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The wavelength range for visible light is \(400 \mathrm{nm}\) to \(700 \mathrm{nm}\) (see Figure 31.10 ) in air. What is the frequency range of visible light?

Short Answer

Expert verified
Answer: The frequency range of visible light is from \(4.29 * 10^{14}\) Hz to \(7.5 * 10^{14}\) Hz.

Step by step solution

01

Convert the wavelength range to meters

The given wavelength range is in nanometers (nm), but we need to convert it to meters (m) for our calculations. Use the conversion factor 1 nm = \(10^{-9}\) m. 400 nm * \(10^{-9}\) m/nm = \(4 * 10^{-7}\) m 700 nm * \(10^{-9}\) m/nm = \(7 * 10^{-7}\) m Now we have the wavelength range in meters: \(4 * 10^{-7}\) m to \(7 * 10^{-7}\) m.
02

Use the speed of light equation to find the frequency range

The speed of light equation is c = λν, where c is the speed of light (approximately \(3 * 10^8\) m/s), λ is the wavelength, and ν is the frequency. We will solve for ν for both ends of the wavelength range: \(4 * 10^{-7}\) m and \(7 * 10^{-7}\) m. For the lower end of the wavelength range: ν = c / λ = \(\frac{3 * 10^8 m/s}{4 * 10^{-7} m}\) = \(7.5 * 10^{14}\) Hz For the higher end of the wavelength range: ν = c / λ = \(\frac{3 * 10^8 m/s}{7 * 10^{-7} m}\) = \(4.29 * 10^{14}\) Hz
03

Write the frequency range of visible light

The frequency range of visible light is from \(4.29 * 10^{14}\) Hz to \(7.5 * 10^{14}\) Hz. The frequency values are rounded to two decimal places.

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Key Concepts

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

Wavelength
In the world of light and electromagnetic waves, the wavelength is a key concept. It measures the distance between two successive peaks of a wave. It is often expressed in meters (m) or nanometers (nm). The visible light spectrum, which you can see with your eyes, typically ranges from 400 nm to 700 nm.
  • 400 nm indicates the start of visible light on the violet end.
  • 700 nm signals the end towards the red side.
The conversion from nanometers to meters involves a simple multiplication. Since 1 nm equals \(10^{-9}\) m, you just multiply the nanometer value by \(10^{-9}\) to convert it to meters. For example:
  • 400 nm becomes \(4 \times 10^{-7}\) m.
  • 700 nm becomes \(7 \times 10^{-7}\) m.
Understanding wavelength is crucial as it helps determine other properties like frequency.
Frequency Range
Frequency tells us how many wave peaks pass a point in one second. It is measured in Hertz (Hz). In the context of visible light, frequency is inversely related to wavelength. When wavelength increases, frequency decreases, and vice versa. This relationship is very important when it comes to identifying the characteristics of different colors within the visible light spectrum.
To find the specific frequency range of visible light, we use the speed of light equation: \[c = \lambda u\]where \(c\) is the speed of light, \(\lambda\) is the wavelength, and \(u\) is the frequency.
For instance:
  • The lower wavelength end, \(4 \times 10^{-7}\) m, corresponds to a frequency of \(7.5 \times 10^{14}\) Hz.
  • The higher end, \(7 \times 10^{-7}\) m, has a frequency of \(4.29 \times 10^{14}\) Hz.
Thus, the frequency range for visible light is from \(4.29 \times 10^{14}\) Hz to \(7.5 \times 10^{14}\) Hz.
Speed of Light
The speed of light is a universal constant that is vitally important in physics. Represented by the symbol \(c\), its approximate value in a vacuum is \(3 \times 10^8\) m/s. This speed is crucial when calculating other properties of light, such as frequency and wavelength.
Using the equation \(c = \lambda u\), the speed of light connects wavelength and frequency. It ensures that:
  • The product of wavelength and frequency equals the constant speed of light.
  • This relationship helps us find the frequency if we know the wavelength, and vice versa.
Understanding the speed and behavior of light not only enhances our comprehension of visible light but also other types of electromagnetic waves.
Electromagnetic Spectrum
The electromagnetic spectrum encompasses all types of electromagnetic radiation, from radio waves to gamma rays. Visible light is only a small part of this spectrum. What makes up the electromagnetic spectrum is the variation in wavelengths and frequencies of different types of waves.
The visible light as we know it is just a small "window" within it, ranging from 400 nm to 700 nm. Each part of the electromagnetic spectrum serves different purposes:
  • Radio waves are used for communication.
  • Ultraviolet light has applications in sterilization.
  • Microwaves are used in cooking and other technologies.
  • X-rays are used in medical imaging.
Recognizing where visible light fits within the electromagnetic spectrum helps us understand its properties and its importance in everyday life.

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Most popular questions from this chapter

Which of the following statements concerning electromagnetic waves are incorrect? (Select all that apply.) a) Electromagnetic waves in vacuum travel at the speed of light. b) The magnitudes of the electric field and the magnetic field are equal. c) Only the electric field vector is perpendicular to the direction of the wave's propagation. d) Both the electric field vector and the magnetic field vector are perpendicular to the direction of propagation. e) An electromagnetic wave carries energy only when \(E=B\).

Which of the following exerts the largest amount of radiation pressure? a) a \(1-\mathrm{mW}\) laser pointer on a \(2-\mathrm{mm}\) -diameter spot \(1 \mathrm{~m}\) away b) a 200-W light bulb on a 4 -mm-diameter spot \(10 \mathrm{~m}\) away c) a 100 -W light bulb on a 2 -mm-diameter spot 4 m away d) a 200 - \(\mathrm{W}\) light bulb on a 2 -mm-diameter spot \(5 \mathrm{~m}\) away e) All of the above exert the same pressure.

Consider an electron in a hydrogen atom, which is \(0.050 \mathrm{nm}\) from the proton in the nucleus. a) What electric field does the electron experience? b) In order to produce an electric field whose root-meansquare magnitude is the same as that of the field in part (a), what intensity must a laser light have?

Determine the distance in feet that light can travel in vacuum during \(1.00 \mathrm{~ns}\).

Alice made a telephone call from her home telephone in New York to her fiancé stationed in Baghdad, about \(10,000 \mathrm{~km}\) away, and the signal was carried on a telephone cable. The following day, Alice called her fiancé again from work using her cell phone, and the signal was transmitted via a satellite \(36,000 \mathrm{~km}\) above the Earth's surface, halfway between New York and Baghdad. Estimate the time taken for the signals sent by (a) the telephone cable and (b) via the satellite to reach Baghdad, assuming that the signal speed in both cases is the same as speed of light, \(c .\) Would there be a noticeable delay in either case?

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