Malu's Law

Etienne-Louis Malus, a French Physicist, derived this law in 1808 in this law, natural Light, which is unpolarised in nature, could be polarised when reflected by a glass surface. The calcite crystal is used to observe this experiment. Malus law deals with the polarisation properties of unpolarized Light. Observing from the experiment, two types of polarisation occurred in natural Light named s and p-type polarisation, which are mutually perpendicular. Malus law is generally used to make polaroids, which may control the intensity of Light.

1.0Definition of Malu's Law

According to this law, transmitted Light through the analyzer varies as the square of the cosine of the angle between the plane of the analyzer and the polariser. This rule is also called the cosine square rule.

2.0Examples of Malu's law in Everyday life

• Polaroid sunglasses

• Window Panes of train and aeroplanes

• Photographic camera

• 3- d movies

• To eliminate dazzle from the headlights of cars, buses, etc.

3.0Experimental setup of  Malu's Law in Polarisation

Note:  I₀ is the intensity of the polariser's transmitted Light, and  I₀ is half the intensity of the incident with unpolarised Light.              

4.0Derivations of Malu’s Law

To derive the Malus Law, let us consider the amplitude' a' of Light transmitted by the polariser and the angle between the polariser's planes and the Analyser.

Now resolving the amplitude'  a' in two components:

(1).”a Cosθ” is parallel to the plane of transmission of the analyzer

(2).”a Sinθ” is perpendicular to the analyzer's transmission plane.

The component that is transmitted through the Analyser is “a cos"

(∵ Intensity ∝ Amplitude² )

I₀ ∝ a²

I ∝ (a Cosθ)²

I=K a² Cos² θ     (∵I₀=Ka²)

I=I₀ Cos² θ

This is Malu's Law, termed as Malu's Formula.

Exceptional cases based on Malu's Law:

(1). When θ=0⁰ or 180⁰ i .e. the plane of the polariser and analyzer are parallel.

Cos θ = cos 0 =1 

I=I₀

In this case, the intensity of the incident plane polarised Light does not change.

(2). When $\theta =\frac{\pi }{2}$, i.e. the plane of the polariser and analyzer are perpendicular to each other, then

$\mathrm{Cos}\theta =\mathrm{Cos}\frac{\pi }{2}=0$

=> I = 0

In this case,  the intensity of the incident plane polarised light becomes zero.

Graph showing the variation  of intensity of transmitted Light on the angle between the polarizer and analyzer:

5.0Solved Examples

Examples:  What do you understand about the term crossed polaroids?

Sol: Two polaroids are said to be crossed polaroids when the plane of polarization of two polaroids is perpendicular(normal) to each other.

Examples: When Light gets polarised, what effect may be observed on the intensity of incident light?

Sol: The intensity of incident light decreases because when the Light gets polarised, the vibration of the electric vector is restricted only in one plane and in the other plane, vibrations of electric vectors are cut off. The intensity of polarised Light is less than that of the unpolarised Light.

Examples: Why does Light from a transparent blue portion of the sky show a rise and fall intensity when viewed through a rotated polaroid?

Sol: The sky appears blue due to the phenomenon of light scattering. Viewing the sky through a Polaroid acts as an analyzer, so the intensity increases or decreases according to the Polaroid's rotation angle.

6.0Solved Questions

Question 1: Prove that when unpolarised Light of intensity I₀ gets polarised on passing through a polaroid, its intensity becomes half, $\mathrm{I}=\frac{I_0}{2}$.

Solution:

In unpolarised Light, the vibration exists in all directions in a plane perpendicular to the direction of propagation. So, the rotation angle can have any value from o to 2.

In this case, we calculate the average value of the square of cosine ${\left[{\mathrm{Cos}}^2\theta \right]}_{avg}$

$\Rightarrow \frac{1}{2\pi }{\int }_0^{2\pi }{\cos }^2\theta \mathrm{d}\theta$

$\Rightarrow \frac{1}{2\pi }{\int }_0^{2\pi }\frac{(1+\cos 2\theta )d\theta }{2}$

$\Rightarrow \frac{1}{2\times 2\pi }{\left(\theta +\frac{\mathrm{Sin}2\theta }{2}\right)}_0^{2\pi }$

$\Rightarrow {\left({\cos }^2\theta \right)}_{\text{Average }}=\frac{1}{2}$

 by using Malus law, $I=I_{0}Co{s}^2\theta$

$\Rightarrow I=\frac{I_0}{2}$

Question 2: Two polaroids are oriented such that the maximum amount of Light is transmitted when the analyzer is rotated (a).30° (b). 60° .To what extent of its maximum value is the intensity of transmitted Light reduced?

Solution:

By using Malu's Law

$\mathrm{I}=I_0{\cos }^2\theta$

(a). $\theta$ =30°

$\mathrm{I}=I_0{\mathrm{Cos}}^{2}30^{\circ }$

$I=I_0{\left(\sqrt{\frac{3}{2}}\right)}^2$

$\mathrm{I}=\frac{3}{4}{I}_0=0.75I_0$

(b). $\theta$ = 60°

$\mathrm{I}=I_0{\cos }^{2}60^{\circ }$

$\mathrm{I}=\frac{I_0}{4}$

Question 3:  A beam of unpolarised Light of intensity I0 is passed through polaroid X and then through another polaroid Y , is oriented such that its plane makes an angle of 45° relative to that of X ; find the intensity of emergent Light.

Solution:

When unpolarised Light passes through polaroid X

By using Malu's law 

The intensity of polarised Light from X, ${\mathrm{I}}_{\mathrm{X}}=\frac{I_0}{2}$

Polaroid Y is oriented at 45° relative to X

The intensity of Light emerges from Polaroid Y, 

${\mathrm{I}}_{\mathrm{Y}}=\frac{I_0}{2}{\cos }^{2}45^{\circ }=\frac{I_0}{2}{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{I_0}{4}$

Hence, the intensity of emergent Light is $\frac{I_0}{4}.$

Question 4: How are the polariser and analyzer arranged so that we get (a) $\mathrm{I}=I_0$  (b) I = 0 ?

Solution:

(a) When polarizer and analyzer are parallel, θ = 0° or 180° so that 

Cos θ = Maximum = 1

$\mathrm{I}=I_0$

(b) When the polarizer and analyzer are perpendicular to each other, θ = 90° or Cos 90⁰ = 0 = Minimum

$I=0$