Three identical polaroids P_1, P_2, and P_3 a | vedclass

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Question

(B) When unpolarized light of intensity $I_0$ passes through the first polaroid $P_1$, the intensity of the transmitted light is $I_1 = \frac{I_0}{2}$

Vedclass · Accepted Answer

$\frac{I_0}{2} \cos^2 60^\circ \cos^2 30^\circ$

Vedclass · Answer

(B) When unpolarized light of intensity $I_0$ passes through the first polaroid $P_1$, the intensity of the transmitted light is $I_1 = \frac{I_0}{2}$.According to Malus's Law, the intensity of light passing through a polaroid is $I = I_{incident} \cos^2 	heta$, where $	heta$ is the angle between the polarization direction of incident light and the pass axis of the polaroid.For $P_2$, the angle between its axis and the axis of $P_1$ is $60^\circ$. Thus, the intensity after $P_2$ is $I_2 = I_1 \cos^2 60^\circ = \frac{I_0}{2} \cos^2 60^\circ$.For $P_3$, the angle between its axis ($90^\circ$ to $P_1$) and the axis of $P_2$ ($60^\circ$ to $P_1$) is $	heta = 90^\circ - 60^\circ = 30^\circ$.Thus, the final intensity after $P_3$ is $I_3 = I_2 \cos^2 30^\circ = (\frac{I_0}{2} \cos^2 60^\circ) \cos^2 30^\circ = \frac{I_0}{2} \cos^2 60^\circ \cos^2 30^\circ$.

Three identical polaroids $P_1, P_2$, and $P_3$ are placed one after another. The pass axes of $P_2$ and $P_3$ are inclined at angles of $60^\circ$ and $90^\circ$ with respect to the axis of $P_1$. The source has an intensity $I_0$. The intensity of light finally coming out is

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