Does this violate energy conservation?

No. The middle sheet absorbs the orthogonal component at each stage; the final intensity is still ≤ the incident intensity. With three ideal sheets the transmitted fraction is at most 1/8 of the incident unpolarized power in the symmetric 0–45–90 configuration.

Why is the “paradox” not seen with two polarizers only?

Two crossed polarizers block because the field after the first is orthogonal to the second. A third at 45° creates a component along the last axis in two steps: first projection onto 45°, then onto 90°.

Three Polarizers (paradox)

P₁–P₂–P₃ Malus chain; crossed P₁⊥P₃ plus P₂ at 45° lets light through.

About this model

Ideal linear polarizers in series: unpolarized light of unit intensity yields I₁ = 1/2 after the first polarizer. Each subsequent polarizer transmits the component along its transmission axis, so I₂ = I₁ cos²(θ₂ − θ₁) and I₃ = I₂ cos²(θ₃ − θ₂). When θ₁ = 0° and θ₃ = 90°, two polarizers alone give extinction, but a middle polarizer at θ₂ = 45° gives I₃ = (I₁/4) sin²(2θ₂), maximal at 45° — not because light is “created,” but because P₂ rotates the polarization state between the crossed outer sheets.

Who it's for: Follows the Malus-law polarizer page; good for lecture demos of quantum vs classical language (this page stays classical).

Key terms

Malus law
linear polarizer
crossed polarizers
projection

How it works

Three linear polarizers in series: P₁ prepares polarization, P₂ is the usual “surprise” insert, P₃ is often crossed with P₁. With only P₁ and P₃ orthogonal, Malus gives I₃ = 0. Adding P₂ at 45° between them gives I₃ = (I₁/4) sin²(2θ₂) when θ₁ = 0 and θ₃ = 90° — nonzero at θ₂ = 45°. The third sheet projects twice; intensities multiply as cos² of successive axis differences (after P₁, I₂ = I₁ cos²(θ₂−θ₁), I₃ = I₂ cos²(θ₃−θ₂)).

Key equations

I₃ = (I₀/2) cos²(θ₂−θ₁) cos²(θ₃−θ₂) · unpolarized incident

θ₁=0, θ₃=90° ⇒ I₃ = (I₀/8) sin²(2θ₂) · max at θ₂ = 45°

Frequently asked questions

Does this violate energy conservation?: No. The middle sheet absorbs the orthogonal component at each stage; the final intensity is still ≤ the incident intensity. With three ideal sheets the transmitted fraction is at most 1/8 of the incident unpolarized power in the symmetric 0–45–90 configuration.
Why is the “paradox” not seen with two polarizers only?: Two crossed polarizers block because the field after the first is orthogonal to the second. A third at 45° creates a component along the last axis in two steps: first projection onto 45°, then onto 90°.

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