Question

The maximum distance up to which the TV transmission from a TV tower of height h can be received is proportional to
a)${{h}^{\dfrac{1}{2}}}$
b)$h$
c)${{h}^{\dfrac{3}{2}}}$
d)${{h}^{2}}$

Answer 1

In this transmission, the reception is possible only if the receiving antenna directly intercepts the signal. If a radio wave is transmitted from an antenna, travelling in a straight line that reaches the receiver depends on the height and the curvature of the Earth. Hence obtaining an expression for maximum distance of transmission in terms of the height of the antenna will enable us to choose the correct alternative.

Complete answer:
From the above diagram we can see that ST forms the curvature of the Earth with radius R. The maximum distance up to which the antenna can transmit a signal from point Q is to point ‘S’ or ‘T’ as clearly seen in the figure.
Let us say the antenna is at a height of ‘h’ from point P. The distances between point SQ and TQ are also equal and let it be equal to ‘d’.
From right angled $\Delta OTQ$, we get
$\begin{aligned} & O{{Q}^{2}}=O{{T}^{2}}+Q{{T}^{2}} \\\ & \because OQ=R+h \\\ & \Rightarrow {{(R+h)}^{2}}={{R}^{2}}+{{d}^{2}} \\\ & \Rightarrow {{R}^{2}}+2Rh+{{h}^{2}}={{R}^{2}}+{{d}^{2}} \\\ & \Rightarrow 2Rh+{{h}^{2}}={{d}^{2}} \\\ & \Rightarrow {{d}^{2}}=2Rh\left[ 1+\dfrac{h}{2R} \right] \\\ & \because h>>R, \\\ & \Rightarrow {{d}^{2}}=2Rh \\\ & \therefore d=\sqrt{2Rh} \\\ \end{aligned}$
Hence the maximum distance up to which the TV transmission from a TV tower of height h can be received up to a distance of $\sqrt{2Rh}$ . Since the height of the tower is raised to half.

The correct answer of the above question is a.

Note:
It is to be noted that we have assumed Earth to be a sphere with uniform radius. Hence the above obtained answer is approximately the maximum distance up to which the signal can be transmitted. More specifically, the maximum distance depends on the line of sight i.e. the distance at which the transmitting and the receiving antenna can see each other.