Consider a step-index fiber for which n1 = 1.475, n2 = 1.460 and a = 25µm. What is the maximum value of θ for which the ray ill be guided by the fiber.
sinθ=n12−n22,\sin \theta=\sqrt{n_1^2-n_2^2},sinθ=n12−n22,
sinθ=1.4752−1.462=0.209,\sin\theta=\sqrt{1.475^2-1.46^2}=0.209,sinθ=1.4752−1.462=0.209,
θ=12.1°.\theta=12.1°.θ=12.1°.
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