The laws of reflection or refraction are the same for sound as for light. The index of refraction of a medium (for sound) is defind as the ratio of the speed of sound in air `343 m//s` to the speed of sound in the medium.
What is the index of refraction (for sound) of water `(v=1498 m//s)`?
What is the critical angle `theta`, for total reflection of sound from water?

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Calculate the Index of Refraction for Sound in Water
The index of refraction (µ) for sound in a medium is defined as the ratio of the speed of sound in air (Va) to the speed of sound in the medium (Vw).

Given:
- Speed of sound in air, Va = 343 m/s
- Speed of sound in water, Vw = 1498 m/s

The formula for the index of refraction is:
\[ \mu_w = \frac{V_a}{V_w} \]

Substituting the values:
\[ \mu_w = \frac{343 \, \text{m/s}}{1498 \, \text{m/s}} \approx 0.229 \]

### Step 2: Calculate the Critical Angle for Total Internal Reflection
The critical angle (θc) is the angle of incidence in the denser medium (water) for which the angle of refraction in the rarer medium (air) becomes 90 degrees.

Using Snell's Law:
\[ \mu_w \sin(\theta_c) = \mu_a \sin(90^\circ) \]

Where:
- µa (index of refraction for air) is approximately 1 (since the speed of sound in air is taken as the reference).

Thus, we can simplify the equation:
\[ \mu_w \sin(\theta_c) = \mu_a \]
\[ \sin(\theta_c) = \frac{\mu_a}{\mu_w} \]

Substituting the values:
\[ \sin(\theta_c) = \frac{1}{0.229} \approx 4.37 \]

Since the sine function cannot exceed 1, we realize that this indicates that total internal reflection cannot occur under normal conditions. However, if we consider the critical angle as:
\[ \sin(\theta_c) = \frac{V_a}{V_w} = 0.229 \]

Now we can find θc:
\[ \theta_c = \sin^{-1}(0.229) \approx 13.29^\circ \]

### Final Answers
1. The index of refraction for sound in water is approximately **0.229**.
2. The critical angle for total internal reflection of sound from water is approximately **13.29 degrees**.

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