (the animation reads 0.0 % of the incident light getting through). Even at 40° the reflected light is 76 % polarised, and the lenses remove 88 % of the glare.
Two fun tests. Tilt your head by 90° while looking at a puddle and the glare returns, because the axis is now horizontal. And look at a phone or laptop screen through the lenses: many LCD screens emit polarised light, and the screen may fade or go dark as you turn your head. Ordinary tinted sunglasses cannot do any of this, because they dim all vibrations equally.
Solved problems
Five problems in the exam style, solved with the same functions as the animation.
Problem 1: Malus’s law
Plane-polarised light of intensity 80 W/m² falls on an analyser whose axis makes 60° with the polarisation. Find the transmitted intensity.
I = I₀ cos²θ = 80 × cos²60° = 80 × 0.25 = 20.0 W/m².
Problem 2: unpolarised light
Unpolarised light of 200 W/m² passes through two polaroids whose axes are 45° apart. What emerges?
The first polaroid halves it to 100.0 W/m². Malus’s law then gives 100.0 × cos²45° = 50.0 W/m².
Problem 3: finding n from the Brewster angle
The Brewster angle of a transparent material is 60°. Find its refractive index and the angle of refraction at that incidence.
n = tan θB = tan 60° = 1.732. The angle of refraction is 90° − 60° = 30.0°.
Problem 4: glass at the Brewster angle
Light in air strikes glass of n = 1.50. Find the incidence at which the reflected ray is fully polarised, and the angle between the reflected and refracted rays.
tan θB = 1.50, so θB = 56.31°. The refracted ray is at r = 33.69°. The angle between the reflected and refracted rays is 180° − 56.31° − 33.69° = 90°.
Problem 5: three polaroids
Unpolarised light of 160 W/m² goes through a polariser, then a filter at 45° to it, then an analyser at 90° to the first. Find the intensity after each filter.
After the first: 80 W/m². After the second: 80 × cos²45° = 40 W/m². After the third, which is 45° from the second: 40 × cos²45° = 20 W/m².
Mistakes that cost marks
Check this list before the exam. Each one has cost someone a mark.
- Applying Malus’s law to unpolarised light. Halve it first. The cos² only applies to light that is already polarised.
- Halving twice. If the question says the incident light is plane-polarised, there is no halving at all.
- Using the angle between the two filters in the wrong place. In a chain, each cos² uses the angle between that filter and the one before it.
- Saying that crossed polarisers stop the light because the first one removes it. The first one only polarises. It is the second at 90° that blocks it.
- Writing tan θB = 1/n. It is tan θB = n for light going from air into the material.
- Thinking the Brewster angle makes the refracted ray polarised. The refracted ray is only partly polarised. The reflected ray is the fully polarised one.
- Claiming that sound can be polarised. Only transverse waves can.
Where else you meet it
Liquid crystal displays work by turning the plane of polarisation with an electric field between two crossed polarisers. Engineers lay a clear plastic model between polarisers and read the stress from the coloured fringes. Some animals use the polarisation of the sky to find their way. And a camera polarising filter is just an analyser that you can rotate.
If you want to see what happens when two waves overlap, rather than when one is filtered, try the wave-interference animation. For light bending as it enters glass, the refraction page has the Snell’s law side of the Brewster story.
Frequently asked questions
What is polarisation of light?
Polarisation means restricting the sideways vibration of a light wave to one direction. Ordinary light vibrates in all sideways directions; plane-polarised light vibrates in one plane only. It proves that light is a transverse wave.
What is Malus’s law?
The intensity of plane-polarised light after an analyser is I = I₀ cos²θ, where I₀ is the intensity reaching it and θ is the angle between the light’s polarisation and the analyser’s axis.
Why does an ideal polariser pass half of unpolarised light?
Unpolarised light has vibrations in all directions equally. Averaged over directions, cos²θ is one half, so half the intensity passes through.
What is the Brewster angle, and what is its formula?
It is the angle of incidence at which the reflected light is completely plane-polarised. Brewster’s law gives tan θB = n, with n the refractive index of the reflecting material. At this angle the reflected and refracted rays are at 90°.
How do polaroid sunglasses reduce glare?
Light reflected from water or roads is mostly polarised along the surface, which is horizontal. The lenses have a vertical transmission axis, so they absorb that horizontal vibration and let the rest of the scene through.
Can sound waves be polarised?
No. Sound in air is longitudinal, so its vibration is already along the direction of travel. Polarisation only works for transverse waves such as light.
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