A sky wave with a frequency of 55 text{ MHz} | vedclass

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(B) The refractive index $n$ of the ionosphere is given by $n = \sqrt{1 - \frac{80.5 N}{f^2}}$, where $N$ is the electron density in $	ext{m}^{-3}$ a

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$45$

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(B) The refractive index $n$ of the ionosphere is given by $n = \sqrt{1 - \frac{80.5 N}{f^2}}$, where $N$ is the electron density in $	ext{m}^{-3}$ and $f$ is the frequency in $	ext{Hz}$.Given: $f = 55 	imes 10^6 	ext{ Hz}$, $N = 400 	ext{ electrons/cm}^3 = 400 	imes 10^6 	ext{ electrons/m}^3$.Calculating the term inside the square root: $\frac{80.5 	imes 400 	imes 10^6}{(55 	imes 10^6)^2} = \frac{32200 	imes 10^6}{3025 	imes 10^{12}} \approx 1.06 	imes 10^{-8}$.Since this value is extremely small compared to $1$, $n \approx \sqrt{1 - 0} = 1$.Using Snell's Law: $n_1 \sin i = n_2 \sin r$. Assuming the wave travels from vacuum $(n_1 \approx 1)$ into the ionosphere $(n_2 \approx 1)$:$1 \cdot \sin(45^\circ) = 1 \cdot \sin r$.Therefore, $r = 45^\circ$.

$A$ sky wave with a frequency of $55 \text{ MHz}$ is incident on the $D$-region of the ionosphere at an angle of $45^\circ$. What is the angle of refraction? (The electron density in the $D$-region is $400 \text{ electrons/cm}^3$).

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