Polarisation and Malus' law
Key idea: H2 Physics lessons on progressive-wave models, standing waves, interference, diffraction and resolution.
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The core idea
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Big question: What does polarisation reveal about a wave?
Only transverse waves can be plane-polarised because their oscillations have a direction perpendicular to travel. An analyser transmits the component parallel to its axis. For initially plane-polarised light, Malus' law I = I₀cos²θ uses the angle between the polarisation and analyser axes.
Use polarisation as evidence
A transverse wave oscillates in directions perpendicular to travel. Plane-polarised light has its electric-field oscillations restricted to one such direction. A longitudinal wave cannot be plane-polarised because its oscillation is already fixed along the travel direction.
An ideal polariser selects one transverse component from unpolarised light. Averaging over all initial directions gives transmitted intensity I₀/2. That half-intensity result applies to unpolarised input, not to every first polariser question.
Check your understanding: What observation provides evidence that light is transverse?
Its transmitted intensity changes as an analyser rotates, including extinction for crossed axes; a longitudinal oscillation has no transverse direction to select.
Project amplitude before squaring
For plane-polarised input of intensity I₀, an analyser at angle θ to the polarisation axis transmits electric-field amplitude E₀ cosθ. Since intensity is proportional to field amplitude squared, Malus' law is I = I₀ cos²θ.
The angle is between the incoming polarisation direction and analyser transmission axis. At 0° transmission is maximum; at 90° it is zero for ideal components.
Check your understanding: Why is Malus' law cos²θ rather than cosθ?
The analyser projects field amplitude by cosθ, while intensity is proportional to amplitude squared.
Key ideas to keep
- Unpolarised light passing one ideal polariser has half its original intensity.
- Use cos²θ, not cosθ, for intensity.
- A 90° crossed analyser gives zero only in the ideal model.
See the reasoning
Worked example
Apply both intensity changes in order
Question: Unpolarised light of intensity 80 W m⁻² passes through a vertical polariser and then an analyser at 35°. Find the final intensity.
Step 1: Treat the unpolarised input
Why: The first ideal polariser selects an average transverse component.
Working: I₁ = 80/2 = 40 W m⁻².
Step 2: Identify the relevant angle
Why: Malus' law uses the angle between polarisation and analyser axes.
Working: θ = 35°.
Step 3: Apply Malus' law
Why: Intensity follows the squared amplitude projection.
Working: I₂ = 40 cos²35° = 26.8 W m⁻².
Answer: The final intensity is 26.8 W m⁻².
Check: It is below 40 W m⁻² but above zero, as expected for an angle between 0° and 90°.
Another worked model
Question
Unpolarised light of intensity 120 W m⁻² passes through an ideal polariser and then an analyser at 60°. Find final intensity and the analyser's amplitude factor.
Check the worked solution
The first polariser transmits 60 W m⁻². The analyser gives I = 60cos²60° = 15 W m⁻². Relative to the already polarised wave, field amplitude is multiplied by cos60° = 0.50.
Use a hint if needed
Practise with support
Try this
Plane-polarised intensity 50 W m⁻² passes through analysers at 20° and then 70° to the original direction. Find the final intensity.
Hint: Apply Malus' law successively; the second relative angle is 70° − 20°.
Check your answer
After the first analyser I₁ = 50cos²20°. The second is 50° from the new polarisation axis, so I₂ = I₁cos²50° = 50cos²20°cos²50° = 18.2 W m⁻².
Now work without the hint
Practise independently
Your turn
Unpolarised light of intensity I enters a polariser followed by an analyser at 45°. Find final intensity and field-amplitude factor relative to the light leaving the first polariser.
Check your answer
The first polariser transmits I/2. The analyser transmits cos²45° = 1/2 of that, so final intensity is I/4. Field amplitude after the analyser is cos45° = 1/√2 of the amplitude leaving the first polariser.
Avoid these traps
Common mistakes
Common mistake
A first ideal polariser leaves unpolarised intensity unchanged.
What is wrong with this reasoning?
Show better thinking
It transmits half the intensity of unpolarised light on average. Malus' law then applies between a known plane-polarisation direction and another axis.
Write for the examiner
Exam guidance
Draw the two transmission axes and mark their angle before applying Malus' law.
Exam-style practice [7 marks]
A student rotates an analyser through 180° after light has passed through a fixed polariser. Describe and explain the intensity variation, including its maxima and minima, and state what the observation shows about light.
Plan before you answer
- Name the selected oscillation direction.
- Use Malus' law over the full rotation.
- Connect the result to transverse waves.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
The first polariser produces plane-polarised light. The analyser transmits the component along its axis, so its output follows I = I₀cos²θ. Intensity is maximum when the axes are aligned, zero at 90° for ideal elements, and maximum again at 180°. Selecting a direction perpendicular to propagation is possible only for transverse oscillations, so the result supports the transverse nature of light.
Come back in three days
Check what stayed with you
Recall question 1
What becomes restricted in plane-polarised light?
Check the answer
The direction of its electric-field oscillation.
Recall question 2
What fraction of unpolarised intensity passes an ideal polariser?
Check the answer
One half.
Recall question 3
State Malus' law.
Check the answer
I = I₀cos²θ for plane-polarised input.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Topic 10 states no explicit exclusions. Topic 11(j) does not require knowledge of spectrometer structure or use. Graph interpretations distinguish time traces at one position from spatial profiles at one instant, and all small-angle equations are used only with their stated geometry.
- GCE A-Level H2 PhysicsTopic 10(i) / Topic 10(j) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027