Question
Assume a transparent rod of diameter $d=2.00 \mu \mathrm{m}$ has an index of refraction of 1.36 Determine the maximum angle $\theta$ for which the light rays incident on the end of the rod in Figure P35.39 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.
Step 1
Step 1: The angle of acceptance is given by the formula $\sin(\theta_A) = \sqrt{n_1^2 - n_2^2}$, where $n_1$ is the index of refraction of the rod and $n_2$ is the index of refraction of air. Show more…
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