Two Polaroids `P_1` and `P_2` are placed with their axis perpendicular to each other. Un-polarized light `I_0` is incident 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. `I_0/4`
B. `I_0/8`
C. `I_0/16`
D. `I_0/2`

Explanation

To solve the problem, we will use Malus's Law, which states that when polarized light passes through a polarizer, the intensity of the transmitted light is given by:

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

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

### Step-by-Step Solution

1. **Incident Light on Polarizer P1**:
- The incident light \( I_0 \) is unpolarized. When it passes through the first polarizer \( P_1 \), the intensity of the transmitted light \( I_1 \) is given by:
\[ I_1 = \frac{I_0}{2} \]
(This is because unpolarized light passing through a polarizer reduces its intensity by half.)

2. **Light Passing Through Polarizer P3**:
- The angle between the axis of \( P_1 \) and \( P_3 \) is \( 45^\circ \). Therefore, we can apply Malus's Law to find the intensity \( I_2 \) after passing through \( P_3 \):
\[ I_2 = I_1 \cos^2(45^\circ) = \frac{I_0}{2} \cdot \cos^2(45^\circ) \]
- Since \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \), we have:
\[ \cos^2(45^\circ) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
- Thus, substituting this value:
\[ I_2 = \frac{I_0}{2} \cdot \frac{1}{2} = \frac{I_0}{4} \]

3. **Light Passing Through Polarizer P2**:
- The angle between the axis of \( P_3 \) and \( P_2 \) is also \( 45^\circ \) (since \( P_2 \) is perpendicular to \( P_1 \)). Therefore, we again apply Malus's Law:
\[ I_3 = I_2 \cos^2(45^\circ) = \frac{I_0}{4} \cdot \cos^2(45^\circ) \]
- Again substituting \( \cos^2(45^\circ) = \frac{1}{2} \):
\[ I_3 = \frac{I_0}{4} \cdot \frac{1}{2} = \frac{I_0}{8} \]

4. **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 \( P_2 \) is \( \frac{I_0}{8} \).