Plane-polarised light of intensity 120 W/m² has its vibration direction at 45° to the transmission axis of a first ideal polariser. It then passes through a second ideal polariser; the two transmission axes differ by 35°. Use I = I₀cos²θ for each filter. (a) Calculate the intensity after each polariser. (b) Predict the final intensity when the two filter axes are 90° apart, leaving the first filter unchanged. (c) Explain what polarisation establishes about light waves.

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After first: 60 W/m²; after second: 40.26 W/m²; crossed axes give zero ideal transmission; light is transverse. The incoming plane-polarised light vibrates at 45° to the first filter's axis. Malus' law gives I₁ = 120 cos²45° = 120/2 = 60 W/m², now polarised along that filter's axis. Apply Malus' law to the second filter using I₁, not the original intensity: I₂ = 60 cos²(35°) = 40.26 W/m². The angle is between the two transmission axes. For crossed axes the angle is 90°, so cos²90° = 0 and the ideal transmitted intensity vanishes. Actual filters can leak some light, which would be a limitation of the ideal-filter assumption. Polarisation selects a vibration direction perpendicular to propagation. This supports the transverse nature of electromagnetic waves: a longitudinal wave has only the propagation direction for particle displacement and cannot be filtered into different transverse vibration directions in this way.

Transmission through two polarisers 5

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