Past exam of the physics 2 paper of the University of Oxford / 2015 / 11 / a / 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 isSnell'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.