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\(\bullet$$\bullet\) A small helium-neon laser emits red visible light with a power of 3.20 \(\mathrm{mW}\) 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 and with the magnetic field? (c) What is the total energy contained in a 1.00 \(\mathrm{m}\) length of the beam?

Short Answer

Expert verified
(a) Calculate fields using intensity; (b) use energy density formulas; (c) multiply total energy density by beam volume.

Step by step solution

01

Calculate the Intensity of the Laser Beam

The power of the laser is given as 3.20 mW, which is 3.20 x 10^{-3} W. The diameter of the beam is 2.50 mm, or 2.50 x 10^{-3} m, making the area of the beam, with a circular cross-section, A = \( \pi (\frac{d}{2})^2 \). Thus, the intensity I of the beam can be calculated using the formula I = \(\frac{P}{A}\). Substitute the values: A = \( \pi ( \frac{2.50 \times 10^{-3}}{2})^2 \) I = \(\frac{3.20 \times 10^{-3}}{A}\)
02

Compute Amplitudes of Electric and Magnetic Fields

The intensity of light is also related to the amplitude of the electric field using the equation I = \(\frac{1}{2} \epsilon_0 c E_0^2 \), where \(\epsilon_0\) is the permittivity of free space \((8.85 \times 10^{-12} \mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2)\), c is the speed of light \((3.00 \times 10^8 \mathrm{m/s})\), and \(E_0\) is the amplitude of the electric field.Rearranging for \(E_0\) gives \(E_0 = \sqrt{\frac{2I}{\epsilon_0 c}} \).Similarly, the magnitude of the magnetic field \(B_0\) relates to the electric field as \(B_0 = \frac{E_0}{c}\).
03

Determine Average Energy Densities

The average energy density of the electric field \(u_E\) can be calculated with \(u_E = \frac{1}{2} \epsilon_0 E_0^2\). Similarly, the average energy density of the magnetic field \(u_B\) is given by \(u_B = \frac{1}{2}B_0^2/\mu_0\), where \(\mu_0\) is the permeability of free space \((4 \pi \times 10^{-7} \mathrm{N}/\mathrm{A}^2)\).
04

Calculate Total Energy in 1.00 meter of Beam

The total energy in a 1.00 meter length of the beam can be found using the energy density, which is given by the sum of the electric \(u_E\) and magnetic field \(u_B\) energy densities and the volume of the beam.The total energy per unit volume is the sum: \( u_{total} = u_E + u_B \). The volume of the beam of length 1 m is Volume = Area x Length = A x 1.00 m.Thus, Total Energy = \( u_{total} \times \text{Volume} \).

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Key Concepts

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

Electric Field Amplitude
The electric field amplitude, symbolized as \( E_0 \), represents the peak value of the electric field's strength in an electromagnetic wave. In laser physics, particularly with a helium-neon laser, determining this amplitude is crucial to understanding the laser's emitted light characteristics. To calculate the electric field amplitude of a laser beam, you can use the formula:
  • \( E_0 = \sqrt{\frac{2I}{\epsilon_0 c}} \)
This equation shows that the electric field amplitude is dependent on:
  • The intensity \( I \) of the laser beam.
  • The vacuum permittivity \( \epsilon_0 \), which is a constant \( 8.85 \times 10^{-12} \mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2 \).
  • The speed of light in a vacuum, \( c = 3.00 \times 10^8 \mathrm{m/s} \).
By substituting the appropriate values into this formula, you gain the maximum electric field strength at a given region within the beam.
Magnetic Field Amplitude
The amplitude of the magnetic field, denoted as \( B_0 \), plays a vital role in the characterization of electromagnetic waves from lasers. In the context of a helium-neon laser, the relationship between the magnetic field amplitude and the electric field amplitude is governed by the equation:
  • \( B_0 = \frac{E_0}{c} \)
Here, \( E_0 \) is the amplitude of the electric field, and \( c \) represents the speed of light. This equation indicates that the magnetic field amplitude is directly proportional to the electric field but inversely proportional to the speed of light.
Understanding the magnetic field amplitude not only helps in grasping how electromagnetic waves work but also illuminates how energy propagates through space in laser beams. Precise calculations allow for better manipulation and application in numerous optical technologies.
Energy Density
Energy density describes how much energy is stored in a given volume of space, crucial for analyzing laser beams. In the case of the helium-neon laser, energy density comprises contributions from both the electric field and magnetic field components. These are expressed as:
  • Electric field energy density: \( u_E = \frac{1}{2} \epsilon_0 E_0^2 \)
  • Magnetic field energy density: \( u_B = \frac{1}{2} \frac{B_0^2}{\mu_0} \)
Here, \( \epsilon_0 \) is vacuum permittivity, \( \mu_0 \) is the permeability of free space \( (4 \pi \times 10^{-7} \mathrm{N}/\mathrm{A}^2) \), and \( E_0 \) and \( B_0 \) are the amplitudes of the electric and magnetic fields, respectively.
The energy density tells us how much energy is available for work or to be transferred in the laser's path. By understanding these densities, you can evaluate how efficiently lasers can deliver energy to a specific point or process.
Helium-Neon Laser
A helium-neon laser, often abbreviated as He-Ne, is one of the most common types of gas lasers known for emitting a reddish-orange beam frequently utilized in research, engineering, and medical applications. These lasers operate by exciting a mixture of helium and neon gases within a glass tube, wherein the electrons transition between energy levels to produce coherent light.
Helium-neon lasers are valued for their:
  • Stable output wavelength, typically around 632.8 nanometers.
  • Low operating costs and ease of maintenance due to long operational life.
  • High-quality beams with minimal divergence and excellent coherence, ideal for interferometry and holography.
Understanding how helium-neon lasers function and their properties allows scientists and engineers to harness these beams effectively for precision tasks, ranging from precision metrology to high-resolution imaging.

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Most popular questions from this chapter

\(\bullet$$\bullet\) A block of glass has a polarizing angle of \(60.0^{\circ}\) for red light and \(70.0^{\circ}\) for blue light, for light traveling in air and reflecting from the glass. (a) What are the indexes of refraction for red light and for blue light? (b) For the same angle of incidence, which color is refracted more on entering the glass?

\(\bullet\) Threshold of vision. Under controlled darkened conditions in the laboratory, a light receptor cell on the retina of a person's eye can detect a single photon (more on photons in Chapter 28 ) of light of wavelength 505 \(\mathrm{nm}\) and having an energy of \(3.94 \times 10^{-19} \mathrm{J} .\) We shall assume that this energy is absorbed by a single cell during one period of the wave. Cells of this kind are called rods and have a diameter of approximately 0.0020 \(\mathrm{mm} .\) What is the intensity (in \(\mathrm{W} / \mathrm{m}^{2} )\) delivered to a rod?

\(\bullet\) The critical angle for total internal reflection at a liquid-air interface is \(42.5^{\circ} .\) (a) If a ray of light traveling in the liquid has an angle of incidence of \(35.0^{\circ}\) at the interface, what angle does the refracted ray in the air make with the normal? (b) If a ray of light traveling in air has an angle of incidence of \(35.0^{\circ}\) at the interface, what angle does the refracted ray in the liquid make with the normal?

\(\bullet\) The indices of refraction for violet light \((\lambda=400 \mathrm{nm})\) and red light \((\lambda=700 \mathrm{nm})\) in diamond are 2.46 and \(2.41,\) respectively. A ray of light traveling through air strikes the diamond surface at an angle of \(53.5^{\circ}\) to the normal. Calculate the angular separation between these two colors of light in the refracted ray.

\(\bullet\) A glass plate having parallel faces and a refractive index of 1.58 lies at the bottom of a liquid of refractive index \(1.70 . \mathrm{A}\) ray of light in the liquid strikes the top of the glass at an angle of incidence of \(62.0^{\circ} .\) Compute the angle of refraction of this light in the glass.

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