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.
I need this answer sir
Given,
n1=1.475n2=1.460n_1=1.475\\n_2=1.460n1=1.475n2=1.460
a=25 μma=25 \ \mu ma=25 μm
Numerical aperture (NA) =n12−n22=(1.475)2−(1.460)2=2.25−2.19=0.210\sqrt{n_1^2-n_2^2}=\sqrt{(1.475)^2-(1.460)^2}=\sqrt{2.25-2.19}=0.210n12−n22=(1.475)2−(1.460)2=2.25−2.19=0.210
θmax=sin−1(0.210)=12.12°\theta_{max}=sin^{-1}(0.210)=12.12\degreeθmax=sin−1(0.210)=12.12°
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