Two thin parallel slits are made in on opaque sheet of film when a monochromatic beam of light is shonw through then at normal incidence. The first bright fringes in the transmitted light occur at `+- 45^(@)` with the original direction of the ligth beam on a distant screen when the apparatus is in air. When the apparatus is immersed in a liquid, the same bright fringe now occur at `+- 30^(@)`. The index of refraction of the liquid is

Options

A. `sqrt 2`
B. `sqrt 3`
C. `(4)/(3)`
D. `(3)/(2)`

Explanation

To solve the problem, we will use the principles of Young's double-slit experiment and the relationship between the angles of the bright fringes and the refractive index of the medium.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two thin parallel slits and a monochromatic light source. The first bright fringes occur at angles of ±45° when in air and at ±30° when in a liquid.

2. **Using the Condition for Bright Fringes**:
- The condition for the first bright fringe in Young's double-slit experiment is given by:
\[ d \sin \theta = n \lambda \]
- Here, \(d\) is the distance between the slits, \(\theta\) is the angle of the bright fringe, \(n\) is the order of the fringe (for the first bright fringe, \(n=1\)), and \(\lambda\) is the wavelength of light.

3. **Setting Up the Equations**:
- For air (where the first bright fringe occurs at ±45°):
\[ d \sin(45^\circ) = \lambda \]
- For the liquid (where the first bright fringe occurs at ±30°):
\[ d \sin(30^\circ) = \lambda' \]
- The wavelength in the liquid \(\lambda'\) is related to the wavelength in air \(\lambda\) by the refractive index \( \mu \):
\[ \lambda' = \frac{\lambda}{\mu} \]

4. **Relating the Two Conditions**:
- From the first condition in air:
\[ d \sin(45^\circ) = \lambda \]
- From the second condition in the liquid:
\[ d \sin(30^\circ) = \frac{\lambda}{\mu} \]

5. **Dividing the Two Equations**:
- Dividing the first equation by the second gives:
\[ \frac{d \sin(45^\circ)}{d \sin(30^\circ)} = \frac{\lambda}{\frac{\lambda}{\mu}} \]
- Simplifying this, we have:
\[ \frac{\sin(45^\circ)}{\sin(30^\circ)} = \mu \]

6. **Calculating the Sines**:
- We know:
\[ \sin(45^\circ) = \frac{\sqrt{2}}{2}, \quad \sin(30^\circ) = \frac{1}{2} \]
- Substituting these values:
\[ \mu = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = \sqrt{2} \]

7. **Conclusion**:
- The index of refraction of the liquid is:
\[ \mu = \sqrt{2} \]