Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 22: Problem 62

Calculate the frequency of an EM wave with a wavelength the size of (a) the thickness of a piece of paper \((60 \mu \mathrm{m}),(\mathrm{b})\) a \(91-\mathrm{m}\) -long soccer field, (c) the diameter of Earth, (d) the distance from Earth to the Sun.

Short Answer

Expert verified
Question: Calculate the frequency of electromagnetic waves for the following wavelengths: (a) the thickness of a piece of paper (60 µm), (b) the length of a soccer field (91 meters), (c) the diameter of Earth (12,742 km), and (d) the distance from Earth to the Sun (149.6 million km). Answer: (a) The frequency of the EM wave with a wavelength equal to the thickness of a piece of paper is approximately 5 * 10^12 Hz. (b) The frequency of the EM wave with a wavelength equal to the length of a soccer field is approximately 3.3 * 10^6 Hz. (c) The frequency of the EM wave with a wavelength equal to the diameter of Earth is approximately 23.6 Hz. (d) The frequency of the EM wave with a wavelength equal to the distance from Earth to the Sun is approximately 2 * 10^(-6) Hz.

Step by step solution

01

Problem 1 - Thickness of a piece of paper; 60 µm

The thickness of the paper is given as 60 micrometers (µm). We need to first convert the micrometers into meters: 60 µm = 60 * 10^(-6) meters Now, we'll calculate the frequency using the formula we derived before: f = c / λ f = (3 * 10^8) / (60 * 10^(-6)) f ≈ 5 * 10^12 Hz So, the frequency of the EM wave with a wavelength equal to the thickness of a piece of paper is approximately 5 * 10^12 Hz.
02

Problem 2 - 91-meter long soccer field

The wavelength of the EM wave is given as the length of a soccer field, which is 91 meters. We'll use our formula to calculate the frequency: f = c / λ f = (3 * 10^8) / 91 f ≈ 3.3 * 10^6 Hz The frequency of the EM wave with a wavelength equal to the length of a soccer field is approximately 3.3 * 10^6 Hz.
03

Problem 3 - Diameter of Earth

For this problem, the wavelength of the EM wave is the diameter of Earth, which is approximately 12,742 kilometers. Before using the formula, we need to convert kilometers to meters: 12,742 km = 12,742,000 meters Now we can calculate the frequency: f = c / λ f = (3 * 10^8) / (12,742,000) f ≈ 23.6 Hz The frequency of the EM wave with a wavelength equal to the diameter of Earth is approximately 23.6 Hz.
04

Problem 4 - Distance from Earth to the Sun

In this case, the wavelength of the EM wave is equal to the average distance from Earth to the Sun, which is approximately 149.6 million kilometers. We will convert it to meters first: 149,600,000 km = 149,600,000,000 meters Now, we can calculate the frequency: f = c / λ f = (3 * 10^8) / (149,600,000,000) f ≈ 2 * 10^(-6) Hz The frequency of the EM wave with a wavelength equal to the distance from Earth to the Sun is approximately 2 * 10^(-6) 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.

Start your free trial

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

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

The electric field in a microwave traveling through air has amplitude $0.60 \mathrm{mV} / \mathrm{m}\( and frequency \)30 \mathrm{GHz}$. Find the amplitude and frequency of the magnetic field.
The electric field in a radio wave traveling through vacuum has amplitude $2.5 \times 10^{-4} \mathrm{V} / \mathrm{m}\( and frequency \)1.47 \mathrm{MHz}$. (a) Find the amplitude and frequency of the magnetic field. (b) The wave is traveling in the \(+x\) -direction. At \(x=0\) and \(t=0,\) the electric field is \(1.5 \times 10^{-4} \mathrm{V} / \mathrm{m}\) in the \- y-direction. What are the magnitude and direction of the magnetic field at \(x=0\) and \(t=0 ?\).
A \(1.0-\mathrm{m}^{2}\) solar panel on a satellite that keeps the panel oriented perpendicular to radiation arriving from the Sun absorbs $1.4 \mathrm{kJ}\( of energy every second. The satellite is located at \)1.00 \mathrm{AU}\( from the Sun. (The Earth Sun distance is defined to be \)1.00 \mathrm{AU} .$. How long would it take an identical panel that is also oriented perpendicular to the incoming radiation to absorb the same amount of energy, if it were on an interplanetary exploration vehicle 1.55 AU from the Sun? ( W) tutorial: sunlight on Io).
The radio telescope in Arecibo, Puerto Rico, has a diameter of $305 \mathrm{m}$. It can detect radio waves from space with intensities as small as \(10^{-26} \mathrm{W} / \mathrm{m}^{2} .\) (a) What is the average power incident on the telescope due to a wave at normal incidence with intensity \(1.0 \times 10^{-26} \mathrm{W} / \mathrm{m}^{2} ?\) (b) What is the average power incident on Earth's surface? (c) What are the rms electric and magnetic fields?
Unpolarized light passes through two polarizers in turn with polarization axes at \(45^{\circ}\) to one another.What is the fraction of the incident light intensity that is transmitted?
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Measurements

Read Explanation

Wave Optics

Read Explanation

Engineering Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Dynamics

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.

Sign-up for free