Question

A 4.7-m-wide swimming pool is filled to the top. The bottom of the pool becomes completely shaded in the afternoon when the sun is 25 degrees above the horizon.

How deep is the pool in meters?

Answer 1

Concepts and reason

The concept needed to solve this question is Snell’s law.

Draw a useful ray diagram. Use Snell’s law, and solve for angle of refraction of a ray which is exactly coming towards the edge of the pool. After that, use a useful trigonometry ratio and solve for the height of the pool.

Fundamentals

The equation of Snell’s law is,

${n_1}\sin i = {n_2}\sin r$

Here, ${n_1}$ is the refractive index of the light incident medium, $i$ is the angle of incidence, ${n_2}$ is the refractive index of light refracted medium, and r is the angle of refraction.

The trigonometric ratio of tangent in a right angle triangle can be write as following:

$\tan \theta = \frac{{{\rm{side opposite to the angle}}}}{{{\rm{side ajacent to the angle}}}}$

Here, $\theta$ is the angle.

The following diagram is useful to calculate the angle of refraction of a ray which is coming towards the edge of the pool:

In the figure, h represents the depth of the pool.

The equation of Snell’s law is,

${n_1}\sin i = {n_2}\sin r$

Rearrange the equation for r.

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

Substitute 1 for ${n_1}$ , $65^\circ$ for $i$ , and1.33 for ${n_2}$ .

$\begin{array}{c}\\r = {\sin ^{ - 1}}\left( {\frac{{\left( 1 \right)\sin 65^\circ }}{{1.33}}} \right)\\\\ = {\rm{43}}^\circ \\\end{array}$

The trigonometric ratio of tangent in a right angle triangle can be write as following:

$\tan \theta = \frac{{{\rm{side opposite to the angle}}}}{{{\rm{side ajacent to the angle}}}}$

Substitute 4.7 m for side opposite to the angle, ${\rm{43}}^\circ$ for $\theta$ , and h for side adjacent to the angle, and solve for h.

$\begin{array}{c}\\\tan {\rm{43}}^\circ = \frac{{{\rm{4}}{\rm{.7 m}}}}{h}\\\\h = \frac{{{\rm{4}}{\rm{.7 m}}}}{{\tan 43^\circ }}\\\\ = {\rm{5}}{\rm{.0 m}}\\\end{array}$

Ans:

The depth of the pool is 5.0 m.