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A rainbow is formed at the boundary between two materials that bend light differently, such as water and air. Using the laws of optics, you can find a formula for the angular size of a rainbow in terms of the refractive index of the ambient material $n_1$ and droplet material $n_2$.

The figure below depicts the rainbow angle as a function of droplet material $n_2$ (assuming $n_1\approx 1$ as in air).

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Question: At around $n_2 \approx 1.12$, the rainbow angle reaches 90 degrees. I'm wondering if it makes physical sense to think of rainbows with angle greater than 90 degrees (perhaps you see such rainbows when facing the sun?), or if the physical derivation breaks down for some reason.





Side derivation: The rainbow angle derivation comes from supposing parallel light rays from the sun travel from the ambient material into a spherical droplet, refracting according to Snell's law. The light reflects just once off the back of the droplet and exits out of the front, refracting again as it exits. The scattering angle---that is, the deviation between entry and exit angle---depends on the materials involved and the angle of entry. The rainbow angle is the minimum achieved angle of deviation, because at this critical point, the rays of light are bunched closest together, forming a bright caustic band.

If light enters at an angle $\alpha$ with respect to a radius of the spherical droplet and refracts into an angle $\beta$, then by Snell's law $n_1 \sin{\alpha} = n_2 \sin{\beta}$, or $\beta = \arcsin{(n_1\sin{\alpha}/n_2)}$.

Given the two refractions and one reflection, we can find the scattering angle as: $D(\alpha,\beta) = \pi + 2\alpha - 4\beta$.

Combining the two formulas, $$D(\alpha) = \pi + 2\alpha - 4 \arcsin{(n_1\sin{\alpha}/n_2)}$$

To find the minimum, we can compute the derivative and set it equal to zero: $$0 = \partial_\alpha D = 2 - \frac{4 (n_1/n_2) \cos{\alpha}}{\sqrt{1-(n_1\sin{\alpha}/n_2)^2}}$$

$${2 (n_1/n_2) \cos{\alpha}} = \sqrt{1-(n_1\sin{\alpha}/n_2)^2}$$
$${2 (n_1/n_2) \cos{\alpha}} = (n_1/n_2) \sqrt{(n_2/n_1)^2-1+1-\sin^2{\alpha}}$$
$$4\cos^2{\alpha} = (n_2/n_1)^2-1+(1-\sin^2{\alpha})$$
$$3\cos^2{\alpha} = (n_2/n_1)^2-1$$
$$\alpha^\star = \arccos\left(\frac{(n_2/n_1)^2 - 1}{3}\right)$$

At this critical angle of entry $\alpha^\star(n_1,n_2)$, the corresponding (minimal) scattering angle is:

$$D(\alpha^\star) = \pi + 2\alpha^\star - 4\arcsin{(n_1\sin{\alpha^\star}/n_2)}$$ which defines the rainbow angle $R(n_1,n_2)$.

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