Unpolarized light of intensity `I` is incident on a system of two polarizers, A followed by B. The intensity of emergent light is `I//2`. If a third polarizer C is placed between A and B the intensity between the polarizers A and C is `theta`, then

Options

A. `cos theta = (2/3)^(1/4)`
B. `cos theta = (1/3)^(1/4)`
C. `cos theta = (1/3)^(1/2)`
D. `cos theta = (2/3)^(1/2)`

Explanation

To solve the problem step by step, we will analyze the situation involving the polarizers and apply Malus's law.

### Step 1: Understand the initial conditions
We have unpolarized light of intensity \( I \) incident on two polarizers A and B. The intensity of the emergent light after passing through both polarizers is given as \( \frac{I}{2} \).

**Hint:** Remember that unpolarized light passing through the first polarizer becomes polarized and its intensity is reduced.

### Step 2: Apply Malus's Law between A and B
According to Malus's law, when unpolarized light passes through a polarizer, the intensity after the first polarizer (A) is given by:
\[ I_A = \frac{I}{2} \]
This is because the intensity of unpolarized light is halved upon passing through the first polarizer.

Now, since the emergent intensity after the second polarizer (B) is also \( \frac{I}{2} \), we can set up the equation:
\[ I_B = I_A \cos^2(\phi) = \frac{I}{2} \cos^2(\phi) \]
where \( \phi \) is the angle between the transmission axes of polarizers A and B.

Since \( I_B = \frac{I}{2} \), we can equate:
\[ \frac{I}{2} \cos^2(\phi) = \frac{I}{2} \]
This implies:
\[ \cos^2(\phi) = 1 \quad \Rightarrow \quad \phi = 0^\circ \]
Thus, the angle between polarizers A and B is \( 0^\circ \).

**Hint:** The angle \( \phi \) between the polarizers affects the intensity; if the angle is zero, the light passes through without further loss.

### Step 3: Introduce the third polarizer C
Now, we place a third polarizer C between A and B. Let the angle between A and C be \( \theta \). The intensity after polarizer C can be calculated using Malus's law again:
\[ I_C = I_A \cos^2(\theta) = \frac{I}{2} \cos^2(\theta) \]

### Step 4: Calculate the intensity after polarizer B
The intensity after polarizer B (which is after C) can be expressed as:
\[ I_B = I_C \cos^2(0^\circ) = I_C \]
Thus, substituting for \( I_C \):
\[ I_B = \frac{I}{2} \cos^2(\theta) \]

### Step 5: Set up the equation with the given emergent intensity
We know from the problem statement that the intensity after polarizer B is \( \frac{I}{3} \):
\[ \frac{I}{2} \cos^2(\theta) = \frac{I}{3} \]

### Step 6: Solve for \( \cos^2(\theta) \)
To find \( \cos^2(\theta) \), we can rearrange the equation:
\[ \cos^2(\theta) = \frac{I/3}{I/2} = \frac{2}{3} \]

### Step 7: Find \( \cos(\theta) \)
Taking the square root gives us:
\[ \cos(\theta) = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} \]

### Final Result
Thus, the value of \( \cos(\theta) \) is:
\[ \cos(\theta) = \sqrt{\frac{2}{3}} \]