Three identical polaroids P_1, P_2 and P_3 ar | vedclass

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(C) According to Malus' law,the intensity of light transmitted through a polarizer is $I = I_{in} \cos^2 	heta$,where $	heta$ is the angle between t

Vedclass · Accepted Answer

$\frac{3 I_0}{32}$

Vedclass · Answer

(C) According to Malus' law,the intensity of light transmitted through a polarizer is $I = I_{in} \cos^2 	heta$,where $	heta$ is the angle between the polarization direction of incident light and the pass axis of the polarizer.$1$. When unpolarized light of intensity $I_0$ passes through the first polaroid $P_1$,the intensity of the transmitted light is $I_1 = \frac{I_0}{2}$.$2$. The light from $P_1$ is now polarized at an angle $0^{\circ}$ relative to $P_1$'s axis. $P_2$ is at $60^{\circ}$ to $P_1$. Thus,the intensity after $P_2$ is:$I_2 = I_1 \cos^2(60^{\circ}) = \frac{I_0}{2} 	imes (0.5)^2 = \frac{I_0}{2} 	imes \frac{1}{4} = \frac{I_0}{8}$.$3$. The light from $P_2$ is polarized at $60^{\circ}$ relative to $P_1$. $P_3$ is at $90^{\circ}$ relative to $P_1$. The angle between the polarization of light from $P_2$ and the pass axis of $P_3$ is $	heta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.$4$. The intensity after $P_3$ is:$I_3 = I_2 \cos^2(30^{\circ}) = \frac{I_0}{8} 	imes \left(\frac{\sqrt{3}}{2}ight)^2 = \frac{I_0}{8} 	imes \frac{3}{4} = \frac{3 I_0}{32}$.

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