Question

Water waves approach an underwater "shelf" where the velocity changes from 2.8 m/s to 2.1 m/s. If the incident wave crests make a 34o angle with the shelf, what will be the angle of refraction?

Answer 1

Answer:

24°

Explanation:

sin(34°)/sin(x)=v2/v1

x=arcsin(2,1*sin(34°)/2,8)=24°

Answer 2

The angle of refraction when water waves approach an underwater "shelf" will be  24 degrees.

What is snell's law?

According to snells law the ratio of the incident angle and refracted angle will be equal to the ratio of the refractive index of the two mediums.Also equal to the ratio of the velocity of the wave in two mediums.

$\dfrac{Sin\ i}{Sin\ r}=\dfrac{V_1}{V_2}$

It is given that

Angle of incident

$i= 34V_1=2.8\ m/s \\\\\\V_2=2.1\ m/s$

putting the values we get

$\dfrac{Sin(34)}{Sin\ r}=\dfrac{2.8}{2.1}r=Sin^{-1}(\dfrac{Sin(34)\times 2.1}{2.8})=24^o$

Thus the angle of refraction when water waves approach an underwater "shelf" will be 24 degrees

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