An unpolarised beam of intensity `I_(0)` is incident on a pair of nicols making an angle of `60^(@)` with each other. The intensity of light emerging from the pair is

Options

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

Explanation

To solve the problem of finding the intensity of light emerging from a pair of nicols (polarizers) making an angle of \(60^\circ\) with each other, we can follow these steps:

### Step-by-Step Solution:

1. **Initial Intensity of Unpolarized Light**:
Let the initial intensity of the unpolarized light be \(I_0\).

2. **First Polarizer**:
When unpolarized light passes through the first polarizer (Nicol), the intensity of the transmitted light is given by:
\[ I_1 = \frac{I_0}{2} \]
This is because unpolarized light can be considered to have equal components in all directions, and the average transmitted intensity through a polarizer is half of the incident intensity.

3. **Second Polarizer**:
The second polarizer is oriented at an angle of \(60^\circ\) with respect to the first polarizer. The intensity of light transmitted through the second polarizer is given by Malus's Law:
\[ I_2 = I_1 \cos^2(\theta) \]
where \(\theta\) is the angle between the transmission axes of the two polarizers. Here, \(\theta = 60^\circ\).

4. **Calculating the Final Intensity**:
Substituting \(I_1\) into the equation for \(I_2\):
\[ I_2 = \left(\frac{I_0}{2}\right) \cos^2(60^\circ) \]
We know that \(\cos(60^\circ) = \frac{1}{2}\), so:
\[ \cos^2(60^\circ) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
Therefore:
\[ I_2 = \left(\frac{I_0}{2}\right) \cdot \frac{1}{4} = \frac{I_0}{8} \]

5. **Final Result**:
The intensity of light emerging from the pair of nicols is:
\[ I_2 = \frac{I_0}{8} \]

### Conclusion:
The intensity of light emerging from the pair of nicols is \(\frac{I_0}{8}\).

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