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Malus's Law (Polarization)

Calculates the intensity of polarized light transmitted through a polarizing filter based on the angle between the light's polarization and the filter's axis.

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Malus's Law CalculatorPolarization

I = I₀ · cos²( θ )
I = transmitted intensity  ·  I₀ = incident intensity  ·  θ = angle between polarization and polarizer axis
⟹ SolveI, I₀, θ
W/m²
W/m²
°
Please fix the errors above.
Solve for:
Presets:
Transmitted Intensity
I: I₀: θ:
I/I₀ = cos²(θ) — the fraction of light transmitted.
✓ Copied!
Transmission Ratio (I/I₀)
Zero (0) Partial (0.25–0.75) Full (1)
I = I₀ · cos²( θ )  ·  θ is the angle between the polarization direction and the polarizer axis.

Interpretation

Malus's law: I = I₀ cos²θ, where I₀ is initial intensity, θ is angle between polarizer axis and light polarization. It describes intensity after passing through a polarizer. Example: θ=45° → I = I₀/2.

I = I0 * cos(theta)^2
Malus's Law (Polarization)

Variables

SymbolQuantityUnit
ITransmitted light intensityW/m2
I0Incident polarized light intensityW/m2
thetaAngle between the incident polarization and the filter's transmission axisdegrees

What it means

Malus’s law gives the intensity of light transmitted through a polarizer when the incident light is linearly polarized and makes an angle θ with the transmission axis. The transmitted intensity is I = I₀ cos²θ. This law is fundamental to the study of polarization and is used in optical instruments, liquid crystal displays (LCDs), and in photography to reduce glare. It also explains how polarizing sunglasses work. The law is derived from the projection of the electric field vector. Understanding Malus’s law is essential for any application involving polarizers and polarization control.

Worked example

Malus's Law – Two Examples

Real‑World
Scenario: Unpolarized light (I₀ = 100 W/m²) passes through a polarizer at 45°. Find transmitted intensity.
ParameterValue
I₀100 W/m²
θ45°
1I = I₀·cos²(θ) = 100 × cos²(45°) = 100 × 0.5 = 50 W/m²
Result 50 W/m² ✓ Half
Scenario: I₀ = 50 W/m², θ = 30°. Find transmitted intensity.
ParameterValue
I₀50 W/m²
θ30°
1I = 50 × cos²(30°) = 50 × 0.75 = 37.5 W/m²
Result 37.5 W/m² ✓ More transmitted
Key insight: Malus's law: I = I₀·cos²θ – intensity depends on the angle between polarizer and polarization direction.

Common mistakes

  • Polarizer angle θ: The angle between the transmission axes of the polarizer and analyzer.
  • Intensity I₀: Incident intensity after the first polarizer (not the original unpolarized light).
  • Cosine squared: I = I₀ cos²θ – do not forget the square.
  • Unpolarized light: After passing through a polarizer, intensity is halved (I₀ = I_original/2).
  • Multiple polarizers: Apply sequentially.

Applications

Malus's law, I = I₀·cos²θ, describes the intensity of polarised light after passing through a polariser at an angle θ. It is used in optical devices such as polarimeters, liquid crystal displays (LCDs), and glare‑reducing sunglasses. Engineers apply it to design optical sensors, to control light intensity in photography, and to analyse stress in transparent materials (photoelasticity). In telecommunications, polarisation is used to reduce signal interference. By understanding Malus's law, professionals can manipulate polarised light for a wide range of applications, from consumer electronics to advanced optical metrology.

  • Design of LCD screens and optical displays
  • Polarimeters for measuring optical activity
  • Photoelastic stress analysis in materials
  • Glare reduction filters in photography
  • Polarisation‑division multiplexing in fibre optics

Frequently Asked Questions

Q01What is Malus's law and what does it describe?
A01

Malus's law gives the intensity of light transmitted through a polarizer when the incident light is already linearly polarized: I = I₀·cos²θ, where I₀ is the initial intensity, and θ is the angle between the polarization direction of the incident light and the transmission axis of the polarizer.

Q02What is the common mistake when applying Malus's law?
A02

Applying it to unpolarized light without first reducing the intensity by 50% (since unpolarized light has equal components in all directions, only half passes through a polarizer). For unpolarized light, I = ½I₀.

Q03What happens when θ = 0° or 90°?
A03

At θ = 0°, I = I₀ (maximum transmission). At θ = 90°, I = 0 (no transmission, crossed polarizers).

Q04How is Malus's law used in optical instruments?
A04

It is used in polarimeters, optical isolators, and variable attenuators. It also explains the operation of liquid crystal displays (LCDs).

Q05What is the intensity if the angle is 45°?
A05

I = I₀·cos²45° = I₀·(1/2) = I₀/2. So half the intensity is transmitted.

Q06What is the difference between a polarizer and an analyzer?
A06

A polarizer is the first element that produces polarized light. An analyzer is the second polarizer used to examine the polarization state of light. Malus's law describes the intensity after the analyzer.

Q07What is the degree of polarization?
A07

It is a measure of how much of the light is polarized: P = (I_max – I_min)/(I_max + I_min). For fully polarized light, P = 1; for unpolarized, P = 0.

Q08How is Malus's law used in stress analysis (photoelasticity)?
A08

When polarized light passes through a stressed transparent material, the polarization changes, and the transmitted intensity varies with stress, allowing visualisation of stress patterns.