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What is the refractive index of the region operating at 16MHz frequency with the electron density 49×10^10/cm^3?(a) 0.518(b) 0.919(c) 0.155(d) 0.845I had been asked this question in an internship interview.This intriguing question comes from Maximum Usable Frequency topic in chapter Virtual Height, Critical Frequency and Muf of Antennas

Answer»

Correct answer is (b) 0.919

The best I can EXPLAIN: Refractive INDEX \(n=\SQRT{(1-\frac{81N}{f^2})}=\sqrt{(1-\frac{81×49×10^{10}}{(16×10^{6})^2})}=\sqrt{(1-0.155)}=\sqrt{0.845}\)

n=0.919



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