Year 12 Physics Module 7 ⏱ ~20 min 5 MC · 2 Short Answer Lesson 5 of 14

Polarisation of Light

You'll be able to calculate transmitted intensity through polariser chains.

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.

  1. When the lenses are aligned the same way, what do you see? When one is rotated 90°, what happens?
  2. Why does light from the sky become partially polarised?
  3. 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

2

Tell unpolarised from plane-polarised light

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

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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.

Unpolarised light, with E oscillating in every direction perpendicular to travel, passes a polarising filter; only the plane along the filter axis remains, at half the intensity.direction of travelUnpolarisedI₀Filteraxis ↕Plane-polarisedI₀/2Only the E component along the axispasses: a single plane remains
Figure 1, Unpolarised light: E-field oscillates in all perpendicular directions. After a polarising filter, only one plane remains
Text description
  • Seen end-on, unpolarised light of intensity I₀ has its electric field oscillating in all directions perpendicular to the direction of travel.
  • A polarising filter with a vertical transmission axis transmits only the component of E along its axis and absorbs the perpendicular component.
  • After the filter the light is plane-polarised: E oscillates in a single vertical plane, and the intensity is I₀/2.

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:

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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.

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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.

Unpolarised light through three polarisers at 0, 45 and 90 degrees: the intensity falls from I0 to I0/2, then I0/4, then I0/8, applying Malus’s law with the angle between adjacent filters.P10°P245°P390°I0I0/2I0/4I0/8Unpolarised in; ° = each filter’s axisApply Malus’s Law sequentially, using theangle between adjacent filters: I = I₀cos²θ
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
Text description
  • Unpolarised light of intensity I₀ passes three polarisers in turn.
  • P₁, axis at 0°: the unpolarised light is halved to I₀/2.
  • P₂, axis at 45°: 45° between P₁ and P₂, so I = (I₀/2)cos²45° = I₀/4.
  • P₃, axis at 90°: 45° between P₂ and P₃, so I = (I₀/4)cos²45° = I₀/8.
  • Apply Malus’s Law sequentially, using the angle between adjacent filters. Without P₂, crossed P₁ and P₃ would pass nothing.
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:

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Solve a three-polariser chain

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

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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).

  1. Calculate the intensity after the first polariser.
  2. Calculate the intensity after the second polariser.
  3. Calculate the intensity after the third polariser.
  4. 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:

5

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 $I = I_0\cos^2\theta$.
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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 skylight is most strongly polarised 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.
Light incident at Brewster’s angle: the reflected ray is completely polarised parallel to the surface, and it is perpendicular to the refracted ray; tan θ_B = n₂/n₁.Air (n₁)Glass / water (n₂)θBUnpolarisedFully polarised(parallel to surface)Refractedtan θB = n₂/n₁; reflected andrefracted rays are at 90°
Figure 3, At Brewster's angle, reflected light is completely polarised parallel to the surface; the reflected and refracted rays are perpendicular
Text description
  • Unpolarised light in air (n₁) strikes the surface of glass or water (n₂) at Brewster’s angle θ_B, measured from the normal.
  • The reflected ray leaves at the same angle θ_B. It is completely polarised parallel to the surface (drawn as dots).
  • The refracted ray bends towards the normal in the denser medium, and it is at 90° to the reflected ray.
  • tan θ_B = n₂/n₁. For n₂/n₁ = 1.5, θ_B ≈ 56° and the refraction angle is ≈ 34°.
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.

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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

  1. Unpolarised light has intensity $I_0 = 200$ W/m². It passes through a polariser (P1). Write the intensity after P1.
  2. An analyser (P2) is at 45° to P1. Calculate the intensity after P2 using Malus's Law.
  3. Rotate P2 to 90°. What is the transmitted intensity now? Explain physically why this happens.
  4. 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°.

  1. Polarising sunglasses worn while fishing from a boat
  2. An LCD screen displaying a white pixel vs a black pixel
  3. 3D cinema glasses
Evaluate7 marks

Evaluate the significance of the polarisation of light for the development of a model of light.

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Unpack the verb

Evaluate means make a judgement based on criteria, weighing what the evidence establishes against what it leaves open. This is the syllabus’s own wording, so expect it in the paper almost verbatim.

7 marks. Plan: the criterion (1), what polarisation proves (2), what it rules out (2), what it fails to settle (1), judgement (1).

Evaluate questions are marked on the weighing. An answer listing everything polarisation shows, with no counterweight, sits in the middle band however accurate it is.
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Analyse6 marks

Two polarising filters are crossed, so that no light is transmitted. Analyse how inserting a third filter between them can allow light through.

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Unpack the verb

Analyse means work the situation through stage by stage. The result looks paradoxical — adding an obstacle increases transmission — so the marks are for identifying which assumption the paradox rests on and dismantling it.

6 marks. Plan: state Malus’ Law and the crossed result (1), name the false assumption (1), apply the law twice (2), find the optimum (1), close (1).

Say early that the middle filter does not “let light through” but rotates the plane of polarisation. That is the whole answer, and everything else is arithmetic.
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