Question #215942

A multimode step-index has a relative refractive index difference of 2% and a core refractive index of 1.5. The Number of modes propagating at a wavelength of 1.3um is 1000. Calculate:

a) The cladding refractive index

b) Numerical Aperture

c) The diameter of the fiber core


1
Expert's answer
2021-07-12T12:15:38-0400

Relative index difference, Δ= 2% = 0.02

λ=1.3  μm=1.3×106  mλ = 1.3 \;\mu m = 1.3 \times 10^{-6} \;m

Core refractive index \mu_1= 1.5

Number of modes M = 1000

a) Proportion:

1.5 – 100 %

μ2\mu_2 – (100 -2) %

μ2=1.5×98100=1.47\mu_2 = \frac{1.5 \times 98 }{100}=1.47

b) Numerical Aperture

NA=μ1(2Δ)12NA=1.5×(2×0.02)12=0.3NA = \mu_1 (2 Δ)^{\frac{1}{2}} \\ NA = 1.5 \times (2 \times 0.02)^{\frac{1}{2}} = 0.3

c) The diameter of the fiber core

d=λ2MπNAd=1.3×1062×10003.14×0.3=61.71×106  m=61.72  μmd = \frac{λ \sqrt{2M}}{\pi NA} \\ d = \frac{1.3 \times 10^{-6} \sqrt{2 \times 1000}}{3.14 \times 0.3} \\ = 61.71 \times 10^{-6} \;m \\ = 61.72 \;\mu m


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