Unpolarised light of intensity I₀ passes through two polarising filters. The first is vertical; the second is rotated 60° from vertical. Find the transmitted intensity. What if a third filter at 30° is placed between them? Step 1: After the first (vertical) filter, the light is vertically polarised with intensity I₁ = I₀/2 (an ideal polariser passes half of unpolarised light). Step 2: Malus's law: I = I₁cos²θ, where θ is the angle between the light's polarisation and the filter axis. θ = 60°, so I₂ = (I₀/2)cos²(60°) = (I₀/2)(0.25) = I₀/8 = 0.125I₀. Step 3: With a middle filter at 30°: after it, I = (I₀/2)cos²(30°) = (I₀/2)(0.75) = 0.375I₀, now polarised at 30°. The last filter is at 60°, so the angle is again 30°: I_final = 0.375I₀ × cos²(30°) = 0.375 × 0.75 I₀ ≈ 0.281I₀. Inserting a middle filter MORE than doubles the transmitted light — a purely transverse-wave effect with no longitudinal analogue.