Two Polaroids `P_(1)` and `P_(2)` are placed with their axis perpendicular to eachother. Unpolarised light `I_(0)` is nicident on `P_(1)`. A third polaroid `P_(3)` is kept in between `P_(1)` and `P_(2)` such that its axis makes an angle `45^(@)` with that of `P_(1)`. The intensity of transmitted light through `P_(2)` is

Options

A. `l_(0)/(2)`
B. `l_(0)/(4)`
C. `l_(0)/(8)`
D. `l_(0)/(16)`

Answer

To solve the problem, we need to analyze the situation step by step, using Malus's law, which states that when polarized light passes through a polarizer, the intensity of transmitted light is given by:

\[ I = I_0 \cos^2(\theta) \]

where \( I_0 \) is the intensity of the incoming light, and \( \theta \) is the angle between the light's polarization direction and the axis of the polarizer.

### Step-by-Step Solution:

1. **Initial Conditions**:
- We have unpolarized light with intensity \( I_0 \) incident on the first polarizer \( P_1 \).
- The axes of \( P_1 \) and \( P_2 \) are perpendicular to each other (90 degrees apart).

2. **Light Passing Through \( P_1 \)**:
- When unpolarized light passes through the first polarizer \( P_1 \), the intensity of the transmitted light is reduced to half:
\[ I_1 = \frac{I_0}{2} \]

3. **Introducing the Third Polarizer \( P_3 \)**:
- The third polarizer \( P_3 \) is placed between \( P_1 \) and \( P_2 \) at an angle of 45 degrees to the axis of \( P_1 \).
- The intensity of light passing through \( P_3 \) can be calculated using Malus's law:
\[ I_2 = I_1 \cos^2(45^\circ) \]
- Since \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ I_2 = \frac{I_1}{2} = \frac{I_0}{2} \cdot \frac{1}{2} = \frac{I_0}{4} \]

4. **Light Passing Through \( P_2 \)**:
- Now, the light from \( P_3 \) passes through \( P_2 \), which is at 90 degrees to \( P_1 \) and thus 45 degrees to \( P_3 \).
- Again applying Malus's law:
\[ I_3 = I_2 \cos^2(45^\circ) \]
- Substituting \( I_2 \):
\[ I_3 = \frac{I_0}{4} \cdot \cos^2(45^\circ) = \frac{I_0}{4} \cdot \frac{1}{2} = \frac{I_0}{8} \]

5. **Final Result**:
- The intensity of the transmitted light through \( P_2 \) is:
\[ I_3 = \frac{I_0}{8} \]

### Conclusion:
The intensity of the transmitted light through the second polarizer \( P_2 \) is \( \frac{I_0}{8} \).