--- title: "Acceptance angle of fiber in a liquid" subject: "Engineering Physics" chapter: 6 key: "semester_1/engineering_physics:fiber-acceptance-angle-liquid-index-change" source: https://www.bctengineer.com/q/semester_1/engineering_physics/fiber-acceptance-angle-liquid-index-change site: BCT Engineering --- # Acceptance angle of fiber in a liquid Engineering Physics · Chapter 6: Laser and Fiber Optics - Asked in: 2070 Chaitra, 5 marks - Video: [Examples of Numerical Aperture, Acceptance Angle, Critical Angle & Refractive Index of Optical Fiber](https://www.youtube.com/watch?v=lGFCPaAI4FM) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/engineering_physics/fiber-acceptance-angle-liquid-index-change - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-6 ## Question 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. ## 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. ## Answer 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 ```text 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 ```text 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 %).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a fiber surrounded by a medium of index n0, which relation gives the acceptance angle θ0? A. sin θ0 = n0 x NA B. n0 sin θ0 = NA C. sin θ0 = NA/n1 D. θ0 is the same as in air whatever n0 is Answer: B. NA is fixed by the fiber; the entrance-face refraction gives n0 sin θ0 = NA, so sin θ0 = NA/n0. (Key point: Acceptance angle in another medium) 2. How does the acceptance angle in the liquid (n0 = 1.25) compare with that in air for this fiber? A. Smaller (about 10.1° against 12.7°), because NA is divided by n0 > 1 B. Larger, because the liquid is denser than air C. Equal, because NA is a fixed property of the fiber D. Zero, because the liquid index is below the core index Answer: A. sin θ0 = 0.22/1.25 = 0.176 gives 10.1°, narrower than sin^-1(0.22) = 12.7° in air. (Key point: Liquid versus air) 3. How is the cladding index found from NA and n1? A. n2 = n1 - NA = 1.40 B. n2 = sqrt(n1^2 + NA^2) ≈ 1.635 C. n2 = n1/NA = 7.4 D. n2 = sqrt(n1^2 - NA^2) ≈ 1.605 Answer: D. Since NA^2 = n1^2 - n2^2, n2^2 = 2.6244 - 0.0484 = 2.5760 and n2 = 1.605. (Key point: Cladding index from NA) 4. What is the fractional index change Δ = (n1 - n2)/n1 here? A. About 9.3 x 10^-2 B. About 0.015 (forgetting to divide by n1) C. About 9.3 x 10^-3 D. About 0.22 Answer: C. Δ = (1.62 - 1.605)/1.62 ≈ 0.0093, which agrees with NA^2/(2 n1^2) = 0.0092. 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