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`

Answer

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

### Step 1: Understand the Initial Conditions
We start with an unpolarized beam of light with intensity \(I_0\). When unpolarized light passes through a polarizer, its intensity is reduced to half.

### Step 2: Calculate the Intensity After the First Nicol
When the unpolarized light passes through the first Nicol (polarizer), the intensity \(I_1\) after the first Nicol can be calculated using Malus's Law. For unpolarized light, the intensity after passing through the first polarizer is:
\[ I_1 = \frac{I_0}{2} \]

### Step 3: Calculate the Intensity After the Second Nicol
The second Nicol is oriented at an angle of \(60^\circ\) to the first Nicol. According to Malus's Law, the intensity after passing through the second polarizer is given by:
\[ I_2 = I_1 \cdot \cos^2(\theta) \]
where \(\theta\) is the angle between the two polarizers. Substituting the values we have:
\[ I_2 = \left(\frac{I_0}{2}\right) \cdot \cos^2(60^\circ) \]

### Step 4: Calculate \(\cos^2(60^\circ)\)
We know that:
\[ \cos(60^\circ) = \frac{1}{2} \]
Thus,
\[ \cos^2(60^\circ) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

### Step 5: Substitute Back to Find \(I_2\)
Now substituting \(\cos^2(60^\circ)\) back into the equation for \(I_2\):
\[ I_2 = \left(\frac{I_0}{2}\right) \cdot \frac{1}{4} = \frac{I_0}{8} \]

### Conclusion
The intensity of light emerging from the pair of nicols is:
\[ I_2 = \frac{I_0}{8} \]