Question #168544

A multimode step index fiber with a core diameter of 80 μm and a relative index difference of 1.5 % is

operating at a wavelength of 0.85 μm. If the core refractive index is 1.48, estimate the normalized

frequency for the fiber and number of guided modes.


1
Expert's answer
2021-03-09T15:34:42-0500

Answer

Here wavelegth

λ=085μm\lambda=085\mu m

Radius of core

a=D/2=40μma=D/2=40\mu m

n1=n_1= core refractive index=1.48

Δ=0.015\Delta=0.015

Putting all value in below equation

normalised frequency of fiber

f=(2π/λ)an1(2Δ)0.5=2×3.14×40μm×1.480.085μm×(2×0.015)0.5=75.8Hzf= (2π/λ) a n_1 (2Δ)^{0.5}\\=\frac{2\times3.14\times40\mu m\times1.48}{0.085\mu m}\times(2\times0.015)^{0.5}\\= 75.8Hz

Now new parameters

V=2πaλ(n12n22)=2×3.14×0.000040.85μm(0.015)=75.80V=\frac{2\pi a}{\lambda}(\sqrt{n_1^2-n_2^2})\\= \frac{2\times3.14\times0.00004}{0.85\mu m}(\sqrt{0.015}) \\ =75.80

the number of guided modes.

MV22=2873M ≈\frac{ V^2} { 2 }= 2873

(i.e. nearly 3000 guided modes!)



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