= Solution
Follow the branch of a <light ray> that undergoes one internal <specular reflection> in the spherical drop. At entry, <refraction> turns its direction by $\alpha-\beta$. The internal <specular reflection> turns it by $\pi-2\beta$, because the radius is the <normal vector> and the incidence and reflection angles are both $\beta$. At exit the second <refraction> contributes another $\alpha-\beta$. Thus the total <deviation angle> is
$$
\boxed{\delta=2(\alpha-\beta)+(\pi-2\beta)=\pi+2\alpha-4\beta.}
$$
<Snell's law> at entry gives $\sin\alpha=n\sin\beta$, taking the surrounding air's <refractive index> as one. The same internal angle at successive encounters follows from the <isosceles triangles> formed by each chord and the two radii. Internal <specular reflection> here need not be <total internal reflection>; the transmitted branch at that encounter does not belong to the one-reflection <trajectory>.
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