Question

Phys 131 HW#2
Q#1 Red laser beam enters hits glass/ water interface with an angle of 30° find the
angle of the refracted beam in the water. (20)
​

Answer 1

Angle of refraction: $34.6^{\circ}$

Explanation:

Refraction is a phenomenon that occurs when light crosses the interface between two mediums.

When this occurs, the ray of light bends, according to Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

$n_1,n_2$ are the index of refraction of the two mediums, respectively

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

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

In this problem, we have:

$n_1 = 1.513$ (index of refraction of glass for red light)

$n_2 = 1.331$ (index of refraction of water for red light)

$\theta_1=30^{\circ}$ (angle of incidence)

Therefore, the angle of refraction is:

$\theta_2 = sin^{-1}(\frac{n_1}{n_2}sin \theta_1)=sin^{-1}(\frac{1.513}{1.331}sin 30^{\circ})=34.6^{\circ}$