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Sky wave propagation involves radio waves being reflected back to Earth by the ionosphere. For this reflection to occur effectively, the frequency of the radio waves must lie within a specific range. If the frequency is too low, the waves get absorbed by the ionosphere, especially in the D layer during the day. If the frequency is too high, the waves penetrate the ionosphere and do not return to Earth.
The critical frequency, denoted as $$f_c$$, is the maximum frequency at which a radio wave can be reflected back to Earth when sent vertically upwards. It is calculated using the formula:
$$f_c = 9 \sqrt{N_{\text{max}}}$$where $$N_{\text{max}}$$ is the maximum electron density in the ionosphere. A typical value for $$N_{\text{max}}$$ is $$10^{12}$$ electrons per cubic meter. Substituting this value:
$$f_c = 9 \times \sqrt{10^{12}}$$Since $$\sqrt{10^{12}} = (10^{12})^{1/2} = 10^{6}$$ (because $$12 \times \frac{1}{2} = 6$$), we get:
$$f_c = 9 \times 10^6 \text{ Hz}$$Converting Hz to MHz (where $$10^6$$ Hz = 1 MHz), this becomes:
$$f_c = 9 \text{ MHz}$$However, for oblique incidence (waves sent at an angle), the maximum usable frequency (MUF) is higher than $$f_c$$ and is given by:
$$\text{MUF} = \frac{f_c}{\cos \theta}$$where $$\theta$$ is the angle of incidence. For long-distance communication, $$\theta$$ is large, so $$\cos \theta$$ is small, allowing MUF to be significantly higher than $$f_c$$. Typically, the MUF can reach up to about 25-30 MHz under favorable conditions.
The effective frequency range for sky wave propagation is generally from 3 MHz to 30 MHz. Frequencies below 3 MHz are prone to absorption, while frequencies above 30 MHz tend to penetrate the ionosphere. Now, examining the options:
Therefore, the correct frequency range for sky wave propagation is given in Option B.
Hence, the correct answer is Option B.
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