Unpolarised light passes through two polaroids, the axis of one is vertical and that of the other is at `45^(@)` to the vertical. Then, the intensity of the transmitted light is

Options

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

Explanation

To solve the problem of unpolarised light passing through two polaroids, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- We have unpolarised light with an initial intensity \( I_0 \).
- The first polaroid is oriented vertically.

2. **First Polaroid Effect**:
- When unpolarised light passes through the first polaroid, it becomes polarised. According to Malus's Law, the intensity of the light after passing through the first polaroid is given by:
\[ I_1 = \frac{I_0}{2} \]
- This is because unpolarised light has equal components in all directions, and the first polaroid only allows the vertical component to pass through.

3. **Second Polaroid Orientation**:
- The second polaroid is oriented at an angle of \( 45^\circ \) to the vertical.

4. **Second Polaroid Effect**:
- The intensity of light after passing through the second polaroid can again be calculated using Malus's Law. The formula is:
\[ I_2 = I_1 \cdot \cos^2(\theta) \]
- Here, \( \theta = 45^\circ \) and \( I_1 = \frac{I_0}{2} \).
- Therefore, substituting the values:
\[ I_2 = \frac{I_0}{2} \cdot \cos^2(45^\circ) \]

5. **Calculate \( \cos^2(45^\circ) \)**:
- We know that \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \).
- Thus, \( \cos^2(45^\circ) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \).

6. **Final Calculation**:
- Substituting \( \cos^2(45^\circ) \) back into the equation for \( I_2 \):
\[ I_2 = \frac{I_0}{2} \cdot \frac{1}{2} = \frac{I_0}{4} \]

7. **Conclusion**:
- The intensity of the transmitted light after passing through both polaroids is:
\[ I_2 = \frac{I_0}{4} \]

### Final Answer:
The intensity of the transmitted light is \( \frac{I_0}{4} \).

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