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Chapter 35: Problem 3

35.3. A radio transmitting station operating at a frequency of 120 \(\mathrm{MHz}\) has two identical antennas that radiate in phase. Antenna \(B\) is 9.00 \(\mathrm{m}\) to the right of antenna \(A\) . Consider point \(P\) between the antennas and along the line connecting them, a horizontal distance \(x\) to the right of antenna \(A\) . For what values of \(x\) will constructive interference occur at point \(P ?\)

Short Answer

Expert verified
Values of \(x\) for constructive interference are \(x = 1.75, 4.0, 6.25, 8.5\) meters within \(0 \leq x \leq 9\).

Step by step solution

01

Identify Wavelength and Recognize Formula

The frequency of the radio wave is given as \(120 \, \text{MHz}\), which is \(120 \times 10^6 \, \text{Hz}\). The speed of light \(c\) is approximately \(3 \times 10^8 \, \text{m/s}\). The wavelength \(\lambda\) can be calculated using the formula:\[ \lambda = \frac{c}{f} = \frac{3 \times 10^8}{120 \times 10^6} = 2.5 \, \text{m} \]
02

Determine Path Difference for Constructive Interference

For constructive interference to occur, the path difference between the waves from antennas \(A\) and \(B\) must be an integer multiple of the wavelength, i.e., \(n\lambda\), where \(n\) is an integer (0, 1, 2, ...). The path difference is given by:\[ \text{Path difference} = (9.00 - x) - x = 9.00 - 2x \]
03

Solve for x with Constructive Interference Condition

Set the path difference to be equal to \(n\lambda\): \[ 9.00 - 2x = n \times 2.5 \]Solve for \(x\):\[ 2x = 9.00 - n \times 2.5 \]\[ x = \frac{9.00 - n \times 2.5}{2} \]
04

Determine Valid Range for x

Since \(x\) is the distance to the right of antenna \(A\) and must be within the 9 meters between the antennas, the valid range for \(x\) is \(0 \leq x \leq 9\). Plug appropriate integer values of \(n\) into the equation to ensure \(x\) lies within this range.

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Key Concepts

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

Constructive Interference
When two or more waves meet and overlap, they can interfere with each other. If the crests and troughs of the waves align perfectly, this results in constructive interference. Constructive interference increases the amplitude of the resulting wave. This occurs when the path difference between the waves is an integer multiple of their wavelength (i.e., 0, 1, 2, and so on). Why does this happen?
  • When the path difference is zero or an exact multiple of the wavelength, the waves are in phase, which means their peaks and troughs align.
  • This alignment allows the waves to add up constructively, leading to a situation where the energy is combined rather than canceled.
Constructive interference is crucial when designing systems that rely on constructive alignment, like certain kinds of antennas or sound systems, ensuring strong and clear signal reception.
Wavelength Calculation
Understanding wavelength is essential for grasping how radio waves function. The wavelength of a wave is the distance between two successive crests. It is calculated using the formula:\[\lambda = \frac{c}{f}\]where \( \lambda \) (lambda) is the wavelength, \( c \) is the speed of light (3 \times 10^8 \text{ m/s} ), and \( f \) is the frequency of the wave.
  • In our exercise, the given frequency is 120 MHz, translating to 120 million cycles per second.
  • By dividing the speed of light by this frequency, we find the wavelength is 2.5 meters.
This calculation helps determine how radio waves of specific frequencies will spread out and interact with the environment, providing a foundational understanding of radio broadcasting technology.
Radio Frequency
Radio frequency (RF) refers to the oscillation rate of electromagnetic waves in the range from around 20 kHz to several hundred GHz, used for communication signals. Mentioned as mega hertz (MHz), it defines the specific band or range of the wave.
  • For example, 120 MHz is within the range of frequencies used in FM radio broadcasting.
  • This frequency dictates the behavior and applications of the specific radio signal, such as broadcasting audio signals to radio receivers.
Radio frequencies are fundamental in daily technology, from radios, televisions, mobile phones, to Wi-Fi. Understanding RF helps design systems that can effectively send and receive information over vast distances with reduced loss of clarity and power.
Path Difference
When two antennas are situated together, like in the problem at hand, the signals they emit can combine constructively or destructively, depending on their path difference. Path difference is crucial in determining whether interference will be constructive. It is calculated by finding the difference in the distance that waves from each antenna travel to a common point.
  • In this scenario, the path difference is calculated as 9.00 - 2x, where x is the position on the line between the two antennas relative to antenna A.
  • For constructive interference, the path difference should equal a multiple of the wavelength, n\(\lambda\), ensuring that the waves meet in phase at the point.
Understanding path difference is essential for configuring devices like antennas, to ensure optimal performance and improved signal qualities through enhanced constructive interference.

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Most popular questions from this chapter

35.29. Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge so that a very thin wedge of air is formed The plates are illuminated at normal incidence by 546 -nm light from a mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge.

35.21. Coherent light with wavelength 500 \(\mathrm{nm}\) passes through narrow slits separated by 0.340 \(\mathrm{mm}\) . At a distance from the slits large compared to their separation, what is the phase difference (in radians) in the light from the two slits at an angle of \(23.0^{\circ}\) from the centerline?

35.8. Young's experiment is performed with light from excited helium atoms \((\lambda=502 \mathrm{nm}) .\) Fringes are measured carefully on a screen 1.20 \(\mathrm{m}\) away from the double slit, and the center of the 20 \(\mathrm{th}\) fringe (not counting the central bright fringe) is found to be 10.6 \(\mathrm{mm}\) from the center of the central bright fringe. What is the separation of the two slits?

35.2. Radio Interference. Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is 120 \(\mathrm{m}\) to the right of antenna \(A\) . Consider point \(Q\) along the extension of the line connecting the antennas, a horizontal distance of 40 \(\mathrm{m}\) to the right of antenna \(B\) . The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q ?\) (b) What is the longest wave-length for which there will be constructive interference at point \(Q\) ?

35.14. Coherent light that contains two wavelengths, 660 \(\mathrm{nm}\) ( red) and 470 \(\mathrm{nm}\) (blue), passes through two narrow slits separated by \(0.300 \mathrm{mm},\) and the interference pattern is observed on a screen 5.00 \(\mathrm{m}\) from the slits. What is the distance on the screen between the first-order bright fringes for the two wavelengths?
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