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 of emergent light is reduced to `I/3`. The angle between the polarizers A and C is `theta`. Then :

Options

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

Answer

To solve the problem step by step, we will use Malus's Law, which states that when polarized light passes through a polarizer, the intensity of the transmitted light is given by:

\[ I' = I \cos^2(\theta) \]

where \( I \) is the intensity of the incident light, \( I' \) is the intensity of the transmitted light, and \( \theta \) is the angle between the light's polarization direction and the axis of the polarizer.

### Step 1: Analyze the first scenario with two polarizers A and B
- Given that unpolarized light of intensity \( I \) is incident on polarizer A.
- After passing through polarizer A, the intensity becomes:
\[ I_A = \frac{I}{2} \]
This is because the intensity of light passing through a polarizer reduces to half when unpolarized light passes through it.

- The light then passes through polarizer B. Since the intensity after passing through B is also \( \frac{I}{2} \), we can conclude that polarizers A and B are aligned (angle between A and B is 0 degrees).

### Step 2: Analyze the scenario with the third polarizer C
- Now, we introduce a third polarizer C between A and B.
- The intensity after passing through polarizer A is still \( I_A = \frac{I}{2} \).
- Let the angle between polarizer A and polarizer C be \( \theta \).

- According to Malus's Law, the intensity after passing through polarizer C is:
\[ I_C = I_A \cos^2(\theta) = \frac{I}{2} \cos^2(\theta) \]

### Step 3: Determine the intensity after passing through polarizer B
- The light that emerges from polarizer C then passes through polarizer B.
- The intensity after passing through polarizer B is given by:
\[ I_B = I_C \cos^2(\phi) \]
where \( \phi \) is the angle between polarizer C and polarizer B. Since A and B are aligned, the angle between C and B can be expressed as \( \phi = 90^\circ - \theta \) (because the total angle between A and B is 90 degrees).

- Therefore, the intensity after passing through polarizer B becomes:
\[ I_B = \left(\frac{I}{2} \cos^2(\theta)\right) \cos^2(90^\circ - \theta) = \frac{I}{2} \cos^2(\theta) \sin^2(\theta) \]

### Step 4: Set up the equation based on the given information
- We know from the problem that the intensity after passing through polarizer B is \( \frac{I}{3} \).
- Setting the two expressions for \( I_B \) equal gives us:
\[ \frac{I}{2} \cos^2(\theta) \sin^2(\theta) = \frac{I}{3} \]

### Step 5: Solve for \( \theta \)
- Cancel \( I \) from both sides:
\[ \frac{1}{2} \cos^2(\theta) \sin^2(\theta) = \frac{1}{3} \]

- Multiply both sides by 6 to eliminate the fraction:
\[ 3 \cos^2(\theta) \sin^2(\theta) = 2 \]

- Use the identity \( \sin^2(\theta) = 1 - \cos^2(\theta) \):
\[ 3 \cos^2(\theta) (1 - \cos^2(\theta)) = 2 \]

- Let \( x = \cos^2(\theta) \):
\[ 3x(1 - x) = 2 \]
\[ 3x - 3x^2 = 2 \]
\[ 3x^2 - 3x + 2 = 0 \]

- Solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 3, b = -3, c = -2 \):
\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3} \]
\[ x = \frac{3 \pm \sqrt{9 + 24}}{6} \]
\[ x = \frac{3 \pm \sqrt{33}}{6} \]

### Step 6: Find \( \theta \)
- Since \( x = \cos^2(\theta) \), we can find \( \theta \):
\[ \cos^2(\theta) = \frac{3 + \sqrt{33}}{6} \text{ or } \frac{3 - \sqrt{33}}{6} \]
- The valid solution will be the one that gives a positive value for \( \cos^2(\theta) \).

### Final Answer
- The angle \( \theta \) can be found using:
\[ \theta = \cos^{-1}\left(\sqrt{\frac{2}{3}}\right) \]