Angle of Rotation of Light by Optically Active Solution

To find the angle of rotation of the light produced by the optically active solution, we apply the principles of polarization and Malus's Law.

1. Intensity after the Polarizer:
   When unpolarized light of initial intensity $I_0$ passes through an ideal linear polarizer, its intensity is halved: $I_1 = \frac{I_0}{2}$

2. Effect of the Optically Active Solution:
   The optically active solution rotates the plane of polarization of the linearly polarized light by an angle $\theta$.

3. Transmission through the Analyzer:
   Since the angle between the transmission axes of the polarizer and the analyzer is $0^\circ$, the transmission axis of the analyzer is aligned with the original plane of polarization. Thus, after rotation by the solution, the angle between the plane of polarization of the light and the pass axis of the analyzer is $\theta$.

   By Malus's Law, the intensity $I$ of light emerging from the analyzer is given by: $I = I_1 \cos^2\theta = \frac{I_0}{2} \cos^2\theta$

4. Solving for the Angle of Rotation ($\theta$):
   We are given that the final intensity emerging from the analyzer is $I = \frac{3}{8}I_0$. Substituting this into the equation: $\frac{I_0}{2} \cos^2\theta = \frac{3}{8} I_0$

   Dividing both sides by $I_0$ and multiplying by $2$: $\cos^2\theta = \frac{3}{4}$

   Taking the square root on both sides: $\cos\theta = \frac{\sqrt{3}}{2}$

   Solving for $\theta$ in degrees: $\theta = \arccos\left(\frac{\sqrt{3}}{2}\right) = 30^\circ$

Thus, the angle of rotation of the light by the solution with respect to the analyzer is 30 degrees.