Question

Derive an expression for covering range of TV transmission tower.

Answer 1

TV signals are high frequency signals. They are not reflected back by atmosphere. These signals can be received only If the receiver antenna directly Intercepts them. Therefore the range of TV depends on the height of TV antenna.
Let $AB$ be a tower of height $h$ and $R_e$. radius of earth. If broadcast is made from the top $B$ of tower $\left(AB\right)$, no reception of direct signals is possible beyond points $C$ and $B$ in figure. The distance upto which signals can be received (i.e., $AC=AD$) is limited due to curvature of earth and is caned the range of transmitter. The signals may be Intercepted In the region $CAD$.
In right angled triangle $BOD,\angle BDO={90}^o$
$\therefore B{O}^2=(OD{)}^2+(BD{)}^2$
i.e, $(R_e+h{)}^2=R_e^2+(BD{)}^2$ ...(i)
As height $h$ of the tower is very small as compared to radius $\left(R_e\right)$ of earth, the point $B$ will be very close to $A$, so that
$BD\approx AD=d$ (say)
$\therefore$ Equation (i) gives $(R_e+h{)}^2=R_e^2+d_2$
Or $d^2=(R^e+h^2)-R_e^2$
$=R_e^2+2h{R}_e+h^2-R_e^2=2R_{e}h+h^2$
Again, $h<<R_e⟹h_2<<2R_{e}h$
$\therefore d^2\approx 2R_{e}h⟹d=\sqrt{2R_{e}h}$
or height of transmitting antenna, $h=\frac{d^2}{2R_e}$
or covering range of TV transmitting tower is $d=\sqrt{2R_{e}h}$
Thus oovering range of TV signal can be increased by Increasing the height of transmission antenna.