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,

               ψ_max= sin^-1(0.076)

                       = 4.411°