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(A) Let $I_P$ be the intensity of the plane-polarized light and $I_0$ be the intensity of the unpolarized light.When the polarizing sheet is rotated,t

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$3/2$

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(A) Let $I_P$ be the intensity of the plane-polarized light and $I_0$ be the intensity of the unpolarized light.When the polarizing sheet is rotated,the transmitted intensity $I$ is given by $I = I_P \cos^2 	heta + \frac{I_0}{2}$,where $	heta$ is the angle between the polarization direction of the incident polarized light and the transmission axis of the sheet.The maximum intensity occurs at $	heta = 0^{\circ}$,so $I_{\max} = I_P + \frac{I_0}{2}$.The minimum intensity occurs at $	heta = 90^{\circ}$,so $I_{\min} = \frac{I_0}{2}$.Given that the ratio $I_{\max} / I_{\min} = 4$,we have $\frac{I_P + I_0/2}{I_0/2} = 4$.This simplifies to $\frac{I_P}{I_0/2} + 1 = 4$,which means $\frac{2I_P}{I_0} = 3$.Therefore,the ratio $\frac{I_P}{I_0} = \frac{3}{2}$.

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