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(D) When natural light of intensity $I_0$ passes through the first polaroid,the transmitted intensity is $I_1 = \frac{I_0}{2}$.For subsequent polaroid

Vedclass · Accepted Answer

$\frac{1}{512}$

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(D) When natural light of intensity $I_0$ passes through the first polaroid,the transmitted intensity is $I_1 = \frac{I_0}{2}$.For subsequent polaroids,Malus' Law states $I_n = I_{n-1} \cos^2 	heta$,where $	heta = 60^{\circ}$.Since $\cos 60^{\circ} = \frac{1}{2}$,then $\cos^2 60^{\circ} = \frac{1}{4}$.For $P_2$: $I_2 = I_1 \cos^2 60^{\circ} = \frac{I_0}{2} 	imes \frac{1}{4} = \frac{I_0}{8}$.For $P_3$: $I_3 = I_2 \cos^2 60^{\circ} = \frac{I_0}{8} 	imes \frac{1}{4} = \frac{I_0}{32}$.For $P_4$: $I_4 = I_3 \cos^2 60^{\circ} = \frac{I_0}{32} 	imes \frac{1}{4} = \frac{I_0}{128}$.For $P_5$: $I_5 = I_4 \cos^2 60^{\circ} = \frac{I_0}{128} 	imes \frac{1}{4} = \frac{I_0}{512}$.The fraction of incident intensity is $\frac{I_5}{I_0} = \frac{1}{512}$.

$A$ beam of natural light falls on a system of $5$ polaroids,which are arranged in succession such that the pass axis of each polaroid is turned through $60^{\circ}$ with respect to the preceding one. The fraction of the intensity of incident light that passes through the system is:

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