Question

The primary rainbow described in Problem 77 is the type commonly seen in regions where rainbows appear. It is produced by light reflecting once inside the drops. Rarer is the secondary rainbow described in Module 33-5, produced by light reflecting twice inside the drops (Fig. 33-68a). (a) Show that the angular deviation of light entering and then leaving a spherical water drop is${\mathit{\theta }}_{\mathbf{d}\mathbf{e}\mathbf{v}}\mathbf{=}\left(180°\right)\mathit{k}\mathbf{+}\mathbf{2}{\mathit{\theta }}_{\mathbf{i}}\mathbf{-}\mathbf{2}\left(k+1\right){\mathit{\theta }}_{\mathbf{r}}$

where, k is the number of internal reflections. Using the procedure of Problem 77, find the angle of minimum deviation for (b) red light and (c) blue light in a secondary rainbow. (d) What is the angular width of that rainbow (Fig. 33-21d)?

The tertiary rainbow depends on three internal reflections (Fig. 33-68b). It probably occurs but, as noted in Module 33-5, cannot be seen with the eye because it is very faint and lies in the bright sky surrounding the Sun. What is the angle of minimum deviation for (e) the red light and (f) the blue light in this rainbow? (g) What is the rainbow’s angular width?

Short answer

a. The angular deviation of light entering and then leaving a spherical water drop is ${\theta }_{dev}=\left(180°\right)k+2{\theta }_i-2\left(k+1\right){\theta }_r$.

b. The angle of minimum deviation for the red light in the secondary rainbow is $230.4°$.

c. The angle of minimum deviation for blue light in the secondary rainbow is $233.48°$.

d. The angular width of the secondary rainbow is $3.1°$.

e. The angle of minimum deviation for the red light in the tertiary rainbow is $317.5°$.

f. The angle of minimum deviation for blue light in the tertiary rainbow is $317.5°$.

g. The angular width of the tertiary rainbow is $4.4°$.

Step-by-step solution

Given Information

Figure 33-68a with the secondary rainbow is given.

Understanding the concept of refraction 

We can use the concept of the primary, secondary, and tertiary rainbow. The primary rainbow is caused due to one total internal reflection and two refractions within the water droplet, while, the secondary rainbow is formed because of two total internal reflections and two refraction of white light by the water droplet. Being brighter secondary rainbow is bigger and more visible. Rays leaving raindrops after three reflections produce a tertiary rainbow.

Formula:

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) Calculation of the formula of angular deviation of the light

According to the first contribution to the overall deviation is at the first refraction, the angle due to this contribution is given by:

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

The angle between the ray light before the reflection and the axis normal to the back surface of the sphere is equal to ${\theta }_r$.

According to the law of reflection, the ray turns by the angle:

$\delta \theta =180°-2{\theta }_r$

For k number of multiple reflections, the second angle is given by:

$\begin{array}{rcl}\delta {\theta }_2& =& k\delta \theta \\ & =& k\left(180°-2{\theta }_r\right)\\ & & \end{array}$

The final contribution is the refraction suffered by the ray upon leaving the sphere is given by:

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

Thus, the total angular deviation while passing through the three different mediums can be given as:

$\begin{array}{rcl}{\theta }_{dev}& =& \delta {\theta }_1+\delta {\theta }_2+\delta {\theta }_3\\ & =& {\theta }_i-{\theta }_r+k\left(180°-2{\theta }_r\right)+{\theta }_i-{\theta }_r\\ & =& 2\left({\theta }_i-{\theta }_r\right)+k\left(180°-2{\theta }_r\right)\\ & =& 2\left({\theta }_i-{\theta }_r\right)+k180°-2k{\theta }_r\\ & =& 2{\theta }_i-2{\theta }_r+k180°-2k{\theta }_r.....................\left(2\right)\\ & =& \left(180°\right)k+2{\theta }_i-2\left(k+1\right){\theta }_r\\ & & \end{array}$

Hence, the value of the angle of deviation of the light is$\left(180°\right)k+2{\theta }_i-2\left(k+1\right){\theta }_r$

b) Calculation of the angle of minimum deviation for the red light in the secondary rainbow

For given values due to secondary reflection, $\text{k}=2$, ${\text{n}}_{\text{2}}\text{(or n)}=1.331$, ${\text{n}}_{\text{1}}=1$and ${\theta }_i=71.90°$

The expression for refracted angle using the above data equation (1) is given as:

${\theta }_r={\sin }^{-1}\left(\frac{1}{n}\sin {\theta }_i\right)$

The expression for minimum deviation after substituting the above value in equation (2) can be given as follows:

$\begin{array}{rcl}{\theta }_{dev}& =& \left(180°\right)k+2{\theta }_i-2\left(k+1\right){\sin }^{-1}\left(\frac{1}{n}\sin {\theta }_i\right)................................\left(3\right)\\ & =& \left(180°\right)2+2\times 71.90°-\left[2\left(2+1\right){\sin }^{-1}\left(\frac{1}{1.331}\sin 71.90°\right)\right]\\ & =& 230.4°\\ & & \end{array}$

Hence, the value of the angle is$230.4°$$230.4°$.

c) Calculation of the angle of minimum deviation for the blue light in the secondary rainbow

For given values due to secondary reflection,$\text{k}=2$ , ${\text{n}}_{\text{2}}\text{(or n)}=1.331$,${\text{n}}_{\text{1}}=1$and${\theta }_i=71.52°$

The angle of deviation for the blue light is as follows:

$\begin{array}{rcl}{\theta }_{dev}& =& \left(180°\right)2+2\times 71.52°-\left[2\left(2+1\right){\sin }^{-1}\left(\frac{1}{1.343}\sin 72.52°\right)\right]\\ & =& 233.48°\end{array}$

Hence, the value of the angle is $233.48°$.

d) Calculation of the angular width for secondary rainbow

The angular width is the difference between the minimum deviation of red and blue light, thus it can be given as follows:

$\begin{array}{rcl}\Delta {\theta }_{dev}& =& 233.48°-230.4°\\ & =& 3.1°\\ & & \end{array}$

Hence, the value of the width is $3.1°$.

e) Calculation of the angle of minimum deviation for the red light in the tertiary rainbow

For given values due to tertiary reflection,$\text{k}=3$,${\text{n}}_{\text{2}}\text{(or n)}=1.331$,${\text{n}}_1=1$and${\theta }_i=76.88$

The expression for refracted angleusing the above data equation (1) is given as:

${\theta }_r={\sin }^{-1}\left(\frac{1}{n}\sin {\theta }_i\right)$

The expression for minimum deviation after substituting the above values in equation (3) can be given as follows:

$\begin{array}{rcl}{\theta }_{dev}& =& \left(180°\right)3+2\times 76.88°-\left[2\left(3+1\right){\sin }^{-1}\left(\frac{1}{1.331}\sin 76.88°\right)\right]\\ & =& 317.5°\\ & & \end{array}$Hence, the value of the angle is $317.5°$.

f) Calculation of the angle of minimum deviation for blue light in the tertiary rainbow

For given values due to tertiary reflection, $\text{k}=2$, ${\text{n}}_{\text{2}}\text{(or n)}=1.343$, ${\text{n}}_{\text{1}}=1$and${\theta }_i=76.62°$

The expression for minimum deviation after substituting the above values in equation (3) can be given as follows:

$\begin{array}{rcl}{\theta }_{dev}& =& \left(180°\right)3+2\times 76.62°-\left[2\left(3+1\right){\sin }^{-1}\left(\frac{1}{1.343}\sin 76.62°\right)\right]\\ & =& 321.9°\\ & & \end{array}$

Hence, the value of the angle is$321.9°$ .

g) Calculation of the angular width for tertiary rainbow

The angular width is the difference between the minimum deviation of red and blue light, thus is given as follows:

$\begin{array}{rcl}\Delta {\theta }_{dev}& =& 321.9°-317.5°\\ & =& 4.4°\\ & & \end{array}$

Hence, the value of the width is$4.4°$.