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Chapter 32: Problem 13

An electromagnetic wave with frequency \(5.70 \times 10^{14} \mathrm{~Hz}\) propagates with a speed of \(2.17 \times 10^{8} \mathrm{~m} / \mathrm{s}\) in a certain piece of glass. Find (a) the wavelength of the wave in the glass; (b) the wavelength of a wave of the same frequency propagating in air; (c) the index of refraction \(n\) of the glass for an electromagnetic wave with this frequency; (d) the dielectric constant for glass at this frequency, assuming that the relative permeability is unity.

Short Answer

Expert verified
The wavelength in the glass is \(3.80 \times 10^{-7}\, \mathrm{m}\), the wavelength in air is \(5.26 \times 10^{-7}\, \mathrm{m}\), the index of refraction is \(1.38\) and the dielectric constant is \(1.90\)

Step by step solution

01

Calculate the wavelength in the glass

We will use the wave equation which states that, for any wave (including electromagnetic), the product of its speed \(v\), its wavelength \(\lambda\) and its frequency \(f\) is constant: \(v = \lambda f\). Solving for the wavelength, we find \(\lambda = \frac{v}{f}\). Substituting the known values, i.e. \(v = 2.17 \times 10^{8} \, \mathrm{m/s}\) and \(f = 5.7 \times 10^{14} \, \mathrm{Hz}\), we can now calculate the wavelength of the wave in the glass.
02

Calculate the wavelength in air

The speed of light in air is very close to the speed of light in vacuum, given by \(c = 3.00 \times 10^{8} \, \mathrm{m/s}\). Repeating the process from the previous step and using \(v = c\) in the wave equation, we can find the wavelength of the wave in air.
03

Calculate the index of refraction in glass

The index of refraction \(n\) is given by the ratio of the speed of light in vacuum to the speed of light in the medium: \(n = \frac{c}{v}\). By substituting \(v\) with the speed of light in the glass, obtained in the first step, we can calculate the index of refraction.
04

Calculate the dielectric constant

For most materials at optical frequencies, the relative permeability is close to 1. In such cases, the dielectric constant \(\varepsilon_r\) of the material can be approximated by the square of the index of refraction: \(\varepsilon_r = n^2\). After finding the index of refraction from the previous step, we can now calculate the dielectric constant.

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Key Concepts

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

Wavelength
Wavelength is a key concept when discussing electromagnetic waves. It refers to the distance between two consecutive peaks or troughs of a wave. In mathematical terms, wavelength (\( \lambda\) ) is calculated as the speed of the wave (\( v\) ) divided by its frequency (\( f\) ).
To find the wavelength of the electromagnetic wave in the given glass, we use the relation:
  • The speed of the wave in the glass: \( v = 2.17 \times 10^{8} \, \mathrm{m/s} \).
  • Frequency of the wave: \( f = 5.7 \times 10^{14} \, \mathrm{Hz} \).
Using the formula \( \lambda = \frac{v}{f} \), substituting these values gives \( \lambda \approx 3.81 \times 10^{-7} \, \mathrm{m} \).

For the wave in air, we know the speed of light \( c = 3.00 \times 10^{8} \, \mathrm{m/s} \), and repeating the calculation for air gives a different wavelength, showcasing how medium changes affect wavelength.
Index of Refraction
The index of refraction, or refractive index, is a measure of how much light bends, or refracts, when entering a material. It is a fundamental concept in optics and describes how light propagates through different materials.
The refractive index (\( n \) ) is calculated by:
  • Dividing the speed of light in a vacuum (\( c = 3.00 \times 10^{8} \, \mathrm{m/s} \) )
  • By the speed of light in the medium (\( v \) )
For the glass in question, we determine \( n = \frac{c}{v} \). Substituting the given speed of light in the glass, we find the refractive index is \( n \approx 1.38 \).

This means light travels slower in this glass than in a vacuum. Light bends as it passes through different media, depending on their indices of refraction. A higher refractive index indicates greater bending.
Dielectric Constant
The dielectric constant, also known as the relative permittivity, is a measure of a material's ability to store electrical energy in an electric field. In the context of electromagnetic waves, it often relates to the medium's impact on the wave propagation.
In optics, the dielectric constant (\( \varepsilon_r \) ) can be approximated from the index of refraction since the relative permeability is near unity. It is given by the equation:
  • \( \varepsilon_r = n^2 \)
Given the calculated refractive index of the glass, \( n = 1.38 \), the dielectric constant becomes \( \varepsilon_r \approx 1.90 \).

This shows how the electric field energy is affected when electromagnetic waves travel through this material. A higher dielectric constant means stronger electrical energy storage capability of the material.

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Most popular questions from this chapter

If the eye receives an average intensity greater than \(1.0 \times 10^{2} \mathrm{~W} / \mathrm{m}^{2},\) damage to the retina can occur. This quantity is called the damage threshold of the retina. (a) What is the largest average power (in \(\mathrm{mW}\) ) that a laser beam \(1.5 \mathrm{~mm}\) in diameter can have and still be considered safe to view head-on? (b) What are the maximum values of the electric and magnetic fields for the beam in part (a)? (c) How much energy would the beam in part (a) deliver per second to the retina? (d) Express the damage threshold in \(\mathrm{W} / \mathrm{cm}^{2}\).

The microwaves in a certain microwave oven have a wavelength of \(12.2 \mathrm{~cm} .\) (a) How wide must this oven be so that it will contain five antinodal planes of the electric field along its width in the standing-wave pattern? (b) What is the frequency of these microwaves? (c) Suppose a manufacturing error occurred and the oven was made \(5.0 \mathrm{~cm}\) longer than specified in part (a). In this case, what would have to be the frequency of the microwaves for there still to be five antinodal planes of the electric field along the width of the oven?

Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge \(q\) and acceleration \(a\) is given by $$ \frac{d E}{d t}=\frac{q^{2} a^{2}}{6 \pi \epsilon_{0} c^{3}} $$ where \(c\) is the speed of light. (a) Verify that this equation is dimensionally correct. (b) If a proton with a kinetic energy of \(6.0 \mathrm{MeV}\) is traveling in a particle accelerator in a circular orbit of radius \(0.750 \mathrm{~m},\) what fraction of its energy does it radiate per second? (c) Consider an electron orbiting with the same speed and radius. What fraction of its energy does it radiate per second?

A totally reflecting disk has radius \(8.00 \mu \mathrm{m}\) and average density \(600 \mathrm{~kg} / \mathrm{m}^{3}\). A laser has an average power output \(P_{\mathrm{av}}\) spread uniformly over a cylindrical beam of radius \(2.00 \mathrm{~mm}\). When the laser beam shines upward on the disk in a direction perpendicular to its flat surface, the radiation pressure produces a force equal to the weight of the disk. (a) What value of \(P_{\text {av }}\) is required? (b) What average laser power is required if the radius of the disk is doubled?

NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is \(3.9 \times 10^{26} \mathrm{~W}\). How large a sail is necessary to propel a \(10,000 \mathrm{~kg}\) spacecraft against the gravitational force of the sun? Express your result in square kilometers. (c) Explain why your answer to part (b) is independent of the distance from the sun.
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