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

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$12$

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(D) When natural (unpolarized) light of intensity $I_0$ passes through the first polaroid,the transmitted intensity becomes $I_1 = \frac{I_0}{2}$.For the subsequent $5$ polaroids,we apply Malus' Law: $I_n = I_{n-1} \cos^2(	heta)$,where $	heta = 30^\circ$.The final intensity $I_6$ after passing through all $6$ polaroids is given by:$I_6 = I_1 	imes (\cos^2 30^\circ)^5 = \frac{I_0}{2} 	imes (\cos^2 30^\circ)^5$.Substituting $\cos 30^\circ = \frac{\sqrt{3}}{2}$:$I_6 = \frac{I_0}{2} 	imes \left(\frac{3}{4}ight)^5 = \frac{I_0}{2} 	imes \frac{243}{1024} = \frac{243}{2048} I_0 \approx 0.1186 I_0$.Converting to percentage: $0.1186 	imes 100 \approx 11.86\% \approx 12\%$.

$A$ beam of natural light falls on a system of $6$ polaroids,which are arranged in succession such that each polaroid is turned through $30^\circ$ with respect to the preceding one. The percentage of incident intensity that passes through the system will be....$\%$

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