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Calculate the required $\Delta$ if a fiber with a 8 $\mu$m core and a 125 $\mu$m cladding is to be single mode at 1300 mm. Assume that the core index is 1.46.

written 9.5 years ago by teamques10 ★ 70k

•   modified 9.5 years ago

Mumbai University > Electronics Engineering > Sem7 > Optical Fiber Communication

Marks: 5M

Year: May 2012

optical fiber communication

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written 9.5 years ago by teamques10 ★ 70k

Given:

Core diameter = 8µm

Core radius (a) = 4 µm

λ = 1300nm

$n_1$ = 1.46

V = 2.405 (Single mode step index)

To find: relative refractive index Δ=?

Solution:

$V = \frac{2πan_1}{λ}\sqrt{2∆}$

$ 2.405 = \frac{2π*4*10^{-6}*1.46*\sqrt{(2*Δ)}}{1300*10^{-9} }$

$Δ = 3.62*10^{-3}$

$\boxed{Δ = 0.362\%}$

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