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 determining the percentage of incident light transmitted through two crossed polaroids when one is rotated by \(60^\circ\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Setup**:
- We have two polaroids, P1 and P2, initially crossed (i.e., at \(90^\circ\) to each other). When unpolarized light (intensity \(I_0\)) hits the first polaroid (P1), only the component of light aligned with the polaroid's axis will pass through.

2. **Calculate the Intensity After the First Polaroid**:
- According to Malus's Law, when unpolarized light passes through a polaroid, the transmitted intensity \(I_1\) is given by:
\[ I_1 = \frac{I_0}{2} \]
- This is because only half of the unpolarized light passes through the first polaroid.

3. **Determine the Angle for the Second Polaroid**:
- The second polaroid (P2) is rotated \(60^\circ\) from the vertical (the axis of P1). Therefore, the angle between the transmitted light from P1 and the axis of P2 is \(60^\circ\).

4. **Calculate the Intensity After the Second Polaroid**:
- Again using Malus's Law, the intensity \(I_2\) after the second polaroid is given by:
\[ I_2 = I_1 \cos^2(\theta) \]
- Here, \(\theta = 60^\circ\), so:
\[ I_2 = I_1 \cos^2(60^\circ) \]
- We know that \(\cos(60^\circ) = \frac{1}{2}\), thus:
\[ I_2 = I_1 \left(\frac{1}{2}\right)^2 = I_1 \cdot \frac{1}{4} \]

5. **Substituting \(I_1\) into the Equation**:
- Now substitute \(I_1 = \frac{I_0}{2}\):
\[ I_2 = \left(\frac{I_0}{2}\right) \cdot \frac{1}{4} = \frac{I_0}{8} \]

6. **Calculate the Percentage of Transmitted Light**:
- To find the percentage of the incident light transmitted through the system, we calculate:
\[ \text{Percentage} = \left(\frac{I_2}{I_0}\right) \times 100 = \left(\frac{I_0/8}{I_0}\right) \times 100 = \frac{1}{8} \times 100 = 12.5\% \]

### Final Answer:
The percentage of incident light transmitted through the system is **12.5%**.