Q. What is the value of frequency at which electromagnetic wave must be propagated for the D-region of atmosphere to have a refractive index of 0.5. Electron density for D-region is 400 electrons/cc
A) 200 kHz
B) 104.2 kHz
C) 208.4 kHz
D) 312.6 kHz
Explanation
(c) Here, $N = 400$ electronic/cc$= 400 \times 10^{6}$ electrons $m^{- 3}$
$\mu = 0.5 = \sqrt{1 - \frac{81.56\ N}{v^{2}}}$
$= \sqrt{1 - \frac{81.56 \times 400 \times 10^{6}}{v^{2}}}$
On solving, $v = 208.4$ kHz.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.