The index of refraction of the core of a piece of fiber optic cable is 1.72.If the index of the surrounding cladding is 1.41,what is the critical angle for total internal reflection for a light ray in the core incident on the core-cladding interface?

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Quizplus · Accepted Answer

The critical angle for total internal reflection is given by the equation:sin(critical angle) = n2/n1where n1 is the index of refraction of the less dense medium (in this case, the cladding) and n2 is the index of refraction of the more dense medium (in this case, the core).Plugging in the given values, we get:sin(critical angle) = 1.41/1.72critical angle = sin^-1(1.41/1.72)critical angle = 55.1°

The Index of Refraction of the Core of a Piece