The third filter

Question

Why can a third polarizer transmit light when the first and last polarizers are crossed?

The first polarizer defines the 0° state. The middle polarizer keeps only its parallel component, so the transmitted fraction after it is cos² α. The surviving field is now aligned with α, not still aligned with 0°. The final polarizer is at 90°, so its relative angle to the middle one is 90° − α. That second projection contributes cos²(90° − α) = sin² α.

Model and variables

α is the middle axis, θ₁ = α is the first relative angle, θ₂ = 90° − α is the second, and T = I/Iref is the final intensity fraction relative to immediately after the first filter. The product rule assumes sequential ideal analyzers and ignores absorption other than the projection loss.

Derivation

Analytical result

Numerical example. With Iref = 100 intensity units immediately after the first 0° filter and α = 30°, the middle passes ≈ 75.00 units and the final filter passes ≈ 18.75 units. At α = 45°, the stages are ≈ 50.00 units and then ≈ 25.00 units, so one quarter remains. This is not energy created by the middle filter: the final analyzer is no longer crossed relative to the state arriving at it.

Reproducible method

I fixed the outer axes at 0° and 90° in my model and calculated the middle and final intensities as the middle angle changes. The interactive model evaluates these projections directly. The calculated curves below sample α from 0° to 90° in 1° steps. For future physical measurements, I would rotate only the middle mount, wait for the detector to settle, take three readings at each angle and record background light separately.

Reading the plot

The final curve is zero at 0° and 90° and peaks at 45°. The middle output falls monotonically from Iref to zero, while the final output is symmetric around 45°. If the measured peak moves away from 45°, check axis alignment and angle calibration. A constant background raises the readings without moving the ideal maximum. Background or source intensity that changes during the sweep can bias the measured peak.

Analytical result

Iref is measured immediately after the first 0° filter. The 25% maximum is 12.5% of an unpolarized source before that first ideal filter.

Differentiating uses radians. Both endpoints give zero; the interior stationary point at 45° gives the maximum. A constant additive detector background raises the curve without shifting this ideal maximum. Drift and angle errors require separate treatment.

Uncertainty and limits

The simple product does not model wavelength dependence, finite extinction ratio, depolarization, or reflections. A detector reading after the first filter is not automatically the same scale as before it unless calibration is checked. The 25% result applies only to ideal crossed analyzers with an aligned linear input.

Further question

Sequential projection and order

The rightmost matrix acts first. Starting with a 0° field, the middle projection gives amplitude Eref cosα along u(α). The last projection contributes sinα. Squaring gives Iref cos²α sin²α. Changing the order changes which state arrives at each filter.

Next question

Can four analyzer readings reconstruct the hidden direction? Finding the hidden direction turns the same projection rule into an inverse problem.