A polarizer-analyser set is adjusted such that the intensity of light coming out of the analyser is just 12.5 % of the original intensity of light incident on the polarizer. Assuming that the polarizer-analyser set does not absorb any light, the angle by which the analyser need to be rotated in any sense to increase the output intensity to double of the previous intensity, is :

Options

A. `15^(@)`
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Explanation

To solve the problem, we need to analyze the situation involving a polarizer-analyzer set and how the intensity of light changes as we adjust the angle of the analyzer.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
The intensity of light coming out of the analyzer is given as 12.5% of the original intensity \( I_0 \). This can be expressed mathematically as:
\[ I = 0.125 I_0 \]

2. **Using Malus's Law**:
According to Malus's Law, the intensity of light after passing through a polarizer-analyzer set is given by:
\[ I = I_0 \cos^2(\theta) \]
where \( \theta \) is the angle between the polarizer and the analyzer.

3. **Setting Up the Equation**:
Since the intensity coming out is \( 0.125 I_0 \), we can set up the equation:
\[ 0.125 I_0 = I_0 \cos^2(\theta) \]
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ 0.125 = \cos^2(\theta) \]

4. **Solving for \( \theta \)**:
Taking the square root:
\[ \cos(\theta) = \sqrt{0.125} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4} \]
Now, we can find \( \theta \):
\[ \theta = \cos^{-1}\left(\frac{\sqrt{2}}{4}\right) \]

5. **Finding the New Intensity**:
We want to double the output intensity. The new intensity \( I' \) should be:
\[ I' = 2 \times 0.125 I_0 = 0.25 I_0 \]
Using Malus's Law again for the new intensity:
\[ 0.25 I_0 = I_0 \cos^2(\theta') \]
Dividing by \( I_0 \):
\[ 0.25 = \cos^2(\theta') \]
Taking the square root:
\[ \cos(\theta') = \sqrt{0.25} = \frac{1}{2} \]
Thus:
\[ \theta' = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

6. **Calculating the Angle of Rotation**:
The angle by which the analyzer needs to be rotated is:
\[ \Delta \theta = \theta' - \theta \]
We previously found \( \theta \) to be \( 60^\circ \) and now we need to find the difference:
\[ \Delta \theta = 60^\circ - 45^\circ = 15^\circ \]

### Final Answer:
The angle by which the analyzer needs to be rotated to double the output intensity is:
\[ \Delta \theta = 15^\circ \]