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Glass-clad fiber numerical aperture

Short answer

Cladding index n₂ = n₁(1−Δ) = 1.549; critical internal reflection angle φ_c = sin⁻¹(n₂/n₁) ≈ 88.1°; numerical aperture NA = n₁√(2Δ) ≈ 0.0514; external acceptance angle θ₀ = sin⁻¹(NA) ≈ 2.95°.

A glass-clad fiber is made with a core glass of refractive index 1.55 and the cladding is doped to give a fractional index difference of 5.5×10⁻⁴. Determine (i) Cladding index (ii) the critical internal reflection angle (iii) the external critical acceptance angle and (iv) numerical aperture (NA).

The fractional index difference is Δ = (n₁ − n₂)/n₁.

Given: n₁ = 1.55, Δ = 5.5 × 10⁻⁴.

(i) Cladding index

n₂ = n₁ (1 − Δ) = 1.55 × (1 − 5.5×10⁻⁴)
   = 1.55 × 0.99945
   = 1.5491

(ii) Critical internal reflection angle (core-cladding)

sin φ_c = n₂/n₁ = 0.99945
φ_c = sin⁻¹(0.99945) = 88.10°

(iii) External critical acceptance angle (from air)

sin θ₀ = NA = 0.0514
θ₀ = sin⁻¹(0.0514) = 2.95°

(iv) Numerical aperture

NA = n₁ √(2Δ) = 1.55 × √(2 × 5.5×10⁻⁴)
   = 1.55 × √(1.1×10⁻³)
   = 1.55 × 0.03317
   = 0.0514

Check: √(n₁² − n₂²) = √(2.4025 − 2.39972) = 0.0514.

Answer: (i) n₂ = 1.549; (ii) φ_c ≈ 88.1°; (iii) θ₀ ≈ 2.95°; (iv) NA ≈ 0.051.

✓ Answer checked by Claude Opus 5.5 (high effort)

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