Question

Consider two antennas separated by 9.00 m that radiate in phase at 120 MHz, as described in Exercise 35.3. A receiver placed 150 m from both antennas measures an intensity I0 . The receiver is moved so that it is 1.8 m closer to one antenna than to the other. (a) What is the phase difference f between the two radio waves produced by this path difference

Answer 1

Answer:

$\phi=4.52 rad$

Explanation:

From the question we are told that

Distance b/e antenna's $d=9.00m$

Frequency of antenna Radiation$F_r=120 MHz \approx 120*10^6Hz$

Distance from receiver $d_r=150m$

Intensity of Receiver $i= 10$

Distance difference of the receiver b/w antenna's $(r^2-r^1)=1.8m$

Generally the equation for Phase difference $\phi$ is mathematically given by

 $\phi=\frac{2\pi}{\frac{c}{f_r}} *(r^2-r^1)$

 $\phi=\frac{2*\pi}{\frac{3*10^{8}}{120*10^6}} *1.8$

 $\phi=\frac{4\pi}{5} *1.8$

 $\phi=4.52 rad$

Therefore phase difference f between the two radio waves produced by this path difference is given as

$\phi=4.52 rad$