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(A) According to Malus's Law,the intensity of transmitted light is given by $I = I_0 \cos^2 	heta$,where $I_0$ is the maximum intensity and $	heta$

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$\frac{I_0}{4}$

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(A) According to Malus's Law,the intensity of transmitted light is given by $I = I_0 \cos^2 	heta$,where $I_0$ is the maximum intensity and $	heta$ is the angle between the transmission axes of the polarizer and the analyzer.Initially,the intensity is maximum,meaning the axes are parallel $(	heta = 0^{\circ})$.When the analyzer is rotated by $60^{\circ}$,the new angle becomes $	heta = 60^{\circ}$.Substituting the values into the formula:$I = I_0 \cos^2(60^{\circ})$$I = I_0 \left( \frac{1}{2} ight)^2$$I = \frac{I_0}{4}$

$A$ polarizer and an analyzer are placed such that the intensity of the transmitted light is maximum. If the analyzer is rotated by $60^{\circ}$,find the intensity of the transmitted light.

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