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/c.c.

Answer

208 . 42 kHz

Explanation

For `D` region , `N = 400` electrons/c.c.
` = 400 xx 10^(6) "electrons"//m^(3)`
`mu //mu_(0) = 0.5`
` mu = 0.5 = sqrt( 1 - ( 81.45 N)/(v^(2))) = sqrt ( 1 - ( 81.45 xx 400 xx 10^(6)) /( v^(2))`
On solving , `v = 208.42 kHz`