/*! 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 38 A television is tuned to a stati... [FREE SOLUTION] | 91Ó°ÊÓ

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A television is tuned to a station broadcasting at a frequency of \(6.60 \times 10^{7} \mathrm{Hz} .\) For best reception, the rabbit-ear antenna used by the TV should be adjusted to have a tip-to-tip length equal to half a wavelength of the broadcast signal. Find the optimum length of the antenna.

Short Answer

Expert verified
The optimum length of the antenna is 2.275 meters.

Step by step solution

01

Identify the Given Values

The problem gives us the frequency of the broadcast signal, which is \( f = 6.60 \times 10^7 \text{ Hz} \).
02

Recall the Speed of Light

The speed of light in a vacuum, which is applicable to electromagnetic waves such as radio waves, is \( c = 3.00 \times 10^8 \text{ m/s} \).
03

Use the Wave Speed Formula

The relationship between wavelength \( \lambda \), frequency \( f \), and speed of light \( c \) is given by the formula: \( c = \lambda f \).
04

Solve for the Wavelength

Rearrange the formula to solve for the wavelength: \[ \lambda = \frac{c}{f} = \frac{3.00 \times 10^8 \text{ m/s}}{6.60 \times 10^7 \text{ Hz}} = 4.55 \text{ m} \].
05

Find the Optimum Length of the Antenna

Because the antenna should be half the wavelength for optimal reception, calculate: \[ \frac{\lambda}{2} = \frac{4.55 \, \text{m}}{2} = 2.275 \, \text{m} \].

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Key Concepts

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

Frequency and Wavelength
When we talk about **frequency and wavelength**, we are discussing fundamental properties of waves. In this case, radio waves. Frequency refers to how often the wave peaks occur over time. This is measured in hertz (Hz), which counts the cycles per second.
Wavelength, denoted by the Greek letter lambda (\( \lambda \)), is the distance between consecutive peaks of a wave. When the frequency of a wave is known, and if the speed of the wave is known, we can find the wavelength using the formula: \( c = \lambda f \).
Here, \( c \) is the speed of light, which is approximately \( 3.00 \times 10^8 \text{ m/s} \). Solving this equation for wavelength gives us \( \lambda = \frac{c}{f} \).
For example, for a frequency of \( 6.60 \times 10^7 \text{ Hz} \), the wavelength can be calculated as follows:
  • Use the formula: \( \lambda = \frac{c}{f} = \frac{3.00 \times 10^8 \text{ m/s}}{6.60 \times 10^7 \text{ Hz}} \).
  • The calculation results in a wavelength of \( 4.55 \text{ m} \).
This equation helps us understand the relationship between frequency and wavelength of any electromagnetic wave, not just radio waves.
Electromagnetic Waves
**Electromagnetic waves** are waves that travel through the vacuum of outer space as well as through air and other media. They include radio waves, microwaves, infrared, visible light, ultraviolet, x-rays, and gamma rays.
All electromagnetic waves move at the speed of light in a vacuum, which is approximately \( 3.00 \times 10^8 \text{ m/s} \). However, their frequencies and wavelengths vary significantly.
Radio waves, used for communication such as in our TV sets, have the longest wavelengths and lowest frequencies of the electromagnetic spectrum. That's why they can travel long distances and penetrate through the atmosphere to reach antennas on Earth.
  • The relationship between wavelength, frequency, and speed (\( c = \lambda f \)) describes all electromagnetic waves.
  • This wave nature is why adjustments such as tuning an antenna involve understanding their geometric and physical properties.
Understanding electromagnetic waves helps us appreciate how signals and information travel over long distances almost instantaneously.
Radio Wave Reception
Receiving a radio signal is not magic but involves understanding **radio wave reception**. For optimal radio wave reception, especially in devices like TVs, antennas must be tuned to the signals they want to pick up.
In our exercise, the *optimum length of the antenna* is key. The length of the antenna should match half the wavelength of the broadcast signal.
This is because the reception is much stronger when the antenna length resonates with the wavelength of the incoming wave.
  • Calculated wavelength for our frequency is \( 4.55 \text{ m} \).
  • For optimal reception, the antenna length should be half the wavelength: \( \frac{4.55 \text{ m}}{2} = 2.275 \text{ m} \).
By aligning these aspects, the signal reception is successfully maximized, ensuring clear sound and picture quality on your TV. Understanding these principles is crucial for anyone interested in radio communications or working in relevant technological fields.

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Most popular questions from this chapter

Experiments show that the ground spider Drassodes cupreus uses one of its several pairs of eyes as a polarization detector, In fact, the two eyes in this pair have polarization directions that are at right angles to one another. Suppose linearly polarized light with an intensity of \(825 \mathrm{W} / \mathrm{m}^{2}\) shines from the sky onto the spider, and that the intensity transmitted by one of the polarizing eyes is \(232 \mathrm{W} / \mathrm{m}^{2}\) (a) For this eye, what is the angle between the polarization direction of the eye and the polarization direction of the incident light? (b) What is the intensity transmitted by the other polarizing eye?

Suppose a VLC unit uses an LED that produces light with an average intensity of \(975 \mathrm{mW} / \mathrm{cm}^{2}\). What is the rms value of the electric field in this beam of light? A. \(606 \mathrm{N} / \mathrm{C}\) B. \(1920 \mathrm{N} / \mathrm{C}\) C. \(3.67 \times 10^{6} \mathrm{N} / \mathrm{C}\) D. \(3.32 \times 10^{7} \mathrm{N} / \mathrm{C}\)

The radiation pressure exerted by beam of light 1 is half the radiation pressure of beam of light \(2 .\) If the rms electric field of beam 1 has the value \(E_{0},\) what is the rms electric field in beam \(2 ?\)

The maximum magnitude of the electric field in an electromagnetic wave is \(0.0400 \mathrm{V} / \mathrm{m}\). What is the maximum magnitude of the magnetic field in this wave?

At the top of Earth's atmosphere, sunlight has an average intensity of \(1360 \mathrm{W} / \mathrm{m}^{2} .\) If the average distance from Earth to the Sun is \(1.50 \times 10^{11} \mathrm{m},\) at what rate does the Sun radiate energy?

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