question_answer

A transmitting antenna of height A 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 \[\sqrt{h}=\frac{d}{\left( \sqrt{2}+\sqrt{\frac{3}{2}} \right)\sqrt{R}}=\frac{\sqrt{2d}}{(2+\sqrt{3})}\times \frac{1}{\sqrt{R}}\] \[=\frac{\sqrt{2}(2-\sqrt{3})d}{\sqrt{R}}\] or \[h=\frac{{{d}^{2}}}{R}{{(2\sqrt{2}-\sqrt{6})}^{2}}=\frac{{{\lambda }^{2}}}{R}{{(2\sqrt{2}-\sqrt{6})}^{2}}\]