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A radio transmitting station operating at a frequency of 120 MHz has two identical antennas that radiate in phase. Antenna B is 9.00 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
Constructive interference occurs at x = 1.25, 4.5, and 7.75 meters.

Step by step solution

01

Understand the Problem

This problem involves calculating positions for constructive interference between two wave sources (antennas). Constructive interference happens when waves from both sources meet in phase. This is determined by the difference in path lengths from each source to point P and the wavelength of the frequency emitted.
02

Calculate the Wavelength

Use the relation between frequency and wavelength: \( \lambda = \frac{c}{f} \) where \( c \) is the speed of light (approximately \( 3 \times 10^8 \) m/s) and \( f \) is the frequency (120 MHz or \( 120 \times 10^6 \) Hz). Calculate \( \lambda \). \[ \lambda = \frac{3 \times 10^8}{120 \times 10^6} = 2.5 \text{ meters} \]
03

Establish Condition for Constructive Interference

For constructive interference, the path difference must be a multiple of the wavelength: \( n \lambda \), where \( n \) is an integer. The path difference here is \( (9 - x) - x = 9 - 2x \). Thus, equate this to \( n \lambda \). \[ 9 - 2x = n \times 2.5 \]
04

Solve for x

Solve the equation derived in Step 3 for \( x \): \[ 9 - 2x = n \times 2.5 \]Divide both sides by 2 to isolate \( x \):\[ x = \frac{9 - n \times 2.5}{2} \]Now substituting different integer values for \( n \), calculate \( x \). Ensure \( x \) lies between 0 and 9 meters (since P is between the antennas).
05

Determine Valid n Values

Calculate possible values for \( n \) such that \( 0 < x < 9 \). For instance, when \( n=0 \), \( x=\frac{9}{2}=4.5 \) meters (valid value within the range). Continue testing other integer values of \( n \) that keep \( x \) within this range.

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Key Concepts

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

Wave Interference
Wave interference occurs when two or more waves meet, affecting the resulting wave's amplitude. There are two types of wave interference: constructive and destructive. - **Constructive interference** happens when waves add up to make a larger amplitude. This occurs when the waves meet in phase, meaning their peaks and troughs align perfectly. - **Destructive interference** happens when waves cancel each other out, resulting in lower amplitude or complete cancellation. Understanding interference is crucial for analyzing wave behavior in various fields, such as acoustics, optics, and radio broadcasting. With radio waves, as in the exercise, the key is to find where along a given path waves amplify each other constructively.
Wavelength Calculation
To calculate the wavelength of a wave, you use the relationship between speed, frequency, and wavelength. The formula is given by:\[ \lambda = \frac{c}{f} \]where:- \( \lambda \) is the wavelength,- \( c \) is the speed of the wave, usually the speed of light for radio waves, approximately \( 3 \times 10^8 \) m/s,- \( f \) is the frequency of the wave.In our specific exercise, the frequency is 120 MHz, which needs to be converted to Hz before performing the calculation.Therefore:\[ \lambda = \frac{3 \times 10^8}{120 \times 10^6} = 2.5 \text{ meters} \] Knowing the wavelength helps determine how waves from different sources interact, as seen in the antennas from the exercise.
Path Difference
Path difference is the difference in distance traveled by two waves from their respective sources to a common point. It's a crucial factor in determining whether waves will interfere constructively or destructively.In the exercise, the path difference is defined as the distance from each antenna to the point P minus each other:\[ (9 - x) - x = 9 - 2x \]For constructive interference, this path difference should be an integer multiple of the wavelength, \( n \lambda \).This requirement ensures that the waves meet in phase, reinforcing each other to create a stronger wave at that point. Calculating the exact path difference is vital for applications involving synchronized wave sources, as in the case of phased array antennas used in radar and communication systems.
Radio Waves
Radio waves are a type of electromagnetic radiation with frequencies ranging from 30 Hz to 300 GHz. They are widely used in communication technologies such as radio, television, and wireless networking.- **Properties of Radio Waves:** - Travel at the speed of light in a vacuum (\(3 \times 10^8\) m/s). - Have longer wavelengths compared to other electromagnetic waves like microwaves and infrared. - Can travel long distances and through various materials, making them ideal for broadcasting.In this exercise, two radio antennas generate signals at 120 MHz, leading to a wavelength of 2.5 meters. The goal is to understand how the waves from both antennas combine at different points, demonstrating key concepts of wave interference and constructive interference in real-world applications.Radio waves' ability to interfere constructively or destructively determines the design and placement of antennas in many telecommunication systems.

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Most popular questions from this chapter

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-nm light from the two slits at a point where the phase difference is 60.0\(^\circ\)?

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of 1.38, while the eyedrops have a refractive index of 1.45. After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 nm has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

Coherent sources \(A\) and \(B\) emit electromagnetic waves with wavelength 2.00 cm. Point \(P\) is 4.86 m from \(A\) and 5.24 m from \(B\). What is the phase difference at \(P\) between these two waves?

Laser light of wavelength 510 nm is traveling in air and shines at normal incidence onto the flat end of a transparent plastic rod that has \(n\) = 1.30. The end of the rod has a thin coating of a transparent material that has refractive index 1.65. What is the minimum (nonzero) thickness of the coating (a) for which there is maximum transmission of the light into the rod; (b) for which transmission into the rod is minimized?

Eyeglass lenses can be coated on the \(inner\) surfaces to reduce the reflection of stray light to the eye. If the lenses are medium flint glass of refractive index 1.62 and the coating is fluorite of refractive index 1.432, (a) what minimum thickness of film is needed on the lenses to cancel light of wavelength 550 nm reflected toward the eye at normal incidence? (b) Will any other wavelengths of visible light be cancelled or enhanced in the reflected light?

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