--- title: "Light propagation; define NA and acceptance angle" subject: "Propagation and Antenna" chapter: 6 key: "semester_6/propagation_and_antenna:light-propagation-define-na-acceptance" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/light-propagation-define-na-acceptance site: BCT Engineering --- # Light propagation; define NA and acceptance angle Propagation and Antenna · Chapter 6: Optical Fibres - Asked in: 2079 Chaitra, 2+2+2 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-define-na-acceptance - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-6 ## Question Explain about the principle of light propagation in optical fiber. Define numerical aperture and acceptance angle with necessary figures and mathematical expressions. ## Short answer Light is guided by total internal reflection at the core-cladding boundary. Acceptance angle θa is the largest input angle that still sustains guiding; Numerical Aperture NA = n0 sin θa = √(n1²−n2²) ≈ n1√(2Δ). Example: for n1=1.48, n2=1.46, NA=0.2425, θa=14.0°. ## 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. #### Acceptance angle The **acceptance angle** θa is the maximum angle to the fibre axis at which light can enter and still be guided by TIR. Rays within the cone of half-angle θa are guided. ```text \ θa \ acceptance cone axis ------*========================== / core / ``` ```text θa = sin⁻¹( √(n1² − n2²) / n0 ) ``` #### Numerical aperture **Numerical aperture (NA)** measures the light-gathering ability of the fibre. It is the sine of the acceptance angle times the index of the outside medium: ```text NA = n0 sin θa = √(n1² − n2²) ≈ n1 √(2Δ) Δ = (n1 − n2)/n1 ``` (This follows from Snell's law n0 sin θa = n1 cos φc and sin φc = n2/n1.) Typical values: 0.1–0.15 for single-mode fibre, 0.2–0.3 for multimode fibre. **Example:** n1 = 1.48, n2 = 1.46: NA = 0.2425, θa = 14.0°. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Light is guided in a fibre core by total internal reflection. What must be true at the core-cladding boundary? A. n1 > n2 and the angle of incidence exceeds the critical angle B. n1 < n2 and the angle of incidence is below the critical angle C. n1 = n2 and the angle of incidence is 90 degrees D. n1 > n2 and the angle of incidence is below the critical angle Answer: A. TIR needs light going from the denser core to the rarer cladding at an angle greater than phi_c = sin^-1(n2/n1). (Key point: TIR condition) 2. Which expression gives the critical angle at the core-cladding boundary? A. sin^-1(n1/n2) B. sin^-1(n2/n1) C. cos^-1(n2/n1) D. tan^-1(n2/n1) Answer: B. Setting the refraction angle to 90 degrees in Snell's law gives sin(phi_c) = n2/n1. (Key point: Critical angle) 3. What is the acceptance angle theta_a? A. The critical angle at the core-cladding boundary B. The minimum angle at which light must enter the fibre C. The maximum angle to the fibre axis at which light can enter and still be guided by TIR D. The angle at which light leaves the cladding Answer: C. Rays inside the cone of half-angle theta_a are guided; rays outside strike the boundary below phi_c and leak into the cladding. (Key point: Acceptance angle) 4. For launch from air, which is the numerical aperture of a step-index fibre? A. n1^2 - n2^2 divided by n0 B. sqrt(n1^2 + n2^2) C. (n1 - n2)/n1 only D. sqrt(n1^2 - n2^2) Answer: D. NA = n0 sin(theta_a) = sqrt(n1^2 - n2^2), obtained from Snell's law and sin(phi_c) = n2/n1. 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