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Most people perceive light having a wavelength between \(630 \mathrm{nm}\) and \(700 \mathrm{nm}\) as red and light with a wavelength between \(400 \mathrm{nm}\) and \(440 \mathrm{nm}\) as violet. Calculate the approximate frequency ranges for (a) violet light and (b) red light.

Short Answer

Expert verified
Violet light: \(6.82 \times 10^{14} \text{ Hz} \) to \(7.50 \times 10^{14} \text{ Hz} \); Red light: \(4.29 \times 10^{14} \text{ Hz} \) to \(4.76 \times 10^{14} \text{ Hz} \).

Step by step solution

01

Understand the Relationship Between Wavelength and Frequency

The frequency \( f \) of light is related to its wavelength \( \lambda \) by the equation \( c = \lambda f \), where \( c \) is the speed of light in a vacuum, approximately \( 3 \times 10^8 \text{ m/s} \). We will use this formula to find the frequency once we know the wavelength.
02

Convert Wavelength from Nanometers to Meters

Given that \(1 \text{ nm} = 10^{-9} \text{ m}\), we need to convert the wavelengths of both violet and red light from nanometers (nm) to meters (m). For example, for violet light at \( 400 \text{ nm} \), this equals \( 400 \times 10^{-9} \text{ m} \). Do the same for all given boundary wavelengths.
03

Calculate Frequency for Violet Light

To find the frequency range of violet light, use the converted wavelengths:\[\lambda_{min} = 400 \times 10^{-9} \text{ m}\] and \[\lambda_{max} = 440 \times 10^{-9} \text{ m}\]. Use the formula \( f = \frac{c}{\lambda} \): - \( f_{max} = \frac{3 \times 10^8 \text{ m/s}}{400 \times 10^{-9} \text{ m}} = 7.50 \times 10^{14} \text{ Hz} \)- \( f_{min} = \frac{3 \times 10^8 \text{ m/s}}{440 \times 10^{-9} \text{ m}} = 6.82 \times 10^{14} \text{ Hz} \).Thus, the frequency range for violet light is \(6.82 \times 10^{14} \text{ Hz} \) to \(7.50 \times 10^{14} \text{ Hz} \).
04

Calculate Frequency for Red Light

Using the same process, calculate the frequency range for red light with:\[\lambda_{min} = 630 \times 10^{-9} \text{ m}\] and \[\lambda_{max} = 700 \times 10^{-9} \text{ m}\]: - \( f_{max} = \frac{3 \times 10^8 \text{ m/s}}{630 \times 10^{-9} \text{ m}} = 4.76 \times 10^{14} \text{ Hz} \)- \( f_{min} = \frac{3 \times 10^8 \text{ m/s}}{700 \times 10^{-9} \text{ m}} = 4.29 \times 10^{14} \text{ Hz} \).The frequency range for red light is \(4.29 \times 10^{14} \text{ Hz} \) to \(4.76 \times 10^{14} \text{ Hz} \).

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Key Concepts

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

Speed of Light
Light travels incredibly fast, making it one of the swiftest phenomena in the universe. The speed of light in a vacuum is a constant value, approximately equal to \(3 \times 10^8 \text{ meters per second (m/s)}\). This means that light can travel 300,000 kilometers, roughly from the Earth to the Moon, in just about a second. Understanding the speed of light is fundamental in physics because it forms the basis for calculating other properties of light, such as frequencies and wavelengths. When it comes to light, any change in wavelength directly affects its frequency, given that the speed of light remains constant. This relationship is concisely expressed in the equation \(c = \lambda f\), where \(c\) is the speed of light, \(\lambda\) is the wavelength, and \(f\) is the frequency.
Violet Light Frequency
Violet light is the most energetic color within the visible light spectrum due to its short wavelengths, ranging from 400 nm to 440 nm. To convert these wavelengths from nanometers to meters, as required by physics equations, we multiply by \(10^{-9}\), giving us \( 400 \times 10^{-9} \text{ m} \) to \( 440 \times 10^{-9} \text{ m} \). Once converted, we use the formula \(f = \frac{c}{\lambda}\) to find the frequency range:
  • Maximum frequency, using 400 nm: \( f_{max} = \frac{3 \times 10^8 \text{ m/s}}{400 \times 10^{-9} \text{ m}} = 7.5 \times 10^{14} \text{ Hz} \)
  • Minimum frequency, using 440 nm: \( f_{min} = \frac{3 \times 10^8 \text{ m/s}}{440 \times 10^{-9} \text{ m}} = 6.82 \times 10^{14} \text{ Hz} \)
Thus, violet light has a frequency range of approximately \(6.82 \times 10^{14} \text{ Hz} \) to \(7.5 \times 10^{14} \text{ Hz}.\)
Red Light Frequency
In contrast to violet light, red light represents the lowest energy portion of the visible spectrum with longer wavelengths ranging from 630 nm to 700 nm. Translating these into meters gives us \( 630 \times 10^{-9} \text{ m} \) and \( 700 \times 10^{-9} \text{ m} \). We apply the same frequency formula to determine the frequency range:
  • Minimum frequency, for 700 nm: \( f_{min} = \frac{3 \times 10^8 \text{ m/s}}{700 \times 10^{-9} \text{ m}} = 4.29 \times 10^{14} \text{ Hz} \)
  • Maximum frequency, for 630 nm: \( f_{max} = \frac{3 \times 10^8 \text{ m/s}}{630 \times 10^{-9} \text{ m}} = 4.76 \times 10^{14} \text{ Hz} \)
Therefore, red light occupies a frequency range from approximately \(4.29 \times 10^{14} \text{ Hz}\) to \(4.76 \times 10^{14} \text{ Hz}.\)
Wavelength Conversion
Wavelengths of light are commonly expressed in nanometers (nm), but physics calculations usually require meters (m). The basic conversion rule is: 1 nm equals \(10^{-9}\) meters.When working with light equations, it's critical to convert all wavelengths to meters to ensure consistency and correctness. For example:
  • A wavelength of \(400 \text{ nm}\) becomes \(400 \times 10^{-9} \text{ m}\).
  • A wavelength of \(700 \text{ nm}\) translates to \(700 \times 10^{-9} \text{ m}\).
Converting wavelength units carefully is important to avoid calculation errors in determining cost/off frequency.
Visible Light Spectrum
The visible light spectrum is a small part of the electromagnetic spectrum that is visible to the human eye, ranging approximately from 380 nm to 750 nm. It represents all the colors we can see, each at a different wavelength. The spectrum starts with violet on the shorter wavelength end and extends to red on the longer wavelength side.
  • Violet wavelengths: 380 - 450 nm
  • Blue wavelengths: 450 - 495 nm
  • Green wavelengths: 495 - 570 nm
  • Yellow wavelengths: 570 - 590 nm
  • Orange wavelengths: 590 - 620 nm
  • Red wavelengths: 620 - 750 nm
Each color has its own unique wavelength that determines its color in the spectrum. This range enables us to see the world in a rainbow of colors, all due to variations in wavelength.

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Most popular questions from this chapter

Laser surgery. Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses of energy absorbed by the retina weld the detached portion back into place. In one such procedure, a laser beam has a wavelength of \(810 \mathrm{nm}\) and delivers \(250 \mathrm{~mW}\) of power spread over a circular spot \(510 \mu \mathrm{m}\) in diameter. The vitreous humor (the transparent fluid that fills most of the eye) has an index of refraction of \(1.34 .\) (a) If the laser pulses are each \(1.50 \mathrm{~ms}\) long, how much energy is delivered to the retina with each pulse? (b) What average pressure does the pulse of the laser beam exert on the retina as it is fully absorbed by the circular spot? (c) What are the wavelength and frequency of the laser light inside the vitreous humor of the eye? (d) What are the maximum values of the electric and magnetic fields in the laser beam?

Scientists are working on a new technique to kill cancer cells by zapping them with ultrahighenergy (in the range of \(10^{12} \mathrm{~W}\) ) pulses of electromagnetic waves that last for an extremely short time (a few nanoseconds). These short pulses scramble the interior of a cell without causing it to explode, as long pulses would do. We can model a typical such cell as a disk \(5.0 \mu \mathrm{m}\) in diameter, with the pulse lasting for \(4.0 \mathrm{~ns}\) with an average power of \(2.0 \times 10^{12} \mathrm{~W}\). We shall assume that the energy is spread uniformly over the faces of 100 cells for each pulse. (a) How much energy is given to the cell during this pulse? (b) What is the intensity (in \(\mathrm{W} / \mathrm{m}^{2}\) ) delivered to the cell? (c) What are the maximum values of the electric and magnetic fields in the pulse?

Radiation falling on a perfectly reflecting surface produces an average pressure \(p .\) If radiation of the same intensity falls on a perfectly absorbing surface and is spread over twice the area, what is the pressure at that surface in terms of \(p ?\)

A beam of polarized light passes through a polarizing filter. When the angle between the polarizing axis of the filter and the direction of polarization of the light is \(30^{\circ}\), the intensity of the emerging beam is \(I\). If you instead want the intensity to be \(I / 2,\) what should be the angle between the filter axis and the original direction of polarization of the light?

Physicians use high-frequency \((f=1 \mathrm{MHz}\) to \(5 \mathrm{MHz}\) ) sound waves, called ultrasound, to image internal organs. The speed of these ultrasound waves is \(1480 \mathrm{~m} / \mathrm{s}\) in muscle and \(344 \mathrm{~m} / \mathrm{s}\) in air. We define the index of refraction of a material for sound waves to be the ratio of the speed of sound in air to the speed of sound in the material. Snell's law then applies to the refraction of sound waves. (a) At what angle from the normal does an ultrasound beam enter the heart if it leaves the lungs at an angle of \(9.73^{\circ}\) from the normal to the heart wall? (Assume that the speed of sound in the lungs is \(344 \mathrm{~m} / \mathrm{s} .)\) (b) What is the critical angle for sound waves in air incident on muscle?

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