Find the core radius necessary for single mode operation at 820 nm of step index fiber with $n_1$=1.482 and $n_2$=1.474.

What is the numerical aperture and acceptance angle of this fiber? Calculate the corresponding solid angle?

Given:

λ =820nm

$n_1$=1.482

$n_2$=1.474

V = 2.405 (single mode step index)

To Find:

1) Core radius (a)

2) Numerical aperture (N.A)

3)acceptance angle ($Ө_a$)

4) solid angle ($τ$) Solution: - N.A = $\sqrt{n_1^2-n_2^2 } \ \ =\sqrt{1.482^2-1.474^2 }$ - N.A= 0.153 $$\boxed{Numerical \ \ aperture \ \ (N.A) =0.153}$$

• $V = \frac{2πa(N.A)}λ$

• $ 2.405 = \frac{2π*a*0.153}{820*10^{-9} }$

• a = 2.051µm

  $\boxed{Core \ \ radius \ \ (a) = 2.051µm}$

  Core radius (a) = 2.051µm

• N.A = sin $Ө_a$

  0.153=sin $Ө_a$

  $Ө_a$ = 8.8°

$$\boxed{Acceptance \ \ angle \ \ (Ө_a) = 8.8°}$$

• $τ =π{(N.A)}^2 \\ = 3.14* (0.153)^2$

• $τ$ =0.073 radians $$\boxed{Solid \ \ angle \ \ (τ) = 0.073 \ \ radians}$$