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 step by step, we will use Malus's Law, which describes how the intensity of polarized light changes as it passes through a polarizer and an analyzer.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- We know that the intensity of light coming out of the analyzer (I') is 12.5% of the original intensity (I0).
- This can be expressed mathematically as:
\[ I' = 0.125 \times I_0 = \frac{I_0}{8} \]

2. **Applying Malus's Law**:
- According to Malus's Law, the intensity of light after passing through a polarizer and an analyzer is given by:
\[ I = I_0 \cdot \frac{1}{2} \cos^2(\theta) \]
- Here, θ is the angle between the light's polarization direction and the axis of the analyzer.

3. **Setting Up the Equation**:
- From the previous step, we know:
\[ I' = \frac{I_0}{2} \cos^2(\theta_1) \]
- Setting \( I' = \frac{I_0}{8} \):
\[ \frac{I_0}{8} = \frac{I_0}{2} \cos^2(\theta_1) \]

4. **Simplifying the Equation**:
- Cancel \( I_0 \) from both sides:
\[ \frac{1}{8} = \frac{1}{2} \cos^2(\theta_1) \]
- Multiply both sides by 2:
\[ \frac{1}{4} = \cos^2(\theta_1) \]
- Taking the square root gives:
\[ \cos(\theta_1) = \frac{1}{2} \]
- Therefore, \( \theta_1 = 60^\circ \).

5. **Finding the New Intensity Condition**:
- We want to find the angle θ2 such that the intensity doubles from the previous intensity:
\[ I'' = 2 \times I' = 2 \times \frac{I_0}{8} = \frac{I_0}{4} \]

6. **Setting Up the New Equation**:
- Using Malus's Law again:
\[ I'' = \frac{I_0}{2} \cos^2(\theta_2) \]
- Setting \( I'' = \frac{I_0}{4} \):
\[ \frac{I_0}{4} = \frac{I_0}{2} \cos^2(\theta_2) \]

7. **Simplifying the New Equation**:
- Cancel \( I_0 \) from both sides:
\[ \frac{1}{4} = \frac{1}{2} \cos^2(\theta_2) \]
- Multiply both sides by 2:
\[ \frac{1}{2} = \cos^2(\theta_2) \]
- Taking the square root gives:
\[ \cos(\theta_2) = \frac{1}{\sqrt{2}} \]
- Therefore, \( \theta_2 = 45^\circ \).

8. **Calculating the Angle of Rotation**:
- The angle of rotation needed to achieve this new intensity is:
\[ \text{Angle rotated} = \theta_2 - \theta_1 = 45^\circ - 60^\circ = -15^\circ \]
- Since we are looking for the magnitude of the rotation, we take the absolute value:
\[ \text{Angle rotated} = 15^\circ \]

### Final Answer:
The angle by which the analyzer needs to be rotated to double the output intensity is **15 degrees**.