Two radio antennas are 120 m apart on a north-south line, and they radiate in phase at a frequency of 3.4 mhz. all radio measurements are made far from the antennas. if the east-west reference line passes midway between the two antennas, what is the smallest angle from the antennas in degrees, measured north of east, at which constructive interference of two radio waves occurs? (c = 3.00 × 10^8 m/s)

a. 22
b. 47
c. 77
d. 63

43degrees

Explanation:

difference in length between the two antennas for an observer in quadrant I is

ΔL = d*Cos(θ)

We can Prove this by letting θ=0 that is the observer is north of the station. Then,

ΔL= d= 120 m.

The wavelength formula is:

λ = c / f = 3*10^8 / 3.4*10^6 = 88.2m

Note that

Constructive interference occurs when the Length difference is equal to an integer multiple of a wavelength.

At the given spacing, and from an observation point north of the stations, they are more than a wavelength apart, so they are not in constructive interference. Looking for the first peak:

d*Cos(θ) = λ

Cos(θ) = λ/d

θ = arcCos(λ/d) = arcCos( 88.2 / 120 ) = 42.7°= 43degrees

For destructive interference it will be

90degrees - 43degrees= 47degrees

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.