Two polaroids are placed in the path of unpolarized beam of intensity `I_(0)` such that no light is emitted from the second polarid. If a third polaroid whose polarization axis makes an angle `theta` with the polarization axis of first polaroid, is placed between these two polariods then the intensity of light emerging from the last polaroid will be

Options

A. `(I_0/8)sin^2(2theta)`
B. `(I_0/4)sin^2(2theta)`
C. `(I_0/2)cos^4(theta)`
D. `I_0cos^4theta`

Answer

To solve the problem step by step, we will analyze the situation with the two initial polaroids and then introduce the third one.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two polaroids, P1 and P2, placed such that no light is emitted from P2. This means that the angle between the polarization axes of P1 and P2 is 90 degrees (θ = 90°).

2. **Intensity of Light After First Polaroid (P1)**:
- The initial intensity of the unpolarized light is \( I_0 \).
- When unpolarized light passes through the first polarizer (P1), the intensity is reduced to half due to averaging over all polarization directions.
- Therefore, the intensity after P1 is:
\[ I_1 = \frac{I_0}{2} \]

3. **Introducing the Third Polaroid (P3)**:
- A third polaroid (P3) is placed between P1 and P2, making an angle \( \theta \) with the polarization axis of P1.
- The angle between P3 and P2 will then be \( 90° - \theta \).

4. **Intensity After Third Polaroid (P3)**:
- The intensity of light after passing through P3 can be calculated using Malus's Law:
\[ I_2 = I_1 \cos^2(\theta) \]
- Substituting \( I_1 \):
\[ I_2 = \left(\frac{I_0}{2}\right) \cos^2(\theta) \]

5. **Intensity After Second Polaroid (P2)**:
- The intensity of light after passing through P2 is given by:
\[ I_3 = I_2 \cos^2(90° - \theta) = I_2 \sin^2(\theta) \]
- Substituting \( I_2 \):
\[ I_3 = \left(\frac{I_0}{2} \cos^2(\theta)\right) \sin^2(\theta) \]

6. **Final Expression for Intensity**:
- Combining the results, we have:
\[ I_3 = \frac{I_0}{2} \cos^2(\theta) \sin^2(\theta) \]
- Using the identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \):
\[ I_3 = \frac{I_0}{2} \cdot \frac{1}{4} \sin^2(2\theta) = \frac{I_0}{8} \sin^2(2\theta) \]

### Final Answer:
The intensity of light emerging from the last polaroid (P2) is:
\[ I_3 = \frac{I_0}{8} \sin^2(2\theta) \]