Question: Rainbows from square drops. Suppose that, on some surreal world, raindrops had a square cross section and always fell with one face horizontal. Figure shows such a falling drop, with a white beam of sunlight incident at $\mathit{\theta }\mathbf{=}\mathbf{70}\mathbf{.}\mathbf{0}\mathbf{°}$at point P. The part of the light that enters the drop then travels to point, where some of it refracts out into the air and the rest reflects. That reflected light then travels to point, where again some of the light refracts out into the air and the rest reflects.(a)What is the difference in the angles of the red light ( n = 1.331) and the blue light(n = 1.343) that emerge atpoint A? and(b) What is the difference in the angles of the red lightn = 1.331 )and the blue light (n = 1.343)that emerge atpoint B? (This angular difference in the light emerging at, say, pointwould be the rainbow’s angular width.)

Figure:

${\theta }_1={70}^0$

1. Angle of incidence
2. Index of refraction for blue$n_{2b}=1.343$
3. Index of refraction for rad$n_{2r}=1.331$

We can use Snell’s law for each color to find the angle of refraction for red and blue colors at point A and at B. From this, the difference in angles of the red light and blue light that emerge at pointand B can be calculated.

Formula:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

According to Snell’s law, for blue color,

$$\begin{gathered} n_{air}\sin {\theta }_1=n_{2b}\sin {\theta }_{2b} \\ {\theta }_{2b}={\sin }^{-1}\left(\frac{n_{air}\sin {\theta }_1}{n_{2b}}\right) \\ {\theta }_{2b}={\sin }^{-1}\left(\frac{\left(1.0\right)\sin {70}^0}{1.343}\right) \\ {\theta }_{2b}=44.{403}^0 \end{gathered}$$

Now, according to Snell’s law, for red color,

$$\begin{gathered} n_{air}\sin {\theta }_1=n_{2r}\sin {\theta }_{2r} \\ {\theta }_{2r}={\sin }^{-1}\left(\frac{n_{air}\sin {\theta }_1}{n_{2r}}\right) \\ {\theta }_{2r}={\sin }^{-1}\left(\frac{\left(1.0\right)\sin {70}^0}{1.331}\right) \\ {\theta }_{2r}=44.{911}^0 \end{gathered}$$

These are the refraction angles at the first surface.

Now, for the second refraction,

According to Snell’s law, for blue color,

$$\begin{gathered} n_{2b}\sin ({90}^0-{\theta }_{2b})=n_{3b}\sin {\theta }_{3b} \\ {\theta }_{3b}={\sin }^{-1}\left(\frac{1.343\sin ({90}^0-{44.403}^0)}{1.0}\right) \\ {\theta }_{3b}=73.{636}^0 \end{gathered}$$

Now, according to Snell’s law, for red color,

$$\begin{gathered} {\theta }_{3r}={\sin }^{-1}\left(\frac{\left(1.331\right)\sin ({90}^0-{44.911}^0)}{1.0}\right) \\ {\theta }_{3r}=70.{497}^0 \end{gathered}$$

Therefore,

$$\begin{gathered} ∆\theta =73.{636}^0-70.{497}^0 \\ ∆\theta =3.1^0 \end{gathered}$$

Therefore, the difference in angles of the red light and blue light that emerge at point A is $3.1^0$.

The angle of incidence and angle of emergence is the same. Therefore, the difference in angles of the red light and blue light that emerge at point B is 0⁰.