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

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 and \( \theta \) is the angle between the light's polarization direction and the axis of the polarizer.

### Step-by-Step Solution:

1. **Initial Conditions**:
- Let the intensity of the unpolarized light incident on Polaroid \( P_1 \) be \( I_0 \).
- The axes of \( P_1 \) and \( P_2 \) are perpendicular to each other (90 degrees).

2. **Intensity after \( P_1 \)**:
- When unpolarized light 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 has equal components in all directions, and a polarizer transmits half the intensity.

3. **Intensity after \( P_3 \)**:
- The third polarizer \( P_3 \) is placed at an angle of \( 45^\circ \) with respect to \( P_1 \).
- Using Malus's Law, the intensity \( I_3 \) after passing through \( P_3 \) is given by:
\[ I_3 = I_1 \cos^2(45^\circ) \]
- Substituting \( I_1 \):
\[ I_3 = \frac{I_0}{2} \cos^2(45^\circ) \]
- Since \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ I_3 = \frac{I_0}{2} \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2} \cdot \frac{1}{2} = \frac{I_0}{4} \]

4. **Intensity after \( P_2 \)**:
- Now, \( P_2 \) is at \( 90^\circ \) to \( P_1 \) and \( 45^\circ \) to \( P_3 \).
- Again applying Malus's Law, the intensity \( I_2 \) after passing through \( P_2 \) is:
\[ I_2 = I_3 \cos^2(45^\circ) \]
- Substituting \( I_3 \):
\[ I_2 = \frac{I_0}{4} \cos^2(45^\circ) \]
- Thus:
\[ I_2 = \frac{I_0}{4} \cdot \frac{1}{2} = \frac{I_0}{8} \]

### Final Result:
The intensity of the transmitted light through \( P_2 \) is:
\[ I_2 = \frac{I_0}{8} \]