The critical frequency is the maximum frequency that can be reflected by a layer of the ionosphere at vertical incidence. It is different for different ionosphere layers and is proportional to the square root of the maximum electron density in that layer. The critical frequency changes throughout the day and with atmospheric conditions, making higher frequencies work best during the day and lower frequencies at night. Given the critical frequency and electron density relationship, examples are provided to calculate critical frequency from electron density and vice versa.

1. Critical Frequency

2.  The maximum frequency which can b
e reflected by a
particular layerofionosphereatvertical incidence.This
highest frequency is called critical frequency for that
particularlayer
.
 Critical frequency is different for different layers of
ionosphere.
2 Definition:-

3.  Criticalfrequencyforparticularlayerisproportionalto
squarerootofthemaxim
u
melectrondensityinthelayer
.
𝒇𝒄 = 𝟗 𝑵𝒎
Where fc is critical frequency (in MHz) and
Nm is maximum electron density (in per cubic meter)
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4. 4
CriticalFrequencychangesasthedaypassesandalsodueto
atmosphericconditions.
Becauseo
f thechangeinIonosphereduringdayandnight,
during daytime higher frequency bands under critical
Frequency work best, but during nighttime the lower
frequencybandsworkbest

5. Derivation:-
from Snell’s Law
𝜇 =
sin i
sin r
angle of incidence= ∠ i = 0 , N=Nm , f=fc
sin r
𝜇 = sin i
= 1 − 81𝑁𝑚
𝑐
𝑓2 = 0
𝑓𝑐 = 9 𝑁𝑚
𝜇 = 1 −
Refractive index of ionized medium is given by
81𝑁
𝑓2
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6. What is the critical frequency for reflection at vertical
incidence if the maximum value of electron density is
1.24 * 106 cm – 3.
Solution:
m – 3
𝑓𝑐 = 9 𝑁𝑚
Nm= 1.24 * 106 cm – 3 = 1.24 * 106 * 10 - 6
𝑓𝑐 = 9 1.24 MHz
𝑓𝑐= 10.026 MHz
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7. 7
The observed critical frequencies of E and F layers at Guwahati at a
particular times are 2.5MHz and 8.4 MHz respectively. Calculate the
maximum electron concentrations of layers.
Solution:
𝑚
𝑓2
𝑐
81
𝑓𝑐 = 9 𝑁𝑚 or 𝑁 =
(i) For E layers, fc= 2.5MHz
𝑁𝑚 =
(2.5∗106)2
81
= 0.07716 * 10 m
12 -3
(ii) For F layers, fc= 8.4 MHz
𝑁𝑚 =
(8.4∗106)2
81
= 0.8711 * 10 m
12 -3

8. Reference:
1. K.D. Prasad, Antenna & wave propagation,
Satya Prakashan, 3/e (2010-11)
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9. Thank YOU
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