recall and use Malus’s law ( I = I0 cos2θ ) to calculate the intensity of a plane-polarised electromagnetic wave after transmission through a polarising filter or a series of polarising filters (calculation of the effect of a polarising filter on the

Polarisation – Cambridge A‑Level Physics 9702 (Syllabus 7.5)

Learning objectives (AO1 & AO2)

• Recall the definition of polarisation and the fact that it is a property of transverse electromagnetic waves.
• State the visible‑light wavelength range (≈ 400 nm – 700 nm) to place polarisation in the electromagnetic‑spectrum context.
• Recall and apply Malus’s law to find the intensity after one polarising filter.
• Apply the “half‑intensity” rule for an unpolarised incident beam and then use Malus’s law for any further filters.
• Calculate the transmitted intensity for a series of polarising filters by successive use of Malus’s law.
• Identify at least three practical applications of polarisation (e.g. LCD screens, polarising sunglasses, stress‑analysis of plastics).
• Discuss the effect of measurement uncertainty in the angle on the calculated intensity.

1. What is polarisation?

• Electromagnetic (EM) waves consist of mutually perpendicular electric (E) and magnetic (B) fields that oscillate perpendicular to the direction of propagation – they are transverse waves.
• In an unpolarised beam the direction of the electric‑field vector changes randomly with time.
• In a plane‑polarised (or linearly polarised) beam the electric field oscillates in a single, fixed direction.
• Visible light occupies the wavelength interval 400 nm – 700 nm; the same polarisation concepts apply across the whole EM spectrum.

2. Polarising filters (analysers)

• A polarising filter has a fixed transmission axis. Only the component of the incident electric field that is parallel to this axis is transmitted; the orthogonal component is absorbed or reflected.
• If the incident light is already plane‑polarised, the transmitted intensity depends on the angle θ between the incident polarisation direction and the filter’s transmission axis.

3. Unpolarised light and the first polariser

When an unpolarised beam encounters the first polariser, the transmitted intensity is reduced to one‑half of the incident intensity because only the average of the two orthogonal components is transmitted:

I₁ = ½ I₀

From this point onward the beam is plane‑polarised, so Malus’s law can be applied to any additional filters.

4. Malus’s law

For a plane‑polarised wave of intensity I₀ incident on a single polarising filter, the transmitted intensity I is

I = I₀ cos²θ

where θ is the angle between the electric‑field direction of the incoming wave and the filter’s transmission axis.

5. Series of polarising filters

When the beam passes through several filters, apply Malus’s law successively. If θ_k‑1,k denotes the angle between the transmission axes of filter k‑1 and filter k, then after n filters

I = I₀ ∏_k=1ⁿ cos²θ_k‑1,k

Special cases:

• Two filters: I = I₀ cos²θ_0,1 cos²θ_1,2
• Three filters: I = I₀ cos²θ_0,1 cos²θ_1,2 cos²θ_2,3

6. Practical applications (AO2)

• Liquid‑crystal displays (LCDs) – each pixel contains two crossed polarisers with a liquid‑crystal layer that rotates the plane of polarisation.
• Polarising sunglasses – block glare by absorbing the horizontally polarised component of reflected sunlight.
• Stress‑analysis (photoelasticity) – stressed plastics become birefringent; polarised light reveals stress patterns.
• 3‑D cinema – orthogonal polarisation for left‑ and right‑eye images.

7. Propagation of angle‑measurement uncertainty

If the angle is measured with an uncertainty Δθ (in radians), the resulting intensity uncertainty can be approximated by

ΔI ≈ 2 I tanθ · Δθ

This formula is useful for AO2 questions that ask for the effect of experimental error.

8. Common pitfalls

• Identifying the correct angle: Use the angle between the *current* polarisation direction and the *next* filter’s axis. Do not simply use the geometric angle between two filters unless the preceding filter has already set the polarisation.
• Unpolarised incident light: Remember the initial ½ I₀ reduction before applying Malus’s law.
• Sign of intensity: Because of the cosine‑squared term, intensity is always ≥ 0; a negative answer indicates a sign error in the angle.
• Units and rounding: Keep angles in degrees for textbook problems (use a calculator set to degrees) but convert to radians when using the uncertainty formula.

9. Worked examples

Example 1 – Single filter (plane‑polarised beam)

Given: I₀ = 120 W m⁻², θ = 30°.

Solution:

1. Apply Malus’s law: I = I₀ cos²θ.
2. cos 30° = √3⁄2.
3. I = 120 (√3⁄2)² = 120 × 3⁄4 = 90 W m⁻².

Example 2 – Two filters, unpolarised incident light

Given: Unpolarised light, I₀ = 200 W m⁻².

First filter aligned with the incident beam (θ₀,₁ = 0°).

Second filter at 45° to the first (θ₁,₂ = 45°).

Solution:

1. First filter: I₁ = ½ I₀ = 100 W m⁻² (beam becomes plane‑polarised).
2. Second filter: I₂ = I₁ cos²45° = 100 × (1/√2)² = 50 W m⁻².

Example 3 – Three filters (crossed polarisers with an intermediate angle)

Given: I₀ = 80 W m⁻², first and third filters are crossed (90° apart).

Middle filter is 45° to each of the outer filters.

Solution:

1. First filter (aligned): I₁ = I₀ = 80 W m⁻².
2. Second filter (θ₁,₂ = 45°): I₂ = 80 cos²45° = 80 × ½ = 40 W m⁻².
3. Third filter (θ₂,₃ = 45°): I₃ = 40 cos²45° = 40 × ½ = 20 W m⁻².

Example 4 – Effect of angle uncertainty

Given: I = 90 W m⁻², θ = 30°, Δθ = ±1° (≈ 0.017 rad).

Solution: ΔI ≈ 2 I tanθ Δθ = 2 × 90 × tan30° × 0.017 ≈ 2 × 90 × 0.577 × 0.017 ≈ 1.8 W m⁻².

10. Summary table

11. Practice questions (AO1 & AO2)

1. A plane‑polarised beam of intensity 150 W m⁻² passes through a filter whose axis is 60° to the polarisation direction. Find the transmitted intensity.
2. Two polarising filters are placed in series. The first is aligned with the incident polarisation. The second is at 30° to the first. If the initial intensity is 100 W m⁻², what intensity emerges from the second filter?
3. Three filters are used. The first and third are crossed (90° apart). The middle filter is at 30° to the first. Starting intensity I₀ = 120 W m⁻². Calculate the final intensity.
4. An unpolarised beam of intensity 80 W m⁻² passes through a single polariser. What intensity emerges? If the angle of the polariser is later measured with an uncertainty of ±2°, estimate the resulting uncertainty in the transmitted intensity.

12. Answers to practice questions

13. Quick reference sheet

• Malus’s law: I = I₀ cos²θ
• Unpolarised → first polariser: I₁ = ½ I₀
• n successive filters: I = I₀ ∏ cos²θ_k‑1,k
• Angle definition: θ is always measured between the *current* electric‑field direction (or the axis of the preceding filter) and the *next* filter’s transmission axis.
• Uncertainty propagation: ΔI ≈ 2 I tanθ Δθ (θ in radians).
• Key applications: LCDs, polarising sunglasses, stress‑analysis (photoelasticity), 3‑D cinema.

14. See also (cross‑references)

• Wave superposition – Syllabus 7.2 (interference of light).
• Diffraction and interference – Syllabus 7.3 (Young’s double‑slit, single‑slit).
• Electromagnetic spectrum – Syllabus 7.1 (properties of EM waves).