Chapter 6.3, Problem 71E

Rainbows Rainbows are created when sunlight of different wavelengths (colors) is refracted and reflected in raindrops. The angle of elevation θ of a rainbow is always the same. It can be shown that θ = 4β − 2α. Where

$\sin \alpha =k\sin \beta$

and α = 59.4° and k = 1.33 is the index of refraction of water. Use the given information to find the angle of elevation θ of a rainbow. [Hint: Find sin β, then use the $\boxed{{\text{SIN}}^{-1}}$ key on your calculator to find β.] (For a mathematical explanation of rainbows see Calculus Early Transcendentals, 7th Edition, by James Stewart, page 282.)

To determine

To find: The value of the angle of elevation, $\theta$ of a rainbow.

Answer to Problem 71E

The value of the angle of elevation, $\theta$ of a rainbow is $43.36°$.

Explanation of Solution

Given:

The angle of elevation, $\theta$ of a rainbow is given by the equation,

$\theta =4\beta -2\alpha$ (I)

Also,

$\sin \alpha =k\sin \beta$ (II)

Calculation:

Substitute $1.33$ for $k$ and $59.4°$ for $\alpha$ in equation (II) to find $\beta$.

$\begin{array}{c}\sin 59.4°=1.33\cdot \sin \beta \\ \sin \beta =\frac{\sin 59.4°}{1.33}\\ =\frac{0.8607}{1.33}\left(\sin 59.4°=0.8607\right)\\ \approx 0.65\end{array}$

Thus, $\sin \beta =0.65$.

Use ${\sin }^{-1}$ key on your calculator to find $\beta$.

$\begin{array}{c}\sin \beta =0.65\\ \beta ={\sin }^{-1}\left(0.65\right)\\ \approx 40.54°\end{array}$

Substitute $59.4$ for $\alpha$ and $40.54°$ for $\beta$ in equation (I),

$\begin{array}{c}\theta =\left(4\times 40.54°\right)-\left(2\times 59.4°\right)\\ =162.16°-118.8°\\ \approx 43.36°\end{array}$

Thus, the value of angle of elevation, $\theta$ of a rainbow is $43.36°$.