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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) Intensity is \(636,619 \text{ W/m}^2\). (b) Max electric field \(E_{max}\) is \(6.92 \times 10^3 \text{ V/m}\), max magnetic field \(B_{max}\) is \(2.31 \times 10^{-5} \text{ T}\). (c) Average energy density is \(2.12 \times 10^{-3} \text{ J/m}^3\).

Step by step solution

01

Calculate the Area of the Laser Beam

The area of the laser beam can be found using the formula for the area of a circle, given the diameter. First, calculate the radius of the beam: \[ r = \frac{d}{2} = \frac{1.00\, \text{mm}}{2} = 0.500 \times 10^{-3} \text{ m} \]The area, \(A\), is then:\[ A = \pi r^2 = \pi (0.500 \times 10^{-3})^2 \approx 7.854 \times 10^{-7} \text{ m}^2 \]
02

Calculate the Intensity of the Laser Beam

Intensity, \(I\), is defined as power per unit area.Using the power \(P = 0.500 \text{ mW} = 0.500 \times 10^{-3} \text{ W}\), we find:\[ I = \frac{P}{A} = \frac{0.500 \times 10^{-3}}{7.854 \times 10^{-7}} \approx 636,619 \text{ W/m}^2 \]
03

Determine Maximum Electric Field

The maximum electric field \(E_{max}\) can be found from the intensity using the relation:\[ I = \frac{1}{2}\epsilon_0 c E_{max}^2 \]Where \( \epsilon_0 \) is the permittivity of free space \(=8.85 \times 10^{-12} \text{ C}^2/\text{N.m}^2\), and \(c\) is the speed of light \(=3 \times 10^8 \text{ m/s}\).Rearrange to solve for \(E_{max}\):\[ E_{max} = \sqrt{\frac{2I}{\epsilon_0 c}} \approx \sqrt{\frac{2 \times 636,619}{8.85 \times 10^{-12} \times 3 \times 10^8}} \approx 6.92 \times 10^3 \text{ V/m} \]
04

Calculate Maximum Magnetic Field

Magnetic field \(B_{max}\) is related to the electric field by:\[ E_{max} = cB_{max} \]So:\[ B_{max} = \frac{E_{max}}{c} \approx \frac{6.92 \times 10^3}{3 \times 10^8} \approx 2.31 \times 10^{-5} \text{ T} \]
05

Calculate the Average Energy Density

Average energy density \(u\) in the beam can be found from:\[ u = \frac{I}{c} \approx \frac{636,619}{3 \times 10^8} \approx 2.12 \times 10^{-3} \text{ J/m}^3 \]

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Key Concepts

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

Wavelength
The wavelength of light is a fundamental property in physics, representing the distance between consecutive crests of a wave. In the context of lasers, the wavelength determines the color of the light emitted.
He-Ne lasers, like the one discussed in the problem, emit light with a wavelength of 633 nanometers (nm), which is in the red part of the visible spectrum.
This wavelength is chosen for He-Ne lasers because it provides a balance between visibility and common applications in optical experiments.
Understanding wavelength is crucial not only for determining the color of light but also for calculating various properties such as energy and frequency. The energy of a photon is inversely proportional to its wavelength, meaning shorter wavelengths have higher energy.
Electric Field
The electric field at any point in space is a measure of the force that would be experienced by a charged particle placed at that point. In the context of a laser beam, the electric field oscillates with the frequency of the light wave.
For a laser beam, such as the He-Ne laser, the intensity of the beam is related to the amplitude of its electric field. The stronger the electric field, the higher the intensity of the light.
In calculations, the maximum electric field is derived from the intensity, using the formula: \( E_{max} = \sqrt{\frac{2I}{\epsilon_0 c}} \). Electric field calculations involve fundamental constants, including the speed of light \(c\) and the permittivity of free space \(\epsilon_0\). Understanding these concepts is essential to predict how light interacts with charged particles.
Magnetic Field
Similar to the electric field, the magnetic field is an important component of an electromagnetic wave, like that produced by a laser.
The magnetic field is perpendicular to the electric field and is created by the movement of the wave through space.
In the context of laser beams, the maximum magnetic field can be calculated from the maximum electric field using the relationship: \( E_{max} = cB_{max} \).
This relationship highlights the intimate connection between electric and magnetic fields in electromagnetic radiation, demonstrating how they propagate through space. Understanding both fields is key to mastering the behavior of light and its applications.
Energy Density
Energy density refers to the amount of energy stored in a given system per unit volume. For a laser beam, this is the energy transmitted by the beam per cubic meter. The average energy density is a critical quantity for understanding the power of a laser beam in physically meaningful terms.
The energy density \( u \) in the laser beam can be calculated by the formula \( u = \frac{I}{c} \), which relies on the intensity of the beam \( I \), divided by the speed of light \(c\).
This value informs us about how much energy is concentrated in the laser, crucial for applications like cutting, medical procedures, and precision measurements.
He-Ne Lasers
He-Ne lasers are a specific type of laser that uses a mixture of helium and neon gases to produce its characteristic red light.
They were among the first lasers created and remain popular for their stability and precision in various applications such as alignment, holography, and spectroscopy.
He-Ne lasers typically emit light at a wavelength of 633 nm, making them suitable for many scientific experiments due to their monochromatic and coherent beam properties.
The technology behind He-Ne lasers involves exciting helium atoms which then transfer energy to neon atoms, creating the laser light. This process is both energy-efficient and reliable, making these lasers a staple in educational and research contexts.

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Most popular questions from this chapter

A space probe \(2.0 \times 10^{10} \mathrm{~m}\) from a star measures the total intensity of electromagnetic radiation from the star to be \(5.0 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2}\). If the star radiates uniformly in all directions, what is its total average power output?

(a) How much time does it take light to travel from the moon to the earth, a distance of \(384,000 \mathrm{~km}\) ? (b) Light from the star Sirius takes \(8.61\) years to reach the earth. What is the distance from earth to Sirius in kilometers?

You want to support a sheet of fireproof paper horizontally, using only a vertical upward beam of light spread uniformly over the sheet. There is no other light on this paper. The sheet measures \(22.0 \mathrm{~cm}\) by \(28.0 \mathrm{~cm}\) and has a mass of \(1.50 \mathrm{~g}\). (a) If the paper is black and hence absorbs all the light that hits it, what must be the intensity of the light beam? (b) For the light in part (a), what are the amplitudes of its electric and magnetic fields? (c) If the paper is white and hence reflects all the light that hits it, what intensity of light beam is needed to support it? (d) To see if it is physically reasonable to expect to support a sheet of paper this way, calculate the intensity in a typical \(0.500-\mathrm{mW}\) laser beam that is \(1.00 \mathrm{~mm}\) in diameter, and compare this value with your answer in part (a).

Fields from a Light Bulb. We can reasonably model a \(75-\mathrm{W}\) incandescent light bulb as a sphere \(6.0 \mathrm{~cm}\) in diameter. Typically, only about \(5 \%\) of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation. (a) What is the visible-light intensity (in \(\mathrm{W} / \mathrm{m}^{2}\) ) at the surface of the bulb? (b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?

Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, if it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an electric field \(\overrightarrow{\boldsymbol{E}}(x, t)=E_{y}(x, t) \hat{\boldsymbol{j}}\) propagating in the \(+x\)-direction within a conductor is $$ \frac{\delta^{2} E_{y}(x, t)}{\delta x^{2}}=\frac{\mu}{\rho} \frac{\delta E_{y}(x, t)}{\delta t} $$ where \(\mu\) is the permeability of the conductor and \(\rho\) is its resistivity. (a) A solution to this wave equation is $$ E_{y}(x, t)=E_{\max } e^{-k_{C} x} \cos \left(k_{C} x-\omega t\right) $$ where \(K_{C}=\sqrt{\omega \mu / 2 \rho}\). Verify this by substituting \(E_{y}(x, t)\) into the above wave equation. (b) The exponential term shows that the electric field decreases in amplitude as it propagates. Explain why this happens. (Hint: The field does work to move charges within the conductor. The current of these moving charges causes \(i^{2} R\) heating within the conductor, raising its temperature. Where does the energy to do this come from?) (c) Show that the electric-field amplitude decreases by a factor of \(1 / e\) in a distance \(1 / k_{\mathrm{C}}=\sqrt{2 \rho / \omega \mu}\), and calculate this distance for a radio wave with frequency \(f=\) \(1.0 \mathrm{MHz}\) in copper (resistivity \(1.72 \times 10^{-8} \Omega \cdot \mathrm{m}\); permeability \(\mu=\mu_{0}\) ). Since this distance is so short, electromagnetic waves of this frequency can hardly propagate at all into copper. Instead, they are reflected at the surface of the metal. This is why radio waves cannot penetrate through copper or other metals, and why radio reception is poor inside a metal structure.

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