Derive numerical aperture of optical fiber

Optical fiber

An optical fiber is a thin, flexible strand of transparent glass or plastic that guides light by total internal reflection. It has a central core (refractive index n₁ = μ_core) surrounded by a cladding of lower refractive index n₂ (n₁ > n₂), covered by a protective jacket.

Acceptance angle and numerical aperture

• Acceptance angle (θ₀): the maximum angle, with respect to the fiber axis, at which light from outside can enter the core and still be guided by total internal reflection. The cone of half-angle θ₀ is the acceptance cone.
• Numerical aperture (NA): n₀ sin θ₀, a measure of the light-gathering ability of the fiber. For air, n₀ = 1, so NA = sin θ₀.

Derivation of NA = n₁ √(2Δ)

          cladding n2
 ------------------------------
        \ phi   /|
 n0  ----\-----/ |  core n1
  theta0  \theta1
 ======axis=========================

1. A ray enters the face of the fiber at angle θ₀ to the axis from a medium of index n₀ (air, 1). It refracts at angle θ₁: n₀ sin θ₀ = n₁ sin θ₁. (1)
2. The ray strikes the core-cladding boundary at angle of incidence φ = 90° − θ₁.
3. For the limiting ray, φ equals the critical angle φ_c: sin φ_c = n₂/n₁.
4. Then sin θ₁ = cos φ_c = √(1 − sin² φ_c) = √(1 − n₂²/n₁²) = √(n₁² − n₂²)/n₁.
5. Substitute in (1): n₀ sin θ₀ = n₁ × √(n₁² − n₂²)/n₁ = √(n₁² − n₂²). For air: NA = sin θ₀ = √(n₁² − n₂²).
6. Define the fractional index difference Δ = (n₁ − n₂)/n₁, so n₂ = n₁(1 − Δ). Then n₁² − n₂² = n₁² [1 − (1 − Δ)²] = n₁² (2Δ − Δ²) ≈ 2Δ n₁² (Δ ≪ 1).
7. Therefore

NA = n₁ √(2Δ) = μ_core √(2Δ)

Any ray entering within the cone of full angle 2θ₀ is guided; rays outside it leak into the cladding.