Subject :- Applied Physics 2.

Topic :- Laser.

Difficulty :- Medium.

The main function of an optical fiber is to accept and transmit as much as light from the source as possible. The light gathering ability of fiber is called as Numerical aperture and it is determined by the acceptance angle. It is measure of amount of light that can be accepted by fiber N.A depends on R.I of core and cladding.

Let us consider the optical fiber into which light is launched. The end at which light enters the fiber is called as launching end. Let $n_1$ & $n_2$ be the refractive indices of core and cladding and no be the refractive medium from which light is launched into the fibre.

Let the light ray enter the fiber at an angle i an angle r and strike the core Cladding interface at an angle φ if ∅≥$∅_c$¸the ray undergoes T.I.R $n_1$ ≥ $n_2$

So long as ∅≥ $∅_c$ , the light stays within the fibre Applying Snell`s law to launching end

               sini / sin⁡r = $n_1$ / $n_0$    -------------------(1)

But $n_0$ = 1

In ∆ ABC ∅=(90-r)

sin∅=sin(90-r)

sin∅=cosr condition for T.I.R $sin∅_c$ ≥ $n_2$ / $n_1$

Therefore cosr ≥ $n_2$ / $n_1$

√(1-$sin^2$ r ) ≥ $n_2$ / $n_1$

1-$sin^2$ r ≥ $n_2^2$ / $n_1^2$

sinr ≤ ($n_1^2$ - $n_2^2$ ) / $n_1$ ...................................(2)

From eq 1 sinr = sini / $n_1$

sin i / $n_1$  ≤ √($n_1^2$ - $n_2^2$ ) / $n_1$

sin⁡i ≤ √($n_1^2$ - $n_2^2$ )

sin $i_m$ = √($n_1^2$ - $n_2^2$ )

The maximum angle of incidence for which the ray propagates down the fiber is called as acceptance angle.

sin $i_m$ = N.A

Where N.A= √($n_1^2$ - $n_2^2$ )

    $i_m$ = $sin^{-1}$ ⁡√($n_1^2$ - $n_2^2$ )

The light rays contained within the cone having full angle 2$i_m$ are accepted and transmitted along the fiber therefore the cone is called acceptance cone.