Question 177: Wave Optics

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.

Accepted Answer

When unpolarized light of intensity $\mathrm{I}_0$ passes through a polarizer, the intensity of the transmitted polarized light $\left(I_1ight)$ is always half of the initial intensity.
$$ \mathrm{I}_1=\frac{\mathrm{I}_0}{2} $$
An optically active solution rotates the plane of polarization by a certain angle. Let the angle of rotation is $	heta$. After passing through the solution, the light is still plane-polarized, but its axis has shifted by $	heta$.
Using Malus's Law when completely plane-polarized light is incident on an analyzer, the intensity I of the light transmitted by the analyzer is directly proportional to the square of the cosine of the angle between the transmission axes of the analyzer and the polarizer.
$$ \mathrm{I}=\mathrm{I}_{	ext {incident }} \cos ^2(\phi) $$
Here, $\phi$ is the final angle between the light's polarization plane (rotated by the solution) and the analyzer's axis.
The initial intensity after polarizer is $\mathrm{I}_1=\frac{\mathrm{I}_0}{2}$ and the final intensity after analyser is $\mathrm{I}=\frac{3}{8} \mathrm{I}_0$.
Since the analyzer and polarizer were originally at $0^{\circ}$, the angle $\phi$ in Malus's law is the rotation angle $	heta$ caused by the solution.
Substituting the values into Malus's Law:
$$ \frac{3}{8} I_0=\left(\frac{I_0}{2}ight) \cos ^2 	heta $$
$$\Rightarrow $$ $\frac{3 / 8}{1 / 2}=\cos ^2(	heta)$
$$\Rightarrow $$ $\frac{3}{4}=\cos ^2(	heta)$
$$\Rightarrow $$ $\cos (	heta)=\frac{\sqrt{3}}{2}$
$$\Rightarrow $$ $$ 	heta=30^{\circ} $$
Therefore, the angle of rotation is 30 degrees.

Wave Optics

Integer / Numeric Answer Type