Question

In space wave propagation, radio waves (having wavelength λ ) are transmitted from transmitter (having height T) and reaches the receiver (having height R) located at a distance, d, by two methods (d >> T, d >> R). The first is the direct transmisstion from transmitter to the antenna. The second is the transmisstion from transmitter to the earth surface, followed by reflection from the earth surface.

Assuming that the reflection, from the earth’s surface, takes place from a point midway between the transmitter and the receiver, the total phase difference between the two waves at the receiver will be

(1) \(\pi\)

(2) \(\frac{4\pi TR}{λd}\)

(3) \(\frac{4\pi TR}{λd}+\pi\)

(4) \(\frac{4\pi TR}{λd}-\pi\)

Answer 1

 (3) \(\frac{4\pi TR}{λd}+\pi\)

The direct distance (AB) traversed by the wave between the transmitter and the receiver can be calculated as,

Similarly, the distance (AC plus CB) traversed by the wave between the transmitter and the receiver will be given by,

Path difference (Δ) between the two wave at the receiver = (AC + CB) – (AB)

Since the second wave gets reflected from the earth surface, which a denser medium, therefore, an additional phase difference is of π will be introduced.

Total phase difference = π + \(\frac{4\pi TR}{λd}\)