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(C) When unpolarized light of intensity $I_0$ passes through the first Nicol prism (polarizer),the intensity of the transmitted light becomes $I_1 = \

Vedclass · Accepted Answer

$12.5$

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(C) When unpolarized light of intensity $I_0$ passes through the first Nicol prism (polarizer),the intensity of the transmitted light becomes $I_1 = \frac{I_0}{2}$.According to Malus' Law,when this polarized light passes through the second Nicol prism (analyzer) inclined at an angle $	heta = 60^o$ to the first,the transmitted intensity $I$ is given by:$I = I_1 \cos^2 	heta = \left(\frac{I_0}{2}ight) \cos^2(60^o)$.Since $\cos(60^o) = \frac{1}{2}$,we have $\cos^2(60^o) = \frac{1}{4}$.Therefore,$I = \left(\frac{I_0}{2}ight) 	imes \left(\frac{1}{4}ight) = \frac{I_0}{8}$.The percentage of incident light transmitted is $\left(\frac{I}{I_0}ight) 	imes 100 = \left(\frac{I_0/8}{I_0}ight) 	imes 100 = \frac{100}{8} = 12.5\%$.

Two Nicol prisms are inclined at an angle of $60^o$ to each other. What percentage of the incident unpolarized light passes through the system?

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