Unpolarized light falls on two polarizing she | vedclass

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Question

(B) Let $I_0$ be the intensity of the unpolarized light.When it passes through the first polarizer,the intensity of the transmitted light is $I_1 = \f

Vedclass · Accepted Answer

$55$

Vedclass · Answer

(B) Let $I_0$ be the intensity of the unpolarized light.When it passes through the first polarizer,the intensity of the transmitted light is $I_1 = \frac{I_0}{2}$.When this light passes through the second polarizer,the intensity of the final transmitted light $I_2$ is given by Malus' Law: $I_2 = I_1 \cos^2 	heta$,where $	heta$ is the angle between the characteristic directions of the sheets.Given that the final intensity $I_2$ is one-third of the maximum intensity of the first transmitted beam $(I_1)$,we have $I_2 = \frac{1}{3} I_1$.Substituting this into the equation: $\frac{1}{3} I_1 = I_1 \cos^2 	heta$.This simplifies to $\cos^2 	heta = \frac{1}{3}$,which means $\cos 	heta = \frac{1}{\sqrt{3}} \approx 0.577$.Calculating the angle: $	heta = \cos^{-1}(0.577) \approx 54.7^o$,which rounds to $55^o$.

Unpolarized light falls on two polarizing sheets placed one on top of the other. What must be the angle between the characteristic directions of the sheets if the intensity of the final transmitted light is one-third the maximum intensity of the first transmitted beam.......$^o$

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