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Polarisation of Light

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Estimated time: 14 minutes
CBSE: Class 12

Definition: Polarisation of Light

Polarisation is the phenomenon of restricting the vibration of a light wave to a particular plane perpendicular to the direction of propagation of the wave, or confining the electric vector vibrations to one direction perpendicular to the direction of propagation.

CBSE: Class 12

Definition: Unpolarised Wave

An unpolarised wave is one in which the plane of vibration changes randomly in very short time intervals.

CBSE: Class 12

Definition: Transverse Wave

A transverse wave is one in which the displacement of particles is perpendicular to the direction of propagation of the wave.

CBSE: Class 12

Definition: Linearly Polarised Wave (Plane Polarised Wave)

A wave in which the electric field vectors are confined in one plane and are parallel to a unique direction is called a linearly polarised wave or plane polarised wave.

CBSE: Class 12

Definition: Polaroid

A Polaroid is a thin film of ultramicroscopic crystals used to produce plane-polarised light.

CBSE: Class 12

Definition: Plane of Vibration

The plane of vibration is the plane in which the electric field vector \[\vec{E}\] vibrates or oscillates.

CBSE: Class 12

Definition: Plane of Polarisation

The plane of polarisation is the plane in which vibrations are present — it is perpendicular to the plane of vibration and contains the direction of propagation.

CBSE: Class 12

Unpolarised vs. Polarised Light

CBSE: Class 12

Law: Malus' Law

Statement

When a beam of plane polarised light is incident on an analyser, the intensity of the transmitted light is directly proportional to the square of the cosine of the angle θ between the pass-axis of the analyser and the plane of polarisation of the incident light.

I = I0 cos⁡2θ

Where:

  • I0​ = intensity of plane-polarised light incident on the analyser
  • I = intensity of the transmitted light
  • θ = angle between the pass axes of the polariser and analyser
Derivation

Step 1: Set up

  • Let plane-polarised light with amplitude a and intensity I0​ be incident on analyser P2. The pass-axis of P2 makes an angle θ with the pass-axis of P1.

Step 2: Resolve the amplitude
The electric field amplitude aaa is resolved into two rectangular components relative to P2's pass-axis:

  • Component parallel to P2's pass-axis: a cos ⁡θ → transmitted
  • Component perpendicular to P2's pass-axis: a sin⁡ θ → absorbed/blocked

Step 3: Calculate transmitted intensity
Since only the parallel component passes through, and intensity ∝ (amplitude)2:

  • I ∝ (a cos ⁡θ)2 = a2 cos⁡2 θ

Step 4: Substitute I0
Since I0 ∝ a2 (the maximum intensity when θ = 0°):

  • I = I0 cos⁡2 θ
This is Malus' Law.
CBSE: Class 12

Special Cases

Angle (θ) cos θ Transmitted Intensity Observation
1 I = I₀ Maximum; pass-axes aligned (parallel)
45° 1/√2 I = I₀/2 Half intensity
90° 0 I = 0 Complete darkness; axes crossed
CBSE: Class 12

Example

Three polaroids are used:

  • P1 polarises the unpolarised light.
  • P2 is placed between P1 and P3 and can be rotated.
  • P3 is crossed with P1 (their pass axes are at 90°), so without P2, no light passes through.

When P2 is rotated by an angle θ:

  • The light intensity after P2 becomes: I = I0 cos⁡2 θ
  • The light then passes through P3, giving the final intensity:
    I = \[I_0\cos^2\theta\sin^2\theta=\frac{I_0}{4}\sin^22\theta\]

The transmitted intensity is maximum when:

θ = 45 (\[\frac {π}{4}\])

Conclusion:
When the middle polaroid (P2) is rotated to 45°, the maximum amount of light passes through all three polaroids, even though the first and third polaroids are crossed.

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