Plane-polarized light of intensity I₀ is incident on a sequence of three ideal polaroids whose transmission axes are at 0°, θ and 90°, respectively. The electric field of the incident light is parallel to the 0° axis. If the final transmitted intensity is I₀ ÷ 16, what is the least positive value of θ?
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Question

Plane-polarized light of intensity I₀ is incident on a sequence of three ideal polaroids whose transmission axes are at 0°, θ and 90°, respectively. The electric field of the incident light is parallel to the 0° axis. If the final transmitted intensity is I₀ ÷ 16, what is the least positive value of θ?

Options

G1 Physics · Accepted Answer

Because the incident light is already polarized along the first axis, the first polaroid does not reduce its intensity. After the second polaroid, I₂ = I₀ cos²θ. The angle between the second and third axes is 90°−θ. Hence, I₃ = I₀ cos²θ sin²θ = I₀ ÷ 4 × sin²2θ. Given, I₃ = I₀ ÷ 16. Therefore, sin²2θ = 1 ÷ 4. For the least positive solution, sin2θ = 1 ÷ 2, so 2θ = 30°. Hence, θ = 15°. Correct option: A.