Answer to Question #189465 in Optics for Ankit

Question #189465

Consider a step-index fiber for which n1 = 1.45, n2 = 1.24 and a = 24µm. What is the maximum value of numerical aperture.

Expert's answer

"\\sin \\theta=\\sqrt{n_1^2-n_2^2},"

"\\sin\\theta=\\sqrt{1.45^2-1.24^2}=0.752,"

"\\theta=48.7\u00b0."

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