An unpolarized light of intensity `I_(0)` passes through three polarizers, sach that the transmission axis of last polarizer is perpendicular to that of first. If the intensity of emergent light is `(3I_(0))/(32)` and the angle between the transmission axis of first two polarizers is `theta`, then

Options

A. `theta=45^(@)`
B. `theta=37^(@)`
C. `theta=30^(@)`
D. `theta=53^(@)`

Answer

To solve the problem, 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:

\[ I = I_0 \cos^2(\theta) \]

where \( I_0 \) is the intensity of the incoming light, \( I \) is the intensity of the transmitted light, and \( \theta \) is the angle between the light's polarization direction and the polarizer's transmission axis.

### Step-by-Step Solution:

1. **Initial Setup:**
- Let the intensity of the unpolarized light be \( I_0 \).
- The first polarizer (P1) will reduce the intensity of the unpolarized light to half:
\[ I_1 = \frac{I_0}{2} \]

2. **Intensity after the Second Polarizer (P2):**
- The angle between the transmission axes of the first polarizer (P1) and the second polarizer (P2) is \( \theta \).
- Using Malus's Law, the intensity after the second polarizer (I2) is:
\[ I_2 = I_1 \cos^2(\theta) = \frac{I_0}{2} \cos^2(\theta) \]

3. **Intensity after the Third Polarizer (P3):**
- The transmission axis of the third polarizer (P3) is perpendicular to that of the first polarizer, which means the angle between P2 and P3 is \( 90^\circ - \theta \).
- Again, applying Malus's Law, the intensity after the third polarizer (I3) is:
\[ I_3 = I_2 \cos^2(90^\circ - \theta) = I_2 \sin^2(\theta) \]
- Substituting \( I_2 \):
\[ I_3 = \left(\frac{I_0}{2} \cos^2(\theta)\right) \sin^2(\theta) = \frac{I_0}{2} \cos^2(\theta) \sin^2(\theta) \]

4. **Using the Given Information:**
- According to the problem, the intensity of the emergent light is given as:
\[ I_3 = \frac{3 I_0}{32} \]
- Setting the two expressions for \( I_3 \) equal to each other:
\[ \frac{I_0}{2} \cos^2(\theta) \sin^2(\theta) = \frac{3 I_0}{32} \]

5. **Canceling \( I_0 \):**
- Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ \frac{1}{2} \cos^2(\theta) \sin^2(\theta) = \frac{3}{32} \]

6. **Multiplying through by 2:**
- This gives:
\[ \cos^2(\theta) \sin^2(\theta) = \frac{3}{16} \]

7. **Using the Identity:**
- We can use the identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \):
\[ \sin^2(2\theta) = 4 \sin^2(\theta) \cos^2(\theta) \]
- Therefore:
\[ \frac{1}{4} \sin^2(2\theta) = \frac{3}{16} \]

8. **Solving for \( \sin^2(2\theta) \):**
- Multiplying both sides by 4:
\[ \sin^2(2\theta) = \frac{3}{4} \]

9. **Finding \( 2\theta \):**
- Taking the square root:
\[ \sin(2\theta) = \frac{\sqrt{3}}{2} \]
- This implies:
\[ 2\theta = 60^\circ \quad \text{or} \quad 2\theta = 120^\circ \]

10. **Finding \( \theta \):**
- Thus:
\[ \theta = 30^\circ \quad \text{or} \quad \theta = 60^\circ \]
- Since the angle between the first two polarizers must be less than \( 90^\circ \), we take:
\[ \theta = 30^\circ \]

### Final Answer:
The angle \( \theta \) between the transmission axes of the first two polarizers is \( 30^\circ \).