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Chapter 35: Problem 21

Consider two antennas separated by \(9.00 \mathrm{~m}\) that radiate in phase at \(120 \mathrm{MHz},\) as described in Exercise \(35.1 .\) A receiver placed \(150 \mathrm{~m}\) from both antennas measures an intensity \(I_{0} .\) The receiver is moved so that it is \(1.8 \mathrm{~m}\) closer to one antenna than to the other. (a) What is the phase difference \(\phi\) between the two radio waves produced by this path difference? (b) In terms of \(I_{0},\) what is the intensity measured by the receiver at its new position?

Short Answer

Expert verified
Part (a): The phase difference between the two radio waves is approximately \(4.536 rad\) or \(260^{\circ}\). Part (b): At its new position, the receiver measures an intensity of \(0.587I_{0}\).

Step by step solution

01

Calculate Wavelength

Before we can calculate the phase difference, we first need to determine the wavelength (\(\lambda\)). We know the frequency (\(f = 120 MHz\)) and the speed of light ( \(c = 3 \times 10^8 m/s\)). We can use the formula \(c = f \times \lambda\) to calculate wavelength. Solving for \(\lambda\) gives: \( \lambda = c/f = 3 \times 10^8 m/s/120 \times 10^6 Hz = 2.5 m\).
02

Calculate Phase Difference

Now we can calculate the phase difference (\(\phi\)). The formula for phase difference is \(\phi=2\pi \times \(\Delta d\) /\(\lambda\)\), where \(\Delta d\) is the path difference and \(\lambda\) is the wavelength. Substituting \(\Delta d = 1.8 m\) and \(\lambda = 2.5 m\), we find \(\phi =2\pi \times (1.8 m) / (2.5 m) = 4.536 rad\) or \(260^{\circ}\).
03

Find Intensity at New Position

Next, let us calculate the intensity at the new position. The formula for the intensity of superposing waves is \(I=4I_{0}\cos^2(\(\phi/2\))\). Substituting the values \(I_{0}\) and \(\phi = 260^{\circ}\), we find \(I = 4I_{0}\cos^2((260/2)^{\circ}) = 0.587I_{0}\).

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Key Concepts

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

Phase Difference
When two waves, such as radio waves from antennas, travel different paths, they can arrive at a point out of sync, creating a phase difference. The phase difference, denoted as \( \phi \), is a measure of how "out of step" two waves are at a given point.
It is expressed in radians or degrees, with one complete wave cycle being \( 2\pi \) radians or \( 360^{\circ} \).
To calculate the phase difference, the formula \( \phi = 2\pi \times (\Delta d) / \lambda \) is used, where:
  • \( \Delta d \) is the path difference, the extra distance one wave travels compared to the other.
  • \( \lambda \) is the wavelength of the waves.
In our specific case, with a path difference of \( 1.8 \) meters and a wavelength of \( 2.5 \) meters, the phase difference can be calculated as \( 4.536 \) radians or \( 260^{\circ} \).
This tells us that the waves are significantly out of phase, impacting how they combine at the receiver's new location.
Intensity of Superposing Waves
The intensity of superposing waves is directly affected by their phase difference. Intensity is a measure of the energy carried by the waves; when two waves meet, they interfere with each other.
Depending on their phase difference, this interference can be constructive or destructive:
  • Constructive interference occurs when the waves are in phase (\( \phi = 0 \)), leading to a combined wave with greater intensity.
  • Destructive interference happens when the waves are out of phase (\( \phi =\pi \)), potentially reducing intensity.
The formula used to determine the intensity of superposing waves is \( I = 4I_{0}\cos^2(\phi/2) \), where \( I_{0} \) is the initial measured intensity.
In scenarios like ours, where the phase difference is \( 260^{\circ} \), the intensity will diminish due to the cosine squared term, leading to an intensity of \( 0.587I_{0} \).
This showcases the impact phase difference can have on wave energies.
Wavelength Calculation
To understand wave behavior, calculating the wavelength is crucial, as it links frequency and speed of light. The speed of light \( c \) is constant at approximately \( 3 \times 10^8 \) meters per second. Wavelength \( \lambda \) is computed using:
  • The speed of light \( c \)
  • Frequency \( f \), given here as \( 120 \) MHz
The formula \( c = f \times \lambda \) allows one to solve for \( \lambda \), rearranging it gives: \( \lambda = c/f \).
For this exercise, the wavelength is \( 2.5 \) meters.
Understanding this characteristic helps in explaining why the waves interact the way they do based on their frequency and medium of travel.
Accurate wavelength calculation is vital for predicting interference patterns when waves overlap.

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Most popular questions from this chapter

A uniform film of \(\mathrm{TiO}_{2}, 1036 \mathrm{nm}\) thick and having index of refraction \(2.62,\) is spread uniformly over the surface of crown glass of refractive index \(1.52 .\) Light of wavelength \(520.0 \mathrm{nm}\) falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels. (a) What is the minimum thickness of \(\mathrm{TiO}_{2}\) that you must add so the reflected light cancels as desired? (b) After you make the adjustment in part (a), what is the path difference between the light reflected off the top of the film and the light that cancels it after traveling through the film? Express your answer in (i) nanometers and (ii) wavelengths of the light in the \(\mathrm{TiO}_{2}\) film.

Two identical horizontal sheets of glass have a thin film of air of thickness \(t\) between them. The glass has refractive index \(1.40 .\) The thickness \(t\) of the air layer can be varied. Light with wavelength \(\lambda\) in air is at normal incidence onto the top of the air film. There is constructive interference between the light reflected at the top and bottom surfaces of the air film when its thickness is \(650 \mathrm{nm} .\) For the same wavelength of light the next larger thickness for which there is constructive interference is \(910 \mathrm{nm}\). (a) What is the wavelength \(\lambda\) of the light when it is traveling in air? (b) What is the smallest thickness \(t\) of the air film for which there is constructive interference for this wavelength of light?

In a two-slit interference pattern, the intensity at the peak of the central maximum is \(I_{0}\). (a) At a point in the pattern where the phase difference between the waves from the two slits is \(60.0^{\circ}\), what is the intensity? (b) What is the path difference for \(480 \mathrm{nm}\) light from the two slits at a point where the phase difference is \(60.0^{\circ} ?\)

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference minima at \(\pm 35.20^{\circ}\) on either side of the central bright spot. You then immerse these slits in a transparent liquid and illuminate them with the same light. Now you find that the first minima occur at \(\pm 19.46^{\circ}\) instead. What is the index of refraction of this liquid?

The professor returns the apparatus to the original setting. She then adjusts the speakers again. All of the students who had heard nothing originally now hear a loud tone, while you and the others who had originally heard the loud tone hear nothing. What did the professor do? (a) She turned off the oscillator. (b) She turned down the volume of the speakers. (c) She changed the phase relationship of the speakers. (d) She disconnected one speaker.
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