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Chapter 23: Problem 7

\(\bullet\) The electric field of a sinusoidal electromagnetic wave obeys the equation \(E=-(375 \mathrm{V} / \mathrm{m}) \sin [(5.97 \times\) \(10^{15} \operatorname{rad} / \mathrm{s} ) t+\left(1.99 \times 10^{7} \mathrm{rad} / \mathrm{m}\right) x ] .\) (a) What are the amplitudes of the electric and magnetic fields of this wave? (b) What are the frequency, wavelength, and period of the wave? Is this light visible to humans? (c) What is the speed of the wave?

Short Answer

Expert verified
The amplitudes are 375 V/m and 1.25 碌T. Frequency: 9.5脳10鹿鈦� Hz, Wavelength: 316 nm, Not visible. Speed: 3脳10鈦� m/s.

Step by step solution

01

Identifying the Amplitude of the Electric Field

The value immediately multiplying the sine function in the equation \( E = -(375 \mathrm{V} / \mathrm{m}) \sin [(5.97 \times 10^{15} \text{ rad} / \mathrm{s} ) t+(1.99 \times 10^{7} \mathrm{rad} / \mathrm{m}) x ] \) is the amplitude of the electric field. Thus, the amplitude of the electric field is \( 375 \text{ V/m} \).
02

Calculating the Amplitude of the Magnetic Field

The amplitude of the magnetic field \( B_0 \) is related to the amplitude of the electric field \( E_0 \) by the equation \( B_0 = \frac{E_0}{c} \), where \( c \) is the speed of light, approximately \( 3 \times 10^8 \text{ m/s} \). Therefore, the amplitude of the magnetic field is \( B_0 = \frac{375}{3 \times 10^8} \approx 1.25 \times 10^{-6} \text{ T} \).
03

Finding the Frequency of the Wave

The frequency \( f \) is obtained from the angular frequency \( \omega = 5.97 \times 10^{15} \text{ rad/s} \) using the formula \( f = \frac{\omega}{2\pi} \). Substituting, we find \( f = \frac{5.97 \times 10^{15}}{2\pi} \approx 9.5 \times 10^{14} \text{ Hz} \).
04

Calculating the Wavelength of the Wave

The wave number \( k = 1.99 \times 10^7 \text{ rad/m} \) is related to the wavelength \( \lambda \) by the equation \( k = \frac{2\pi}{\lambda} \). Solving for \( \lambda \), we have \( \lambda = \frac{2\pi}{1.99 \times 10^7} \approx 3.16 \times 10^{-7} \text{ m} \).
05

Determining the Period of the Wave

The period \( T \) is the inverse of the frequency: \( T = \frac{1}{f} \). Using the previously calculated frequency, \( T = \frac{1}{9.5 \times 10^{14}} \approx 1.05 \times 10^{-15} \text{ s} \).
06

Checking the Visibility of Light

The wavelength of visible light ranges from about \( 400 \text{ nm} \) to \( 700 \text{ nm} \). Our calculated wavelength of \( 316 \text{ nm} \) (or \( 3.16 \times 10^{-7} \text{ m} \)) falls outside this range, indicating that this light is not visible to humans.
07

Finding the Speed of the Wave

The speed \( v \) of a wave is given by \( v = f \times \lambda \). Using the calculated frequency and wavelength, \( v = 9.5 \times 10^{14} \times 3.16 \times 10^{-7} \approx 3 \times 10^8 \text{ m/s} \), which is the speed of light.

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Key Concepts

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

Electric Field Amplitude
The electric field amplitude in an electromagnetic wave represents the maximum strength of the electric field at any point in the wave. In the given equation \(E = -(375 \mathrm{V} / \mathrm{m}) \sin [(5.97 \times 10^{15} \mathrm{rad} / \mathrm{s} ) t+(1.99 \times 10^{7} \mathrm{rad} / \mathrm{m}) x ] \), the term directly preceding the sine function, \(-375 \mathrm{V} / \mathrm{m}\), indicates the electric field amplitude. This negative sign simply indicates the direction of the field, while the amplitude itself is \(375 \mathrm{V/m}\). This amplitude reflects the peak electric force that acts on charges in the field.
Magnetic Field Amplitude
The magnetic field amplitude is linked directly to the electric field amplitude via the speed of light \(c\). Using the relation \(B_0 = \frac{E_0}{c}\), where \(E_0\) is the electric field amplitude and \(c\) is approximately \(3 \times 10^8 \mathrm{m/s}\), we can compute the magnetic field amplitude. In this exercise, substituting \(E_0 = 375 \mathrm{V/m}\) gives \( B_0 = \frac{375}{3 \times 10^8} \approx 1.25 \times 10^{-6} \mathrm{T}\). This value is the highest intensity that the magnetic field reaches and provides information about the magnetic component of the wave.
Wave Frequency
Wave frequency is a fundamental characteristic that describes how often wave cycles pass a point within one second. We determine it from the angular frequency using \(f = \frac{\omega}{2\pi}\). Given \(\omega = 5.97 \times 10^{15} \mathrm{rad/s}\), the frequency is \(f \approx \frac{5.97 \times 10^{15}}{2\pi} = 9.5 \times 10^{14} \mathrm{Hz}\). This high frequency indicates the wave oscillates extremely rapidly and is characteristic of the electromagnetic spectrum such as ultraviolet or other non-visible light.
Wave Period
The wave period is the duration of time for one complete cycle of the wave to pass a point. It is the reciprocal of the frequency, defined as \(T = \frac{1}{f}\). With the previously calculated frequency of \(9.5 \times 10^{14} \mathrm{Hz}\), we find \(T \approx \frac{1}{9.5 \times 10^{14}} = 1.05 \times 10^{-15} \mathrm{s}\). This extremely short period is typical of high-frequency electromagnetic waves, showing how quickly these waves repeat.
Wavelength
Wavelength is the distance between consecutive crests (or cycles) of a wave. It's determined from the wave number \(k\) with \(\lambda = \frac{2\pi}{k}\). Given \(k = 1.99 \times 10^7 \mathrm{rad/m}\), the wavelength \(\lambda \approx \frac{2\pi}{1.99 \times 10^7} = 3.16 \times 10^{-7} \mathrm{m}\). This wavelength is in the range of ultraviolet light, which is not perceptible by the human eye, thus confirming the non-visibility conclusion.
Speed of Light
The speed of light \(c\) is a fundamental constant in physics, defined as \(3 \times 10^8 \mathrm{m/s}\). For electromagnetic waves in a vacuum, like light, the speed \(v\) is computed by multiplying frequency \(f\) and wavelength \(\lambda\), \[ v = f \times \lambda. \]Substituting \(f = 9.5 \times 10^{14} \mathrm{Hz}\) and \(\lambda = 3.16 \times 10^{-7} \mathrm{m}\) gives \(v \approx 3 \times 10^8 \mathrm{m/s}\), which matches the constant speed of light. This consistency confirms that such waves are electromagnetic in nature and behave predictably according to established physical laws.

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Most popular questions from this chapter

\(\bullet$$\bullet\) A light beam is directed parallel to the axis of a hollow cylindrical tube. When the tube contains only air, it takes the light 8.72 ns to travel the length of the tube, but when the tube is filled with a transparent jelly, it takes the light 2.04 ns longer to travel its length. What is the refractive index of this jelly?

\(\bullet\) A glass plate having parallel faces and a refractive index of 1.58 lies at the bottom of a liquid of refractive index \(1.70 . \mathrm{A}\) ray of light in the liquid strikes the top of the glass at an angle of incidence of \(62.0^{\circ} .\) Compute the angle of refraction of this light in the glass.

\(\bullet\) Light inside the eye. The vitreous humor, a transparent, gelatinous fluid that fills most of the eyeball, has an index of refraction of 1.34 . Visible light ranges in wavelength from 400 \(\mathrm{nm}\) (violet) to \(700 \mathrm{nm}(\mathrm{red}),\) as measured in air. This light travels through the vitreous humor and strikes the rods and cones at the surface of the retina. What are the ranges of (a) the wavelength, (b) the frequency, and (c) the speed of the light just as it approaches the retina within the vitreous humor?

\(\bullet\) The indices of refraction for violet light \((\lambda=400 \mathrm{nm})\) and red light \((\lambda=700 \mathrm{nm})\) in diamond are 2.46 and \(2.41,\) respectively. A ray of light traveling through air strikes the diamond surface at an angle of \(53.5^{\circ}\) to the normal. Calculate the angular separation between these two colors of light in the refracted ray.

\(\bullet\) Threshold of vision. Under controlled darkened conditions in the laboratory, a light receptor cell on the retina of a person's eye can detect a single photon (more on photons in Chapter 28 ) of light of wavelength 505 \(\mathrm{nm}\) and having an energy of \(3.94 \times 10^{-19} \mathrm{J} .\) We shall assume that this energy is absorbed by a single cell during one period of the wave. Cells of this kind are called rods and have a diameter of approximately 0.0020 \(\mathrm{mm} .\) What is the intensity (in \(\mathrm{W} / \mathrm{m}^{2} )\) delivered to a rod?
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