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(C) Let $I_0$ be the initial intensity of unpolarized light.$1$. Without the third polarizer: The first polarizer reduces intensity to $I_1 = I_0/2$.

Vedclass · Accepted Answer

$9$/$4$

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(C) Let $I_0$ be the initial intensity of unpolarized light.$1$. Without the third polarizer: The first polarizer reduces intensity to $I_1 = I_0/2$. The second polarizer at $90^\circ$ relative to the first (which is at $30^\circ$) has an angle $	heta = 90^\circ - 30^\circ = 60^\circ$. The output intensity is $I_{final} = I_1 \cos^2(60^\circ) = (I_0/2) 	imes (1/4) = I_0/8$.$2$. With the third polarizer at $60^\circ$: $I_1 = I_0/2$. Intensity after the third polarizer (angle $60^\circ - 30^\circ = 30^\circ$): $I_2 = I_1 \cos^2(30^\circ) = (I_0/2) 	imes (3/4) = 3I_0/8$. Intensity after the second polarizer (angle $90^\circ - 60^\circ = 30^\circ$): $I_{final}' = I_2 \cos^2(30^\circ) = (3I_0/8) 	imes (3/4) = 9I_0/32$.Ratio = $(9I_0/32) / (I_0/8) = 9/4$.

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