written 8.5 years ago by | • modified 8.5 years ago |
Mumbai University > Electronics Engineering > Sem7 > Optical Fiber Communication
Marks: 5M
Year: May2014, Dec2013, Dec2012
written 8.5 years ago by | • modified 8.5 years ago |
Mumbai University > Electronics Engineering > Sem7 > Optical Fiber Communication
Marks: 5M
Year: May2014, Dec2013, Dec2012
written 8.5 years ago by | • modified 8.5 years ago |
$\frac{sin{Ө_1 }}{sinӨ_2 } = \frac{n_1}{n_o} =n_1 \ \ as \ \ n_0= 1$
In right angled Δabc,
$θ_2 = \frac{π}2 - ∅_c$
$sin θ_1= n_1sin (\frac{π}2 - ∅_c) = n_1cos ∅_c$
$cos ∅_c = (1-sin∅_c ^2)^\frac{1}2$
From the above equation
$sin θ_1=n_1 (1-sin∅_c ^2)^\frac{1}2$
when the TIR takes place, $\phi_c=\theta_c \ \ and \ \ \theta_1 =\theta_a$
$$sin\theta_a=n_1\big[1-(\frac{n_2}{n_1})^2\big]^{1/2}$$
$$N.A.=sin\theta_a$$
sin $θ_a$ represents all the light rays within cone of $θ_a$ , which maintain the condition of TIR inside the fiber.
The NA is always chosen so as to accept maximum incident light, satisfying other requirements.
$$NA = sin θ_a= \sqrt{(n_1^2-n_2^2 )}$$
The core transmits the optical signal while the cladding guides the light within the core and if the cladding is removed then there is a loss of signal.
For a step index fiber with a constant refractive index core, the wave equation is Bessel differential equation and solutions are cylindrical functions, therefore if cladding is removed the step index fiber behaves like a cylindrical circular optical fiber with $n_2$=1 of air.
Thus, NA increases if the cladding is removed.