Consider two antennas separated by a distance of 168 m which are each generating

Question

Consider two antennas separated by a distance of 168 m which are each generating radio waves with a frequency of 7.4 MHz in phase with each other. The two antennas are located with one directly west of the other one. What are the two farthest points directly north of one of the antenna where there will be destructive interference, so that a radio receiver placed at that poit would not pick up any signal? (Hint: Consider the difference in path length for the two signals that will give destructive interference.)

Farthest y?
Next farthest y?

Explanation / Answer

wavelength = L = c/f = 3*10^8/6.1*10^6 = 49.2 m

at the first one maxPD = nL/2

194 = n*24.6

n = 7

s1 = y

s2 = sqrt(x^2+y^2)

ds = s2-s1 = sqrt(x^2+y^2) -y = 5L/2 = 123

sqrt(x^2+y^2) = 123 + y

x^2 +y^2 = 15129 + y^2 + 246y

194^2 = 15129 + 246y

y = 91.5 m

-------------------

ds = s2-s1 = sqrt(x^2+y^2) -y = 3L/2 = 73.8

sqrt(x^2+y^2) = 73.8 + y

x^2 +y^2 = 5446.44 + y^2 + 147.6y

194^2 = 5446.44 + 147.6y

y = 218.08 m

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