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Acceptance angle of fiber in water

Short answer

Acceptance angle in water: θ₀ = sin⁻¹(NA/n₀) = sin⁻¹(0.2/1.33) ≈ 8.65° (core index n₁ ≈ 1.60).

An optical fiber has a NA of 0.2 and a cladding refractive index of 1.59. Determine the acceptance angle for the fiber in water which has a refractive index of 1.33.

For a medium of index n₀ outside the fiber, the acceptance angle θ₀ satisfies n₀ sin θ₀ = NA, where NA = √(n₁² − n₂²) is the value measured from air.

Given: NA = 0.2, n₂ = 1.59, n₀ (water) = 1.33.

Core index (for reference)

n₁² = NA² + n₂² = 0.04 + 2.5281 = 2.5681
n₁  = 1.6025

Acceptance angle in water

sin θ₀ = NA / n₀ = 0.2 / 1.33 = 0.1504
θ₀ = sin⁻¹(0.1504) = 8.65°

(In air, θ₀ = sin⁻¹ 0.2 = 11.54°; the cone narrows in water because the refractive-index contrast at the entrance face is smaller.)

Answer: the acceptance angle in water is about 8.65° (core index ≈ 1.60).

✓ Answer checked by Claude Opus 5.5 (high effort)

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