A cordless telephone is designed to operate at a frequency of \(8.5 \times 10^{8} \mathrm{~Hz}\). Determine the wavelength of these telephone waves.

Short Answer

Expert verified
Answer: The wavelength of the telephone waves is approximately 0.353 meters.

Step by step solution

01

Note down given information

The given frequency of the cordless telephone is \(8.5 \times 10^{8}\) Hz.
02

Write down the formula connecting frequency, speed of light, and wavelength

The formula for the relationship between frequency, speed of light, and wavelength is: $$c = f \times \lambda$$
03

Plug in the given values and constants

We know the speed of light, c, is approximately \(3.0 \times 10^{8} \mathrm{~m/s}\). We can now plug in the values and solve for λ. $$3.0 \times 10^{8} \mathrm{~m/s} = (8.5 \times 10^{8} \mathrm{~Hz}) \times \lambda$$
04

Solve for the wavelength, λ

To find the wavelength λ, we need to rearrange the equation and solve for it: $$\lambda = \frac{c}{f} = \frac{3.0 \times 10^{8} \mathrm{~m/s}}{8.5 \times 10^{8} \mathrm{~Hz}}$$ Now, divide the numbers: $$\lambda = \frac{3.0}{8.5} \times \frac{10^{8} \mathrm{~m/s}}{10^{8} \mathrm{~Hz}}$$ $$\lambda \approx 0.353 \mathrm{~m}$$ So, the wavelength of the telephone waves is approximately 0.353 meters.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Frequency of Waves
When discussing the characteristics of waves, the term 'frequency' plays a pivotal role. The frequency of a wave, usually denoted by the symbol 'f', refers to the number of full cycles of the wave that pass a given point within one second. It is measured in units called hertz (Hz).

For instance, if a cordless telephone operates at a frequency of 8.5 x 108 Hz, it means that 850 million cycles of that wave occur in just one second. High-frequency waves have shorter wavelengths and typically carry more energy, while low-frequency waves have longer wavelengths and carry less energy.

Understanding frequency is crucial as it directly affects the radiation's ability to transmit information over distance without attenuation, which is vital for devices like cordless telephones that rely on electromagnetic waves to function.
Speed of Light
The speed of light, symbolized as 'c', is a universal physical constant that is vital in many areas of physics. In a vacuum, light travels at an incredible speed of approximately 3.0 x 108 meters per second (m/s). This speed can vary based on the medium through which the light is traveling.

In materials like glass or water, light slows down, and this slowing effect is associated with the material's refractive index. The speed of light is not just important for understanding how fast light travels, but it is also a key component in the wave equation that relates frequency and wavelength, and hence is integral when addressing problems related to wave propagation and communications technology.
Wave Equation
The wave equation is a fundamental relationship that connects frequency, wavelength, and the speed at which the wave travels. It is elegantly represented as \( c = f \times \lambda \), where 'c' is the speed of light, 'f' is the frequency of the waves, and '\(\lambda\)' (lambda) is the wavelength, or the distance between successive crests of the wave.

Using this equation, we have a powerful tool to determine unknown properties of a wave when given enough information. In the example of the cordless telephone, once you know the operation frequency, you can utilize the wave equation to solve for the wavelength. If the telephone operates at a frequency of 8.5 x 108 Hz and you use the constant speed of light, you can rearrange the formula to \( \lambda = \frac{c}{f} \). This simplicity and universality make the wave equation a cornerstone concept in physics and engineering, serving as the basis for understanding and designing all kinds of wave-based technologies.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For a surface, how is irradiation defined? For diffusely incident radiation, how is irradiation on a surface related to the intensity of incident radiation?

What is thermal radiation? How does it differ from the other forms of electromagnetic radiation?

A long metal bar \(\left(c_{p}=450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=\right.\) \(7900 \mathrm{~kg} / \mathrm{m}^{3}\) ) is being conveyed through a water bath to be quenched. The metal bar has a cross section of \(30 \mathrm{~mm} \times 15 \mathrm{~mm}\), and it enters the water bath at \(700^{\circ} \mathrm{C}\). During the quenching process, \(500 \mathrm{~kW}\) of heat is released from the bar in the water bath. In order to prevent thermal burn on people handling the metal bar, it must exit the water bath at a temperature below \(45^{\circ} \mathrm{C}\). A radiometer is placed normal to and at a distance of \(1 \mathrm{~m}\) from the bar to monitor the exit temperature. The radiometer receives radiation from a target area of \(1 \mathrm{~cm}^{2}\) of the bar surface. Irradiation signal detected by the radiometer is used to control the speed of the bar being conveyed through the water bath so that the exit temperature is safe for handing. If the radiometer detects an irradiation of \(0.015 \mathrm{~W} / \mathrm{m}^{2}\), determine the speed of the bar bar can be approximated as a blackbody.

Why is radiation usually treated as a surface phenomenon?

For a surface, how is radiosity defined? For diffusely emitting and reflecting surfaces, how is radiosity related to the intensities of emitted and reflected radiation?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free