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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

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?

A plastic film with index of refraction 1.70 is applied to the surface of a car window to increase the reflectivity and thus to keep the car's interior cooler. The window glass has index of refraction 1.52. (a) What minimum thickness is required if light of wavelength 550 nm in air reflected from the two sides of the film is to interfere constructively? (b) Coatings as thin as that calculated in part (a) are difficult to manufacture and install. What is the next greater thickness for which constructive interference will also occur?

Coherent light with wavelength 450 nm falls on a pair of slits. On a screen 1.80 m away, the distance between dark fringes is 3.90 mm. What is the slit separation?

Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, 2.04 \(\mu\)m apart, and in line with an observer, so that one source is 2.04 \(\mu\)m farther from the observer than the other. (a) For what visible wavelengths (380 to 750 nm) will the observer see the brightest light, owing to constructive interference? (b) How would your answers to part (a) be affected if the two sources were not in line with the observer, but were still arranged so that one source is 2.04 \(\mu\)m farther away from the observer than the other? (c) For what visible wavelengths will there be \(destructive\) interference at the location of the observer?

Coherent light that contains two wavelengths, 660 nm (red) and 470 nm (blue), passes through two narrow slits that are separated by 0.300 mm. Their interference pattern is observed on a screen 4.00 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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