Question

A diver shines a flashlight upward from beneath the water at a 42.5? angle to the vertical. The index of refraction of water is 1.33.
(a.) Determine the angle at which light is reflected at the water-air surface.
(b.) Calculate the angle at which the light leaves the water.

Answer 1

Concepts and reason

The main concept required to solve this problem is the law of reflection, refraction and Snell’s law.

Initially, write the equation for the law of reflection and the Snell’s law. Use these equations and find the angle at which the light is reflected at the water-air surface, and the angle at which the light leaves the water.

Fundamentals

Law of reflection states that the angle of incident ray is equal to the angle of reflected ray on the smooth surface, the equation for the law of reflection is,

$i = r$

Here, i is the angle of incident ray and r is the angle of reflected ray.

The equation for the Snell’s law is,

${n_1}\sin {i_1} = {n_2}\sin {r_1}$

Here, ${n_1}$ is the refractive index of the first medium, ${n_2}$ is the refractive index of the second medium, ${i_1}$ is the angle of incidence in first medium and ${r_1}$ is the angle of refracted ray in the second medium.

The angle of incidence of the flashlight on the water-air surface is,

$i = {42.5^{\rm{o}}}$

According to the law of reflection, the angle of incidence of the flash light should be equal to the angle of reflection of the flashlight, that is,

$i = r$

Therefore, the angle of reflection on the water-air surface is,

$r = {42.5^{\rm{o}}}$

Use the Snell’s law for the water-air interface as follows,

${n_1}\sin {i_1} = {n_2}\sin {r_1}$

Here, ${n_1}$ is the refractive index of the water, ${i_1}$ is the incident angle, ${n_2}$ is the refractive index of the air, and ${r_1}$ if the angle of refraction in air.

Rearrange the above equation for ${r_1}$ .

$\begin{array}{c}\\\sin {r_1} = \frac{{{n_1}\sin {i_1}}}{{{n_2}}}\\\\{r_1} = {\sin ^{ - 1}}\left( {\frac{{{n_1}\sin {i_1}}}{{{n_2}}}} \right)\\\end{array}$

Substitute $1.33$ for ${n_1}$ , ${42.5^{\rm{o}}}$ for ${i_1}$ , and 1 for ${n_2}$ in above equation.

$\begin{array}{c}\\{r_1} = {\sin ^{ - 1}}\left( {\frac{{\left( {1.33} \right)\sin {{42.5}^{\rm{o}}}}}{{\left( 1 \right)}}} \right)\\\\ = {\sin ^{ - 1}}\left( {\left( {1.33} \right)\left( {0.67559} \right)} \right)\\\\ = {\sin ^{ - 1}}\left( {0.898534} \right)\\\\ = {63.96^{\rm{o}}}\\\end{array}$

Ans:

The angle at which the light leaves the water is ${63.96^{\rm{o}}}$ .