---
title: "MUF range numerical: 400 km reflection height"
subject: "Propagation and Antenna"
chapter: 4
key: "semester_6/propagation_and_antenna:muf-range-numerical-400km"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/muf-range-numerical-400km
site: BCT Engineering
---

# MUF range numerical: 400 km reflection height

Propagation and Antenna · Chapter 4: Propagation and Radio Frequency Spectrum

- Asked in: 2070 Magh, 5 marks
- Video: [Maximum Usable Frequency (MUF) (Definition, Basics, Derivation, Formula & Calculation) Explained](https://www.youtube.com/watch?v=0Houq2d-Jtk) — 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/muf-range-numerical-400km
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-4

## Question

Assume that reflection takes place at a height of 400 km and that the maximum density in the ionosphere corresponds to a 0.8 refractive index at 15 MHz. What will be the range (assume flat earth) for which the MUF is 20 MHz?

## Short answer

Using n=0.8 at 15 MHz, maximum electron density Nmax=10^12 m^-3 gives critical frequency fc=9 MHz. For MUF=20 MHz, sec(theta_i)=20/9, tan(theta_i)=1.9845. With flat earth at height 400 km, range d=2*h*tan(theta_i)=1588 km.

## Answer

The MUF over a flat-earth path is the critical frequency multiplied by the secant of the angle of incidence (the **secant law**). So we first find the maximum electron density from the given refractive index, then the critical frequency, then the angle and the range.

#### Given

- Virtual height of reflection, h = 400 km
- At the layer maximum, n = 0.8 at f = 15 MHz
- Required MUF = 20 MHz

#### Step 1: Maximum electron density

The refractive index of the ionosphere is n² = 1 − 81N/f² (N in electrons/m³, f in Hz).

```text
81·Nmax = f²(1 − n²)
        = (15×10⁶)² × (1 − 0.8²)
        = 2.25×10¹⁴ × 0.36
        = 8.1×10¹³
Nmax    = 1.0×10¹² electrons/m³
```

#### Step 2: Critical frequency

```text
fc = 9√Nmax = 9 × √(10¹²) = 9×10⁶ Hz = 9 MHz
```

#### Step 3: Angle of incidence for MUF = 20 MHz

```text
MUF = fc·sec θi
sec θi = 20/9 = 2.222
cos θi = 0.45  →  θi = 63.26°
tan θi = √(1 − 0.45²)/0.45 = 1.9845
```

#### Step 4: Range (flat earth)

```text
     ionosphere ~~~~~~~~~~~ P ~~~~~~~~~~~ (reflection point)
                           /|\
                          / | \
                      θi /  |  \ θi
                        /   | h = 400 km
                       /    |    \
                     Tx ----M---- Rx      (flat earth)
                     <----- d ----->
```

For a flat earth, half the range is d/2 = h·tan θi, so:

```text
d = 2h·tan θi = 2 × 400 × 1.9845 = 1587.6 km
```

Check with the combined formula MUF = fc·√(1 + (d/2h)²):

```text
d = 2h·√((MUF/fc)² − 1) = 800 × √(4.938 − 1)
  = 800 × 1.9845 = 1587.6 km
```

**Answer: Nmax = 10¹² /m³, fc = 9 MHz, θi = 63.26°, range d ≈ 1588 km (about 1590 km).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is Nmax if n = 0.8 at 15 MHz?
   A. 10^12 /m^3
   B. 10^10 /m^3
   C. 2.25 x 10^14 /m^3
   D. 8.1 x 10^13 /m^3
   Answer: A. 81N = f^2(1 - n^2) = 2.25e14 x 0.36 = 8.1e13, so N = 1e12. (Key point: Electron density)

2. The critical frequency for this layer is:
   A. 9 MHz
   B. 15 MHz
   C. 20 MHz
   D. 12 MHz
   Answer: A. fc = 9 sqrt(1e12) = 9 MHz. (Key point: Critical frequency)

3. For MUF = 20 MHz, cos(theta_i) equals:
   A. 0.8
   B. 0.45
   C. 0.9
   D. 2.22
   Answer: B. sec theta = 20/9 = 2.222, so cos theta = 0.45 (theta = 63.26 degrees). (Key point: Angle)

4. The flat-earth range is:
   A. About 1588 km
   B. About 794 km
   C. About 400 km
   D. About 3176 km
   Answer: A. d = 2h tan theta = 2 x 400 x 1.9845 = 1587.6 km. (Key point: Range)

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