Mumbai University > Electronics and telecommunication > Sem 7 > optical communication and networks

Marks: 05

Years: MAY 2012

Given: Core diameter $= 8µm$

Core radius (a) $= 4 µm$

$ λ = 1300nm \\ n_1 = 1.46 \\ V = 2.405 \hspace{2cm}$ (Single mode step index)

To find: relative refractive indexΔ=?

Solution:

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