n polarizing sheets are arranged such that ea | vedclass

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(B) When unpolarized light of intensity $I$ passes through the first polarizing sheet,the intensity becomes $I_1 = \frac{I}{2}$.For subsequent sheets,

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$6$

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(B) When unpolarized light of intensity $I$ passes through the first polarizing sheet,the intensity becomes $I_1 = \frac{I}{2}$.For subsequent sheets,we use Malus' Law: $I_{k} = I_{k-1} \cos^2(	heta)$,where $	heta = 45^{\circ}$.Since $\cos^2(45^{\circ}) = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$,the intensity after each subsequent sheet is halved.After the first sheet,the intensity is $I_1 = \frac{I}{2}$.After the second sheet,$I_2 = I_1 \cos^2(45^{\circ}) = \frac{I}{2} \cdot \frac{1}{2} = \frac{I}{2^2}$.After the $n^{	ext{th}}$ sheet,the intensity is $I_n = \frac{I}{2} \cdot (\frac{1}{2})^{n-1} = \frac{I}{2^n}$.Given $I_n = \frac{I}{64}$,we have $\frac{I}{2^n} = \frac{I}{64}$.$2^n = 64 = 2^6$.Therefore,$n = 6$.

$n$ polarizing sheets are arranged such that each makes an angle $45^{\circ}$ with the preceding sheet. An unpolarized light of intensity $I$ is incident on this arrangement. The output intensity is found to be $\frac{I}{64}$. The value of $n$ will be:

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