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Find the core radius necessary for single mode operation at 820nm of step index fiber with $n_1=1.482$ and $n_2=1.474$
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Given: $ λ =820nm \\ n_1=1.482 \\ n_2=1.474 \\ V = 2.405 \text {(single mode step index)} $

To Find:

1) Core radius (a)

2) Numerical aperture (N.A)

3) acceptance angle (Өa)

4) solid angle (τ)

Solution:

$\Rightarrow N.A = \sqrt{n_1^2-n_2^2 } \\ =\sqrt{1.482^2-1.474^2 } \\ N.A= 0.153 $

Numerical aperture $(N.A) =0.153$

$\Rightarrow V = \dfrac {2πa(N.A)}λ \\ 2.405 = \dfrac {2π*a*0.153}{820*10^{-9}} \\ a = 2.051µm$

Core radius $(a) = 2.051µm$

Core radius $(a) = 2.051µm \\ \Rightarrow N.A = \sin Өa \\ 0.153=\sin Өa \\ Өa = 8.8°$

Acceptance angle $(Өa) = 8.8°$

$\Rightarrow τ =π(N.A)^2 \\ = 3.14* (0.153)^2 \\ τ =0.073\text { radians}$

Solid angle $(τ) = 0.073$ radians

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