Question

An unpolarized light of intensity $I_0$ 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 $0^\circ$ and intensity of light emerged from analyser is $\frac{3}{8} I_0$, the angle of rotation of the light by the solution with respect to analyser is ________ degrees.

Hint:

Recall that light intensity becomes $I_0/2$ after the first polarizer. Then use Malus's Law $I = I_{polarised} \cos^2 \theta$, where $\theta$ is the total angle between the plane of polarization and the analyser axis.

Answer 1

Correct Answer: 30

To solve this problem, we track the polarization state of the light. Initially, we have unpolarized light with intensity $I_0$.

1. After the polarizer: The light becomes linearly polarized in a plane defined by the polarizer. The intensity becomes $I_P = \frac{1}{2} I_0$.

2. Rotation by the solution: The optically active solution rotates the plane of polarization by an angle $\phi$. The intensity does not change at this step, so $I_{sol} = I_P = \frac{I_0}{2}$.

3. Passage through the Analyser: The light incident on the analyser has its plane of polarization at an angle $\phi$ relative to the original polarizer axis. Since the analyser and polarizer are parallel (angle between them is $0^\circ$), the angle between the light's polarization and the analyser's transmission axis is simply $\phi$.

Applying Malus's Law for the final intensity $I_f$:
$$I_f = I_{sol} \cos^2 \phi$$
$$\frac{3}{8} I_0 = \frac{I_0}{2} \cos^2 \phi$$
$$\cos^2 \phi = \frac{3}{4}$$
$$\cos \phi = \frac{\sqrt{3}}{2}$$
$$\phi = \arccos\left(\frac{\sqrt{3}}{2}\right) = 30^\circ$$
Therefore, the rotation angle is 30 degrees.