Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A 1 ns pulse of electromagnetic waves would be \(30 \mathrm{~cm}\) Iong. (a) Consider such a pulse of \(633 \mathrm{~nm}\) wavelength laser light. Find its central wave number and the range of wave numbers it comprises. (b) Repeat part (a), bue for a 1 ns pulse of \(100 \mathrm{MHz}\) radio waves.

Short Answer

Expert verified
For the laser light with a wavelength of 633 nm, the central wave number is approximately \(9.93 x 10^{6} m^{-1}\) and the range of wave numbers is approximately \(20.94 m^{-1}\). For the radio waves with a frequency of 100 MHz, their central wave number is about \(0.00209 m^{-1}\) and the wave number range is the same as the laser light, \(20.94 m^{-1}\).

Step by step solution

01

Compute for the Central Wave Number for the Laser Light

To find the central wave number of the laser light with wavelength of 633 nm, convert the wavelength into meters, then substitute it into the wave number formula \(k = 2π/λ\). This will result in \( k = 2π/(633 x 10^{-9})\)
02

Calculate the Range of Wave Numbers for the Laser Light

The range of wave numbers for the laser light can be found using the formula Δk=2π/Δx, where Δx = 30 cm or 0.3 m. Thus, Δk= 2π/(0.3).
03

Compute for the Central Wave Number for the Radio Waves

Finding the central wave number for the radio waves requires converting frequency into wavelength then use the wave number formula. Using the speed of light \(c = 3 x 10^{8} m/s\), convert the frequency of 100 MHz (1 MHz = \(10^6 Hz\)) to wavelength by using the formula \(λ = c/f = 3 x 10^{8}/(100 x 10^6)\). After determining the wavelength, substitute it into the formula to get the wave number \( k = 2π/λ\).
04

Calculate the Range of Wave Numbers for the Radio Waves

The range of wave numbers can be found using the formula Δk=2π/Δx, where Δx = 30 cm or 0.3 m, the same as in step 2. Thus, Δk= 2π/(0.3) as well.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Wave Number Calculation
Understanding the wave number helps grasp the nature of different electromagnetic waves. The wave number is inversely related to the wavelength, with its unit being reciprocal meters \( m^{-1} \). Generally, it's defined as \( k = \frac{2\pi}{\lambda} \), where \( \lambda \) is the wavelength and \( \pi \) is the mathematical constant pi, approximately 3.14159. This relationship highlights that shorter wavelengths have higher wave numbers and vice versa.

For example, in a 1 ns pulse of electromagnetic waves, calculating the wave number provides us with valuable information about the spatial frequency of the wave—essentially telling us how many wave cycles can exist over a certain distance. It's a fundamental concept in understanding the behavior of waves, from the light of lasers to the transmission of radio signals.
Laser Light Wavelength
Laser light is often defined by its wavelength because it falls within the visible to infrared range of the electromagnetic spectrum. The wavelength of laser light affects its color and energy. For instance, a 633 nm wavelength corresponds to the red portion of the spectrum.

To connect this to our exercise, imagining a pulse of laser light as a physical entity 30 cm long helps us visualize the compactness of that light. The central wave number, which we calculate by inverting the wavelength, represents the most common number or the 'average' spatial frequency of the wave within that pulse.

Real-World Applications

  • Laser Precision: The precise wavelength of laser light is crucial for applications like cutting, medical procedures, and optical communications.
  • Wavelength and Energy: Shorter wavelengths, like those of blue or ultraviolet laser light, carry more energy, which is important in applications like laser imaging or spectroscopy.
Radio Waves Frequency
Radio waves are a type of electromagnetic radiation with frequencies as low as 3 kHz to as high as 300 GHz. A frequency of 100 MHz, which falls in the radio spectrum, is commonly used for FM radio broadcasting. Unlike the laser light wavelength, radio waves have much longer wavelengths and thus lower frequencies.

In our problem, converting from frequency to wavelength allows us to apply the same wave number concept to these much larger and longer-pulse radio waves. The central wave number gives us a measure of the number of wave cycles per unit distance within the radio pulse, which is crucial for understanding signal propagation and antenna design.

Importance in Communication

  • Frequency Bandwidth: Different frequencies are used for various communication services, with certain frequencies penetrating buildings and traveling longer distances better than others.
  • Antenna Design: The wavelength of the radio waves determines the size and type of the antenna needed to transmit or receive those waves effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Equation \((3-1)\) expresses Planck's spectral energy density as an energy per range \(d f\) of frequencies. Quite of ten, it is more convenient to express it as an energy per range \(d \lambda\) of wavelengths, By differentiating \(f=c / \lambda,\) we find that df \(=-c / \lambda^{2} d \lambda\). Ignoring the minus sign (we are interested only in relating the magnitudes of the ranges \(d f\) and \(d \lambda\) ). show that, in terms of wavelength. Planck's formula is \(\frac{d U}{d \lambda}=\frac{8 \pi V h c}{e^{h c / \lambda k_{B} T}-1} \frac{1}{\lambda^{5}}\)

A photon has the same momentum as an electron moving at \(10^{6} \mathrm{~m} / \mathrm{s}\). (a) Determine the photon's wavelength \(\mathrm{N}\). (b) What is the ratio of the kinetic energies of the two?

A bedrock topic in quantum mechanics is the uncertainty principle. It is discussed mostly for massive objects in Chapter \(4,\) but the idea also applies to light: Increasing certainty in knowledge of photon position implies increasing uncertainty in knowledge of its momentum, and vice versa. A single-slit pattern that is developed (like the double-slit pattern of Section 3.6 ) one photon at a time provides a good example. Depicted in the accompanying figure, the pattern shows that photons emerging from a narrow slit are spread allover; a photon's \(x\) -component of momentum can be any value over a broad range and is thus uncertain. On the other hand, the \(x\) -coordinate of position of an emerging photon covers a fairly small range, for \(w\) is small. Using the single-slit diffraction formula \(n \lambda=w \sin \theta,\) show that the range of likely values of \(p_{x}\), which is roughly \(p \sin \theta\), is inversely proportional to the range \(w\) of likely position values. Thus, an inherent wave nature implies that the precisions with which the particle properties of position and momentum can be known are inversely proportional.

You are conducting a photoelectric effect experiment by shining light of \(500 \mathrm{nm}\) wavelength a a piece of metal and determining the stopping potential. If, unbeknownst to you, your 500 nm light source actually contained a small amount of ultraviolet light, would it throw off your results by a small amount or by quite a bit? Explain.

A stationary muon \(\mu^{-}\) annihilates with a stationary antimuon \(\mu^{+}\) (same mass. \(1.88 \times 10^{\circ} 28 \mathrm{~kg}\), but opposite charge). The two disappear, replaced by electromagnetic radiation. (a) Why is it not possible for a single photon to result? (b) Suppose two photons result. Describe their possible directions of motion and wavelengths.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free