--- title: "Multimode vs single mode; NA derivation" subject: "Propagation and Antenna" chapter: 6 key: "semester_6/propagation_and_antenna:multimode-vs-single-na-derivation" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/multimode-vs-single-na-derivation site: BCT Engineering --- # Multimode vs single mode; NA derivation Propagation and Antenna · Chapter 6: Optical Fibres - Asked in: 2070 Magh, 3+7 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/multimode-vs-single-na-derivation - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-6 ## Question What are advantages and disadvantages of multimode fiber over single mode fiber? Derive expression of Numerical Aperture of a stepped index optical fiber. ## Short answer Multimode fibre has larger core, easier LED launch, cheaper connectors and lower system cost, but suffers high intermodal dispersion, high attenuation, and short reach. NA for step-index fibre is derived from Snell's law and TIR condition: NA = n0 sin θa = sqrt(n1^2 - n2^2) ≈ n1 sqrt(2Δ). ## Answer #### Multimode fibre compared with single mode fibre **Advantages of multimode fibre** - Larger core (50–200 µm) and larger NA, so light can be launched from cheap LEDs. - Easier and cheaper splicing, connectors and alignment. - Lower total system cost for short links (LANs, buildings, campuses). - Can be used with low-cost sources at 850 nm. **Disadvantages of multimode fibre** - Large intermodal dispersion, so low bandwidth (MHz·km to about 1 GHz·km). - Higher attenuation than single mode fibre. - Short maximum link length; not suitable for long-haul or high-speed WDM systems. #### Derivation of NA of a step-index fibre Consider a step-index fibre with core index $n_1$, cladding index $n_2$ ($n_1 > n_2$), and outside medium index $n_0$ (air, $n_0 = 1$). A meridional ray enters the core end at angle $\theta_0$ to the axis. ```text n0 | cladding n2 ray \ |__________________ \th0| /\phi axis ---*--+--/--\----------- core n1 |th1 \ |__________________ cladding n2 ``` **Step 1: refraction at the input face.** By Snell's law: $$ n_0\sin\theta_0 = n_1\sin\theta_1 $$ **Step 2: angle at the core–cladding boundary.** The refracted ray meets the boundary at angle $\phi$ to the normal. From the right-angled triangle: $$ \phi = 90^\circ - \theta_1 \quad\Rightarrow\quad \sin\theta_1 = \cos\phi $$ **Step 3: condition for total internal reflection.** The ray is guided only if $\phi \ge \phi_c$, where the critical angle is given by $$ \sin\phi_c = \frac{n_2}{n_1} $$ **Step 4: maximum acceptance angle.** The largest input angle $\theta_0 = \theta_a$ corresponds to $\phi = \phi_c$: $$ \begin{aligned} n_0\sin\theta_a &= n_1\sin\theta_1 = n_1\cos\phi_c \\ &= n_1\sqrt{1 - \sin^2\phi_c} \\ &= n_1\sqrt{1 - \frac{n_2^2}{n_1^2}} \\ &= \sqrt{n_1^2 - n_2^2} \end{aligned} $$ **Step 5: definition of NA.** $$ \text{NA} = n_0\sin\theta_a = \sqrt{n_1^2 - n_2^2} $$ For air, $\text{NA} = \sin\theta_a$ and $\theta_a = \sin^{-1}\sqrt{n_1^2 - n_2^2}$. **Step 6: in terms of relative index difference.** With $\Delta = \dfrac{n_1^2 - n_2^2}{2n_1^2} \approx \dfrac{n_1 - n_2}{n_1}$ (since $n_1 \approx n_2$): $$ n_1^2 - n_2^2 = 2n_1^2\Delta \quad\Rightarrow\quad \text{NA} = n_1\sqrt{2\Delta} $$ **Example:** $n_1 = 1.48$, $n_2 = 1.46$: $$ \begin{aligned} \text{NA} &= \sqrt{1.48^2 - 1.46^2} = \sqrt{0.0588} = 0.2425 \\ \theta_a &= \sin^{-1}(0.2425) = 14.03^\circ \end{aligned} $$ NA depends only on the refractive indices, not on the core diameter. The full acceptance cone angle is $2\theta_a$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is multimode fibre preferred for short links such as LANs? A. It has lower attenuation B. It has higher bandwidth C. Its large core and NA allow cheap LEDs and easy, cheap splicing D. It has no dispersion Answer: C. The large core and NA make coupling and alignment easy and let low-cost 850 nm sources be used, lowering system cost. (Key point: Advantage of multimode) 2. What is the main limitation of multimode fibre? A. Large intermodal dispersion giving low bandwidth B. Core too small to couple light C. Cannot use total internal reflection D. Needs expensive lasers Answer: A. Many modes arrive at different times, which limits bandwidth and link length. (Key point: Disadvantage of multimode) 3. In the NA derivation, the limiting ray at the acceptance angle has the angle phi at the core-cladding boundary equal to: A. 90 degrees B. The critical angle, with sin(phi_c) = n2/n1 C. Zero D. The angle theta_0 Answer: B. The largest guided input angle corresponds to the ray just meeting the TIR limit, phi = phi_c. (Key point: Derivation: TIR condition) 4. Which step gives NA = sqrt(n1^2 - n2^2)? A. n0 sin(theta_a) = n1 sin(phi_c) = n2 B. n0 sin(theta_a) = n1 cos(phi_c) = n1 sqrt(1 - n2^2/n1^2) C. n0 sin(theta_a) = n2 cos(phi_c) D. n0 sin(theta_a) = n1 tan(phi_c) Answer: B. Since sin(theta_1) = cos(phi), at the limit n0 sin(theta_a) = n1 cos(phi_c), and cos(phi_c) = sqrt(1 - n2^2/n1^2). 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