Two antennas located at points A and B are broadcasting radio waves of frequency
ID: 779086 • Letter: T
Question
Two antennas located at points A and B are broadcasting radio waves of frequency 94.0 MHz, perfectly in phase with each other. The two antennas are separated by a distance d= 6.20 m. An observer, P, is located on the x axis, a distance x= 68.0 m from antenna A, so that APB forms a right triangle with PB as hypotenuse. What is the phase difference between the waves arriving at P from antennas A and B?
b) Now observer P walks along the x axis toward antenna A. What is P's distance from A when he first observes fully destructive interference between the two waves? I attempted it by solving for the path difference for the first destructive interference and using that number in a pythagoreum equation but however I was doing it is wrong and I'm not sure why.
c) If observer P continues walking until he reaches antenna A, at how many places along the x axis (including the place you found in the previous problem) will he detect minima in the radio signal, due to destructive interference?
Thanks for the help
5.553x1o-1 rad You are correct. Your receipt no. is 159-5737 Previous Tries Wwnohe nhrst obwerlves tuny destru@tive tonerferentee Detveen the twg wistvesse from A As P gets closer A, the path length difference gets larger. What's the smallest path length Submit Answer Incorrect. Tries 2/5 Previous Tries If observer P continues walking until, he reaches antenna A, at how many places along the radios(andu.dine thedesu tove IoterferethePrevious problem) wil he detect minima ih the when he first observes fully destructive interference between the two waves? 66.35 m difference that gives destructive interference?Explanation / Answer
b)for destructive interference
x2 - x1 = lambda/2
x2 = sqrt (d^2 + x1^2)
sqrt (d^2 + x1^2) - x1 = lambda/2
d^2 + x1^2 = (x1^2 + lambda/2)^2
d^2 = x1lambda + lamba^2/4
x1 = d^2/lambda - lambda/4
lambda = 3 x 10^8/94 x 10^6 = 3.19
x1 = 6.2^2/3.19 - 3.19/4 = 11.25 m
Hence, x1 = 11.25 m
pd = sqrt (68^2 + 6.2^2) - 68 = 0.28 m
x1' = 0.28/3.19 = 0.088
x2 = 6.2/3.19 = 1.94
for minimas, permitted values are : 0.5, 1.5, 2.5...
between 0.088 and 1.94
0.5, 1.5