Chapter 38, Problem 1PQ

The Sun appears at an angle of 53.0° above the horizontal as viewed by a dolphin swimming underwater. What angle does the sunlight striking the water actually make with the horizon?

To determine

The location of sun above the horizon.

Answer to Problem 1PQ

The location of the sun above the horizon is $36.7°$.

Explanation of Solution

Write the expression for the angle of incidence on the air-water interface.

    ${\theta }_{\text{i}}={\sin }^{-1}\left(\frac{n_{\text{r}}}{n_{\text{i}}}\sin {\theta }_{\text{r}}\right)$        (I)

Here, $n_{\text{i}}$ is the refractive index of the first medium, $n_{\text{r}}$ refractive index of second medium, ${\theta }_{\text{i}}$ is the angle of incidence, and ${\theta }_{\text{r}}$ is the angle of refraction.

Write the expression to calculate the location of the sun above the horizon.

    $\varphi =90°-{\theta }_{\text{i}}$        (II)

Here, $\varphi$ is location of sun above the horizon.

Conclusion:

Calculate the angle of refraction of the light rays.

    $\begin{array}{c}{\theta }_{\text{r}}=90°-53.0°\\ =37.0°\end{array}$

Substitute $1.333$ for $n_{\text{r}}$, $1.0$ for $n_{\text{i}}$  and $37.0°$ for ${\theta }_{\text{r}}$ in equation (I) to calculate ${\theta }_{\text{i}}$.

    $\begin{array}{c}{\theta }_{\text{i}}={\sin }^{-1}\left(\frac{1.333}{1.0}\sin 37.0°\right)\\ =53.34°\end{array}$

Substitute $53.34°$ for ${\theta }_{\text{i}}$ in equation (II) to calculate $\varphi$.

    $\begin{array}{c}\varphi =90°-53.33°\\ =36.7°\end{array}$

Therefore, the location of the sun above the horizon is $36.7°$.