Question

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) 0.845

I 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 1

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