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 due to atmospheric conditions, making higher frequencies better during the day and lower frequencies better at night. Given the maximum electron density, the critical frequency can be calculated using the formula: fc = 9√Nm, where fc is the critical frequency in MHz and Nm is the maximum electron density in electrons per cubic meter.

1. Critical Frequency

2. Definition:-
 The maximum frequency which can be reflected by a
particular layer of ionosphere at vertical incidence. This
highest frequency is called critical frequency for that
particularlayer.
 Critical frequency is different for different layers of
ionosphere.
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3.  Criticalfrequencyforparticularlayerisproportionalto
squarerootofthemaximumelectrondensityinthelayer.
Where fc is critical frequency (in MHz) and
Nm is maximum electron density (in per cubic meter)
𝒇 𝒄 = 𝟗 𝑵 𝒎
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4. 4
Critical Frequency changes as the day passes and also due to
atmosphericconditions.
Because of the change in Ionosphere during day and night,
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
𝜇 =
sin i
sin r
= 1 −
81𝑁 𝑚
𝑓𝑐
2 = 0
𝑓𝑐 = 9 𝑁 𝑚
Refractive index of ionized medium is given by
𝜇 = 1 −
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:
𝑓𝑐 = 9 𝑁 𝑚
Nm= 1.24 * 106 cm – 3 = 1.24 * 106 * 10 - 6 m – 3
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:
𝑓𝑐 = 9 𝑁 𝑚 or 𝑁 𝑚 =
𝑓𝑐
2
81
(i) For E layers, fc= 2.5MHz
𝑁 𝑚 =
(2.5∗106)2
81
= 0.07716 * 1012 m-3
(ii) For F layers, fc= 8.4 MHz
𝑁 𝑚 =
(8.4∗106)2
81
= 0.8711 * 1012 m-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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