Answer:
[tex]\phi=4.52 rad[/tex]
Explanation:
From the question we are told that
Distance b/e antenna's [tex]d=9.00m[/tex]
Frequency of antenna Radiation[tex]F_r=120 MHz \approx 120*10^6Hz[/tex]
Distance from receiver [tex]d_r=150m[/tex]
Intensity of Receiver [tex]i= 10[/tex]
Distance difference of the receiver b/w antenna's [tex](r^2-r^1)=1.8m[/tex]
Generally the equation for Phase difference [tex]\phi[/tex] is mathematically given by
[tex]\phi=\frac{2\pi}{\frac{c}{f_r}} *(r^2-r^1)[/tex]
[tex]\phi=\frac{2*\pi}{\frac{3*10^{8}}{120*10^6}} *1.8[/tex]
[tex]\phi=\frac{4\pi}{5} *1.8[/tex]
[tex]\phi=4.52 rad[/tex]
Therefore phase difference f between the two radio waves produced by this path difference is given as
[tex]\phi=4.52 rad[/tex]
