Question

Light passes from air into water at an angle of incidence of 42.3 determine the angle of refraction in the water

Answer 1

Answer:

$30.4^{\circ}$

Explanation:

Refraction occurs when a ray of light passes from a medium into another medium. In this situation, the ray of light bends and changes speed, according to Snell's Law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

$n_1$ is the index of refraction of the 1st medium

$n_2$ is the index of refraction of the 2nd medium

$\theta_1$ is the angle of incidence (the angle between the incident ray and the normal to the boundary)

$\theta_2$ is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

In this problem, we have:

$n_1=1.00$ is the index of refraction of air

$n_2=1.33$ is the index of refraction of water

$\theta_1=42.3^{\circ}$ is the angle of incidence in air

Solving for $\theta_2$, we find the angle of refraction in water:

$\theta_2=sin ^{-1}(\frac{n_1}{n_2}sin \theta_1)=sin^{-1}(\frac{1.00}{1.33}sin42.3^{\circ})=30.4^{\circ}$