Question

Rainbow Figure 33-67 shows a light ray entering and then leaving a falling, spherical raindrop after one internal reflection (see Fig. 33-21a). The final direction of travel is deviated (turned) from the initial direction of travel by angular deviation
${\mathit{\theta }}_{\mathbf{d}\mathbf{e}\mathbf{v}}$. (a) Show that localid="1664200532112" ${\mathit{\theta }}_{\mathbf{d}\mathbf{e}\mathbf{v}}$ is localid="1664200226807" ${\mathit{\theta }}_{\mathbf{d}\mathbf{e}\mathbf{v}}\mathbf{=}\mathbf{180}\mathbf{°}\mathbf{+}\mathbf{2}{\mathit{\theta }}_{\mathbf{i}}\mathbf{-}\mathbf{4}{\mathit{\theta }}_{\mathbf{r}}$ , where localid="1664200612169" ${\mathit{\theta }}_{\mathbf{i}}$is the angle of incidence of the ray on the drop and localid="1664200615282" ${\mathit{\theta }}_{\mathbf{r}}$is the angle of refraction of the ray within the drop. (b) Using Snell’s law, substitute for localid="1664200618431" ${\mathit{\theta }}_{\mathbf{r}}$in terms of localid="1664200621396" ${\mathit{\theta }}_{\mathbf{i}}$and the index of refraction n of the water. Then, on a graphing calculator or with a computer graphing package, graph localid="1664200624361" ${\mathit{\theta }}_{\mathbf{d}\mathbf{e}\mathbf{v}}$versus localid="1664200627334" ${\mathit{\theta }}_{\mathbf{i}}$for the range of possible localid="1664200636137" ${\mathit{\theta }}_{\mathbf{i}}$values and for localid="1664200630531" $\mathit{n}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{333}$for red light (at one end of the visible spectrum) and localid="1664200633245" $\mathit{n}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{331}$for blue light (at the other end). The red-light curve and the blue-light curve have a different minimum, which means that there is a different angle of minimum deviation for each color. The light of any given color that leaves the drop at that color’s angle of minimum deviation is especially bright because rays bunch up at that angle. Thus, the bright red light leaves the drop at one angle and the bright blue light leaves it at another angle minimum deviation from the localid="1664200639414" ${\mathit{\theta }}_{\mathbf{d}\mathbf{e}\mathbf{v}}$curve for (c) red light and (d) blue light. (e) If these colors form the inner and outer edges of a rainbow (Fig. 33-21a), what is the angular width of the rainbow?

Short answer

a. It is proved that the angular deviation of light entering and then leaving a spherical water drop is${\theta }_{dev}=180°+2{\theta }_i-4{\theta }_r$

b. The graph ${\theta }_{dev}$ versus is plotted.

c. The angle of minimum deviation for red light in rainbow is$137.6°$.

d. The angle of minimum deviation for blue light in rainbow is$139.4°$

e. The angular width of the rainbow is$1.72°$ .

Step-by-step solution

Given Information

a. The refractive index of red light,$n=1.331$

b. The refractive index of blue light,$n=1.333$

Determining the formula for the Snell’s law

The Snell’s law of refraction of the light between two mediums of two materials,

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$................(1)

a) Determining the formula of the angle of deviation

The figure shows the light ray entering and then leaving a falling, spherical raindrop after one internal reflection. This is a primary rainbow.

The first contribution to the overall deviation at the first refraction is given as:

$\delta {\theta }_1={\theta }_i-{\theta }_r$

The next contribution to the overall deviation is the reflection.

Letbe the angle between the ray light before reflection and the axis normal to the back surface of the sphere. Thus, the second angle is given by:

$\delta {\theta }_2={180}^0-2{\theta }_r$

The final contribution is the refraction suffered by the ray upon leaving the surface and the angle is given by:

$\delta {\theta }_3={\theta }_i-{\theta }_r$

Thus, the total angular deviation of the light passing through three phase change is given as:

$\begin{array}{rcl}{\theta }_{dev}& =& \delta {\theta }_1+\delta {\theta }_2+\delta {\theta }_3\\ & =& {\theta }_i-{\theta }_r+180-2{\theta }_r+{\theta }_i-{\theta }_r\\ & =& 180+2{\theta }_i-4{\theta }_r\\ & & \end{array}$ …… (ii)

Hence, the angle of deviation of the refracted light is $180+2{\theta }_i-4{\theta }_r$.

b) Plotting the graph between angular deviation and angle of incidence

Here, refractive index of the material medium is given by, $n_2=n$and the refractive index of air medium is given by, $n_1=1$

Thus, using equation (i), the angle of refraction can be given as:

$\begin{array}{rcl}\sin {\theta }_i& =& n\sin {\theta }_r\\ {\theta }_r& =& {\sin }^{-1}\left(\frac{\sin {\theta }_i}{n}\right)\\ & & \end{array}$

Thus, using the above value in equation (ii), determine the angular deviation of the light as:

${\theta }_{dev}=180+2{\theta }_i-4\left({\sin }^{-1}\left(\frac{\sin {\theta }_i}{n}\right)\right)$

For red light, the refractive index is given by:$n=1.331$

For blue light, the refractive index is given by:$n=1.333$

c) Determining the angle of minimum deviation for red light in the rainbow

From the graph, we get the angle of minimum deviation for the red light as,

$\begin{array}{rcl}{\theta }_{dev}& =& 137.63°\\ & & ~137.6°\\ & & \end{array}$

Hence, the value of minimum angle is $137.6°$.

d) Determining the angle of minimum deviation for blue light in the rainbow

From the graph, determine the angle of minimum deviation for blue light as,

$\begin{array}{rcl}{\theta }_{dev}& =& 139.35°\\ & & ~139.4°\\ & & \end{array}$

Hence, the value of minimum angle is $139.4°$.

e) Determining the angular width of the rainbow

The angular width of the rainbow is the difference between the angle of minimum deviation for red and blue light and that is given by:

$\begin{array}{rcl}\Delta {\theta }_{dev}& =& 139.35°-137.63°\\ & =& 1.72°\\ & & \end{array}$

Hence, the value of the angular width is $1.72°$.