Question

A transmitting antenna of height $h$ and the receiving antenna of height $\frac{3}{4}h$ are separated by a distance of $d$ for satisfactory communication in line-of-sight mode. Then, the value of $h$ is
[Given, radius of the earth is $R$].

• A. $\frac{d^2}{2R}(2\sqrt{2}-\sqrt{6}{)}^2$
• B. $\frac{d^2}{4R}(2\sqrt{2}-\sqrt{6}{)}^2$
• C. $\frac{d^2}{R}(2\sqrt{2}-\sqrt{6}{)}^2$
• D. $\frac{d^2}{8R}(2\sqrt{2}-\sqrt{6}{)}^2$

Answer 1

The correct option is C $\frac{d^2}{R}(2\sqrt{2}-\sqrt{6}{)}^2$

Given,

$h_T=h$ where $h_T$ is the height of the transmitting antenna

$h_R=\frac{3h}{4}$ where $h_R$ is the height of the receiving antenna
$Radiusofearth=R$
The formula used is:
$d_m=\sqrt{2R{h}_T}+\sqrt{2R{h}_R}$
where $d_m$ is the maximum line of sight distance between the two antennae.
Now, we have
$d_m=d$ (Given)

$d=\sqrt{2Rh}+\sqrt{2R\frac{3h}{4}}d=\sqrt{2R}\times (\sqrt{h}+\sqrt{\frac{3h}{4}})$
$\frac{d}{\sqrt{2R}}=\sqrt{h}(1+\sqrt{\frac{3}{4}})\frac{d}{\sqrt{2R}}=\sqrt{h}\left(\frac{2+\sqrt{3}}{2}\right)$
$\frac{2d}{\sqrt{2R}(2+\sqrt{3})}=\sqrt{h}$
Factorizing the denominator, we have
$\sqrt{h}=\frac{2d(2-\sqrt{3})}{\sqrt{2R}}\sqrt{h}=\frac{2\sqrt{2}d(2-\sqrt{3})}{2\sqrt{R}}$

$\sqrt{h}=\frac{\sqrt{2}d(2-\sqrt{3})}{\sqrt{R}}\sqrt{h}=\frac{d(2\sqrt{2}-\sqrt{6})}{\sqrt{R}}$

Squaring both sides, we have
$h=\frac{d^2{(2\sqrt{2}-\sqrt{6})}^2}{R}$

Hence, the answer is OPTION C.