/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Two radio-transmitting antennas ... [FREE SOLUTION] | 91Ó°ÊÓ

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Two radio-transmitting antennas are separated by half the broadcast wavelength and are driven in phase with each other. In which directions are (a) the strongest and (b) the weakest signals radiated?

Short Answer

Expert verified
The strongest signals are radiated in directions parallel to the line connecting the antennas, while the weakest signals are radiated in directions perpendicular to this line.

Step by step solution

01

Understand Superposition

The principle of superposition states that when two waves interact, the resultant wave will be the sum of the amplitudes of the individual waves. If the waves are in phase (that is, they reach their peak and trough at the same time), then constructive interference occurs and the amplitudes add together to create a more intense wave. If the waves are out of phase (that is, while one wave is at peak, the other is at trough), destructive interference occurs and the amplitudes subtract from each other, yielding a weaker wave or no wave at all. This will be relevant for both parts of the exercise.
02

Analyzing Strongest Signal

In the case of this problem, the two antennas are in phase and separated by half of a wavelength. This means that at certain points, the waves emitted by the two antennas will meet when they are both at the same point in their cycle, creating constructive interference. This yields the strongest signal. This signal is radiated in directions parallel to the line connecting the antennas, as the paths of the radio waves in these directions are at the perfect distance apart to add together.
03

Analyzing Weakest Signal

Destructive interference, yielding the weakest signal, occurs between waves traveling in directions perpendicular to the line connecting the antennas. Because the two antennas are a half wavelength apart, waves emitted from one antenna will meet waves emitted from the other antenna a half cycle later, when they are at opposite points in their cycles, creating destructive interference.

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Key Concepts

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

Superposition
Wave superposition describes how two or more waves overlap in space and time. Imagine the surface of a calm lake: if you toss in two pebbles, ripples spread outward, and where these ripples meet, they create new patterns. Waves on the lake, like electromagnetic waves, interfere with each other based on this concept of superposition. The resultant wave pattern is simply the sum of the amplitudes of the individual waves at each point.

With superposition, waves can add together to form a larger, more intense wave or cancel each other out to create a less intense one. This concept is foundational in understanding both constructive and destructive interference, which dictate how waves like sound, light, or radio waves behave in the world around us.
Constructive Interference
Constructive interference occurs when waves align perfectly in phase. This means that their peaks and troughs match up exactly. When this happens, the amplitudes (or heights) of the waves add together.

  • The result is a new wave that is much stronger, or higher, than either of the original waves alone. Imagine two people pushing a swing at the same time; the swing goes higher because the pushes add together.
  • In the problem of the radio antennas, when the waves are in phase, the strongest signal is produced in directions parallel to the line connecting the antennas. Here, the waves perfectly reinforce one another.
Understanding where and why constructive interference occurs helps in designing and using technologies such as radios and other communication devices effectively.
Destructive Interference
Destructive interference is the phenomenon that occurs when waves are out of phase, which means they do not align at all. One wave is at its peak while the other is at its trough—effectively cancelling each other out.

  • This results in a wave that is weaker or even non-existent, as the opposing amplitudes subtract from one another. Think of two people trying to pull a rope in opposite directions with equal force; the rope doesn't move because the forces cancel each other out.
  • In the case of the radio antennas, the weakest signals are emitted in directions perpendicular to the line connecting the antennas. Here, the half-wavelength separation ensures that waves from each antenna are out of sync or out of phase, leading to destructive interference.
Grasping the mechanisms of destructive interference is crucial for avoiding broadcast interference in various technological applications.

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Most popular questions from this chapter

A plane electromagnetic wave varies sinusoidally at \(90.0 \mathrm{MHz}\) as it travels along the \(+x\) direction. The peak value of the electric field is \(2.00 \mathrm{mV} / \mathrm{m},\) and it is directed along the \(\pm y\) direction. (a) Find the wavelength, the period, and the maximum value of the magnetic field. (b) Write expressions in SI units for the space and time variations of the electric field and of the magnetic field. Include numerical values and include subscripts to indicate coordinate directions. (c) Find the average power per unit area that this wave carries through space. (d) Find the average energy density in the radiation (in joules per cubic meter). (e) What radiation pressure would this wave exert upon a perfectly reflecting surface at normal incidence?

\- Write down expressions for the electric and magnetic fields of a sinusoidal plane electromagnetic wave having a frequency of \(3.00 \mathrm{GHz}\) and traveling in the positive \(x\) direction. The amplitude of the electric field is \(300 \mathrm{V} / \mathrm{m}\).

Consider a small, spherical particle of radius \(r\) located in space a distance \(R\) from the Sun. (a) Show that the ratio \(F_{\mathrm{rad}} / F_{\mathrm{grav}}\) is proportional to \(1 / r,\) where \(F_{\mathrm{rad}}\) is the force exerted by solar radiation and \(F_{\mathrm{grav}}\) is the force of gravitational attraction. (b) The result of part (a) means that, for a sufficiently small value of \(r,\) the force exerted on the particle by solar radiation exceeds the force of gravitational attraction. Calculate the value of \(r\) for which the particle is in equilibrium under the two forces. (Assume that the particle has a perfectly absorbing surface and a mass density of \(1.50 \mathrm{g} / \mathrm{cm}^{3} .\) Let the particle be located \(3.75 \times 10^{11} \mathrm{m}\) from the Sun, and use \(214 \mathrm{W} / \mathrm{m}^{2}\) as the value of the solar intensity at that point.)

The intensity of solar radiation at the top of the Earth's atmosphere is \(1340 \mathrm{W} / \mathrm{m}^{2} .\) Assuming that \(60 \%\) of the incoming solar energy reaches the Earth's surface and assuming that you absorb \(50 \%\) of the incident energy, make an order-of-magnitude estimate of the amount of solar energy you absorb in a \(60-\) min sunbath.

A very long, thin rod carries electric charge with the linear density \(35.0 \mathrm{nC} / \mathrm{m}\). It lies along the \(x\) axis and moves in the \(x\) direction at a speed of \(15.0 \mathrm{Mm} / \mathrm{s} .\) (a) Find the electric field the rod creates at the point \((0,20.0 \mathrm{cm}, 0)\) (b) Find the magnetic field it creates at the same point. (c) Find the force exerted on an electron at this point, moving with a velocity of \((240 \hat{\mathbf{i}}) \mathrm{Mm} / \mathrm{s}\)

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