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Find the core radius necessary for single mode operation at 820nm of step index fiber with n1=1.482 and n2=1.474
1 Answer
written 8.1 years ago by |
Given: λ=820nmn1=1.482n2=1.474V=2.405(single mode step index)
To Find:
1) Core radius (a)
2) Numerical aperture (N.A)
3) acceptance angle (Өa)
4) solid angle (τ)
Solution:
⇒N.A=√n21−n22=√1.4822−1.4742N.A=0.153
Numerical aperture (N.A)=0.153
⇒V=2πa(N.A)λ2.405=2π∗a∗0.153820∗10−9a=2.051µm
Core radius (a)=2.051µm
Core radius (a)=2.051µm⇒N.A=sinӨa0.153=sinӨaӨa=8.8°
Acceptance angle (Өa)=8.8°
⇒τ=π(N.A)2=3.14∗(0.153)2τ=0.073 radians
Solid angle (τ)=0.073 radians