Destructive Interference when walking toward an antenna

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SUMMARY

The discussion centers on calculating the points along the x-axis where an observer detects minima in radio signals due to destructive interference from two antennas broadcasting at 96.0 MHz. The wavelength (λ) is determined to be 3.125 m using the formula λ = c/f, where c is the speed of light. The correct condition for destructive interference is that the difference in distances from the observer to the two antennas must equal 0.5λ, 1.5λ, 2.5λ, etc. The initial calculation of 18 minima was incorrect because it did not account for the distance to both antennas.

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Homework Statement


Two antennas located at points A and B are broadcasting radio waves of frequency 96.0 MHz, perfectly in phase with each other. The two antennas are separated by a distance d=12.40m. An observer, P, is located on the x axis, a distance x=55.0m from antenna A, so that APB forms a right triangle with PB as hypotenuse.

If observer P starts walking until he reaches antenna A, at how many places along the x-axis will he detect minima in the radio signal, due to destructive interference?


Homework Equations


Destructive interference = lambda / 2
lambda = c/f



The Attempt at a Solution


Since lambda is 3.125 m (3E8 / 96MHz), I thought the total number of times the observer would experience destructive interference would be once every wavelength (where it's lambda / 2) so I took 55 / 3.125 and got 17.6, which would produce 18 destructive interferences. This is incorrect, not sure what I'm doing wrong. Picture is attached. Thanks!
 
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The picture is not attached, but the description is pretty clear so we can get the idea.

Your idea of when destructive interference happens is a little mixed up. The difference in the distances to each speaker must equal 0.5λ, 1.5λ, 2.5λ, etc. (It's not just the distance to antenna A, and ignoring antenna B, as you are saying.)
 

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