A sky wave with a frequency 55 , MHz is incid | vedclass

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(B) The refractive index $n$ of the ionosphere is given by $n = \sqrt{1 - \frac{81N}{f^2}}$,where $N$ is the electron density in $m^{-3}$ and $f$ is t

Vedclass · Accepted Answer

$45$

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(B) The refractive index $n$ of the ionosphere is given by $n = \sqrt{1 - \frac{81N}{f^2}}$,where $N$ is the electron density in $m^{-3}$ and $f$ is the frequency in $Hz$.Given $N = 400 \, 	ext{electrons}/cm^3 = 400 	imes 10^6 \, 	ext{electrons}/m^3$ and $f = 55 	imes 10^6 \, Hz$.Calculating the term $\frac{81N}{f^2} = \frac{81 	imes 400 	imes 10^6}{(55 	imes 10^6)^2} = \frac{32400 	imes 10^6}{3025 	imes 10^{12}} \approx 1.07 	imes 10^{-8}$.Since this value is extremely small,$n = \sqrt{1 - 1.07 	imes 10^{-8}} \approx 1$.According to Snell's Law,$n_1 \sin i = n_2 \sin r$. Assuming the wave travels from vacuum $(n_1 \approx 1)$ into the $D$-region $(n_2 \approx 1)$,$1 \cdot \sin(45^\circ) = 1 \cdot \sin r$.Therefore,$\sin r = \sin(45^\circ)$,which implies $r = 45^\circ$.

$A$ sky wave with a frequency $55 \, MHz$ is incident on the $D$-region of the Earth's atmosphere at $45^\circ$. The angle of refraction is ......... $^\circ$ (electron density for the $D$-region is $400 \, \text{electrons}/cm^3$).

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