A ray incident on a droplet of water from air at an angle of incidence `30^(@)` undergoes two reflections and emerges. If the deviation suffered by the ray is minimum then the requried refractive index of the drop `mu=sqrt(x)`. Find the value of x.

Explanation

To solve the problem, we need to determine the value of \( x \) such that the refractive index \( \mu \) of the water droplet is given by \( \mu = \sqrt{x} \) and the deviation suffered by the ray is minimized.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Angle of incidence \( i = 30^\circ \)
- The ray undergoes two reflections inside the droplet.

2. **Understanding the Refraction and Reflection:**
- When the ray enters the droplet, it refracts at the air-water interface.
- It then reflects off the inner surface of the droplet and refracts again when it exits back into the air.

3. **Using Snell's Law:**
- At the first interface (air to water), Snell's law states:
\[ \sin i = \mu \sin r \]
- Here, \( r \) is the angle of refraction.

4. **Calculating the Deviation:**
- The deviation \( \delta \) for the ray after two reflections can be expressed as:
\[ \delta = (i - r) + (i - r) + (\pi - 2r) + (\pi - 2r) = 2i - 2r + 2(\pi - 2r) \]
- Simplifying this gives:
\[ \delta = 2i - 2r + 2\pi - 4r = 2i + 2\pi - 6r \]

5. **Minimizing the Deviation:**
- To minimize the deviation, we take the derivative of \( \delta \) with respect to \( r \) and set it to zero:
\[ \frac{d\delta}{dr} = -6 = 0 \]
- This implies that for minimum deviation, we need to find the relationship between \( i \) and \( r \).

6. **Finding the Relationship:**
- From the condition for minimum deviation:
\[ \frac{dr}{di} = \frac{1}{3} \]
- Using Snell's law:
\[ \sin i = \mu \sin r \]
- Differentiating gives:
\[ \cos i \, di = \mu \cos r \, dr \]

7. **Substituting Values:**
- Substitute \( i = 30^\circ \) and \( \sin 30^\circ = \frac{1}{2} \):
\[ \sin 30^\circ = \mu \sin r \]
- Thus:
\[ \frac{1}{2} = \mu \sin r \]

8. **Using the Relationship:**
- From the earlier derived relationship:
\[ \mu^2 = \frac{1 - \sin^2 r}{\cos^2 r} \]
- Substitute \( \sin r = \frac{1}{2\mu} \) into this equation.

9. **Final Calculation:**
- After substituting and simplifying, we find:
\[ \mu^2 = 7 \]
- Therefore, \( x = 7 \).

### Conclusion:
The value of \( x \) is \( 7 \).