Acceptance angle and NA of step-index fibre

Acceptance angle

The acceptance angle θa is the maximum angle, measured from the fibre axis, at which a ray can enter the fibre core and still be guided by total internal reflection. Rays inside the cone of half-angle θa (the acceptance cone) are trapped; rays outside it are refracted into the cladding and lost.

Derivation of NA for a step-index fibre

          n0 (air)  |  n2 cladding
                    |-----------------------------
        θa  ray     |   φ  \  /  φ  (φ ≥ φc: TIR)
   ---------------->*-------\/----------------->
     (acceptance)   |θ1  n1 core   axis
                    |-----------------------------
                    |      n2 cladding

Let a ray enter the core from a medium of index n0 at angle θa to the axis, refract to angle θ1 inside the core, and meet the core–cladding boundary at angle φ (measured from the normal), where φ = 90° − θ1.

1. Snell's law at the input face:

n0 sin θa = n1 sin θ1

2. For total internal reflection at the core–cladding boundary, φ must be at least the critical angle φc, where

sin φc = n2 / n1

3. The limiting (maximum) entry angle θa corresponds to φ = φc, i.e. θ1 = 90° − φc:

n0 sin θa = n1 sin(90° − φc) = n1 cos φc
          = n1 √(1 − sin²φc)
          = n1 √(1 − n2²/n1²)
          = √(n1² − n2²)

4. The numerical aperture is defined as NA = n0 sin θa, so

NA = n0 sin θa = √(n1² − n2²)

5. With the relative index difference Δ = (n1² − n2²)/(2n1²) ≈ (n1 − n2)/n1:

NA = n1 √(2Δ)

For launching from air (n0 = 1): θa = sin⁻¹(NA), and the full acceptance cone angle is 2θa.

Example: n1 = 1.48, n2 = 1.46 gives NA = √(1.48² − 1.46²) = 0.2425 and θa = 14.0°.