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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

Two flat plates of glass with parallel faces are on a table, one plate on the other. Each plate is \(11.0 \mathrm{~cm}\) long and has a refractive index of \(1.55 .\) A very thin sheet of metal foil is inserted under the end of the upper plate to raise it slightly at that end, in a manner similar to that discussed in Example 35.4 . When you view the glass plates from above with reflected white light, you observe that, at \(1.15 \mathrm{~mm}\) from the line where the sheets are in contact, the violet light of wavelength \(400.0 \mathrm{nm}\) is enhanced in this reflected light, but no visible light is enhanced closer to the line of contact. (a) How far from the line of contact will green light (of wavelength \(550.0 \mathrm{nm}\) ) and orange light (of wavelength \(600.0 \mathrm{nm}\) ) first be enhanced? (b) How far from the line of contact will the violet, green, and orange light again be enhanced in the reflected light? (c) How thick is the metal foil holding the ends of the plates apart?

Red light with wavelength \(700 \mathrm{nm}\) is passed through a two-slit apparatus. At the same time, monochromatic visible light with another wavelength passes through the same apparatus. As a result, most of the pattern that appears on the screen is a mixture of two colors; however, the center of the third bright fringe \((m=3)\) of the red light appears pure red, with none of the other color. What are the possible wavelengths of the second type of visible light? Do you need to know the slit spacing to answer this question? Why or why not?

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 \(60.0^{\circ} ?\)

A researcher measures the thickness of a layer of benzene \((n=1.50)\) floating on water by shining monochromatic light onto the film and varying the wavelength of the light. She finds that light of wavelength \(575 \mathrm{nm}\) is reflected most strongly from the film. What does she calculate for the minimum thickness of the film?

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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