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Calculate the cut-off parameter & the number of modes for a fiber of core diameter = 50$\mu m$, core RI = 1.50 & cladding RI = 1.48, operating at a wavelength of 1 $\mu m$.

written 8.3 years ago by zuzu.zz299 • 0

modified 4.3 years ago by krithikkm200 • 10

Subject: Applied Physics 2

Topic: Fiber Optics

Difficulty: Medium

engineering physics

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1 Answer

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written 8.2 years ago by zuzu.zz299 • 0

modified 8.2 years ago by arnavyadav0002 • 0

$d =50*10^{-6} m ,\\ n_1 = 1.50 ,\\ n_2 = 1.48, \\ λ = 1*10^{-6} m ,\\ V =? ,\\ N_m = ? \\V = 2 П a NA / λ \\ 2a == d , NA \\ NA = 0.2441 \\ V = 50 x10^{-6}* 0.2411 / 1*10^{-6} \\ V = 12.055 \\ \text {For sep index fiber N_m = } V^2 / 2 \\ N_m = 72.66 \\ N_m = 72 $

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