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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written 8.5 years ago by

teamques10 ★ 68k

• modified 8.4 years ago

Mumbai University > Electronics and Telecommunication > Sem7 > Optical Communication and Networks

Marks: 5M

Year: May 2012

optical networks

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written 8.4 years ago by

teamques10 ★ 68k

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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