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.
Optically Active Solution Rotation Angle Calculation
This problem involves calculating the angle of rotation caused by an optically active solution using principles of polarization and Malus's Law.
Step-by-Step Solution
- Initial Intensity: The unpolarized light has an initial intensity of $I_0$.
- After First Polarizer: When unpolarized light passes through a polarizer, its intensity reduces to half. So, the intensity after the first polarizer is $I_1 = \frac{I_0}{2}$. The light is now linearly polarized.
- Rotation by Solution: The optically active solution rotates the plane of polarization of the light by an angle $\theta$.
- Malus's Law: The light then passes through an analyzer. Let the angle between the transmission axis of the analyzer and the plane of polarization of the light emerging from the solution be $\phi$. According to Malus's Law, the final intensity $I_f$ is given by $I_f = I_{incident} \cos^2 \phi$.
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Applying Malus's Law:
- The intensity incident on the analyzer is $I_1 = \frac{I_0}{2}$.
- The angle between the analyzer and the polarizer is given as $0^\circ$. This means the analyzer's axis is aligned with the first polarizer's axis.
- The solution rotates the light by $\theta$. Therefore, the angle between the rotated plane of polarization and the analyzer's axis is $\phi = \theta$.
- The final intensity is given as $I_f = \frac{3}{8} I_0$.
- Calculation: Substitute the values into Malus's Law: $ \frac{3}{8} I_0 = \frac{I_0}{2} \cos^2 \theta $ Divide both sides by $I_0$: $ \frac{3}{8} = \frac{1}{2} \cos^2 \theta $ Solve for $\cos^2 \theta$: $ \cos^2 \theta = \frac{3}{8} \times 2 = \frac{3}{4} $ Solve for $\cos \theta$: $ \cos \theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} $ Find the angle $\theta$: $ \theta = \arccos\left(\frac{\sqrt{3}}{2}\right) $ The angle $\theta$ is $30^\circ$.
The angle of rotation of the light by the solution with respect to the analyser is $30^\circ$.
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