--- title: "Light propagation; NA and acceptance angle" subject: "Propagation and Antenna" chapter: 6 key: "semester_6/propagation_and_antenna:light-propagation-na-acceptance-relation" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/light-propagation-na-acceptance-relation site: BCT Engineering --- # Light propagation; NA and acceptance angle Propagation and Antenna · Chapter 6: Optical Fibres - Asked in: 2080 Chaitra, 5+6 marks - Video: [Acceptance Angle & Numerical Aperture NA (Basics, Definition, Derivation & Formula) Explained](https://www.youtube.com/watch?v=q37CpocWZyw) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/light-propagation-na-acceptance-relation - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-6 ## Question Explain the principle of light propagation through a fiber with necessary diagram. Derive the relationship between numerical aperture and acceptance angle. ## Short answer Light propagates via total internal reflection when the incident angle at the core-cladding interface exceeds the critical angle φc=sin⁻¹(n2/n1). The maximum launch angle θa (acceptance angle) satisfies NA = n0 sin θa = √(n1² − n2²). For n1=1.50, n2=1.48, NA=0.244 and θa=14.1°. ## Answer #### Principle of light propagation in an optical fibre Light travels in an optical fibre by **total internal reflection (TIR)**. The core has a higher refractive index (n1) than the cladding (n2). When a ray inside the core meets the core–cladding boundary at an angle (from the normal) greater than the **critical angle** φc = sin⁻¹(n2/n1), it is totally reflected back into the core with no refraction loss. Repeated TIR guides the ray along the fibre, even around gentle bends. ```text cladding n2 ___________________________________ \ /\ /\ / core n1 ray \ φ / \ / \ / ----->\ / \ / \ / ---> \ / \ / \ / cladding n2 ____\/________\/________\/_________ φ > φc at every reflection (TIR) ``` Conditions for TIR: 1. Light must travel from a denser medium to a rarer medium (n1 > n2). 2. The angle of incidence at the core–cladding boundary must exceed the critical angle φc = sin⁻¹(n2/n1). To meet condition 2, light must enter the fibre end within the **acceptance angle** θa. Rays entering at larger angles strike the boundary below φc, partly refract into the cladding and are lost after a short distance. In a step-index fibre the rays follow zig-zag paths; in a graded-index fibre the index decreases gradually from the axis, so the rays bend continuously (curved paths) instead of reflecting sharply. In a single-mode fibre the core is so small that only one mode, travelling almost along the axis, exists. #### Relationship between numerical aperture and acceptance angle ```text n0 (air) | n2 cladding |----------------------------- θa ray | φ \ / φ (φ ≥ φc: TIR) ---------------->*-------\/-----------------> (acceptance) |θ1 n1 core axis |----------------------------- | n2 cladding ``` Let a ray enter the core from a medium of index n0 at angle θa to the axis, refract to angle θ1 inside the core, and meet the core–cladding boundary at angle φ (measured from the normal), where φ = 90° − θ1. 1. Snell's law at the input face: ```text n0 sin θa = n1 sin θ1 ``` 2. For total internal reflection at the core–cladding boundary, φ must be at least the critical angle φc, where ```text sin φc = n2 / n1 ``` 3. The limiting (maximum) entry angle θa corresponds to φ = φc, i.e. θ1 = 90° − φc: ```text n0 sin θa = n1 sin(90° − φc) = n1 cos φc = n1 √(1 − sin²φc) = n1 √(1 − n2²/n1²) = √(n1² − n2²) ``` 4. The numerical aperture is defined as NA = n0 sin θa, so ```text NA = n0 sin θa = √(n1² − n2²) ``` 5. With the relative index difference Δ = (n1² − n2²)/(2n1²) ≈ (n1 − n2)/n1: ```text NA = n1 √(2Δ) ``` For launching from air (n0 = 1): θa = sin⁻¹(NA), and the full acceptance cone angle is 2θa. So the NA is the sine of the acceptance angle (for launching from air) and depends only on the core and cladding indices, not on the fibre diameter. A larger NA gives a wider acceptance cone and easier coupling, but more modes and more modal dispersion. **Example:** n1 = 1.50, n2 = 1.48: NA = √(2.25 − 2.1904) = 0.244, θa = 14.1°. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. TIR requires: A. n1 < n2 B. incidence angle below φc C. n1 = n2 D. n1 > n2 and incidence angle above φc Answer: D. Both conditions needed. (Key point: TIR condition) 2. The critical angle is: A. sin^-1(n2/n1) B. sin^-1(n1/n2) C. cos^-1(n2/n1) D. tan^-1(n2/n1) Answer: A. sin φc = n2/n1. (Key point: Critical angle) 3. For launch from air, NA equals: A. 2θa B. sin θa C. cos θa D. tan θa Answer: B. NA = n0 sin θa with n0 = 1. (Key point: NA and θa) 4. A larger NA gives: A. No change B. Less coupling C. Wider acceptance cone but more modal dispersion D. Narrower cone Answer: C. Easier coupling, more modes. (Key point: Larger NA) 5. For n1 = 1.50, n2 = 1.48, NA is about: A. 0.02 B. 1.0 C. 0.5 D. 0.244 Answer: D. sqrt(2.25 − 2.1904) = 0.244. 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