An unpolarized light of intensity I_o passes through polarizer and...

Intensity of light after passing through the first polarizer is:

$I_1 = \dfrac{I_o}{2}$

Let the optically active solution rotate the plane of polarization by an angle $\theta$. Since the angle between the polarizer and the analyser is $0^\circ$, the angle between the plane of polarization of the light emerging from the solution and the transmission axis of the analyser is also $\theta$.

According to Malus's law, the intensity of light emerging from the analyser is:

$I_2 = I_1 \cos^2 \theta = \dfrac{I_o}{2} \cos^2 \theta$

Given that the final intensity is $\dfrac{3}{8}I_o$, we have:

$\dfrac{I_o}{2} \cos^2 \theta = \dfrac{3}{8}I_o$

$\cos^2 \theta = \dfrac{3}{4}$

$\cos \theta = \dfrac{\sqrt{3}}{2}$

$\theta = 30^\circ$

Answer: $30$