Unpolarized light of intensity I₀ passes through a polarizer, then through an optically active solution, and finally through an analyser. The transmission axes of the polarizer and analyser are parallel. The intensity emerging from the analyser is 3I₀/8. If the solution rotates the plane of polarization by θ relative to the analyser axis, find θ in degrees.
Physics · Wave Optics · polarisation
JEE Mains 2026 6 April Morning Shift

Question

Unpolarized light of intensity I₀ passes through a polarizer, then through an optically active solution, and finally through an analyser. The transmission axes of the polarizer and analyser are parallel. The intensity emerging from the analyser is 3I₀/8. If the solution rotates the plane of polarization by θ relative to the analyser axis, find θ in degrees.

Physics · Wave Optics · polarisation

JEE Mains 2026 6 April Morning Shift

G1 Physics · Accepted Answer

After passing through the first polarizer, the intensity of initially unpolarized light becomes I' = I₀ ÷ 2. The optically active solution rotates the plane of polarization by θ. The analyser axis is parallel to the original polarizer axis, so Malus' law gives I = I'cos²θ. Therefore, 3I₀ ÷ 8 = (I₀ ÷ 2)cos²θ. Cancel I₀: 3 ÷ 8 = (1 ÷ 2)cos²θ. Hence, cos²θ = 3 ÷ 4, cosθ = √3 ÷ 2. For the acute rotation angle, θ = 30°. Therefore the answer is 30.