Editing Openai/6897769e-4ee4-800f-aba5-69cca34f701c (section)

=== For a ray entering a spherical water droplet with external incidence angle θi\theta_iθi (measured from the droplet normal at the entry point), Snell’s law inside the drop gives: ===

θt=arcsin⁡ ⁣(sin⁡θin(λ)).\theta_t = \arcsin\!\Big(\frac{\sin\theta_i}{n(\lambda)}\Big).θt=arcsin(n(λ)sinθi).
For the primary rainbow (one internal reflection) the total deviation angle D (between incoming and outgoing rays) is:

D(θi;λ)  =  π+2θi−4θt(θi,λ).D(\theta_i;\lambda) \;=\; \pi + 2\theta_i - 4\theta_t(\theta_i,\lambda).D(θi;λ)=π+2θi−4θt(θi,λ).
(Units: radians; convert to degrees for reporting.)

The observed rainbow corresponds to the stationary (extremal) ray: the deviation is extremal when

∂D∂θi=0.\frac{\partial D}{\partial \theta_i}=0.∂θi∂D=0.
This yields the θi\theta_iθi that produces Dmin⁡(λ)D_{\min}(\lambda)Dmin(λ). Numerically we minimize D(θi;λ)D(\theta_i;\lambda)D(θi;λ) over θi∈[0,π2]\theta_i\in[0,\tfrac{\pi}{2}]θi∈[0,2π] to get Dmin⁡(λ)D_{\min}(\lambda)Dmin(λ).

Remark: the visible rainbow angle from the antisolar point is 180∘−Dmin⁡180^\circ - D_{\min}180∘−Dmin. If Dmin⁡≈138∘D_{\min}\approx 138^\circDmin≈138∘, then 180−138=42∘180-138=42^\circ180−138=42∘.