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Polarisation

Key idea: Explain what polarisation is, why it only occurs for transverse waves, and how polarisers affect intensity (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Explain polarisation and apply Malus’ law to amplitude and intensity.

1. Definitions (Must Know)

A. Polarisation

Polarisation is the restriction of a wave’s oscillations to one plane.

B. Plane-polarised wave

A plane-polarised wave is a transverse wave whose oscillations are in a single plane.

C. Unpolarised wave

An unpolarised wave has oscillations in many planes (randomly distributed).

For light, “unpolarised” means the electric field direction varies rapidly and randomly.

D. Polariser / analyser (transmission axis)

A polariser (or analyser) has a transmission axis. It only transmits the component of a transverse wave along that axis.

2. Key Ideas (What Earns Marks)

  • Polarisation is a phenomenon associated with transverse waves only.
  • Showing that a wave can be polarised is evidence it is transverse.
  • A polariser reduces transmitted intensity because it only passes one component of the oscillation.
  • For unpolarised light passing through an ideal polariser: I = (1/2)I₀
  • For plane-polarised light through a polariser at angle θ, use Malus’ law: I = I₀ cos² θ
Link: Malus’ law

Full calculation method: Malus’ Law.

3. Detailed Explanations

A. Why polarisation only applies to transverse waves

To “restrict oscillations to one plane”, the oscillation must be able to point in different directions perpendicular to the wave travel direction.

That is true for transverse waves.

For a longitudinal wave, oscillations are along the direction of travel, so there is no “plane of oscillation” to restrict.

B. How a polariser works (idea)

For light, the oscillation is the electric field vector E.

A polariser transmits only the component of vector E along its transmission axis. That reduces the amplitude and therefore the intensity.

C. How to test whether light is polarised

Place a second polariser (the analyser) after the first and rotate it.

  • If the intensity changes with rotation, the light is polarised.
  • If the intensity stays constant for all angles, the light is unpolarised (or the setup is not sensitive enough).

4. Common Mistakes

  • Mixing up unpolarised with “weak” (unpolarised can still be intense).
  • Saying “all laser light is perfectly plane-polarised” (some lasers are strongly polarised, but it depends on the laser and optics).
  • Claiming longitudinal waves can be polarised.

5. Exam Tips

  • Use the exact phrase: “polarisation is a phenomenon associated with transverse waves”.
  • If the question gives angles between polarisers, it is usually a Malus’ law question.
  • If the question is qualitative, focus on “component along the transmission axis”.

6. Worked Examples

Modelled example 1

Why polarisation implies a transverse wave

Core

Problem

Explain why observing polarisation shows that a wave is transverse.
Study the worked solution

  1. State what polarisation does

    Method

    Polarisation restricts oscillations to one plane.

    Reason

    A selectable plane requires more than one possible oscillation direction.

    Working

    many transverse directions → one plane

  2. Connect to transverse waves

    Method

    A transverse oscillation is perpendicular to propagation and can have different orientations in that perpendicular plane.

    Reason

    A polariser can therefore select one component.

    Working

    oscillation ⊥ propagation

  3. Exclude longitudinal waves

    Method

    A longitudinal wave cannot be polarised in this way.

    Reason

    Its oscillation is fixed along the propagation direction, leaving no transverse plane to select.

    Working

    longitudinal: oscillation ∥ propagation

Guided practice 2

Unpolarised light through one polariser

About 3 min

Problem

Unpolarised light of intensity I₀ passes through one ideal polariser. Find the transmitted intensity.

Try this before viewing the solution

Hints

Hint 1: distinguish unpolarised input
Malus’ law needs a known input polarisation direction; use the average-component result for the first polariser.
View solution step by step

  1. Average over input directions

    Method

    The ideal polariser transmits half the incident intensity.

    Reason

    Unpolarised light has rapidly varying transverse directions whose mean squared projection on any fixed axis is one half.

    Working

    I = (1/2)I₀

  2. State the output state

    Method

    The transmitted light is plane-polarised along the transmission axis.

    Reason

    Only that field component passes.

    Working

    output direction = polariser axis

Common misconception 3

Two polarisers at 90°

Find and correct the mistake

Learner claim

Plane-polarised light passes through two ideal polarisers with crossed axes. A learner says each polariser halves intensity, so one quarter remains. Diagnose the claim.

Try this before viewing the solution

View solution step by step

  1. Reject repeated averaging

    Method

    The half-intensity rule is not applied again to plane-polarised light.

    Reason

    The second analyser sees a definite polarisation direction, not random directions.

    Working

    known direction → Malus’ law

  2. Use the relative angle

    Method

    The analyser axis is 90° from the incident polarisation.

    Reason

    Crossed axes have no transmitted field component along the second axis.

    Working

    cos 90° = 0

  3. Find intensity

    Method

    The ideal final intensity is zero.

    Reason

    Intensity follows the squared field projection.

    Working

    I = I₀ cos² 90° = 0

Examiner practice 4

Unpolarised light through two polarisers

4 marks

Examination question

Unpolarised light of intensity I₀ passes through a polariser and then an analyser at 60° to the first. Find the final intensity. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Treat the unpolarised input

    1 mark

    Method

    The first polariser transmits I₁ = I₀/2.

    Reason

    It averages the squared projections of random input directions.

    Working

    I₁ = (1/2)I₀

  2. Set the analyser angle

    1 mark

    Method

    The plane-polarised input to the analyser is at 60° to its axis.

    Reason

    Angles for Malus’ law are measured between polarisation and transmission directions.

    Working

    θ = 60°

  3. Apply Malus' law

    1 mark

    Method

    I₂ = I₁ cos² 60°.

    Reason

    Intensity is proportional to the square of the transmitted field component.

    Working

    I₂ = (I₀/2)(0.50)²

  4. Simplify

    1 mark

    Method

    The final intensity is I₀/8.

    Reason

    One half multiplied by one quarter is one eighth.

    Working

    I₂ = (1/8)I₀

Challenge 5

Three polarisers (step-by-step)

Minimal support

Independent transfer

Unpolarised light of intensity I₀ passes through ideal polarisers whose axes are at 0°, 30°, and 90°. Find the final intensity.

Try this before viewing the solution

Hints

Hint 1: use successive relative angles
After the first half rule, use 30° and then 90°-30°, not 90° again.
View solution step by step

  1. Pass the first polariser

    Method

    I₁ = I₀/2.

    Reason

    The incident light is unpolarised.

    Working

    I₁ = (1/2)I₀

  2. Pass the second polariser

    Method

    I₂ = 3I₀/8.

    Reason

    The second axis is 30° from the current polarisation direction.

    Working

    I₂ = (I₀/2) cos² 30° = (3/8)I₀

  3. Use the final relative angle

    Method

    The last angle is 60°, giving I₃ = 3I₀/32.

    Reason

    After the second polariser, the light is polarised at 30°, so compare 90° with that direction.

    Working

    I₃ = (3I₀/8) cos² 60° = (3/32)I₀

7. Mind Stretchers

Mind stretcher 1: Adding a third polariserExtension

Two polarisers are crossed (90° apart), so no light is transmitted.

Explain qualitatively why inserting a third polariser at 45° between them allows some light through.

Show Answer

The middle polariser changes the polarisation direction step-by-step: the first polariser sets one direction, the middle one transmits the component along 45°, and the final one transmits the component of that along its axis. Because the “rotation” happens in two stages, the final transmitted intensity is not zero.

Mind stretcher 2: Combining the “half” rule with Malus’ lawExtension

Show that unpolarised light of intensity I₀ passing through a polariser, then an analyser at angle θ, has final intensity I = 1/2 I₀ cos² θ.

Show Answer

After the first polariser, unpolarised light transmits with half intensity: I₁ = 1/2 I₀ After the analyser at angle θ: I = I₁ cos² θ = (1/2 I₀) cos² θ

Mind stretcher 3: Optional (Enrichment)Extension

A. Polarisation by reflection (glare)

Reflected light from a surface can be partially polarised, which is why some sunglasses reduce glare. This typically goes beyond what most exam questions require unless explicitly mentioned.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027