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Malus' Law

Key idea: Use Malus’ law (I ∝ cos²θ) to calculate intensity and amplitude of plane-polarised light after a polarising filter (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. Plane-polarised wave

A plane-polarised wave is a transverse wave whose oscillations are restricted to one plane.

B. Polarising filter (polariser / analyser)

A polarising filter has a transmission axis. It only transmits the component of the wave’s electric field along this axis.

C. Angle, θ

θ is the angle between the incident polarisation direction and the polariser’s transmission axis.

D. Intensity, I (W m⁻²)

Intensity, I, is power per unit area carried by the wave.

2. Key Ideas (What Earns Marks)

  • After transmission through a polariser:
    • amplitude scales as A = A₀ cos θ
    • intensity scales as I = I₀ cos² θ
  • This is Malus’ law.
  • Quick checks:
    • θ = 0°: I = I₀ (maximum transmission)
    • θ = 90°: I = 0 (crossed polarisers)
    • θ = 45°: I = (1/2)I₀

Malus’ law curve

Transmitted intensity follows cos squared of the angle between polarisation direction and polariser axis.

Scroll across the graph to read all labels.

Transmitted intensity follows cos squared of the angle between polarisation direction and polariser axis.Transmitted intensity follows cos squared of the angle between polarisation direction and polariser axis.
At θ = 0°, I = I₀; at θ = 90°, I = 0. The curve follows I = I₀ cos² θ.
Open full-size graph
View figure data
Values for Malus’ law curve
Angle, θ (°)Series 1
01
150.933
300.75
450.5
600.25
750.067
900

3. Detailed Explanations

A. Why the cosine appears

A polariser only transmits the component of the electric field along its transmission axis.

If the incident field amplitude is A₀ and the axis makes angle θ, the transmitted component is the projection: A = A₀ cos θ

B. From amplitude to intensity

For electromagnetic waves, intensity is proportional to the square of the field amplitude: I ∝ A²

So:

I/I₀ = (A/A₀)² = (cos θ)²; I = I₀ cos² θ

4. Common Mistakes

  • Using I = I₀ cos θ instead of I = I₀ cos² θ.
  • Mixing degrees and radians on a calculator (use degrees unless specified).
  • Measuring θ from the wrong reference line (it must be between polarisation direction and transmission axis).

5. Exam Tips

  • Write the correct form immediately: I = I₀ cos² θ.
  • If the question asks for amplitude after the polariser, use A = A₀ cos θ (not the squared version).
  • If multiple polarisers are in series, apply Malus’ law step-by-step to each stage.

6. Worked Examples

Modelled example 1

Intensity after a polariser

Core

Problem

Plane-polarised light of intensity 8.0 W m⁻² passes through a polariser at 60°. Find the transmitted intensity.
Study the worked solution

  1. Choose the intensity relation

    Method

    Use I = I₀ cos² θ.

    Reason

    The incident light is already plane-polarised and the angle is relative to the transmission axis.

    Working

    I = I₀ cos² θ

  2. Project and square

    Method

    cos 60° = 0.50, so the intensity factor is 0.25.

    Reason

    Field amplitude is projected by cosine and intensity depends on amplitude squared.

    Working

    cos² 60° = (0.50)² = 0.25

  3. Calculate

    Method

    The transmitted intensity is 2.0 W m⁻².

    Reason

    One quarter of 8.0 is 2.0.

    Working

    I = 8.0(0.25) = 2.0 W m⁻²

Guided practice 2

Angle for a given fraction of intensity

About 4 min

Problem

The transmitted intensity is 0.25I₀. Find the acute angle between incident polarisation and the polariser axis.

Try this before viewing the solution

Unit: °

Hints

Hint 1: undo the square first
cos θ = square root of (I/I₀) for an acute axis angle.
View solution step by step

  1. Form the intensity fraction

    Method

    cos² θ = 0.25.

    Reason

    Malus’ law directly relates the transmitted fraction to the squared cosine.

    Working

    I/I₀ = cos² θ

  2. Undo the square

    Method

    cos θ = 0.50.

    Reason

    The angle is taken in the physical 0° to 90° range.

    Working

    square root of 0.25 = 0.50

  3. Find the angle

    Method

    θ = 60°.

    Reason

    cos⁻¹ (0.50) = 60°.

    Working

    θ = 60°

Common misconception 3

Amplitude after a polariser

Find and correct the mistake

Learner claim

The transmitted intensity is 0.36I₀. A learner says the field amplitude is therefore 0.36A₀. Diagnose the claim.

Try this before viewing the solution

View solution step by step

  1. Relate the ratios

    Method

    I/I₀ = (A/A₀)².

    Reason

    Electromagnetic intensity depends on the square of field amplitude.

    Working

    0.36 = (A/A₀)²

  2. Recover amplitude

    Method

    A/A₀ = 0.60.

    Reason

    Amplitude is the positive magnitude of the field projection.

    Working

    A/A₀ = square root of 0.36 = 0.60

Examiner practice 4

Angle for 75% transmission

3 marks

Examination question

Plane-polarised light is transmitted at 0.75I₀. Find the acute polariser angle. [3 marks]

Try this before viewing the solution

Unit: °

View solution step by step

  1. Apply Malus' law

    1 mark

    Method

    cos² θ = 0.75.

    Reason

    The given quantity is an intensity fraction.

    Working

    I/I₀ = cos² θ

  2. Find the cosine

    1 mark

    Method

    cos θ = 0.866.

    Reason

    Take the positive square root for an acute angle.

    Working

    square root of 0.75 = 0.866

  3. Find the angle

    1 mark

    Method

    θ ≈ 30°.

    Reason

    Use inverse cosine in degree mode.

    Working

    θ = cos⁻¹ (0.866) ≈ 30°

Challenge 5

Find the incident intensity

Minimal support

Independent transfer

At 45°, the transmitted intensity is 2.0 W m⁻². Find the incident plane-polarised intensity.

Try this before viewing the solution

Unit: W m^-2

Hints

Hint 1: reverse Malus' law
Rearrange I = I₀ cos² θ for I₀.
View solution step by step

  1. Evaluate the transmission factor

    Method

    cos² 45° = 1/2.

    Reason

    The 45° field projection has magnitude 1/square root of 2.

    Working

    I = I₀/2

  2. Solve backwards

    Method

    The incident intensity is 4.0 W m⁻².

    Reason

    The measured transmitted value is half of the incident value.

    Working

    I₀ = 2I = 4.0 W m⁻²

7. Mind Stretchers

Mind stretcher 1: Two polarisers in seriesExtension

Plane-polarised light of intensity I₀ passes through two polarisers. The first is aligned with the incident polarisation. The second is at 30° to the first. Find the final intensity.

Show Answer

After the first polariser (aligned): I₁ = I₀.

After the second polariser: I₂ = I₁ cos² 30° = I₀((square root of 3)/2)² = (3/4)I₀

Mind stretcher 2: Adding a third polariserExtension

Plane-polarised light of intensity I₀ passes through three ideal polarisers with axes at 0°, 45°, and 90°. Find the final intensity.

Show Answer

First polariser is aligned: I₁ = I₀.

Second polariser at 45°: I₂ = I₁ cos² 45° = I₀(1/(square root of 2))² = (1/2)I₀

Third polariser is 45° relative to the second: I₃ = I₂ cos² 45° = ((1/2)I₀)(1/2) = (1/4)I₀

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