/*! 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 104 Calculate the ratio of the highe... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate the ratio of the highest to lowest 104. frequencies of electromagnetic waves the eye can see, given the wavelength range of visible light is from 380 to \(760 \mathrm{nm} .\) (Note that the ratio of highest to lowest frequencies the ear can hear is \(1000 .\) )

Short Answer

Expert verified
The ratio of the highest to lowest frequencies of electromagnetic waves the eye can see within the visible light wavelength range is 2.0.

Step by step solution

01

Convert the wavelength range to meters

We need to convert the given wavelength range of visible light from nanometers (nm) to meters (m). To do this, we will multiply the given wavelengths by \(10^{-9}\). For 380 nm wavelength: \[\lambda_1 = 380 \times 10^{-9} \mathrm{m}\] For 760 nm wavelength: \[\lambda_2 = 760 \times 10^{-9} \mathrm{m}\]
02

Relationship between wavelength, frequency, and speed of light

Using the formula \(c = \lambda \cdot f\), we will find the frequencies for both wavelengths. Since we have the wavelengths and the speed of light (c = \(3 \times 10^8 \mathrm{m/s}\)), we can solve for the frequencies. For \(\lambda_1\), the frequency calculation is: \[f_1 = \frac{c}{\lambda_1}\] For \(\lambda_2\), the frequency calculation is: \[f_2 = \frac{c}{\lambda_2}\]
03

Calculate the frequencies

Now, we will plug the wavelengths and speed of light into the frequency calculation formulas. For \(\lambda_1\): \[f_1 = \frac{3 \times 10^8 \mathrm{m/s}}{380 \times 10^{-9} \mathrm{m}}\] For \(\lambda_2\): \[f_2 = \frac{3 \times 10^8 \mathrm{m/s}}{760 \times 10^{-9} \mathrm{m}}\]
04

Simplify the frequencies

Simplify and calculate the results of the frequencies. For \(\lambda_1\): \[f_1 = \frac{3 \times 10^8 \mathrm{m/s}}{380 \times 10^{-9} \mathrm{m}} = 7.89 \times 10^{14} \mathrm{Hz}\] For \(\lambda_2\): \[f_2 = \frac{3 \times 10^8 \mathrm{m/s}}{760 \times 10^{-9} \mathrm{m}} = 3.95 \times 10^{14} \mathrm{Hz}\]
05

Calculate the ratio of the frequencies

Finally, calculate the ratio of the highest to lowest frequencies by dividing \(f_1\) by \(f_2\). Ratio = \(\frac{f_1}{f_2}\) = \(\frac{7.89 \times 10^{14} \mathrm{Hz}}{3.95 \times 10^{14} \mathrm{Hz}}\)
06

Simplify the ratio

Simplify the result of the ratio. Ratio = \(\frac{7.89 \times 10^{14} \mathrm{Hz}}{3.95 \times 10^{14} \mathrm{Hz}} = 2.0\) The ratio of the highest to lowest frequencies of electromagnetic waves the eye can see is 2.0.

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Key Concepts

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

Visible Light
Visible light is the portion of the electromagnetic spectrum that is visible to the human eye. It consists of the range of wavelengths that our eyes can perceive as different colors. This range falls between approximately 380 nanometers (nm) to 760 nm.

These wavelengths correspond to the colors we perceive as violet to red respectively:
  • Violet has the shortest wavelength, near 380 nm.
  • Red has the longest wavelength, near 760 nm.
Our eyes perceive color depending on how these different wavelengths interact with the cones in our retinas. Visible light is vital for life on Earth, as it plays a crucial role in processes like photosynthesis in plants, let's not forget its essential role in human and animal vision.
Wavelength
Wavelength is a critical characteristic of waves, especially in the context of light and electromagnetic waves. It is defined as the distance between consecutive peaks (or troughs) in the wave. Wavelength is usually denoted by the Greek letter lambda (\( \lambda \)).

In the context of visible light, wavelength is vital in defining the color of light, with different wavelengths corresponding to different colors. Because visible light wavelengths are so small, they are often measured in nanometers (nm), where 1 nm equals \(10^{-9}\) meters.

Wavelengths play a significant role when calculating other properties of waves, such as their frequency, by using the relationship with the speed of light. Adjustments to the wavelength directly impact calculations for frequency, influencing the perceived energy and color of the electromagnetic waves. Understanding wavelengths helps us grasp how light interacts with matter.
Frequency
Frequency refers to the number of wave cycles that pass a given point per second. It is denoted by the letter \( f \) and is measured in Hertz (Hz), where one Hertz is one cycle per second. In the context of light, frequency is inversely proportional to wavelength, according to the equation \( c = \lambda \cdot f \), where \( c \) is the speed of light.

For visible light, higher frequencies correspond to shorter wavelengths, resulting in colors toward the violet end of the spectrum. Conversely, lower frequencies correspond to longer wavelengths, appearing nearer to the red end:
  • Higher frequency: Violet light.
  • Lower frequency: Red light.
Frequency is essential in understanding light because it relates to the energy it carries. Higher frequency waves carry more energy, which influences applications like lasers, telecommunications, and even biological processes. By knowing the frequency, we can better understand the behavior and application of light in various technologies.
Speed of Light
The speed of light is a fundamental constant in physics, denoting how fast light travels through a vacuum. Represented by the letter \( c \), it is approximately \( 3 \, \times \, 10^8 \, \text{m/s} \). This value is crucial when calculating other properties of electromagnetic waves, such as wavelength and frequency.

The equation \( c = \lambda \cdot f \) is a key expression that links the speed of light to wavelength and frequency. When solving problems related to electromagnetic waves, this equation allows us to find the missing variable if two are known.

The speed of light is significant not just for visible light, but across the entire electromagnetic spectrum. It plays a pivotal role in understanding phenomena like refraction, reflection, and interference. Acknowledging the speed of light helps us in designing optical systems, understanding cosmological distances, and advancing technologies like fiber optics.

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Most popular questions from this chapter

A 150-W lightbulb emits 5\% of its energy as electromagnetic radiation. What is the magnitude of the average Poynting vector \(10 \mathrm{m}\) from the bulb?

A radio station broadcasts at a frequency of \(760 \mathrm{kHz}\) At a receiver some distance from the antenna, the maximum magnetic field of the electromagnetic wave detected is \(2.15 \times 10^{-11} \mathrm{T}\) (a) What is the maximum electric field? (b) What is the wavelength of the electromagnetic wave?

The magnetic field of a plane electromagnetic wave \(\overrightarrow{\mathbf{B}}=B_{0}(\cos k z+\omega t) \hat{\mathbf{j}},\) where \(B_{0}=5.00 \times 10^{-10} \mathrm{T}\) and \(k=3.14 \times 10^{-2} \mathrm{m}^{-1}\) (a) Write an expression for the electric field associated with the wave. (b) What are the frequency and the wavelength of the wave? (c) What is its average Poynting vector?

During normal beating, the heart creates a maximum \(4.00-\mathrm{mV}\) potential across \(0.300 \mathrm{m}\) of a person's chest, creating a \(1.00-\mathrm{Hz}\) electromagnetic wave. (a) What is the maximum electric field strength created? (b) What is the corresponding maximum magnetic field strength in the electromagnetic wave? (c) What is the wavelength of the electromagnetic wave?

The Poynting vector describes a flow of energy whenever electric and magnetic fields are present. Consider a long cylindrical wire of radius \(r\) with a current \(I\) in the wire, with resistance \(R\) and voltage \(V\). From the expressions for the electric field along the wire and the magnetic field around the wire, obtain the magnitude and direction of the Poynting vector at the surface. Show that it accounts for an energy flow into the wire from the fields around it that accounts for the Ohmic heating of the wire.

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