Unpolarised light falls on two polarizing she | vedclass

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(B) Let the intensity of the unpolarised light be $I_0$. After passing through the first polariser,the intensity of the transmitted light is $I_1 = \f

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$

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(B) Let the intensity of the unpolarised light be $I_0$. After passing through the first polariser,the intensity of the transmitted light is $I_1 = \frac{I_0}{2}$.This $I_1$ is the maximum intensity of the first transmitted beam.According to Malus's Law,the intensity of light transmitted through the second polariser is $I_2 = I_1 \cos^2 	heta$,where $	heta$ is the angle between the characteristic directions of the two sheets.We are given that $I_2 = \frac{1}{3} I_1$.Substituting this into the equation: $\frac{1}{3} I_1 = I_1 \cos^2 	heta$.$\cos^2 	heta = \frac{1}{3}$.$\cos 	heta = \frac{1}{\sqrt{3}}$.$	heta = \cos^{-1}\left(\frac{1}{\sqrt{3}}ight)$.

Unpolarised 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?

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