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Ocean waves moving at a speed of 4.0 m/s are approaching a beach at angle θ1=30°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 θ2between 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.

Short Answer

Expert verified

(a) The angle between direction of motion and the normal is22° .

(b) Refraction causes the angle of incidence to steadily decrease, as demonstrated in the math above (from 30° to 22°). After multiple refractions, θ2will eventually approach zero. Because of this, most waves arrive at a shore normally.

Step by step solution

01

Given in the question.

Speed of deep water isvd=4.0m/s.

Speed of shallow water isvs=3.0m/s .

02

 Step 2: Law of refraction.

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.

n1sinθ1=n2sinθ2

Here, n1is refractive index in rarer medium, n2is refractive index in denser medium.

03

(a) The angle between direction of motion and the normal.

Apply the law of refraction,

sinθ2sinθ1=vsvd

Substitute all the value in the above equation.

Consequently solve as follows:

θ2=sin-1vssin30°vd=sin-13.0m/s×sin30°4.0m/s=22°

Hence, the angle between direction of motion and the normal is22° .

04

(b) Explanation of more waves comes in normal. 

Refraction causes the angle of incidence to steadily decrease, as demonstrated in the math above (from 30° to 22°). After multiple refractionsθ2 ,will eventually approach zero. Because of this, most waves arrive at a shore normally.

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