question_answer

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)  \[\frac{{{d}^{2}}}{8R}{{(2\sqrt{2}-\sqrt{6})}^{2}}\]          

Correct Answer: C

Solution :

The situation can be shown in the diagram Here,     \[d={{d}_{1}}+{{d}_{2}}=\sqrt{2hR}+\sqrt{2\times \frac{3}{4}hR}\]                 \[=\sqrt{2hR}+\sqrt{\frac{3}{2}hR}\] \[\Rightarrow \]               \[d=\sqrt{R}\left( \sqrt{2}+\sqrt{\frac{3}{2}} \right)\sqrt{h}\] or       \[\sqrt{h}=\frac{d}{\left( \sqrt{2}+\sqrt{\frac{3}{2}} \right)\sqrt{R}}=\frac{\sqrt{2}d}{(2+\sqrt{3})}\times \frac{1}{\sqrt{R}}\]                 \[=\frac{\sqrt{2}(2-\sqrt{3})d}{\sqrt{R}}\] or            \[h=\frac{{{d}^{2}}}{R}{{(2\sqrt{2}-\sqrt{6})}^{2}}=\frac{{{\lambda }^{2}}}{R}{{(2\sqrt{2}-\sqrt{6})}^{2}}\]