Question

A diver shines a flashlight upward from beneath the water (n=1.33) at an angle of 40.0° to the vertical. At what angle does the light refract through the air above the surface of the water?

Answer 1

Given:

The refractive index of water, n₁=1.33

The angle of incidence, θ₂=40.0°

To find:

The angle of refraction.

Explanation:

The refractive index of the air is n₂=1

From the snell's law,

$n_1\sin\theta_1=n_2\sin\theta_2$

Where θ₂ is the angle of refraction.

On substituting the known values,

$\begin{gathered} 1.33\sin40.0\degree=1\sin\theta_2 \\ \implies\theta_2=sin^{-1}(\frac{1.33\sin40.0\degree}{1}) \\ =58.75\degree \end{gathered}$

Final answer:

The angle of refraction is 58.75°