Deduce an expression for the distance from which the T.V. signals can directly be received from a T.V. tower of height h.

Explanation

Let `AB` be a T.V. tower of height `h` above the earth. `T.V.` signal is received within a circle of `CQ=QD` on the surface of earth. Let `R` be the radius of the earth.
Let `CQ=QD=d` and `AB=h`.
`:. OB=R+h`
In rt. `/_d` triangle `OCB,`
`OB^(2)=OC^(2)+BC^(2)`
`(R+h)^(2)=R^(2)+d^(2)`
`d^(2)=(R+h)^(2)-R^(2)`
`=R^(2)+h^(2)+2Rh-R^(2)`
or `d^(2)=h^(2)+2Rh`
Since `h^(2)` can be neglected as compared to `2hR`
`:.d^(2)=2Rh` or `d=sqrt(2hR)`
or `h=(d^(2))/(2R).`

Area covered by T.V. signal `=pid^(2)`
`=pi2hR=2pihR`
Population covered = Area covered `xx` Population density
Since `dpropsqrt(h,)( :.2R` is constant )
So greater the height of T.V. transmitting antenna, grater is its range.