Two polaroids are kept crossed to each other . If one of them is rotated an angle `60^@` , the percentage of incident light now transmitted through the system is

Options

A. `10%`
B. `20%`
C. `25%`
D. `12.5%`

Answer

To solve the problem of finding the percentage of incident light transmitted through a system of two crossed polaroids when one is rotated by an angle of \(60^\circ\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- We have two polaroids, initially crossed, meaning they are at \(90^\circ\) to each other. When unpolarized light passes through the first polaroid, it becomes polarized.

2. **Intensity After First Polaroid**:
- The intensity of the light after passing through the first polaroid is given by Malus's Law:
\[ I_1 = I_0 \cdot \frac{1}{2} \]
- Here, \(I_0\) is the intensity of the incident unpolarized light, and \(\frac{1}{2}\) represents the fraction of light transmitted through the first polaroid.

3. **Rotating the Second Polaroid**:
- The second polaroid is rotated by \(60^\circ\) from its original position (which was at \(90^\circ\) to the first). The angle between the light emerging from the first polaroid and the second polaroid is now \(30^\circ\) (since \(90^\circ - 60^\circ = 30^\circ\)).

4. **Intensity After Second Polaroid**:
- The intensity of the light after passing through the second polaroid is again given by Malus's Law:
\[ I_2 = I_1 \cdot \cos^2(30^\circ) \]
- Substituting \(I_1\):
\[ I_2 = \left(I_0 \cdot \frac{1}{2}\right) \cdot \cos^2(30^\circ) \]

5. **Calculating \(\cos^2(30^\circ)\)**:
- We know that \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\), thus:
\[ \cos^2(30^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

6. **Final Intensity Calculation**:
- Now substituting \(\cos^2(30^\circ)\) back into the equation:
\[ I_2 = \left(I_0 \cdot \frac{1}{2}\right) \cdot \frac{3}{4} = I_0 \cdot \frac{3}{8} \]

7. **Calculating the Percentage of Incident Light Transmitted**:
- The percentage of incident light transmitted through the system is given by:
\[ \text{Percentage} = \left(\frac{I_2}{I_0}\right) \times 100 = \left(\frac{3/8}{1}\right) \times 100 = 37.5\% \]

### Final Answer:
The percentage of incident light transmitted through the system is **37.5%**.