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Polarisation, Malus's Law and Brewster's Law

Polarisation is the restriction of light vibrations to one plane perpendicular to the direction of propagation. It proves that light waves are transverse.

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Student-friendly explanation

Ordinary light has electric-field vibrations in many planes perpendicular to the direction of travel. A polariser transmits vibrations along its transmission axis and blocks the perpendicular component. When polarised light passes through an analyser, the transmitted intensity depends on the angle between their transmission axes. Reflection can also produce plane-polarised light at Brewster's angle.

How to write this in exams

  1. 1

    Start with the exact idea

    Polarisation is the restriction of light vibrations to one plane perpendicular to the direction of propagation. It proves that light waves are transverse.

  2. 2

    Then show how to use it

    First identify whether the incident light is unpolarised or already plane-polarised. For unpolarised light, use I0/2 after the first polariser because only one vibration plane is selected. For plane-polarised light through an analyser, resolve the field component along the analyser axis and use I = I0 cos^2 theta. For Brewster-angle questions, use tan theta_B = n and state the final geometry cue: reflected and refracted rays are at 90 degree to each other.

  3. 3

    Add one concrete example

    Polaroid sunglasses reduce glare from horizontal surfaces because reflected glare is strongly polarised and the lenses block much of that vibration direction.

  4. 4

    Avoid this incomplete answer

    A common wrong answer uses sin^2 theta in Malus's law. The transmitted component is along the analyser axis, so the intensity varies as cos^2 theta.

Definition

Polarisation is the restriction of light vibrations to one plane perpendicular to the direction of propagation. It proves that light waves are transverse.

Example

Polaroid sunglasses reduce glare from horizontal surfaces because reflected glare is strongly polarised and the lenses block much of that vibration direction.

Rule to remember

Malus's law assumptions: the incident light on the analyser is plane-polarised and the analyser is ideal. Key step: resolve the electric-field amplitude along the analyser axis, E = E0 cos theta; since intensity is proportional to E^2, the final result is I = I0 cos^2 theta. Brewster's law result: tan theta_B = n for incidence from air, with reflected and refracted rays perpendicular at theta_B. Intensities are in W m^-2 and refractive index has no SI unit.

Memory hook

Polariser selects a plane; analyser measures the cosine-squared component.

Examples and method

Worked example

Plane-polarised light of intensity 80 W m^-2 enters an analyser with its axis at 30 degree to the vibration direction. I = I0 cos^2 theta = 80 x cos^2 30 degree = 80 x 3/4 = 60 W m^-2. The analyser transmits 60 W m^-2.

Method to apply

First identify whether the incident light is unpolarised or already plane-polarised. For unpolarised light, use I0/2 after the first polariser because only one vibration plane is selected. For plane-polarised light through an analyser, resolve the field component along the analyser axis and use I = I0 cos^2 theta. For Brewster-angle questions, use tan theta_B = n and state the final geometry cue: reflected and refracted rays are at 90 degree to each other.

Diagram support

A diagram is useful when showing polariser and analyser axes, unpolarised light becoming plane-polarised light, and Brewster-angle reflection. Labels should include transmission axis, analyser angle theta, reflected ray, refracted ray and 90 degree angle between reflected and refracted rays at Brewster's angle.

How CBSE asks it

It is asked as proof of transverse nature, Malus's-law numerical, Brewster-angle calculation, or conceptual questions on glare reduction and crossed polaroids.

Avoid common mistakes

Common confusion

Students often apply Malus's law directly to unpolarised light without first halving the intensity after the first polariser. For unpolarised light through one ideal polariser, transmitted intensity becomes I0/2.

Common wrong answer

A common wrong answer uses sin^2 theta in Malus's law. The transmitted component is along the analyser axis, so the intensity varies as cos^2 theta.

Exam tip

In Malus's law, theta is the angle between the transmission axes of the polariser and analyser, not the angle of incidence. For Brewster's law, tan theta_B = n is used for light incident from air into the medium.

Quick check

What is the transmitted intensity when plane-polarised light of intensity I0 passes through an analyser at 60 degree to its vibration direction?

Using Malus's law, I = I0 cos^2 60 degree = I0 x (1/2)^2 = I0/4. So one-fourth of the incident plane-polarised intensity is transmitted.

Answer writing and exam use

1-mark answer

Polarisation is the restriction of light vibrations to one plane perpendicular to the direction of propagation. It proves that light waves are transverse.

2-mark answer

Polarisation is the restriction of light vibrations to one plane perpendicular to the direction of propagation. It proves that light waves are transverse. Malus's law assumptions: the incident light on the analyser is plane-polarised and the analyser is ideal. Key step: resolve the electric-field amplitude along the analyser axis, E = E0 cos theta; since intensity is proportional to E^2, the final result is I = I0 cos^2 theta. Brewster's law result: tan theta_B = n for incidence from air, with reflected and refracted rays perpendicular at theta_B. Intensities are in W m^-2 and refractive index has no SI unit. Polaroid sunglasses reduce glare from horizontal surfaces because reflected glare is strongly polarised and the lenses block much of that vibration direction.

3-mark answer

Ordinary light has electric-field vibrations in many planes perpendicular to the direction of travel. A polariser transmits vibrations along its transmission axis and blocks the perpendicular component. When polarised light passes through an analyser, the transmitted intensity depends on the angle between their transmission axes. Reflection can also produce plane-polarised light at Brewster's angle. Malus's law assumptions: the incident light on the analyser is plane-polarised and the analyser is ideal. Key step: resolve the electric-field amplitude along the analyser axis, E = E0 cos theta; since intensity is proportional to E^2, the final result is I = I0 cos^2 theta. Brewster's law result: tan theta_B = n for incidence from air, with reflected and refracted rays perpendicular at theta_B. Intensities are in W m^-2 and refractive index has no SI unit. Plane-polarised light of intensity 80 W m^-2 enters an analyser with its axis at 30 degree to the vibration direction. I = I0 cos^2 theta = 80 x cos^2 30 degree = 80 x 3/4 = 60 W m^-2. The analyser transmits 60 W m^-2. It is asked as proof of transverse nature, Malus's-law numerical, Brewster-angle calculation, or conceptual questions on glare reduction and crossed polaroids. A common wrong answer uses sin^2 theta in Malus's law. The transmitted component is along the analyser axis, so the intensity varies as cos^2 theta.
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