I = I₀ cos²θ, where θ is the angle between the incident light's polarisation direction and the polariser's transmission axis.

What happens at θ = 90°?

I = 0. Crossed polarisers block all the light. This is the basis of LCD displays and photographic polarising filters.

Does Malus's law work for unpolarised light?

Unpolarised light passing through the first polariser becomes polarised at intensity I₀/2 (the factor of cos²θ averaged over all angles). After that first polariser, subsequent polarisers obey Malus's law.

Why does cos² appear?

Only the component of the electric field along the transmission axis passes through, giving E_transmitted = E₀ cos θ. Intensity is proportional to the square of the field, so I/I₀ = cos²θ.

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NSW · HSCModule 7

Malus's law calculator

I = I₀ cos²θ. Enter the incident intensity (polarised parallel to the first polariser) and the angle of the second polariser.

Inputs

Incident intensity I₀ (W/m² (or any unit))Angle between polariser axes θ (degrees)

Assumes the incident light is already plane-polarised parallel to the first polariser's transmission axis.

Result

Transmitted intensity I

75.00same unit as I₀

I / I₀ = cos²θ

0.7500

Malus's law: I = I₀ cos²θ. At θ = 0, I = I₀; at θ = 90°, I = 0 (axes crossed).

How this calculator works

Polarised light through a polariser at angle θ has intensity I = I₀ cos²θ. At θ = 0 (axes aligned), all the light passes. At θ = 90° (axes crossed), none passes.

Common questions

What is Malus's law?: I = I₀ cos²θ, where θ is the angle between the incident light's polarisation direction and the polariser's transmission axis.
What happens at θ = 90°?: I = 0. Crossed polarisers block all the light. This is the basis of LCD displays and photographic polarising filters.
Does Malus's law work for unpolarised light?: Unpolarised light passing through the first polariser becomes polarised at intensity I₀/2 (the factor of cos²θ averaged over all angles). After that first polariser, subsequent polarisers obey Malus's law.
Why does cos² appear?: Only the component of the electric field along the transmission axis passes through, giving E_transmitted = E₀ cos θ. Intensity is proportional to the square of the field, so I/I₀ = cos²θ.