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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.

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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.

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Most popular questions from this chapter

A blackbody plate \(\left(A_{1}=5 \mathrm{~cm}^{2}\right)\) is subjected to a uniform heat flux of \(1000 \mathrm{~W} / \mathrm{m}^{2}\) on the bottom, while the top surface is exposed to an ambient surrounding at \(5^{\circ} \mathrm{C}\). The heat transfer coefficient due to natural convection on the plate surface is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). A radiometer is placed \(1 \mathrm{~m}\) above the plate normal to the direction of viewing from the plate. Determine the irradiation that the radiometer would detect.

The reflectivity of aluminum coated with lead sulfate is \(0.35\) for radiation at wavelengths less than \(3 \mu \mathrm{m}\) and \(0.95\) for radiation greater than \(3 \mu \mathrm{m}\). Determine the average reflectivity of this surface for solar radiation \((T \approx 5800 \mathrm{~K})\) and radiation coming from surfaces at room temperature \((T \approx 300 \mathrm{~K})\). Also, determine the emissivity and absorptivity of this surface at both temperatures. Do you think this material is suitable for use in solar collectors?

An opaque horizontal plate is well insulated on the edges and the lower surface. The irradiation on the plate is \(3000 \mathrm{~W} / \mathrm{m}^{2}\), of which \(500 \mathrm{~W} / \mathrm{m}^{2}\) is reflected. The plate has a uniform temperature of \(700 \mathrm{~K}\) and has an emissive power of \(5000 \mathrm{~W} / \mathrm{m}^{2}\). Determine the total emissivity and absorptivity of the plate.

The variation of the spectral transmissivity of a \(0.6\)-cm-thick glass window is as given in Fig. P12-80. Determine the average transmissivity of this window for solar radiation \((T \approx 5800 \mathrm{~K})\) and radiation coming from surfaces at room temperature \((T \approx 300 \mathrm{~K})\). Also, determine the amount of solar radiation transmitted through the window for incident solar radiation of \(650 \mathrm{~W} / \mathrm{m}^{2}\).

What is the greenhouse effect? Why is it a matter of great concern among atmospheric scientists?

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