Question

A diver shines a flashlight upward from beneath the water (n=1.33) at an angle of 43.4° to the vertical. At what angle does the light refract through the air above the surface of the water?

Answer 1

To determine the angle of refraction we can use Snell's law:

`n_1\sin\theta_1=n_2\sin\theta_2`

For the problem given we know that:

`\begin{gathered} n_1=1.33 \\ \theta_1=43.4 \\ n_2=1\text{ \lparen The refactive index of air is one\rparen} \end{gathered}`

Plugging these values and solving for the second angle we have:

`\begin{gathered} 1.33\sin43.4=1\sin\theta_2 \\ \theta_2=\sin^(-1)(1.33\sin43.4) \\ \theta_2=66.04 \end{gathered}`

Therefore, the light refracts through the air with an angle 66.04°