--- title: "Numerical aperture and dispersion" subject: "Propagation and Antenna" chapter: 6 key: "semester_6/propagation_and_antenna:na-and-dispersion" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/na-and-dispersion site: BCT Engineering --- # Numerical aperture and dispersion Propagation and Antenna · Chapter 6: Optical Fibres - Asked in: 2071 Magh, 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/na-and-dispersion - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-6 ## Question Describe the following terms: Numerical Aperture and Dispersion in Optical Fiber. ## Short answer Numerical aperture is the light-gathering ability of a fibre, NA = n0 sin θa = sqrt(n1^2 - n2^2) ≈ n1 sqrt(2Δ); it defines the acceptance cone. Dispersion is pulse spreading in time, caused by intermodal, chromatic (material and waveguide), and polarization mode dispersion, limiting bit rate and bandwidth. ## Answer #### Numerical aperture (NA) **Numerical aperture** is the light-gathering ability of a fibre. It is the sine of the acceptance angle $\theta_a$ (the largest angle to the axis at which an incoming ray is still guided by total internal reflection) multiplied by the index of the launching medium: $$ \text{NA} = n_0\sin\theta_a = \sqrt{n_1^2 - n_2^2} \approx n_1\sqrt{2\Delta}, \qquad \Delta = \frac{n_1 - n_2}{n_1} $$ ```text acceptance cone \ cladding n2 \ ____________________ ---*<)th_a core n1 --> / ____________________ / ``` - Rays inside the cone of half-angle $\theta_a$ are guided; rays outside it are lost into the cladding. - A large NA makes it easy to couple light from an LED, but also increases modal dispersion. - Typical NA: 0.2–0.3 for multimode, about 0.1 for single mode fibre. - Example: $n_1 = 1.48$, $n_2 = 1.46$ gives NA $= 0.2425$ and $\theta_a = 14.03^\circ$. #### Dispersion **Dispersion** is the spreading of a light pulse in time as it travels along the fibre. Spread pulses overlap with neighbours (inter-symbol interference), which limits the bit rate and the bandwidth–distance product. ```text input _|^|_ output _/^^\_ (wider, lower) ``` Types: 1. **Intermodal (modal) dispersion**: in multimode fibre, different modes travel different path lengths and arrive at different times. For step-index fibre $\Delta T \approx \dfrac{L n_1 \Delta}{c}$. Reduced by graded-index fibre; absent in single mode fibre. 2. **Intramodal (chromatic) dispersion**: the source emits a range of wavelengths that travel at different speeds. - **Material dispersion**: refractive index of silica changes with wavelength. - **Waveguide dispersion**: the mode's speed depends on core size relative to wavelength. - Chromatic dispersion is near zero at about 1310 nm in standard fibre. 3. **Polarization mode dispersion**: the two polarizations travel at slightly different speeds due to fibre asymmetry; important at very high bit rates. Dispersion is reduced by using single mode fibre, narrow-linewidth lasers, operation near 1310 nm, or dispersion-shifted fibre at 1550 nm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Numerical aperture of a fibre is best defined as: A. n0 sin(theta_a) = sqrt(n1^2 - n2^2) B. n1 - n2 C. The core diameter divided by wavelength D. The critical angle of the core-cladding boundary Answer: A. NA is the sine of the acceptance angle times the launching-medium index, which equals sqrt(n1^2 - n2^2). (Key point: Definition of NA) 2. What is the effect of a large NA? A. Harder LED coupling and less modal dispersion B. Easier LED coupling but more modal dispersion C. No change in coupling but lower loss D. Zero chromatic dispersion Answer: B. A wider acceptance cone collects more light, but it allows more modes and path differences, so modal dispersion rises. (Key point: Effect of large NA) 3. For n1 = 1.48 and n2 = 1.46 in air, what is NA? A. 0.02 B. 0.1 C. 0.2425 D. 0.99 Answer: C. NA = sqrt(1.48^2 - 1.46^2) = sqrt(0.0588) = 0.2425. (Key point: NA calculation) 4. Which dispersion is absent in single mode fibre and reduced by graded-index fibre? A. Intermodal (modal) dispersion B. Material dispersion C. Waveguide dispersion D. Polarization mode dispersion Answer: A. Modal dispersion arises from different modes taking different path lengths; a single mode fibre has only one mode. 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