Two polaroids are crossed. If now one of them is rotated through `30^(@)` and unpolarised light of intensity `l_(0)` is incident on the first polaroid, then the intensity of transmitted light will be

Options

A. `(I_(0))/(4)`
B. `(3I_(0))/(4)`
C. `(3I_(0))/(8)`
D. `(I_(0))/(8)`

Answer

To solve the problem of finding the intensity of transmitted light when two polaroids are crossed and one is rotated through \(30^\circ\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- Initially, we have two polaroids that are crossed, meaning the angle between their transmission axes is \(90^\circ\).
- Unpolarized light with intensity \(I_0\) is incident on the first polaroid.

2. **Intensity After First Polaroid**:
- When unpolarized light passes through a polaroid, the intensity of the transmitted light is given by:
\[ I_1 = \frac{I_0}{2} \]
- This is because the average value of \(\cos^2 \theta\) for unpolarized light is \(\frac{1}{2}\).

3. **Rotating the Second Polaroid**:
- The second polaroid is rotated by \(30^\circ\). Therefore, the angle between the first polaroid (which is fixed) and the second polaroid (after rotation) is:
\[ \theta = 90^\circ - 30^\circ = 60^\circ \]

4. **Intensity After Second Polaroid**:
- Now, we apply Malus's Law for the second polaroid. The intensity after passing through the second polaroid is given by:
\[ I_2 = I_1 \cos^2 \theta \]
- Substituting \(I_1\) and \(\theta\):
\[ I_2 = \left(\frac{I_0}{2}\right) \cos^2(60^\circ) \]
- Since \(\cos(60^\circ) = \frac{1}{2}\), we have:
\[ I_2 = \left(\frac{I_0}{2}\right) \left(\frac{1}{2}\right)^2 = \left(\frac{I_0}{2}\right) \left(\frac{1}{4}\right) = \frac{I_0}{8} \]

5. **Final Result**:
- The intensity of the transmitted light after passing through both polaroids is:
\[ I_2 = \frac{I_0}{8} \]

### Conclusion:
The intensity of the transmitted light when one of the crossed polaroids is rotated through \(30^\circ\) is \(\frac{I_0}{8}\).

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