An unpolarized light beam is incident on the polarizer of a polarization experiment and the intensity of light beam emerging from the analyzer is measured as 100 Lumens. Now, if the analyzer is rotated around the horizontal axis (direction of light) by `30^@` in clockwise direction, the intensity of emerging light will be _____ Lumens.

Answer

To solve the problem, we will use the principles of polarization and the formula for the intensity of light passing through a polarizer and analyzer.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- We have an unpolarized light beam incident on a polarizer.
- The intensity of the light emerging from the analyzer is given as 100 Lumens.

2. **Intensity After the Polarizer**:
- When unpolarized light passes through a polarizer, the intensity of the transmitted light is halved.
- Therefore, if the initial intensity of the unpolarized light is \( I_0 \), the intensity after the polarizer \( I' \) is:
\[ I' = \frac{I_0}{2} \]
- Since the intensity after the analyzer is given as 100 Lumens, we can set:
\[ I' = 100 \text{ Lumens} \]
- Thus, we have:
\[ \frac{I_0}{2} = 100 \implies I_0 = 200 \text{ Lumens} \]

3. **Analyzing the Effect of Rotation**:
- The analyzer is rotated by \( 30^\circ \) clockwise.
- The intensity of light emerging from the analyzer after rotation is given by Malus's Law:
\[ I = I' \cos^2(\theta) \]
- Here, \( I' = 100 \text{ Lumens} \) and \( \theta = 30^\circ \).

4. **Calculating the New Intensity**:
- We need to calculate \( \cos^2(30^\circ) \):
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \implies \cos^2(30^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]
- Now substituting back into the intensity formula:
\[ I = 100 \cdot \cos^2(30^\circ) = 100 \cdot \frac{3}{4} = 75 \text{ Lumens} \]

5. **Final Result**:
- The intensity of the emerging light after the analyzer is rotated by \( 30^\circ \) is:
\[ \boxed{75} \text{ Lumens} \]