Question

Two radio antennas are 130 m apart on a north-south line. The two antennas radiate in phase at a frequency of 4.1 MHz. All radio measurements are made far from the antennas. The smallest angle, reckoned north of east from the antennas, at which destructive interference of the two radio waves occurs, is closest to:

Two radio antennas are 10 km apart on a north-south axis on a seacoast. The antennas broadcast identical AM radio signals, in phase, at a frequency of 5.9 MHz. A steamship, 200 km offshore, travels due north at a speed of 15 km/hr and passes east of the antennas. A radio on board the ship is tuned to the broadcast frequency. The reception of the radio signal on the ship is a maximum at a given instant. The time interval until the next occurrence of maximum reception is closest to:

Answer 1

wavelength = 3e8 / 4.1e6

wavelength = 73.17 m

now,

for destructive interference

d*sin = ( m + 1/2)wavelength

m = 0

d*sin = wavelength / 2

sin = wavelength / 2d

= 16.345^o

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wavelength = 3e8 / 5.9e6

wavelength = 50.847 m

so,

m * wavelength = d * sin

or

m * wavelength = d * y/L

y = m * wavelength * L / d

y = 1016.95 m

or y = 1.01695 km

time = 4.0677 minutes.