/*! 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 1 A certain helium-neon laser emit... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 33: Problem 1

A certain helium-neon laser emits red light in a narrow band of wavelengths centered at \(632.8 \mathrm{~nm}\) and with a "wavelength width" (such as on the scale of Fig. \(33-1\) ) of \(0.0100 \mathrm{~nm}\). What is the corresponding "frequency width" for the emission?

Short Answer

Expert verified
The frequency width is approximately \(7.51 \times 10^{9} \text{ Hz}\).

Step by step solution

01

Understand the Relationship between Frequency and Wavelength

In order to find the 'frequency width', we need to use the relationship between frequency \( f \) and wavelength \( \lambda \), given by the formula \( c = \lambda f \), where \( c \) is the speed of light (approximately \( 3 \times 10^8 \) m/s).
02

Wavelength Conversion to Meters

Convert the central wavelength and the wavelength width from nanometers to meters. Since \(1 \text{ nm} = 10^{-9} \text{ m} \), the central wavelength is \(632.8 \text{ nm} = 632.8 \times 10^{-9} \text{ m} \), and the wavelength width is \(0.0100 \text{ nm} = 0.0100 \times 10^{-9} \text{ m} \).
03

Calculate the Central Frequency

Using the formula \( f = \frac{c}{\lambda} \), calculate the central frequency for \( \lambda = 632.8 \times 10^{-9} \text{ m} \). This yields:\[ f = \frac{3 \times 10^8 \text{ m/s}}{632.8 \times 10^{-9} \text{ m}} \approx 4.74 \times 10^{14} \text{ Hz} \].
04

Calculate the Change in Frequency

Use the derivative of the frequency with respect to wavelength, given by \( \frac{df}{d\lambda} = -\frac{c}{\lambda^2} \), to find the frequency width \( \Delta f \). The change in wavelength is \( \Delta \lambda = 0.0100 \times 10^{-9} \text{ m} \) and:\[ \Delta f \approx \left| \frac{-c}{\lambda^2} \right| \Delta \lambda \approx \frac{3 \times 10^8}{(632.8 \times 10^{-9})^2} \times 0.0100 \times 10^{-9} \approx 7.51 \times 10^{9} \text{ Hz} \].

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 91影视!

Key Concepts

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

Speed of Light
In the realm of physics, the speed of light is an essential constant, playing a crucial role in the understanding of electromagnetic waves. Denoted as \( c \), the speed of light is approximately \( 3 \times 10^8 \) meters per second (m/s). This constant signifies how fast light travels in a vacuum, which is fundamental when working with concepts related to electromagnetic radiation.
For any kind of light, whether seen or unseen, this speed remains unchanged. This makes it a key factor in formulas that link frequency and wavelength. When you know two of these variables, with the speed of light, you can find the third.
  • Example: Light from the sun takes about 8 minutes to reach the earth.
  • In Context: For any light or electromagnetic wave, the relationship between frequency (\( f \)) and wavelength (\( \lambda \)) is defined by \( c = \lambda f \).
Nanometer to Meter Conversion
Wavelengths in the electromagnetic spectrum are often expressed in nanometers (nm) due to the small size of these waves. A nanometer is one billionth of a meter or \(1 \, \text{nm} = 10^{-9} \, \text{m}\). Converting nanometers to meters is crucial for precise calculations when dealing with universal physical constants.
Consider the case where you need to find the frequency of a wavelength measured in nanometers. You must first convert it to meters.
  • Example: If you have a wavelength of 632.8 nm, you convert it by multiplying by \(10^{-9}\) to get \(632.8 \times 10^{-9} \, \text{m}\).
This conversion ensures consistency in units, allowing you to use formulas involving the speed of light without any discrepancies.
Frequency Calculation
Understanding how to calculate frequency from a given wavelength is fundamental when working with wave equations. The frequency of a wave indicates how often the wave oscillates in a second and is typically measured in Hertz (Hz).
Using the relationship \( c = \lambda f \), you can rearrange the formula to solve for frequency \( f \): \[ f = \frac{c}{\lambda} \]
  • Steps:
  • Convert wavelength from nanometers to meters (if necessary).
  • Use the speed of light \( c \) as \( 3 \times 10^8 \text{ m/s} \).
  • Plug the values into the formula to find \( f \).

In our example, by converting the wavelength from nanometers to meters and substituting into the formula, we were able to calculate a central frequency of approximately \( 4.74 \times 10^{14} \text{ Hz} \). This calculation is integral to understanding the characteristics of electromagnetic waves.
Derivative in Physics
In physics, derivatives are used to determine how a quantity changes in relation to another. When dealing with wave properties, applying derivatives helps us understand how changes in one aspect, like wavelength, affect another, such as frequency.
The formula \( \frac{df}{d\lambda} = -\frac{c}{\lambda^2} \) represents the derivative of frequency with respect to wavelength. This equation helps find how a tiny change in wavelength affects the frequency's change, known as frequency width \( \Delta f \).
  • Conceptualization:
  • The negative sign indicates that frequency decreases as the wavelength increases. This inverse relationship is a principal characteristic of wave behavior.
  • Example: A small change in wavelength \( \Delta \lambda = 0.0100 \times 10^{-9} \text{ m} \) results, in our scenario, in a frequency change \( \Delta f \approx 7.51 \times 10^9 \text{ Hz} \).
By employing derivatives, one gains insight into the dynamics of wave interactions and can predict changes within these systems.

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

Frank D. Drake, an investigator in the SETI (Search for Extra-Terrestrial Intelligence) program, once said that the large radio telescope in Arecibo, Puerto Rico (Fig. 33-36), "can detect a signal which lays down on the entire surface of the earth a power of only one picowatt." (a) What is the power that would be received by the Arecibo antenna for such a signal? The antenna diameter is \(300 \mathrm{~m}\). (b) What would be the power of an isotropic source at the center of our galaxy that could provide such a signal? The galactic center is \(2.2 \times 10^{4}\) ly away. A light-year is the distance light travels in one year.

A plane electromagnetic wave has a maximum electric field magnitude of \(3.20 \times 10^{-4} \mathrm{~V} / \mathrm{m} .\) Find the magnetic field amplitude.

An electromagnetic wave with frequency \(4.00 \times 10^{14} \mathrm{~Hz}\) travels through vacuum in the positive direction of an \(x\) axis. The wave has its electric field oscillating parallel to the \(y\) axis, with an amplitude \(E_{m}\). At time \(t=0\), the electric field at point \(P\) on the \(x\) axis has a value of \(+E_{m} / 4\) and is decreasing with time. What is the distance along the \(x\) axis from point \(P\) to the first point with \(E=0\) if we search in (a) the negative direction and (b) the positive direction of the \(x\) axis?

We want to rotate the direction of polarization of a beam of polarized light through \(90^{\circ}\) by sending the beam through one or more polarizing sheets. (a) What is the minimum number of sheets required? (b) What is the minimum number of sheets required if the transmitted intensity is to be more than \(60 \%\) of the original intensity?

A beam of partially polarized light can be considered to be a mixture of polarized and unpolarized light. Suppose we send such a beam through a polarizing filter and then rotate the filter through \(360^{\circ}\) while keeping it perpendicular to the beam. If the transmitted intensity varies by a factor of \(5.0\) during the rotation, what fraction of the intensity of the original beam is associated with the beam's polarized light?
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Dynamics

Read Explanation

Rotational Dynamics

Read Explanation

Electromagnetism

Read Explanation

Linear Momentum

Read Explanation

Electric Charge Field and Potential

Read Explanation
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.