Assume a transparent rod of diameter d = 3.19 µm has an index of refraction of 1.33. Determine the maximum angle θ (in degrees) 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.
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Critical angle is sin^–1 (1/µ) = 48.75˚
Angle of refraction that will make the ray strike at the critical angle is 90 – 48.75 = 41.25˚
sin i = 1.33 sin 41.25
i = 61.3˚
Angle of refraction that will make the ray strike at the critical angle is 90 – 48.75 = 41.25˚
sin i = 1.33 sin 41.25
i = 61.3˚