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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

Sagot :

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]