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Short note: NA for meridional rays

Short answer

For meridional rays, which pass through the fibre axis, NA = n0 sin θa = sqrt(n1² − n2²) ≈ n1√(2Δ). Only rays inside the acceptance cone of half-angle θa are guided; NA depends only on core and cladding indices. Example: n1=1.48, n2=1.46 gives NA=0.2425, θa=14.03°.

Write a short note on NA for meridional rays in optical fiber.

Numerical aperture (NA) of a fibre measures its light-gathering ability. For meridional rays (rays that pass through the fibre axis and stay in one plane), it is the sine of the maximum acceptance angle multiplied by the index of the outside medium.

        n0       cladding n2
   \  ________________________
 th0\/  phi /\      /\
 ----*-----/--\----/--\---- axis  core n1
   ________________________

A ray entering at angle θ0\theta_0 is refracted to θ1\theta_1 (Snell's law n0sin⁡θ0=n1sin⁡θ1n_0\sin\theta_0 = n_1\sin\theta_1) and meets the core–cladding boundary at ϕ=90∘−θ1\phi = 90^\circ - \theta_1. It is guided only if ϕ≥ϕc\phi \ge \phi_c where sin⁡ϕc=n2/n1\sin\phi_c = n_2/n_1. At the limit:

n0sin⁡θa=n1cos⁡ϕc=n11−n22n12NA=n0sin⁡θa=n12−n22≈n12Δ\begin{aligned} n_0\sin\theta_a &= n_1\cos\phi_c = n_1\sqrt{1-\frac{n_2^2}{n_1^2}} \\ \text{NA} = n_0\sin\theta_a &= \sqrt{n_1^2 - n_2^2} \approx n_1\sqrt{2\Delta} \end{aligned}

where Δ=(n1−n2)/n1\Delta = (n_1 - n_2)/n_1 is the relative index difference.

Key points

  • NA depends only on n1n_1 and n2n_2, not on core size.
  • Rays inside the cone of half-angle θa\theta_a are guided; others leak into the cladding.
  • Typical values: 0.1–0.15 (single mode), 0.2–0.3 (multimode).
  • Example: n1=1.48n_1 = 1.48, n2=1.46n_2 = 1.46 gives NA =0.2425= 0.2425 and θa=14.03∘\theta_a = 14.03^\circ in air.
  • Skew rays (not passing through the axis) can be accepted at slightly larger angles, so the meridional NA is the standard, conservative value.

✓ Answer checked by Claude Opus 5.5 (high effort)

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