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Consider two identical antennas separated by 9.00 m that radiate in phase at 120 MHz. 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.What is the phase difference ? between the two radio waves produced by this path difference?

In terms of I0, what is the intensity measured by the receiver at its new position?

User Bim
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Final answer:

The phase difference between two radio waves caused by a path difference affects the measured intensity at a receiver's new position. The phase difference can be calculated and then used to determine the new intensity with respect to the initial intensity I0.

Step-by-step explanation:

When two identical antennas radiate in phase and the receiver is moved closer to one of the antennas, the path difference causes a phase difference (δ) between the two incoming waves, which in turn affects the intensity measured at the new position of the receiver. The phase difference can be calculated using the formula δ = (2π / λ) ∆d, where ∆d is the path difference and λ is the wavelength of the radiation. Knowing the frequency (f) of the antennas allows us to find the wavelength using λ = c/f, where c is the speed of light.

The intensity, I, at the new receiver position will depend on the phase difference and can be expressed using an intensity interference pattern formula. If I0 is the initial intensity when both paths are equal, the new intensity can be found by using the intensity interference formula, which takes into account the path difference and the amplitude of the waves.

User Sebastian The Crab
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