Sequential application of Malus's law Steps:

• Vertically polarized wave hits filter A at θ=30° to vertical; I₁ = I₀ cos²(30°) = I₀ (√3/2)² = (3/4)I₀.
• Transmitted light polarized at 30° to vertical.
• Second filter at 90° to vertical; angle to first transmission axis is 60°.
• I₂ = I₁ cos²(60°) = (3/4)I₀ (1/2)² = (3/4)I₀ (1/4) = (3/16)I₀ ≈ 0.19 I₀. Why C is correct:
• Malus's law I = I_in cos²θ applied twice yields (3/16)I₀ = 0.19 I₀ after both filters. Why the others are wrong:
• A: Assumes zero transmission for crossed axes, but first filter rotates polarization.
• B: Equals I after first filter only (cos²30° ≈0.75, but ignores second).
• D: Equals cos²60°=0.25 of original, but skips first filter's 3/4 factor.

Final answer: C