An unpolarized light of intensity \(I_o\) passes through polarizer and then through a certain optically active solution and finally it goes to analyser. If the angle between analyser and polariser is \(0°\) and intensity of light emerged from analyser is \(\dfrac{3}{8}I_o\), the angle of rotation of the light by the solution with respect to analyser is _______ degrees.

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: \(30\)