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?

Mumbai University > Electronics and Telecommunication > Sem7 > Optical Communication and Networks

Marks: 10M

Year: May 2013

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 (Өa)

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 = \left( \frac {2πa(N.A)}λ\right)$

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

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

Core radius (a) = 2.051µm

• N.A = sin $Ө_a$

0.153=sin $Ө_a$

$Ө_ a = 8.8°$

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

• $τ =π(N.A)^2$

  = $3.14* (0.153)^2$

$τ =0.073 $ radians

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