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(D) According to Malus' Law,the intensity of transmitted light $I$ through a polaroid is given by $I = I_0 \cos^2(\phi)$,where $I_0$ is the maximum in

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$45$

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(D) According to Malus' Law,the intensity of transmitted light $I$ through a polaroid is given by $I = I_0 \cos^2(\phi)$,where $I_0$ is the maximum intensity and $\phi$ is the angle between the polarizing directions of the two polaroids.Initially,the polarizing directions are parallel,so $\phi = 0^\circ$ and $I = I_0$.We want the new intensity $I'$ to be half of the initial intensity,so $I' = \frac{I_0}{2}$.Substituting this into Malus' Law: $\frac{I_0}{2} = I_0 \cos^2(\phi)$.This simplifies to $\cos^2(\phi) = \frac{1}{2}$,which means $\cos(\phi) = \frac{1}{\sqrt{2}}$.Therefore,$\phi = 45^\circ$.

Two polaroids have their polarizing directions parallel so that the intensity of the transmitted light is maximum. The angle through which either polaroid must be turned so that the intensity of the transmitted light is dropped to one-half of the previous transmitted intensity is $.....^\circ$.

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