B

To solve this problem, we will use Snell's law, which states that \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \).

• Let \( n_1 \) be the refractive index of air (approximately 1.00).
• Let \( n_2 \) be the refractive index of water (approximately 1.33).
• The angle of incidence \( \theta_1 = 60^\circ \).

Now, substitute the values into Snell's law:
\( 1.00 \cdot \sin(60^\circ) = 1.33 \cdot \sin(\theta_2) \).
The sine of 60 degrees is approximately 0.866. Therefore:
\( 0.866 = 1.33 \cdot \sin(\theta_2) \).
Solving for \( \sin(\theta_2) \):
\( \sin(\theta_2) = \frac{0.866}{1.33} \approx 0.651 \).
Thus, \( \theta_2 = \sin^{-1}(0.651) \approx 40.5^\circ \).
Since 40.5º is less than 60º, the sound wave is indeed refracted towards the normal (not away from it). Therefore, the statement in the question is incorrect.