--- title: "Refractive index, critical frequency and MUF" subject: "Propagation and Antenna" chapter: 4 key: "semester_6/propagation_and_antenna:refractive-index-fc-muf" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/refractive-index-fc-muf site: BCT Engineering --- # Refractive index, critical frequency and MUF Propagation and Antenna · Chapter 4: Propagation and Radio Frequency Spectrum - Asked in: 2071 Bhadra, 8 marks - Video: [Critical Frequency (Definition, Basics, Derivation, Formula & Calculation) Explained](https://www.youtube.com/watch?v=A7WNqtKj-YA) — 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/refractive-index-fc-muf - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-4 ## Question With a mathematical relation of refractive index of ionospheric layer derive a relation of critical frequency and maximum usable frequency (MUF) of radio waves with necessary explanation. Consider the earth is not curved. ## Short answer The refractive index of an ionospheric layer is n = sqrt(1 - 81N/f^2). Critical frequency fc = 9*sqrt(Nmax) is the highest frequency reflected vertically. Maximum usable frequency MUF = fc * sec(theta_i) = fc * sqrt(1 + D^2/(4h^2)) for flat earth, where theta_i is incidence angle, D is distance, h is virtual height. ## Answer #### Refractive index of an ionospheric layer An ionised layer contains N free electrons per m³ (charge e, mass m). Under a wave field E sin ωt, each electron moves with velocity v: ```text m dv/dt = e E sin ωt v = −(eE / mω) cos ωt ``` The electron current J = Nev lags E by 90° (an inductive current), so it is opposite to the displacement current ε₀∂E/∂t. Total current density: ```text J_total = ε₀ ∂E/∂t + Nev = ωε₀E cos ωt − (Ne²/mω) E cos ωt = ωε₀ [1 − Ne² / (ε₀ m ω²)] E cos ωt ``` So the effective relative permittivity is εr = 1 − Ne²/(ε₀mω²), and the refractive index n = √εr. Putting e = 1.602 × 10⁻¹⁹ C, m = 9.109 × 10⁻³¹ kg, ε₀ = 8.854 × 10⁻¹² F/m and ω = 2πf: ```text e² / (4π² ε₀ m) = 80.6 ≈ 81 n = √(1 − 81N / f²) (N per m³, f in Hz) ``` Since n < 1 and falls as N rises with height, an upward ray bends away from the normal and returns to earth. #### Critical frequency At vertical incidence the wave returns where n = 0 at the level of maximum density Nmax: ```text 1 − 81Nmax / fc² = 0 fc = 9 √Nmax ``` **Critical frequency** is the highest frequency reflected at vertical incidence. #### Maximum usable frequency (flat earth) ```text ionosphere (virtual height h) ~~~~~~~~~~~~~~~~~~ P ~~~~~~~~~~~~~~~~~~ / | \ /θi| \ / |h \ / | \ flat earth T-----M-----R <--- D ----> TM = MR = D/2, θi = angle of incidence ``` Refractive index of an ionised layer (electron density N per m³, frequency f in Hz): ```text n = √(1 − 81N / f²) ``` By Snell's law, a wave entering at angle θi turns back where the ray becomes horizontal (refraction angle 90°): ```text sin θi = n = √(1 − 81Nmax / f²) cos² θi = 81Nmax / f² f = 9√Nmax / cos θi ``` For vertical incidence (θi = 0) this gives the critical frequency fc = 9√Nmax, so ```text f_MUF = fc sec θi (secant law) ``` From the flat-earth geometry (triangle TMP): ```text tan θi = (D/2) / h sec θi = √(1 + tan² θi) = √(1 + D² / 4h²) f_MUF = fc √(1 + D² / 4h²) = fc √(4h² + D²) / (2h) ``` Solving for the skip distance: ```text (f_MUF / fc)² = 1 + D² / 4h² D = 2h √[(f_MUF / fc)² − 1] ``` **Explanation:** fc depends only on the layer's peak electron density; the MUF depends also on the angle of incidence, so a longer path (larger θi) can use a higher frequency. With flat earth, MUF = fc sec θi = fc√(1 + D²/4h²), always ≥ fc. Frequencies above the MUF escape the layer; in practice OWF ≈ 0.85 MUF is used. **Example:** Nmax = 10¹² /m³ gives fc = 9 × 10⁶ Hz = 9 MHz; for D = 600 km and h = 300 km, f_MUF = 9 × √(1 + 1) = 12.73 MHz. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is n less than 1 in an ionised layer? A. The electron current lags E by 90 degrees and opposes the displacement current B. Electrons have no charge C. The layer has high permittivity D. The wave speeds are lower Answer: A. Electron current is inductive, reducing effective permittivity: er = 1 - Ne^2/(e0 m w^2). (Key point: Electron current) 2. Which gives the refractive index? A. n = sqrt(1 + 81N/f^2) B. n = sqrt(1 - 81N/f^2) C. n = 1 - 81N/f D. n = 81N/f^2 Answer: B. Substituting constants gives e^2/(4 pi^2 e0 m) = 80.6 approx 81. (Key point: Index formula) 3. Which step leads to f = fc sec(theta_i)? A. cos^2 theta_i = 81Nmax/f^2, so f = 9 sqrt(Nmax)/cos theta_i B. sin theta_i = 1 C. n = 0 at all angles D. tan theta_i = f/fc Answer: A. Snell's law gives sin theta = n; then cos^2 theta = 81Nmax/f^2. (Key point: Secant law derivation) 4. For Nmax = 10^12 /m^3, D = 600 km, h = 300 km, the MUF is: A. 9 MHz B. 12.73 MHz C. 18 MHz D. 6.4 MHz Answer: B. fc = 9 MHz and sec theta = sqrt(1 + 1) = 1.414, so MUF = 12.73 MHz. 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