A diver at a depth of 12 m in water `(mu=4 //3)` sees the sky in a cone of semi-vertical angle

Options

A. `sin^(-1)(4/3)`
B. `tan^(-1)(4/3)`
C. `sin^(-1)(3/4)`
D. `90^(@)`

Explanation

To solve the problem, we need to determine the semi-vertical angle of the cone in which a diver at a depth of 12 meters in water sees the sky. The refractive index of water is given as \( \mu = \frac{4}{3} \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The diver is at a depth of 12 meters in water.
- The refractive index of water \( \mu_{\text{water}} = \frac{4}{3} \).
- The refractive index of air \( \mu_{\text{air}} = 1 \).
- We need to find the semi-vertical angle \( \theta \) of the cone of light that the diver can see.

2. **Using Snell's Law**:
- According to Snell's law, we have:
\[ \mu_{\text{water}} \sin(\theta) = \mu_{\text{air}} \sin(r) \]
- Here, \( r \) is the angle of refraction when light exits the water into the air. Since the light emerges at the critical angle, we can consider \( r = 90^\circ \).

3. **Substituting Values**:
- Substituting the known values into Snell's law:
\[ \frac{4}{3} \sin(\theta) = 1 \cdot \sin(90^\circ) \]
- Since \( \sin(90^\circ) = 1 \), we can simplify this to:
\[ \frac{4}{3} \sin(\theta) = 1 \]

4. **Solving for \( \sin(\theta) \)**:
- Rearranging the equation gives:
\[ \sin(\theta) = \frac{3}{4} \]

5. **Finding \( \theta \)**:
- Now, we can find \( \theta \) by taking the inverse sine:
\[ \theta = \sin^{-1}\left(\frac{3}{4}\right) \]

6. **Conclusion**:
- The semi-vertical angle \( \theta \) of the cone in which the diver sees the sky is \( \sin^{-1}\left(\frac{3}{4}\right) \).

### Final Answer:
\[ \theta = \sin^{-1}\left(\frac{3}{4}\right) \]