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Q.
Plane polarised light is passed through an analyser and the intensity of emerging light is reduced by $75 \%$. Optical vibrations make an angle $\theta$ with the axis of analyser. Then $\theta$ is
Suppose intensity of light incident on the polariser be $I_0$. When it passes through polariser, its intensity is reduced to $\frac{I_0}{2}$
$ \, \, \, \, \, I = \frac{I_0}{2} \cos^2 \alpha \, \, \, \, \, \, \, \, $ (Malus's law)
where a is the angle between the plane of transmission analyser and polariser
According to the question,
$I = 75 \% $ of $\frac{I_0}{2} = \frac{3}{4} \times \frac{I_0}{2}$
Hence, $\cos^2 \alpha = \frac{3}{4}$ or $\cos \alpha = \frac{\sqrt{3}}{2} = \cos 30^{\circ}$
$ \therefore \, \alpha = 30^{\circ}$