Question

1) A diver shines a flashlight upward from beneath the water at an angle of 46.5 degrees to the vertical.
At what angle does the light leave the water?

Answer 1

\[\mu=\frac{\sin \theta _{i}}{\sin \theta _{r}}=>\theta _{i}=\sin^{-1} (\mu \sin \theta _{r})=\sin^{-1} (1.33\times \sin 46.5)=74.7degrees\]