/*! 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 29 Laboratory Lasers. He-Ne lasers ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Laboratory Lasers. He-Ne lasers are often used in physics demonstrations. They produce light of wavelength 633\(\mathrm { nm }\) and a power of 0.500\(\mathrm { mW }\) spread over a cylindrical beam 1.00\(\mathrm { mm }\) in diameter (although these quantities can vary). (a) What is the intensity of this laser beam? (b) What are the maximum values of the electric and magnetic fields? (c) What is the average energy density in the laser beam?

Short Answer

Expert verified
(a) 637 W/m², (b) 69.3 V/m for electric field and 2.31 \times 10^{-7} T for magnetic field, (c) 2.12 \times 10^{-6} J/m³.

Step by step solution

01

Calculate the Intensity of the Laser Beam

The intensity of a laser beam can be determined using the formula for intensity, which is given by \[I = \frac{P}{A}\]where \(I\) is the intensity in watts per square meter (W/m²), \(P\) is the power of the laser in watts, and \(A\) is the area of the beam in square meters. First, convert the power from milliwatts to watts:\[P = 0.500 \text{ mW} = 0.500 \times 10^{-3} \text{ W} = 5.00 \times 10^{-4} \text{ W}\]Next, calculate the area \(A\) of the circular beam using the formula for the area of a circle \(A = \pi \left( \frac{d}{2} \right)^2\):\[A = \pi \left( \frac{1.00 \times 10^{-3}}{2} \right)^2 \text{ m}^2 = \pi \left( 0.5 \times 10^{-3} \right)^2 \text{ m}^2 = \pi \times 0.25 \times 10^{-6} \text{ m}^2 = 0.7854 \times 10^{-6} \text{ m}^2\]Finally, calculate the intensity:\[I = \frac{5.00 \times 10^{-4}}{0.7854 \times 10^{-6}} \text{ W/m}^2 = 637 \text{ W/m}^2\]
02

Determine the Maximum Electric Field

The maximum value of the electric field \(E_{max}\) can be calculated using the relationship between intensity \(I\) and the electric field in a wave:\[I = \frac{1}{2} \cdot c \cdot \varepsilon_0 \cdot E_{max}^2\]where \(c\) is the speed of light \((3 \times 10^8 \text{ m/s})\) and \(\varepsilon_0\) is the permittivity of free space \((8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2)\).Rearranging to solve for \(E_{max}\):\[E_{max} = \sqrt{\frac{2I}{c \cdot \varepsilon_0}} = \sqrt{\frac{2 \times 637}{3 \times 10^8 \times 8.85 \times 10^{-12}}} = \sqrt{4.807 \times 10^3} \approx 69.3 \text{ V/m}\]
03

Calculate the Maximum Magnetic Field

The maximum magnetic field \(B_{max}\) can be determined using the relationship between \(E_{max}\) and \(B_{max}\):\[B_{max} = \frac{E_{max}}{c}\]Substituting the value of \(E_{max}\):\[B_{max} = \frac{69.3}{3 \times 10^8} \approx 2.31 \times 10^{-7} \text{ T}\]
04

Compute the Average Energy Density

The average energy density \(u\) in the laser beam can be determined using the intensity and the speed of light with the formula:\[u = \frac{I}{c}\]Substituting the known values:\[u = \frac{637}{3 \times 10^8} \text{ J/m}^3 \approx 2.12 \times 10^{-6} \text{ J/m}^3\]

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 91Ó°ÊÓ!

Key Concepts

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

Intensity
In the context of a He-Ne laser beam, intensity is a crucial concept. Intensity refers to the power per unit area carried by the beam. For this laser, we calculate it using the formula \( I = \frac{P}{A} \), where \( P \) is power and \( A \) is the area of the beam.
First, the power (0.500 mW) is converted to watts, resulting in \( 5.00 \times 10^{-4} \) W.
The beam's cross-sectional area is circular, with a diameter of 1.00 mm (which is 0.001 m), so we calculate the area using \( A = \pi \left( \frac{d}{2} \right)^2 \). This gives us \( A \approx 0.7854 \times 10^{-6} \) m².
Finally, dividing the power by the calculated beam area results in an intensity of approximately 637 W/m².
Electric Field
Understanding the electric field's role in a laser beam is essential since it's directly linked to the wave's intensity. The peak, or maximum electric field strength, denoted as \( E_{max} \), can be found with \( I = \frac{1}{2} \cdot c \cdot \varepsilon_0 \cdot E_{max}^2 \), where \( c \) is the speed of light and \( \varepsilon_0 \) is the vacuum permittivity.
Solving for \( E_{max} \), we use \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \) and \( c = 3 \times 10^8 \text{ m/s} \). Inserting the intensity (637 W/m²), we find that \( E_{max} \approx 69.3 \text{ V/m} \).
This value represents the field's strength at its highest point, a critical parameter in describing the laser beam's characteristics.
Magnetic Field
Magnetic fields are integral to electromagnetic waves, such as light from a He-Ne laser. They're related to the electric field through the speed of light. The maximum (peak) magnetic field \( B_{max} \) can be calculated using the max electric field value with \( B_{max} = \frac{E_{max}}{c} \).
Given the speed of light \( c = 3 \times 10^8 \text{ m/s} \) and the previously calculated \( E_{max} \approx 69.3 \text{ V/m} \), we find \( B_{max} \approx 2.31 \times 10^{-7} \text{ T} \).
This value helps describe how the magnetic component of the electromagnetic wave contributes to energy propagation in the laser beam.
Energy Density
Energy density allows us to quantify how much energy the laser beam contains per unit volume. For electromagnetic waves, like those in a laser, energy density \( u \) is expressed as \( u = \frac{I}{c} \), where \( I \) is intensity and \( c \) is the speed of light.
Taking the known intensity (637 W/m²) and \( c = 3 \times 10^8 \text{ m/s} \), we find \( u \approx 2.12 \times 10^{-6} \text{ J/m}^3 \).
This tells us how densely packed the energy is within the beam, a useful measure for understanding its power and effect over a given area.

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

A plane sinusoidal electromagnetic wave in air has a wavelength of 3.84\(\mathrm { cm }\) and an \(\vec { \boldsymbol { E } }\) -field amplitude of 1.35\(\mathrm { V } / \mathrm { m }\) . (a) What is the frequency? (b) What is the \(\vec { \boldsymbol { B } }\) -field amplitude? (c) What is the intensity? (d) What average force does this radiation exert on a totally absorbing surface with area 0.240\(\mathrm { m } ^ { 2 }\) perpendicular to the direction of propagation?

Interplanetary space contains many small particles referred to as interplanetary dust. Radiation pressure from the sun sets a lower limit on the size of such dust particles. To see the origin of this limit, consider a spherical dust particle of radius \(R\) and mass density \(\rho .\) (a) Write an expression for the gravitational force exerted on this particle by the sun (mass \(M\) ) when the particle is a distance \(r\) from the sun. (b) Let \(L\) represent the luminosity of the sun, equal to the rate at which it emits energy in electromagnetic radiation. Find the force exerted on the (totally absorbing) particle due to solar radiation pressure, remembering that the intensity of the sun's radiation also depends on the distance \(r .\) The relevant area is the cross-sectional area of the particle, not the total surface area of the particle. As part of your answer, explain why this is so. (c) The mass density of a typical interplanetary dust particle is about 3000\(\mathrm { kg } / \mathrm { m } ^ { 3 } .\) Find the particle radius \(R\) such the gravitational and radiation forces acting on the particle are equal in magnitude. The luminosity of the sun is \(3.9 \times 10 ^ { 26 }\) W. Does your answer depend on the distance of the particle from the sun? Why or why not? (d) Explain why dust particles with a radius less than that found in part (c) are unlikely to be found in the solar system. LHint: Construct the ratio of the two force expressions found in parts (a) and (b).]

In the 25 -ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead are lamps can produce light of intensity 2500\(\mathrm { W } / \mathrm { m } ^ { 2 }\) at the floor of the facility. This simulates the intensity of sunlight near the planet Venus.) Find the average radiation pressure (in pascals and in atmospheres) on (a) a totally absorbing section of the floor and (b) a totally reflection section of the floor. (c) Find the average momentum densit (momentum per unit volume) in the light at the floor.

A small helium-neon laser emits red visible light with a power of 4.60\(\mathrm { mW }\) in a beam that has a diameter of 2.50\(\mathrm { mm }\) . (a) What are the amplitudes of the electric and magnetic fields of the light? (b) What are the average energy densities associated with the electric field are the magnetic field? (c) What is the total energy contained in a \(1.00 - \mathrm { m }\) length of the beam?

Testing a Space Radio Transmitter. You are a NASA mission specialist on your first flight aboard the space shuttle. Thanks to your extensive training in physics, you have been assigned to evaluate the performance of a new radio transmitter on board the International Space Station (ISS). Perched on the shuttle's movable arm, you aim a sensitive detector at the ISS, which is 2.5\(\mathrm { km }\) away. You find that the electric-field amplitude of the radio waves coming from the ISS transmitter is 0.090\(\mathrm { V } / \mathrm { m }\) and that the frequency of the waves is 244\(\mathrm { MH } z\) . Find the following: (a) the intensity of the radio wave at your location; (b) the magnetic-field amplitude of the wave at your location; (c) the total power output of the ISS radio transmitter. (d) What assumptions, if any, did you make in your calculations?

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.