Question

The electron density of a layer of ionosphere at a height $150km$ from the Earth's surface is $9\times {10}^{10}/m^3$. For the sky transmission from this layer up to a range of $250km$.

Angle of incidence of this layer is

• A. $0^{\circ }$
• B. ${38}^{\circ }$
• C. ${31.6}^{\circ }$
• D. ${42}^{\circ }$

Answer 1

The correct option is C ${31.6}^{\circ }$

$f/f_c=\sec {\varphi }_i$

$f_c=9\sqrt{N_m}=9\times \sqrt{9\times {10}^{10}}=2.7\times {10}^{6}Hz=2.7MHz$

$f=f_c\sqrt{1+\frac{D^2}{4h^2}}=2.7\times {10}^6\times \sqrt{1+\frac{{(250\times {10}^3)}^2}{{4(150\times {10}^3)}^2}}=3.51\times {10}^{6}Hz$
$\sec {\varphi }_i=\frac{f}{f_c}=\frac{3.51\times {10}^6}{2.7\times {10}^6}=1.174$
${\varphi }_i={\sec }^{-1}\left(1.174\right)={31.6}^{\circ }$