Two Nicols are oriented with their principal planes making an angle of `60^@`. The percentage of incident unpolarised light which passes through the system is

Options

A. `50%`
B. `100%`
C. `12.5%`
D. `37.5%`

Answer

To solve the problem of finding the percentage of incident unpolarized light that passes through two Nicol prisms oriented at an angle of 60 degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- We have unpolarized light incident on the first Nicol prism (Nicol 1).
- The intensity of the incident light is denoted as \( I_0 \).

2. **First Nicol Prism**:
- When unpolarized light passes through the first Nicol prism, the intensity of the transmitted light is reduced to half.
- Therefore, the intensity after Nicol 1, denoted as \( I_1 \), is given by:
\[ I_1 = \frac{I_0}{2} \]

3. **Second Nicol Prism**:
- The second Nicol prism (Nicol 2) is oriented at an angle of 60 degrees with respect to the first Nicol prism.
- The intensity of light transmitted through Nicol 2 can be calculated using Malus's Law, which states:
\[ I_2 = I_1 \cos^2(\theta) \]
- Here, \( \theta = 60^\circ \). Thus, we can substitute \( I_1 \) into the equation:
\[ I_2 = \left(\frac{I_0}{2}\right) \cos^2(60^\circ) \]

4. **Calculating \( \cos(60^\circ) \)**:
- We know that:
\[ \cos(60^\circ) = \frac{1}{2} \]
- Therefore, \( \cos^2(60^\circ) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).

5. **Substituting Back**:
- Now substituting \( \cos^2(60^\circ) \) back into the equation for \( I_2 \):
\[ I_2 = \left(\frac{I_0}{2}\right) \cdot \frac{1}{4} = \frac{I_0}{8} \]

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

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

---