Chapter 30, Problem 35P

To determine

The horizontal distance of the diver from the edge of the lake.

Answer to Problem 35P

The horizontal distance of the diver from the edge of the lake is $5.1\text{m}$.

Explanation of Solution

Given Info: The distance from the edge is $2.3\text{m}$, the refractive index for water is 1.333. the refractive index for air is 1.000, the height of the eyes from the water surface is $1.7\text{m}$, the depth of the water is $4.5\text{m}$ and angle of incidence of the light inside the water is $42°$.

Explanation:

The figure shows the direction of light ray and angle subtended by the light ray at each surface. The angle ${\alpha }_2$ is $42°$.

Write the expression for the Snell’s law.

$\frac{\sin {\alpha }_2}{\sin {\alpha }_1}=\frac{n_1}{n_2}$

${\alpha }_1$ is the angle of refraction

${\alpha }_2$ is the angle of incidence

$n_1$ is the refractive index of air

$n_2$ is the refractive index of water

Substitute $42°$ for ${\alpha }_2$, 1.333 for $n_2$ and 1.000 for $n_1$ in the above equation to calculate ${\alpha }_1$.

$\begin{array}{c}\frac{\sin 42°}{\sin {\alpha }_1}=\frac{1.000}{1.333}\\ \sin {\alpha }_1=\sin 42°\left(1.333\right)\\ {\alpha }_1={\sin }^{-1}\left(\sin 42°\left(1.333\right)\right)\\ =63.1°\end{array}$

Write the expression for ${\alpha }_1$.

$\tan {\alpha }_1=\frac{2.3\text{m}+l_1}{1.7\text{m}}$

Substitute $63.1°$ for ${\alpha }_1$ in the above equation to calculate $l_1$.

$\begin{array}{c}\frac{2.3\text{m}+l_1}{1.7\text{m}}=\tan 63.1°\\ 2.3\text{m}+l_1=\tan 63.1°\left(1.7\text{m}\right)\\ l_1=3.35\text{m}-2.3\text{m}\\ =\text{1}\text{.05}\text{m}\end{array}$

Write the expression for ${\alpha }_2$.

$\tan {\alpha }_2=\frac{l_2}{4.5\text{m}}$

Substitute $42°$ in the above equation to calculate $l_2$.

$\begin{array}{c}\frac{l_2}{4.5\text{m}}=\tan 42°\\ l_2=\left(4.5\text{m}\right)\tan 42°\\ =4.05\text{m}\end{array}$

Write the formula to calculate the horizontal distance of the diver from the edge.

$x=l_1+l_2$

x is the horizontal distance of the diver from the edge of the lake

Substitute $\text{1}\text{.05}\text{m}$ for $l_1$ and $\text{4}\text{.05}\text{m}$ for $l_2$ in the above equation to calculate x.

$\begin{array}{c}x=\text{1}\text{.05}\text{m}+\text{4}\text{.05}\text{m}\\ =5.1\text{m}\end{array}$

Conclusion:

Therefore, the horizontal distance of the diver from the edge of the lake is $5.1\text{m}$.