Polarisation
Section: Waves | Syllabus: Cambridge AS Level Physics (9702)
Polarisation
Polarisation is a phenomenon unique to transverse waves that restricts oscillations to a single plane.
Key Definitions
Polarisation: The process of restricting the oscillations of a transverse wave to a single plane. Only transverse waves can be polarised. Unpolarised wave: A transverse wave in which oscillations occur in all planes perpendicular to the direction of wave travel. Plane-polarised wave: A transverse wave in which oscillations occur in only one plane (which contains the direction of wave travel). Polarising filter (polariser): A device that transmits oscillations in only one plane while absorbing oscillations in perpendicular planes. Transmission axis: The plane of oscillation that a polariser allows through.Core Principles
Why Only Transverse Waves Can Be Polarised
- In transverse waves, oscillations are perpendicular to the direction of travel
- These perpendicular oscillations can occur in any plane
- A polariser can select just one of these planes
- In longitudinal waves, oscillations are parallel to wave travel-there is only one possible direction, so polarisation is not applicable
Unpolarised Light
- Light from most sources (sun, bulbs) is unpolarised
- Contains oscillations in all planes perpendicular to the direction of travel
- Can be represented as oscillations in all directions in a plane perpendicular to the ray
Effect of a Single Polariser on Unpolarised Light
- The transmitted light is plane-polarised
- The intensity is reduced to half: I = I₀/2
- (This halving is not required for calculations at AS Level)
Effect of a Polariser on Plane-Polarised Light
- The transmitted intensity depends on the angle between the polarisation plane and the transmission axis
- This is described by Malus's law
Show the concept of polarisation: (a) Unpolarised light represented as arrows oscillating in all planes perpendicular to the direction of travel. (b) Light passing through a polariser showing only vertical oscillations transmitted. (c) Two polarisers: when aligned (parallel), light passes through; when crossed (90°), no light passes. Label the transmission axis of each polariser.
Malus's Law
When plane-polarised light of intensity I₀ passes through a polarising filter (analyser), the transmitted intensity is:
I = I_0 cos^2θwhere:
- I = transmitted intensity (W m⁻²)
- I₀ = incident intensity of plane-polarised light (W m⁻²)
- θ = angle between the plane of polarisation and the transmission axis of the filter
Key Cases
| Angle θ | Transmitted Intensity | Result |
|---|---|---|
| 0° | I = I₀ cos²(0°) = I₀ × 1 | I₀ (maximum transmission) |
| 90° | I = I₀ cos²(90°) = I₀ × 0 | 0 (no transmission-crossed polarisers) |
| 45° | I = I₀ cos²(45°) = I₀ × 0.5 | 0.5I₀ (half transmitted) |
| 60° | I = I₀ cos²(60°) = I₀ × 0.25 | 0.25I₀ (quarter transmitted) |
Concept Insight: Rope and Fence Analogy
Imagine sending waves along a rope that passes through a vertical fence slot:
- Vertical oscillations pass through unaffected
- Horizontal oscillations are blocked completely
- Oscillations at an angle have only their vertical component transmitted
The fence slot acts like a polarising filter:
- Its orientation determines which oscillation plane is transmitted
- Only the component of oscillation parallel to the slot passes through
With two fence slots:
- If both are vertical (parallel): waves pass through both
- If one is vertical and one horizontal (perpendicular/crossed): waves are blocked
- If at some angle between: partial transmission occurs
Series of Polarisers
When light passes through multiple polarisers in series, apply Malus's law at each stage:
Two polarisers at angle θ to each other:I_final = I_0 cos^2θ Three polarisers (first and third crossed at 90°, middle at angle θ to first):
- After first polariser: light is plane-polarised with intensity I₁
- After second polariser (at angle θ): I₂ = I₁ cos²θ
- After third polariser (at angle (90° − θ) to second): I₃ = I₂ cos²(90° − θ) = I₂ sin²θ
With three polarisers, light CAN pass through crossed polarisers if an intermediate polariser is inserted at an angle.
Worked Examples
Worked Example 1: Single PolariserQuestion: Plane-polarised light of intensity 20 W m⁻² passes through a polariser with its transmission axis at 30° to the plane of polarisation. Calculate the transmitted intensity.
SolutionUsing Malus's law:
I = I_0 cos^2θ = 20 × cos^2(30°)
I = 20 × (0.866)^2 = 20 × 0.75 = 15 W m^-2
Worked Example 2: Crossed PolarisersQuestion: Two polarising filters are placed with their transmission axes perpendicular to each other (crossed). Plane-polarised light of intensity 50 W m⁻² (with polarisation parallel to the first filter) is incident on the first filter. Calculate the intensity after: (a) the first filter, (b) the second filter.
Solution(a) The incident light is already polarised parallel to the first filter (θ = 0°):
I_1 = I_0 cos^2(0°) = 50 × 1 = 50 W m^-2
(b) The second filter is at 90° to the first, so θ = 90°:
I_2 = I_1 cos^2(90°) = 50 × 0 = 0 W m^-2
No light passes through crossed polarisers.
Worked Example 3: Finding the AngleQuestion: Plane-polarised light passes through a polariser. The transmitted intensity is one-third of the incident intensity. Find the angle between the polarisation plane and the transmission axis.
SolutionI = I_0 cos^2θ
(I/I_0) = cos^2θ = (1/3)
cosθ = √(1/3) = (1/√(3)) = 0.577
θ = cos^-1(0.577) = 54.7°
Worked Example 4: Three Polarisers in SeriesQuestion: Three polarisers are arranged in series. The first and third have their transmission axes perpendicular (crossed). The second polariser is placed between them with its axis at 45° to the first. Plane-polarised light of intensity I₀ (polarised parallel to the first polariser) enters the system. Find the final transmitted intensity.
SolutionAfter first polariser (θ = 0°):
I_1 = I_0 cos^2(0°) = I_0
After second polariser (at 45° to first):
I_2 = I_1 cos^2(45°) = I_0 × (0.707)^2 = 0.5I_0
After third polariser (at 45° to second, since third is at 90° to first):
I_3 = I_2 cos^2(45°) = 0.5I_0 × 0.5 = 0.25I_0
The middle polariser allows 25% of the original intensity to pass through what were originally crossed polarisers.
Worked Example 5: Intensity ReductionQuestion: A polariser reduces the intensity of plane-polarised light to 40% of the incident value. Calculate the angle between the plane of polarisation and the transmission axis.
Solutioncos^2θ = 0.40
cosθ = √(0.40) = 0.632
θ = cos^-1(0.632) = 50.8°
Applications of Polarisation
Interactive revision notes, videos and practice questions load below.