Unpolarized light of intensity I₀ is sent through an arrangement of 4 polarizers. Each polarizer (after the first) is rotated 30° relative to the preceding one, so that the last polarizer is perpendicular to the first. What is the intensity transmitted by this system? A) 0.75 I₀ B) 0.50 I₀ C) 0.42 I₀ D) 0.21 I₀ E) 0

Question

Quizplus · Accepted Answer

Each polarizer cuts the intensity of the light in half. After passing through the first polarizer, the intensity will be $I_0/2$. After passing through the second polarizer, which is rotated 30° relative to the first, the intensity will be $(I_0/2) \cdot \cos^2(30^\circ) = (I_0/4)$. After passing through the third polarizer, which is rotated 30° relative to the second, the intensity will be $(I_0/4) \cdot \cos^2(30^\circ) = (I_0/8)$. Finally, after passing through the fourth polarizer, which is rotated 30° relative to the third, the intensity will be $(I_0/8) \cdot \cos^2(30^\circ) = (I_0/16)$. Therefore, the transmitted intensity is 0.5 * 0.5 * 0.5 * 0.5 = 0.0625 times the original intensity, or 0.21 $I_0$.

Unpolarized Light of Intensity I0 Is Sent Through an Arrangement

Unpolarized Light of Intensity I₀ Is Sent Through an Arrangement