Question

Many times, radio antennas occur in pairs. They then produce constructive interference in one direction while producing destructive interference in another direction-acting as a directional antenna-so that their emissions don't overlap with nearby stations. How far apart at a minimum should a local radio station, operating at \(88.1 \mathrm{MHz},\) place its pair of antennas operating in phase so that no emission occurs along a line \(45.0^{\circ}\) from the line joining the antennas?

Short answer

Answer: The minimum distance between the antennas should be approximately 2.4 meters.

Step-by-step solution

Find the wavelength of the radio antennas

We know the operating frequency of the radio antennas, let's find the wavelength. We can use the formula: \(λ= \dfrac{c}{f}\) where \(λ\) is the wavelength, \(c\) is the speed of light (\(3×10^8 m/s\)), and \(f\) is the frequency of the radio waves (\(88.1 MHz\)). \(λ = \dfrac{3×10^8 m/s}{88.1×10^6 Hz}\) \(λ ≈ 3.4 m\)

Calculate the path difference

The path difference must be half of the wavelength for destructive interference to occur: \(Δd = \dfrac{λ}{2}\) \(Δd = \dfrac{3.4 m}{2}\) \(Δd = 1.7 m\)

Calculate the minimum distance between the antennas

Now, we will use the angle \(θ = 45.0 ^{\circ}\) between the line joining the antennas and the direction with no interference. The path difference is related to the distance \(d\) between the antennas and the sin of the angle \(θ\) like this: \(Δd = d \sinθ\) We need to find the minimum distance \(d\) between the antennas: \(d = \dfrac{Δd}{\sinθ}\) \(d = \dfrac{1.7 m}{\sin{45.0^{\circ}}}\) \(d \approx 2.4 m\) So, the minimum distance between the pair of antennas should be about \(2.4\) meters for no emission along a line \(45.0^{\circ}\) from the line joining the antennas.