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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.

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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:

$V = \frac{2πan_1}{λ}(\sqrt{2∆}) \\ 2.405 = \frac{2π*4*10^{-6}*1.46*\sqrt{2*Δ}}{1300*10^{-9}} \\ Δ = 3.62*10^{-3} \\ Δ = 0.362 \%$

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