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

`\phi=4.52 rad`

Step-by-step 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=(2\pi)/((c)/(f_r)) *(r^2-r^1)`

`\phi=(2*\pi)/((3*10^(8))/(120*10^6)) *1.8`

`\phi=(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`