Ocean waves moving at a speed of 4.0 m/s are approaching a beach at angle ${\theta }_{\mathit{1}}\mathit{=}\mathit{30}\mathit{°}$to the normal, as shown from above in Fig. 35-55. Suppose the water depth changes abruptly at a certain distance from the beach and the wave speed there drops to 3.0 m/s. (a) Close to the beach, what is the angle ${\mathrm{\theta }}_2$between the direction of wave motion and the normal? (Assume the same law of refraction as for light.) (b) Explain why most waves come in normal to a shore even though at large distances they approach at a variety of angles.

Speed of deep water is$v_d=4.0 \text{m/s}$.

Speed of shallow water is$v_s=3.0 \text{m/s}$ .

All three rays—the incident, normal at the point of incidence, and refracted—lie in the same plane. According to the law of refraction, this is the case.

${\mathbf{n}}_{\mathbf{1}}{\mathbf{sin\theta }}_{\mathbf{1}}\mathbf{=}{\mathbf{n}}_{\mathbf{2}}{\mathbf{sin\theta }}_{\mathbf{2}}$

Here, $n_1$is refractive index in rarer medium, $n_2$is refractive index in denser medium.

Apply the law of refraction,

$\frac{\sin {\theta }_2}{\sin {\theta }_1}=\frac{v_s}{v_d}$

Substitute all the value in the above equation.

Consequently solve as follows:

$$\begin{gathered} {\theta }_2={\sin }^{-1}\left(\frac{v_s\sin 30°}{v_d}\right) \\ ={\sin }^{-1}\left(\frac{3.0 \text{m/s}\times \sin 30°}{4.0 \text{m/s}}\right) \\ =22° \end{gathered}$$

Hence, the angle between direction of motion and the normal is$22°$ .

Refraction causes the angle of incidence to steadily decrease, as demonstrated in the math above (from 30° to 22°). After multiple refractions${\theta }_2$ ,will eventually approach zero. Because of this, most waves arrive at a shore normally.