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Find the core radius necessary for a single mode operation at 1320nm of a step index fibre with n1=1.48 and n2=1.478. What are the N.A and maximum acceptance angle for this fibre?
Ans:
For a single mode step index fibre, V≤2.405
Take V= 2.4
Given Λ= 1320nm
n1=1.48
n2=1.478
We have V= Πd (n12- n22)½ / λ
Substituting the values of V, n1, n2, d and solving for Λ, we get
d= 1.311 x 10-5 m
Hence, core radius is given by
r=d/2= 6.555µm
Numerical aperture
Sinψ = (n12- n22)½= (1.482- 1.4782)½ =0.076
Maximum acceptance angle,
ψmax= sin-1(0.076)
= 4.411°
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