Saved Bookmarks
| 1. |
On a particular day, the maximum frequency reflected from the ionosphere is 8 MHz. On another day, it was found to increase to 9 MHz. Calculate the ratio of maximum electron densities of the ionosphere on the two days. Point out a plausible explanations for this. |
|
Answer» Solution :Here, `v_c = 8 MHz , v'_c = 9 MHz` Now, `v_c = 9 (N_(max))^(1//2) and v'_c = 9(N'_(max))^(1//2) :. (N'_(max))/(N_(max)) = ((v'_(c))/v_c)^2 = (9/8)^2 = 1.266` The increase in the maximum number density of electron `(N_(max))`in ionosphere shows more IONISATION DUE to high energy radiations from sun reaching there, which is possible when sun rays are falling normally. For a layer of ionosphere, the refractive index of layer is given by ` mu = mu_0 ( 1 - (81.45N)/v^2)^(1//2)` As we go deep into the ionosphere, N increase, so `mu` decreases. Due to this, the bending of the incident beam of e.m. waves away from the normal CONTINUES till it reaches the critical angle stage, after which the total internal reflection takes place and electromagnetic beam is reflected back. If frequency v is too high, then N may not be so high to produce enough bending for attainment of critical angle or CONDITIONS of total reflection. |
|