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Many times, radio antennas occur in pairs. The effect is that they will then produce constructive interference in one direction while producing destructive interference in another direction - 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 antennae operating in phase such that no emission occurs along a line \(45.0^{\circ}\) from the line joining the antennae?

Short Answer

Expert verified
Answer: The minimum distance between the two radio antennas is approximately 2.41 m.

Step by step solution

01

Calculate the wavelength of the radio wave

First, we need to find the wavelength (λ) of the radio wave. To calculate the wavelength, we use the formula: \(λ = \dfrac{c}{f}\) Where c is the speed of light in a vacuum (approximately \(3*10^8 m/s\)) and f is the frequency of the radio wave. In this case, the frequency is \(88.1 \mathrm{MHz}\): \(λ = \dfrac{3*10^8 m/s}{88.1 * 10^6 Hz}\)
02

Solve for the wavelength

Now, we will calculate the wavelength of the radio wave: \(λ = \dfrac{3*10^8 m/s}{88.1 * 10^6 Hz} = 3.41 m\) The wavelength of the radio wave is 3.41 m.
03

Apply the destructive interference condition

The destructive interference occurs when the path difference between the two radio waves is equal to an odd multiple of half wavelength (n(λ/2)), where n is an odd integer: Path difference = \(n * \cfrac{λ}{2}\) In this problem, we want to find the minimum distance, which means n = 1. Therefore, the path difference between the two radio waves should be: Path difference = \(\cfrac{λ}{2} = \cfrac{3.41 m}{2} = 1.705 m\)
04

Apply trigonometry to find the minimum distance between the antennae

We can use the formula for sine function in a right triangle: \(\sin (θ) = \cfrac{\textrm{opposite side}}{\textrm{hypotenuse}}\) Since we know the angle (45 degrees) and the path difference (opposite side) is 1.705 m, we can use the sine function to find the minimum distance between the antennae (hypotenuse): \(\sin(45^{\circ}) = \cfrac{1.705 m}{\textrm{hypotenuse}}\) Now, solve for the hypotenuse: Hypotenuse = \(\cfrac{1.705 m}{\sin(45^{\circ})}\)
05

Calculate the minimum distance between the antennae

Finally, we compute the minimum distance between the antennae: Hypotenuse = \(\cfrac{1.705 m}{\sin(45^{\circ})} = 2.41 m\) The minimum distance between the two antennas to produce destructive interference at an angle of 45 degrees from the line joining the antennae is approximately 2.41 m.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radio Frequency
Radio frequency (RF) refers to the range of electromagnetic frequencies used for communication by radio waves. These waves are immensely useful as they can travel long distances and transmit data effectively. In this case, the frequency we are dealing with is 88.1 MHz, a typical frequency for FM radio signals. Radio frequency is crucial for determining the characteristics of the radio wave, such as its wavelength and its interference pattern.
Understanding the radio frequency is essential, as it affects how radio waves interact with each other. The different behaviors, such as constructive and destructive interference, depend directly on this frequency. Radio frequencies are measured in hertz (Hz), which indicates the number of cycles a wave completes in one second.
Through this frequency, we can determine the wavelength, which is significant when designing systems like antennas to either maximize or minimize interference.
Wavelength Calculation
Wavelength calculation is a fundamental step when working with radio waves, especially when we need to manage interference. To calculate the wavelength (\(λ\)), we use the formula:\(λ = \frac{c}{f}\), where \(c\) is the speed of light (approximately \(3 \times 10^8 \text{ m/s}\)), and \(f\) is the radio frequency (in this case, 88.1 MHz).
Wavelength signifies the distance between repeating units of a wave, like from crest to crest or trough to trough. For this exercise, the wavelength is calculated as 3.41 meters after implementation of the given formula. This calculation helps us in understanding how waves might interact with each other when emitted from different sources.
Therefore, accurate wavelength calculation guides the design and layout of radio infrastructure to ensure optimal signal clarity and strength while reducing unwanted interference.
Destructive Interference
Destructive interference occurs when two waves of the same frequency meet while being out of phase, effectively canceling each other out. This happens when the path difference between the waves is a half-integer multiple of the wavelength. The waves combine to form a reduction in amplitude, which can result in no signal being detected.
In practical applications like designing radio antennas, creating destructive interference in specific directions can be useful. For instance, the exercise shows that in order to ensure that emissions do not interfere with nearby stations, antennas are strategically placed. By ensuring the path difference is an odd multiple of half the wavelength, we can achieve this effect. Hence, the relationship between wavelength and antenna placement distance plays a critical role.
This principle of destructive interference is used when it is essential to prevent radio waves from overlapping and causing unwanted noise in specific directions.
Directional Antennas
Directional antennas are designed to focus radio wave emissions in particular directions, enhancing signal quality and reducing interference with other devices. This is achieved by manipulating the way waves interfere with each other.
An antenna pair can direct the radio waves by creating a pattern of constructive and destructive interference. In areas where we want the signal to be strong, constructive interference is optimized; alternatively, areas where no signal is desired involve setting up conditions for destructive interference.
This concept is critical in densely packed environments, like urban areas, where different radio signal sources need to coexist without issues. Directing an antenna's emission path ensures efficiency and clarity for users relying on those signals, while also minimizing interference with adjacent stations or equipment.
  • Maximizes signal strength in desired directions.
  • Minimizes interference with adjacent stations.
  • Utilizes principles of interference for optimal design.
By understanding and designing for interference, an effectively placed directional antenna setup can greatly enhance communication capabilities.

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Most popular questions from this chapter

A Young's interference experiment is performed with monochromatic green light \((\lambda=540 \mathrm{nm}) .\) The separation between the slits is \(0.100 \mathrm{~mm},\) and the interference pattern on a screen shows the first side maximum \(5.40 \mathrm{~mm}\) from the center of the pattern. How far away from the slits is the screen?

Why are radio telescopes so much larger than optical telescopes? Would an \(X\) -ray telescope also have to be larger than an optical telescope?

In a double-slit experiment, He-Ne laser light of wavelength \(633 \mathrm{nm}\) produced an interference pattern on a screen placed at some distance from the slits. When one of the slits was covered with a thin glass slide of thickness \(12.0 \mu \mathrm{m},\) the central fringe shifted to the point occupied earlier by the 10 th dark fringe (see figure). What is the refractive index of the glass slide? (a) Without the glass slide (b) With glass slide

A diffraction grating with exactly 1000 lines per centimeter is illuminated by a He-Ne laser of wavelength \(633 \mathrm{nm}\). a) What is the highest order of diffraction that could be observed with this grating? b) What would be the highest order if there were exactly 10,000 lines per centimeter?

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