Malus Law

Malus law is crucial if we want to learn or understand the polarisation properties of light. The law helps us to study the light intensity relation of the polariser-analyser. Malus law is named after Étienne-Louis Malus, who, in the year 1808, discovered that natural incident light could be polarised when it was reflected by a glass surface. He used calcite crystal for his experiment.

After observing the results, he further put forth a concept that natural light consisted of the s- and p-polarisation and that they were perpendicular to each other. Today, this law is used to define the intrinsic connection between optics and electromagnetism as well as demonstrate the transverse nature of electromagnetic waves.

Table of Contents

• What Is Malus Law?
• Malus Law Formula
• Solved Problems and Questions

What Is Malus Law?

Malus law states that the intensity of plane-polarised light that passes through an analyser varies as the square of the cosine of the angle between the plane of the polariser and the transmission axes of the analyser.

Also Read: Electromagnetic Spectrum and Electromagnetic Waves

Malus Law Formula

The law helps us quantitatively verify the nature of polarised light. Let us understand the expression of Malus law.

Point 1 – When unpolarised light is incident on an ideal polariser, the intensity of the transmitted light is exactly half that of the incident unpolarised light, no matter how the polarising axis is oriented.

Point 2 – An ideal polarising filter passes 100% of incident unpolarised light, which is polarised in the direction of the filter’s (Polariser) polarising axis.

From point (1) and point (2), we can assume I = I_o cos² ϕ

The average value of I (< I >):

We know

Which satisfies point (2) mentioned above.

To show point (1), let us consider ϕ = 0

That implies cos²ϕ = 1

I = I₀

To determine the direction of polarisation, we need one polariser, which is known as analyser oriented, making an angle (p) with the polariser.

Also Read: Polarisation of Light

What happens when the linearly polarised light emerging from a polariser passes through a second polariser (analyser)? In general, the polarising axis of the second polariser (analyser) makes an angle (d) with the polarising axis of the first polariser.

Since the intensity of an electromagnetic wave is proportional to the square of the amplitude of the wave, the ratio of transmitted to incident amplitude is cos ϕ, so the ratio transmitted to incident intensity is cos²ϕ

Solved Problems and Questions

1. What is the difference between unpolarised light and plane-polarised light?

Solution:

Unpolarised light	Plane polarised light
The orientation of electric field vectors \(\begin{array}{l}\left( \overrightarrow{E} \right)\ \text{will be in all possible directions}\end{array} \)	The orientation of electric field vectors \(\begin{array}{l}\left( \overrightarrow{E} \right)\ \text{will be in a direction on a plane}\end{array} \) perpendicular to the direction of propagation of light.

2. An unpolarised light with intensity (I) is passing through a polariser. What happens to the intensity of incident light?

Solution:

If an unpolarised light of intensity (I) passes through a polariser, outcoming light intensity becomes half of its initial value (I/2).

3. An unpolarised light passes through two successive polaroids (P₁ and P₂); the polaroid P₁ makes an angle θ with the axis of the polaroid P₂. Find out the intensity final outcoming light. And if θ is varied from 0 to 27, plot the intensity variation graph.

Solution:

An unpolarised light passes through two successive polaroids.

We know that when an unpolarised light passes through a polaroid, its intensity becomes half of its initial intensity. If it passes again through any other polaroids, then its intensity given by Malus law is,

Here, θ is the angle between the axis of the polaroid’s intensity variation with respect to 0 to 2π. It is nothing (cos² θ) curve.

4. How does an unpolarised light of intensity (I₀) get plane-polarised when passing through a polaroid?

(i) An unpolarised light of intensity (I₀) passes through two successive polaroids (P₁ and P₂), and corresponding intensities of light coming out from them (I₁ and I₂) clearly distinguish the difference between (I₁) and (I₂).

(ii) What is the necessary condition to get maximum intensity after passing through two successive polaroids?

Solution:

Where θ – is the angle between the plane of the polariser and the analyser.

The necessary condition for intensity maximum,

θ should be 0 or π.

The angle (θ) between the plane of transmission of the polariser and analyser must be zero (or) the polariser and analyser must be parallel to each other.

When cos θ = ±1,

I = I₀

When cos θ = 0,

I = 0

Intensity coming out from polaroid P₁ and P₂ (I₁ and I₂)

From Malus law,

We know the average value of

When (I₀) intensity of unpolarised light passes through a polaroid (P₁), its intensity becomes (I₁), and unpolarised light has an electric field vector in all possible directions; when they pass through a polaroid, they get polarised along a direction from Malus law

Then, the same unpolarised light of intensity (I₀) after passing through a polaroid, its intensity becomes (I₁), and if this (I₁) passes through the second polaroid (Analyser), then its intensity becomes (I₂).

It is a function of the square of the cosine angle (θ) between the arms of polarisation of the polaroid and analyser.

5. An Unpolarised The light of intensity (I₀) passes through three successive polaroids P₁, P₂ and P₃. The polarisation axes of P₁ make an angle of 60° with P_2, and P₂ makes an angle of 30° with P₃. Then, what are the corresponding intensity outcomes from P₁, P₂ and P₃ (I₁, I₂ and I₃)? 

Solution: