Year 12 Physics Module 7 ⏱ ~40 min 5 MC · 2 Short Answer Lesson 5 of 14

Polarisation of Light

In 1929 Edwin Land patented the first practical sheet polariser, aligning tiny iodo-quinine sulphate crystals in a stretched polymer film. He founded the Polaroid Corporation in 1937, and sheet polarisers went on to be manufactured at large scale for sunglasses, camera filters and glare-reducing optics. The technology works because light is a transverse wave: only transverse waves can be restricted to a single plane of oscillation. Longitudinal waves, like sound, cannot be polarised at all. Brewster's angle, $\theta_B = \arctan n$, is the incidence angle at which the reflected light is completely plane-polarised. For a glass of refractive index 1.5 that is about 56°; the exact value depends on the glass.

Today's hook: In 1929 Edwin Land at Harvard made the first sheet polariser by stretching polymer films containing iodine crystals. When two such sheets are crossed at 90°, zero light passes through, yet if you slip a third sheet at 45° between them, light reappears. Why does adding an obstacle let light through?

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Orient and predict

Predict what crossed sunglasses do, then set your goals and vocabulary for polarised light.

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response Master Mastery tasks Exam Exam-style questions

Before you read, predict

You hold two pairs of polarising sunglasses in front of you, one behind the other, and look through both at a bright screen.

When the lenses are aligned the same way, what do you see? When one is rotated 90°, what happens?
Why does light from the sky become partially polarised?
Can sound waves be polarised? Why or why not?

Write your predictions before reading on, you will revisit them at the end.

Warm-up, which type of wave can be polarised?

Learning Intentions

goals

Know, Polarisation Concepts

Polarisation as a property of transverse waves
Unpolarised vs plane-polarised light
Polarising filters transmit one plane of oscillation

Understand, Malus's Law

$I = I_0 \cos^2\theta$ for plane-polarised incident light
Crossed polarisers transmit zero intensity
A middle polariser at 45° lets some light through, counterintuitive but explainable

Can Do, Apply Polarisation

Calculate transmitted intensity through polariser chains
Explain applications: sunglasses, LCDs, 3D cinema
Relate polarisation to the transverse wave nature of light

Scan these before reading

vocab

PolarisationThe restriction of the oscillations of a transverse wave to a single plane.

Plane-polarised lightLight in which the electric field oscillates in only one direction perpendicular to propagation.

Polarising filterA material that transmits only one plane of polarisation and absorbs the perpendicular component.

Malus's Law$I = I_0 \cos^2\theta$: transmitted intensity through an analyser depends on the square of the cosine of the angle between the polariser axes.

Brewster's angleThe angle of incidence at which reflected light is completely polarised: $\tan\theta_B = n_2/n_1$.

Cross-lesson links: L04 confirmed diffraction, wave bending around obstacles. L05 confirms transverse polarisation: only transverse waves can be restricted to a single oscillation plane, which Edwin Land's 1929 sheet polariser took from the laboratory to mass manufacture. Together, L01–L05 build an airtight classical wave model. From L09 onward, phenomena like atomic spectra and the photoelectric effect show where that model fails, requiring a quantum description.

Core Content

Tell unpolarised from plane-polarised light

See why only transverse waves can be polarised, and what one ideal filter does to unpolarised light.

What Is Polarisation?

+5 XP

Why only transverse waves can be polarised

Put on polarising sunglasses and look at the glare bouncing off the road on a sunny day. Rotate the lenses 90°: the glare brightens dramatically, then vanishes again as you complete the rotation. The glare is not just bright, it is vibrating in a preferred direction, and the lens either aligns with or blocks that direction. This selective transmission is polarisation. In an unpolarised electromagnetic wave, the electric field oscillates in all directions perpendicular to the direction of propagation; in plane-polarised light, it oscillates in only one direction.

Polarisation is possible only for transverse waves. In a longitudinal wave (like sound in air), oscillations occur parallel to the direction of propagation, there is no "perpendicular plane" to restrict. This is why sound cannot be polarised, but light can.

Figure 1, Unpolarised light: E-field oscillates in all perpendicular directions. After a polarising filter, only one plane remains

When unpolarised light passes through a polarising filter (such as Polaroid), the filter absorbs the component of the electric field perpendicular to its transmission axis and transmits the parallel component. The transmitted light is plane-polarised, with intensity equal to half the incident intensity (for an ideal polariser):

Unpolarised through one polariser

$I_{\text{transmitted}} = \dfrac{1}{2}I_0$

Stop & Check

Explain why polarisation is impossible for sound waves travelling through air, but possible for light. What fundamental difference between these waves makes polarisation possible for one but not the other?

Polarisation is the restriction of a transverse wave's electric field oscillations to a single plane. Only transverse waves can be polarised, longitudinal waves (sound) cannot because oscillations are parallel to propagation. A polarising filter transmits the parallel E-field component only; transmitted intensity = $I_0/2$ for unpolarised incident light.

Pause, copy the highlighted definition into your book before moving on.

Unpolarised light of intensity $I_0$ passes through a single ideal polarising filter. The transmitted intensity is:

Apply Malus's Law to an analyser

Use $I = I_0\cos^2\theta$ for the key angles, then handle the crossed-plus-middle-filter case.

Malus's Law

+5 XP

How intensity changes when polarisers cross

We just saw that a single polariser halves the intensity of unpolarised light. That raises a question: if you add a second polariser at angle $\theta$, how does intensity depend on that angle? This card answers it → Malus's Law $I = I_0\cos^2\theta$.

When already plane-polarised light passes through a second polarising filter (an analyser), the transmitted intensity depends on the angle $\theta$ between the transmission axes of the two filters.

Malus's Law

$I = I_0 \cos^2\theta$

Key cases:

$\theta = 0°$: filters aligned, maximum transmission ($I = I_0$)
$\theta = 45°$: $I = I_0/2$ (half the intensity)
$\theta = 90°$: filters crossed, zero transmission ($I = 0$)

For unpolarised light passing through two polarisers at angle $\theta$ to each other, the combined result is:

Unpolarised through polariser then analyser

$I = \dfrac{1}{2}I_0 \cos^2\theta$

When unpolarised light passes through two crossed polarisers ($\theta = 90°$), no light emerges. But if a third polariser is inserted between them at 45°, some light does emerge. The first polariser reduces intensity to $I_0/2$ and polarises the light. The middle polariser at 45° transmits $(I_0/2)\cos^2 45° = I_0/4$. The final polariser at 45° to the middle transmits $(I_0/4)\cos^2 45° = I_0/8$. This counterintuitive result is a favourite HSC exam question.

Figure 2, Applying Malus's Law sequentially through a three-polariser chain; the middle polariser re-polarises light so some can pass the final filter

HSC Tip, The Three-Polariser Trap

Two polarisers at 90° with a third in between do not block all light. Apply Malus's Law sequentially, using the angle between adjacent polarisers, not the total angle. Also remember: the first polariser always halves intensity of unpolarised light.

Stop & Check

Unpolarised light of intensity $I_0$ passes through two polarisers whose transmission axes are at 30° to each other. Calculate the final intensity. Then a third polariser at 30° to the second is inserted between them. Calculate the new final intensity.

Malus's Law: $I = I_0\cos^2\theta$ (plane-polarised incident light, $\theta$ = angle between axes). Unpolarised through two polarisers: $I = \tfrac{1}{2}I_0\cos^2\theta$. Three-polariser chain at 45°–45°: final $I = I_0/8$. Apply the law sequentially, using the angle between adjacent polarisers.

Add the highlighted law and key cases to your notes before the check below.

Plane-polarised light of intensity 40 W/m² passes through an analyser at 60° to the polarisation direction. The transmitted intensity is:

Solve a three-polariser chain

Work a full worked example filter by filter, and see why a middle polariser can raise the transmitted intensity.

Worked Example, Malus's Law and Polariser Chains

+5 XP

Intensity through multiple polarisers

Problem

Unpolarised light of intensity 80 W/m² passes through three polarising filters. The first has its transmission axis vertical. The second is rotated 30° from vertical. The third is rotated 60° from vertical (30° from the second).

Calculate the intensity after the first polariser.
Calculate the intensity after the second polariser.
Calculate the intensity after the third polariser.
What would the intensity be after the third polariser if the second polariser were removed?

Step 1, After first polariser

$I_1 = \dfrac{1}{2}I_0 = \dfrac{1}{2}(80) = $ 40 W/m²

The light is now plane-polarised vertically.

Step 2, After second polariser

$I_2 = I_1 \cos^2 30° = 40 \times (\sqrt{3}/2)^2 = 40 \times 0.75 = $ 30 W/m²

Step 3, After third polariser

The angle between the second and third polarisers is 30°.

$I_3 = I_2 \cos^2 30° = 30 \times 0.75 = $ 22.5 W/m²

Step 4, Without the second polariser

The angle between first and third is 60°.

$I_3' = I_1 \cos^2 60° = 40 \times (0.5)^2 = 40 \times 0.25 = $ 10 W/m²

The middle polariser increases the transmitted intensity from 10 to 22.5 W/m²! It provides an intermediate step, preventing the stronger cancellation that occurs when jumping directly from 0° to 60°.

Stop & Check

Unpolarised light passes through two crossed polarisers, so no light emerges. A third polariser at 45° is inserted between them. Show that the final transmitted intensity is $I_0/8$, where $I_0$ is the initial intensity.

Multi-polariser method: first polariser gives $I_0/2$ from unpolarised light; then apply $I = I_\text{prev}\cos^2\theta$ for each subsequent filter using the angle between adjacent axes. A middle polariser can increase total transmission by bridging a large angle gap.

Add the highlighted multi-polariser method to your notes before the check below.

Two crossed polarisers block all light. A third polariser is inserted between them at 45°. The final transmitted intensity as a fraction of the original unpolarised intensity is:

Explain natural sources and applications

Link reflection, scattering and birefringence to Brewster's angle, sunglasses, LCDs and 3D cinema.

Beyond the syllabus. Polarisation evidence and quantitative Malus's Law — including multiple-polariser chains — are core. Brewster's Law calculations, stress birefringence and LCD/3D display engineering are extension — exam marks come from $I = I_0\cos^2\theta$.

Natural Sources and Applications of Polarisation

+5 XP

Reflection, scattering, birefringence, and technology

We just saw how to calculate intensity through polariser chains. That raises a question: where does naturally polarised light come from, and how is polarisation exploited in everyday technology? This card answers it → reflection, scattering, and applications from sunglasses to LCD screens.

Polarised light occurs naturally and is exploited in many technologies. Recognising the mechanism behind each source is essential for HSC questions.

Natural sources of polarised light:

Reflection: Light reflected from non-metallic surfaces (water, glass, roads) is partially polarised parallel to the surface. At Brewster's angle, reflected light is completely polarised. $\tan\theta_B = n_2/n_1$.
Scattering: Sunlight scattered by air molecules becomes partially polarised. This is why polarising sunglasses reduce glare from the sky, and why the sky is brightest at 90° to the Sun.
Birefringence: Certain crystals (like calcite) split unpolarised light into two polarised beams travelling at different speeds.

Applications of polarisation:

Polarising sunglasses: Block horizontally polarised glare from reflective surfaces (water, wet roads).
LCD screens: Use crossed polarisers with liquid crystals that rotate polarisation when voltage is applied, controlling which pixels emit light.
Photoelasticity: Stress analysis in transparent materials, stressed regions rotate polarisation, revealing stress patterns in colour.
3D cinema: Two images projected with orthogonal polarisations; each eye sees only one image through matching polarised glasses.

Figure 3, At Brewster's angle, reflected light is completely polarised parallel to the surface; the reflected and refracted rays are perpendicular

Stop & Check

A photographer uses a polarising filter to reduce glare from water. Explain why rotating the filter changes the amount of glare removed, and in which orientation is the filter most effective?

Reflected light from non-metallic surfaces is partially polarised parallel to the surface; at Brewster's angle ($\tan\theta_B = n_2/n_1$) it is fully polarised. Polarising sunglasses (vertical transmission axis) block horizontal glare. LCDs use liquid crystals between crossed polarisers to control pixel brightness via voltage.

Pause, write the highlighted applications and Brewster's angle formula into your book before the check below.

Polarising sunglasses work best when the transmission axis is vertical, to block horizontally polarised reflected glare.

LCD screens project two images with different colours to create the 3D effect.

At Brewster's angle, the reflected and refracted rays are perpendicular to each other.

Apply and consolidate

Run your own polariser calculations, then explain the mechanism behind three real applications.

Activity 1, Malus's Law Calculations

ApplyBand 4

Apply the law step-by-step to a polariser chain

Unpolarised light has intensity $I_0 = 200$ W/m². It passes through a polariser (P1). Write the intensity after P1.
An analyser (P2) is at 45° to P1. Calculate the intensity after P2 using Malus's Law.
Rotate P2 to 90°. What is the transmitted intensity now? Explain physically why this happens.
A third polariser at 45° to P1 is inserted between P1 and P2 (which is at 90° to P1). Calculate the intensity after each filter and explain why light now passes through P2.

Activity 2, Polarisation Applications Analysis

UnderstandBand 5

Explain the mechanism behind each real-world use

For each application below, identify: (i) the source of polarised light, (ii) how the polarisation is exploited, and (iii) what would happen if the polariser were rotated 90°.

Polarising sunglasses worn while fishing from a boat
An LCD screen displaying a white pixel vs a black pixel
3D cinema glasses

Independent practice

Show what you can do without prompts

Answer a shuffled question-bank set, then write two HSC-style responses on polarisation and Malus's Law.

Quick recall, polarisation of light

+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 7 marks

+5 XP

ApplyBand 4(3 marks) 1. Unpolarised light of intensity 100 W/m² passes through two polarising filters whose transmission axes are at 40° to each other. (a) Calculate the intensity after the first filter. (b) Calculate the final intensity after the second filter. (c) A third filter is inserted between the first two, at 20° to the first. Calculate the new final intensity and explain why it differs from part (b).

1 mark: correct intensity after first filter (50 W/m²) · 1 mark: correct final intensity with Malus's Law · 1 mark: correct new final intensity and clear explanation

EvaluateBand 6(4 marks) 2. (a) Explain why polarisation provides evidence that light is a transverse wave and not a longitudinal wave. (b) A student claims that two crossed polarisers will always block all light, regardless of what is inserted between them. Evaluate this claim. Support your answer with a worked calculation using a polariser inserted at 45° between the crossed pair, with initial unpolarised intensity $I_0$.

1 mark: transverse nature explanation (oscillations perpendicular to propagation, restricting to one plane is possible) · 1 mark: correct identification that the claim is false · 1 mark: correct calculation showing $I = I_0/8$ · 1 mark: clear physical explanation of re-polarisation

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1 (a): After first filter: $I_1 = I_0/2 = 100/2 = 50$ W/m² (1 mark).

Q1 (b): $I_2 = I_1\cos^2 40° = 50 \times (0.766)^2 = 50 \times 0.587 = 29.3$ W/m² (1 mark).

Q1 (c): With middle at 20° to first: after middle: $I_m = 50\cos^2 20° = 50 \times 0.883 = 44.2$ W/m². The second original filter is at 40° to the first, so it is 20° to the middle filter. After final: $I_f = 44.2\cos^2 20° = 44.2 \times 0.883 = 39.0$ W/m². This is higher than 29.3 W/m² because the middle polariser provides an intermediate step, reducing the total angular jump and allowing more light through at each stage (1 mark).

Q2 (a): Transverse waves have oscillations perpendicular to the direction of propagation. A polarising filter restricts these oscillations to one plane, this is only possible when there is a perpendicular plane to restrict. Longitudinal waves oscillate parallel to propagation and have no "perpendicular direction", so cannot be polarised. Since light can be polarised, it must be transverse (1 mark).

Q2 (b): The claim is incorrect. A middle polariser at 45° allows some light through (1 mark). After first polariser (vertical): $I_1 = I_0/2$. After middle (at 45° to first): $I_m = (I_0/2)\cos^2 45° = I_0/4$. After crossed polariser (at 45° to middle): $I_f = (I_0/4)\cos^2 45° = I_0/8$ (1 mark). The middle polariser "re-polarises" the light at 45°, so the final polariser at 90° to the original actually sees light at only 45° to its axis. Some component of this re-polarised light passes through (1 mark).

Final step

Retrieve, reflect and finish

Check what stuck, revisit your opening predictions and record the rule you will reuse.

Check what actually stuck

Take the full module quiz

quiz

A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.

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How did your thinking change?

Look back at your Think First answers:

Did you predict that aligned polarising sunglasses let light through, while crossed ones block it completely? Malus's Law gives $I = I_0\cos^2 90° = 0$.
Did you explain that sunlight scattered by air molecules becomes partially polarised perpendicular to the scattering plane? This is why polarising sunglasses are most effective when looking at right angles to the Sun.
Did you correctly state that sound waves in air cannot be polarised because they are longitudinal?

The hook answer: inserting a third polariser at 45° between two crossed polarisers allows light through, the middle sheet re-polarises the light at 45°, so the final filter is only 45° from the new polarisation direction. Final intensity = $I_0/8$.

The deeper historical connection: Edwin Land's 1929 sheet polariser, aligned iodo-quinine sulphate crystals in a stretched polymer film and manufactured at scale by the Polaroid Corporation he founded in 1937, is the technological embodiment of Malus's Law and of Brewster's condition, about 56° for a glass of refractive index 1.5. Land's work confirmed that light is a transverse wave at an industrial, everyday scale.

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