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CIE AS/A Level Physics 7.5 Polarisation Study Notes- 2025-2027 Syllabus

CIE AS/A Level Physics 7.5 Polarisation Study Notes – New Syllabus

CIE AS/A Level Physics 7.5 Polarisation Study Notes at  IITian Academy  focus on  specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:

  1. understand that polarisation is a phenomenon associated with transverse waves
  2. 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 intensity of an unpolarised wave is not required)

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Polarisation — a Property of Transverse Waves

Polarisation is the phenomenon in which the oscillations of a transverse wave are restricted to a particular direction or pattern. It is a property associated only with transverse waves, where the oscillating field or displacement is perpendicular to the direction of propagation.

Key Points:

  • Polarisation applies to transverse waves because their oscillations have directions perpendicular to propagation that can be oriented (e.g., the electric field vector of an electromagnetic wave).
  • Longitudinal waves (e.g., sound in a fluid) cannot be polarised because their oscillations are parallel to the direction of propagation and have no transverse direction to restrict.
  • For electromagnetic waves, polarisation describes the direction of the electric field vector \( \mathrm{\vec{E}} \). Common types include plane (linear) polarisation, circular polarisation, and elliptical polarisation.
  • An unpolarised light wave is a superposition of waves with electric field vectors in all transverse directions; a polariser selects one direction (or component) of the \( \mathrm{\vec{E}} \)-field.

Representation (plane-polarised EM wave):

\( \mathrm{\vec{E}(x,t) = E_0 \hat{u} \cos(kx – \omega t)} \)

where \( \mathrm{\hat{u}} \) is a unit vector giving the direction of the electric field (the polarisation direction).

Example

State whether the following waves can be polarised:

(a) water surface waves,

(b) sound waves in air,

(c) light from a laser that has passed through a polariser.

▶️ Answer / Explanation

(a) Water surface waves: Yes — surface (transverse) components can be polarised because particle motion has transverse directions; practical polarisation concepts apply to transverse components.

(b) Sound waves in air: No — sound in fluids is longitudinal, so it cannot be polarised.

(c) Light from a laser after a polariser: Yes — passing through a polariser produces plane-polarised light with the electric field oscillating in a specific direction.

Malus’s Law — Intensity of Plane-Polarised Light After Passing Through a Polarising Filter

Malus’s law describes how the intensity of a plane-polarised electromagnetic wave changes as it passes through a polarising filter (also called an analyser). It relates the transmitted intensity to the angle between the light’s polarisation direction and the filter’s transmission axis.

\( \mathrm{I = I_0 \cos^2 \theta} \)

Where:

  • \( \mathrm{I_0} \): initial intensity of the incident plane-polarised light
  • \( \mathrm{I} \): transmitted intensity after passing through the polariser
  • \( \mathrm{\theta} \): angle between the light’s plane of polarisation and the transmission axis of the polarising filter

Explanation:

  • When a plane-polarised wave passes through a polarising filter, only the component of the electric field parallel to the filter’s transmission axis passes through.
  • The amplitude of this transmitted component is \( \mathrm{E = E_0 \cos \theta} \).
  • Since intensity \( \mathrm{I} \) is proportional to the square of the amplitude (\( \mathrm{I \propto E^2} \)), the transmitted intensity becomes \( \mathrm{I = I_0 \cos^2 \theta} \).

Key Observations:

  • If \( \mathrm{\theta = 0°} \): the polarisation direction is aligned with the filter full transmission (\( \mathrm{I = I_0} \)).
  • If \( \mathrm{\theta = 90°} \): the polarisation direction is perpendicular to the filter — complete extinction (\( \mathrm{I = 0} \)).
  • If \( \mathrm{\theta = 45°} \): \( \mathrm{I = \tfrac{1}{2} I_0} \).

Series of Polarisers:

  • When light passes through multiple polarising filters, each filter transmits intensity according to Malus’s law, relative to the angle between successive filters.
  • Thus, if \( \mathrm{I_1 = I_0 \cos^2 \theta_1} \) and \( \mathrm{I_2 = I_1 \cos^2 \theta_2} \), then overall \( \mathrm{I_2 = I_0 \cos^2 \theta_1 \cos^2 \theta_2} \).

Example

A plane-polarised light of intensity \( \mathrm{I_0 = 20 \ W/m^2} \) passes through a polarising filter whose axis makes an angle of \( \mathrm{30°} \) with the light’s polarisation direction. Calculate the transmitted intensity.

▶️ Answer / Explanation

Using \( \mathrm{I = I_0 \cos^2 \theta} \):

\( \mathrm{I = 20 \times \cos^2 30° = 20 \times (0.866)^2 = 20 \times 0.75 = 15 \ W/m^2.} \)

Therefore, the transmitted intensity is \( \mathrm{15 \ W/m^2.} \)

Example

Plane-polarised light of intensity \( \mathrm{I_0 = 12 \ W/m^2} \) passes successively through two polarising filters. The first filter is aligned at \( \mathrm{0°} \), and the second at \( \mathrm{60°} \). Determine the transmitted intensity after the second filter.

▶️ Answer / Explanation

Using Malus’s law for the second filter relative to the first: \( \mathrm{I = I_0 \cos^2 60°.} \)

\( \mathrm{I = 12 \times (0.5)^2 = 12 \times 0.25 = 3 \ W/m^2.} \)

Hence, the transmitted intensity after both filters is \( \mathrm{3 \ W/m^2.} \)

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