Surface waves travelling in deep water at 15ms-1 are incident at a shallow water boundary. If the angles of incidence and refraction are 45o and 30o respect...

Answer Details

When surface waves travelling in deep water meet a shallow water boundary, they experience a change in speed and direction. This is known as refraction. The speed of the waves is dependent on the medium they are travelling through, so when they enter shallow water, their speed will change. We can use Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the velocities in the two media. In this case, we have: \begin{align*} \frac{\sin i}{\sin r} &= \frac{v_i}{v_r} \\ \frac{\sin 45}{\sin 30} &= \frac{15}{v_r} \\ \frac{\sqrt{2}}{0.5} &= \frac{15}{v_r} \\ v_r &= \frac{15}{\frac{\sqrt{2}}{0.5}} \\ v_r &= 10.6\text{ms}^{-1} \end{align*} Therefore, the speed of the waves in shallow water is approximately 10.6ms^-1. The correct option is (c) 10.6ms^-1.