A plane weve of sound travelling in air is incident upon a plane water surface. The angle of incidence is `60^(@)`. If velocity of sound in air and water are `330 m//s` and `1400 m//s`, then the wave undergoes

Options

A. refraction only
B. reflection only
C. Both reflection and refraction
D. neither reflection nor refraction

Explanation

To solve the problem, we need to analyze the behavior of a sound wave as it travels from air into water at an angle of incidence of \(60^\circ\). We will determine whether the wave undergoes reflection, refraction, or both.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Angle of incidence, \(I = 60^\circ\)
- Velocity of sound in air, \(V_{air} = 330 \, m/s\)
- Velocity of sound in water, \(V_{water} = 1400 \, m/s\)

2. **Use Snell's Law for Refraction:**
Snell's Law relates the angles of incidence and refraction to the velocities of sound in the two media:
\[ \frac{\sin r}{\sin i} = \frac{V_{water}}{V_{air}} \]
Where \(r\) is the angle of refraction and \(i\) is the angle of incidence.

3. **Substitute Known Values:**
Substitute \(I\) and the velocities into Snell's Law:
\[ \frac{\sin r}{\sin 60^\circ} = \frac{1400}{330} \]

4. **Calculate \(\sin 60^\circ\):**
The value of \(\sin 60^\circ\) is \(\frac{\sqrt{3}}{2} \approx 0.866\).

5. **Calculate the Right Side:**
Calculate the right side of the equation:
\[ \frac{1400}{330} \approx 4.24 \]

6. **Set Up the Equation:**
Now we can set up the equation:
\[ \sin r = \sin 60^\circ \cdot \frac{1400}{330} \]
\[ \sin r = 0.866 \cdot 4.24 \approx 3.67 \]

7. **Analyze the Result:**
Since \(\sin r\) cannot be greater than 1, we conclude that:
\[ \sin r > 1 \quad \text{(which is not possible)} \]

8. **Conclusion:**
Since refraction is not possible (as \(\sin r\) cannot exceed 1), the wave will only reflect off the water surface.

### Final Answer:
The wave undergoes reflection only.

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