The polaroids `P_(1), P_(2) & P_(3)` are arranged coaxially. The angle between `P_(1)` and `P_(2)` is `37^(@)`. The angle between `P_(2)` and `P_(3)` is, if intensity of emerging light is a quarter of intensity of unploarized light

Options

A. `theta = cos^(-1) ((5)/(4))`
B. `theta = cos^(-1) ((4)/(5))`
C. `theta = cos^(-1) ((4)/(5sqrt(2)))`
D. `theta = cos^(-1)((5)/(4sqrt(2)))`

Explanation

To solve the problem involving three polaroids arranged coaxially, we will apply Malus's Law, which relates the intensity of polarized light transmitted through a polaroid to the angle between the light's polarization direction and the polaroid's axis.

### Step-by-Step Solution

1. **Understanding the Initial Setup**:
- We have three polaroids: P1, P2, and P3.
- The angle between P1 and P2 is given as \(37^\circ\).
- The light incident on P1 is unpolarized.

2. **Intensity After P1**:
- When unpolarized light passes through the first polaroid (P1), the intensity of the transmitted light is halved.
- Therefore, if \(I_0\) is the intensity of the unpolarized light incident on P1, the intensity after P1 (\(I_1\)) is:
\[ I_1 = \frac{I_0}{2} \]

3. **Intensity After P2**:
- The intensity after passing through P2 can be calculated using Malus's Law:
\[ I_2 = I_1 \cdot \cos^2(\theta) \]
- Here, \(\theta\) is the angle between P1 and P2, which is \(37^\circ\).
- Substituting the values:
\[ I_2 = \frac{I_0}{2} \cdot \cos^2(37^\circ) \]

4. **Calculating \(\cos^2(37^\circ)\)**:
- We know that \(\cos(37^\circ) \approx 0.8\) (exact value is \(\frac{4}{5}\)).
- Therefore:
\[ \cos^2(37^\circ) = \left(\frac{4}{5}\right)^2 = \frac{16}{25} \]
- Now substituting this back into the equation for \(I_2\):
\[ I_2 = \frac{I_0}{2} \cdot \frac{16}{25} = \frac{16 I_0}{50} = \frac{8 I_0}{25} \]

5. **Intensity After P3**:
- We know that the intensity of the emerging light from the system is given as a quarter of the initial intensity:
\[ I_3 = \frac{I_0}{4} \]
- Using Malus's Law again for P3:
\[ I_3 = I_2 \cdot \cos^2(\phi) \]
- Here, \(\phi\) is the angle between P2 and P3, which we need to find.

6. **Setting Up the Equation**:
- We can equate the two expressions for \(I_3\):
\[ \frac{I_0}{4} = \frac{8 I_0}{25} \cdot \cos^2(\phi) \]
- Canceling \(I_0\) from both sides:
\[ \frac{1}{4} = \frac{8}{25} \cdot \cos^2(\phi) \]

7. **Solving for \(\cos^2(\phi)\)**:
- Rearranging gives:
\[ \cos^2(\phi) = \frac{25}{32} \]

8. **Finding \(\phi\)**:
- Taking the square root:
\[ \cos(\phi) = \sqrt{\frac{25}{32}} = \frac{5}{\sqrt{32}} = \frac{5}{4\sqrt{2}} \]
- Finally, to find \(\phi\):
\[ \phi = \cos^{-1}\left(\frac{5}{4\sqrt{2}}\right) \]

### Final Answer
The angle between P2 and P3 is \(\phi = \cos^{-1}\left(\frac{5}{4\sqrt{2}}\right)\).