Malus's Law — Physica

Equivalent forms

\frac{I}{I_0} = \cos^2\theta

I = I_0 (\hat{e}_1 \cdot \hat{e}_2)^2

A single cosine-squared captures every interaction between linearly polarized light and an ideal polarizer.

Symbol	Name	SI unit	Dimension	Range
$I_0$	Incident intensity Intensity of polarized light before the analyzer	W/m^2	M T^-3	0 – 1000
$theta$	Angle between axes Angle between incoming polarization and polarizer transmission axis	rad	1	0 – 3.14159
$I$	Transmitted intensityoutput Intensity after passing through the analyzer	W/m^2	M T^-3	0 – 1000

Unit systems

Symbol	SI	CGS	Imperial
$I_0$	W/m^2	erg/(s*cm^2)	W/m^2
$I$	W/m^2	erg/(s*cm^2)	W/m^2
$theta$	rad	rad	deg

Where it holds

Valid for ideal linear polarizers and monochromatic linearly polarized input. Real polarizers have finite extinction ratios and absorption losses.

Discovery

Étienne-Louis Malus · 1809

Malus observed sunlight reflecting off the windows of the Luxembourg Palace through a calcite crystal and noticed the intensity varied with crystal orientation. He formulated the cosine-squared law to describe the relationship.

Try this

Why do polarized sunglasses cut glare from a wet road?

Unpolarized light of intensity 100 W/m^2 passes through two polarizers whose axes differ by 30°. Find the transmitted intensity.

Research status: stable

Real-world applications

Polarized sunglasses reduce horizontal glare from water and roads.
LCD screens use crossed polarizers with switchable liquid crystals.
Photographic polarizing filters control reflections and deepen sky contrast.
Optical stress analysis uses photoelasticity through crossed polarizers.

Common misconceptions

Polarizers 'rotate' light — they actually project the field onto their axis.
Crossed polarizers must transmit zero — a third polarizer between them at $45^{\circ}$ transmits 1/8 of original intensity.
Malus's law applies to unpolarized light — only after the first polarizer makes it polarized.

Experimental verification

Verified using laser beams through Polaroid sheets; rotating a second polarizer produces a clean cos^2 intensity curve measurable with a photodiode.

Derivation

An electric field E_0 polarized at angle theta to the transmission axis decomposes into E_0*cos(theta) (transmitted) and E_0*sin(theta) (blocked).

Since intensity scales with |E|^2, the transmitted intensity is

I = I_0 * \cos ^2(\theta )

.

Limiting cases

\theta = 0

⟶

I = I_0

Aligned polarizers transmit fully.

\theta = pi/2

⟶

I = 0

Crossed polarizers block all light.

Unpolarized input⟶

I = I_0 / 2

Random orientations average cos^2 to 1/2.

What if…

What if the input is unpolarized?

The first polarizer halves the intensity (I_0/2); Malus's law applies for subsequent polarizers.

What if a third polarizer is inserted between crossed polarizers at

45^{\circ}

?

Each $\cos ^2(45^{\circ}) = 0.5$ , so transmitted intensity $= I_0 * 0.5 * 0.5 = I_0/8$ — light reappears from nowhere.

1

Light through two polarizers at 30°

Given ·

I 0:: 100
theta:: 0.5236

Find · I

Steps

Apply Malus's law: $I = I_0 * \cos ^2(\theta )$
$\cos (30^{\circ}) = \cos (0.5236) = 0.866$
$\cos ^2(30^{\circ}) = 0.75$
$I = 100 * 0.75 = 75\,\mathrm{W/m}^2$

Result ·

I = 100 * \cos ^2(30^{\circ}) = 100 * (0.866)^2 = 100 * 0.75 = 75\,\mathrm{W/m}^2

Brewster's Angle→Intensity of an Electromagnetic Wave→