Answer to Question #335608 in Optics for asra

Question

Assume a transparent rod of diameter d = 6.06 𝜇m has an index of refraction of 1.25. Determine the maximum angle 𝜃 for which the light rays incident on the end of the rod in the figure below are subject to total internal reflection along the walls of the rod. Your answer defines the size of the cone of acceptance for the rod.

Answer 1

$\beta=\arcsin \frac 1n=53.1°,$

$\theta=90°-\beta=36.9°,$

$\alpha=\arcsin(n\sin\theta)=48.6°.$