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Acceptance angle of fiber in a liquid

Short answer

Acceptance angle in the liquid: θ₀ = sin⁻¹(NA/n₀) = sin⁻¹(0.22/1.25) ≈ 10.1°. Fractional refractive index change Δ = (n₁−n₂)/n₁ with n₂ = √(n₁²−NA²) ≈ 1.605, giving Δ ≈ 9.3×10⁻³.

An optical fiber has a numerical aperture of 0.22 and core refractive index 1.62. Determine the acceptance angle for the fiber in a liquid which has a refractive index of 1.25. Also, determine the fractional refractive index change.

The numerical aperture is defined for light entering from air: NA = √(n₁² − n₂²) = 1 × sin θ_air. For another surrounding medium of index n₀, n₀ sin θ₀ = NA.

Given: NA = 0.22, n₁ = 1.62, n₀ (liquid) = 1.25.

Acceptance angle in the liquid

sin θ₀ = NA / n₀ = 0.22 / 1.25 = 0.176
θ₀ = sin⁻¹(0.176) = 10.14°

(In air it would be sin⁻¹ 0.22 = 12.7°.)

Fractional refractive index change Δ

n₂² = n₁² − NA² = 2.6244 − 0.0484 = 2.5760
n₂  = 1.6050

Δ = (n₁ − n₂)/n₁ = (1.62 − 1.6050)/1.62 = 0.00926

Using the approximation NA = n₁√(2Δ): Δ = NA²/(2n₁²) = 0.0484/5.2488 = 0.00922, which agrees.

Answer: the acceptance angle in the liquid is about 10.1°; the fractional index change is Δ ≈ 9.3 × 10⁻³ (about 0.93 %).

✓ Answer checked by Claude Opus 5.5 (high effort)

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