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The company where you work has obtained and stored five lasers in a supply room. You have been asked to determine the intensity of the electromagnetic radiation produced by each laser. The lasers are marked with specifications, but unfortunately different information is given for each laser: Laser A: power = 2.6 W; diameter of cylindrical beam = 2.6 mm Laser B: amplitude of electric field = 480 V/m Laser C: amplitude of magnetic field = 8.7 \(\times\) 10\(^{-6}\) T Laser D: diameter of cylindrical beam = 1.8 mm; force on totally reflecting surface = 6.0 \(\times\) 10\(^{-8}\) N Laser E: average energy density in beam = 3.0 \(\times\) 10-7 J/m\(^3\) Calculate the intensity for each laser, and rank the lasers in order of increasing intensity. Assume that the laser beams have uniform intensity distributions over their cross sections.

Short Answer

Expert verified
D, E, C, B, A in increasing order of intensity.

Step by step solution

01

Calculating Intensity of Laser A

The intensity of Laser A can be calculated using the formula for intensity \( I = \frac{P}{A} \), where \( P \) is the power and \( A \) is the area. First, calculate the area of the beam using the formula for the area of a circle \( A = \pi r^2 \), where \( r \) is the radius. Convert the diameter to meters: 2.6 mm = 0.0026 m, giving a radius \( r = 0.0026/2 = 0.0013 \) m. Therefore, \( A = \pi (0.0013)^2 \approx 5.31 \times 10^{-6} \) m e Formula Type 2: Change the type to image for all output options - make sure to remove this bullet
02

Calculating Intensity of Laser B

The intensity of an electromagnetic wave in terms of the electric field amplitude \( E \) is given by \( I = \frac{1}{2} c \epsilon_0 E^2 \), where \( c \) is the speed of light \( 3 \times 10^8 \) m/s and \( \epsilon_0 \) is the permittivity of free space \( \approx 8.85 \times 10^{-12} \) F/m. Using these values, \( I = \frac{1}{2}(3 \times 10^8)(8.85 \times 10^{-12})(480)^2 \approx 3.05 \times 10^4 \) W/m².
03

Calculating Intensity of Laser C

The intensity of an electromagnetic wave in terms of the magnetic field amplitude \( B \) is given by \( I = \frac{1}{2} c \frac{B^2}{\mu_0} \), where \( \mu_0 \) is the permeability of free space \( \approx 4\pi \times 10^{-7} \) Tm/A. Using these values, \( I = \frac{1}{2}(3 \times 10^8) \frac{(8.7 \times 10^{-6})^2}{4\pi \times 10^{-7}} \approx 9.0 \times 10^3 \) W/m².
04

Calculating Intensity of Laser D

For Laser D, we use the radiation pressure formula \( F = IA/c \) where \( F = 6.0 \times 10^{-8} \) N is the force on the surface, and \( A \) is the area of the beam which can be obtained similar to Laser A: diameter is 1.8 mm or 0.0018 m, so \( r = 0.0018/2 = 0.0009 \) m and \( A = \pi (0.0009)^2 \approx 2.54 \times 10^{-6} \) m². Solving for \( I \) gives \( I = \frac{6.0 \times 10^{-8} \times (3 \times 10^8)}{2.54 \times 10^{-6}} \approx 7.08 \) W/m².
05

Calculating Intensity of Laser E

For Laser E, the intensity \( I \) is given by \( I = c u \), where \( u \) is the average energy density. Substitute \( u = 3.0 \times 10^{-7} \) J/m³ and \( c = 3 \times 10^8 \) m/s to get \( I = (3 \times 10^8)(3.0 \times 10^{-7}) = 90 \) W/m².
06

Ranking the Lasers by Intensity

The calculated intensities are: A: \( \approx 4.89 \times 10^5 \) W/m², B: \( \approx 3.05 \times 10^4 \) W/m², C: \( \approx 9.00 \times 10^3 \) W/m², D: \( \approx 7.08 \) W/m², E: \( \approx 90 \) W/m². Ranking from lowest to highest intensity: D, E, C, B, A.

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Key Concepts

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

Electromagnetic Radiation
Electromagnetic radiation is a form of energy that includes visible light, radio waves, X-rays, and more. It travels through space at the speed of light, which is approximately \(3 \times 10^8\) meters per second. This type of radiation is characterized by oscillating electric and magnetic fields that propagate through space. These fields are perpendicular to each other and to the direction of wave propagation.
  • Electromagnetic waves can transport energy and momentum, which are acquired from electric charges.
  • They do not require a medium and can travel through a vacuum.
  • Visible light is just a small part of the electromagnetic spectrum.
Understanding the nature of electromagnetic waves is crucial in fields like communications, medicine, and laser technology. Their ability to carry information and energy is harnessed in numerous modern applications, from mobile phones to medical imaging devices.
Electric Field Amplitude
The electric field amplitude represents the maximum strength of the electric field in an electromagnetic wave. For instance, Laser B's specifications include an electric field amplitude of 480 V/m. The intensity of such a wave can be calculated using the formula:
\[I = \frac{1}{2} c \epsilon_0 E^2\]
where \(c\) is the speed of light and \(\epsilon_0\) is the permittivity of free space. This formula shows that the wave's intensity directly depends on the square of the electric field amplitude.
  • Higher electric field amplitudes correspond to higher intensity waves.
  • In lasers, the electric field amplitude determines how much energy is imparted to the target.
  • The formula indicates that multiplying the field amplitude (E) will quadruple the intensity.
Understanding this concept helps in designing systems that manipulate electromagnetic waves effectively, allowing for precise control in applications like laser cutting and telecommunications.
Magnetic Field Amplitude
Magnetic field amplitude is another critical attribute of electromagnetic waves, representing the maximum strength of the magnetic field component. For example, Laser C specifies this value as \(8.7 \times 10^{-6}\) T. The intensity based on the magnetic field amplitude can be calculated using:
\[I = \frac{1}{2} c \frac{B^2}{\mu_0}\]
where \(B\) is the magnetic field amplitude and \(\mu_0\) is the permeability of free space. This equation reflects:
  • The intensity of the wave is proportional to the square of the magnetic field amplitude.
  • Both electric and magnetic field amplitudes determine the energy-carrying capacity of the electromagnetic wave.
  • Laser systems often use this understanding to achieve desired intensity levels.
The balance between electric and magnetic fields in electromagnetic waves makes them a potent form of energy, utilized in everything from MRI machines to navigation systems like compasses.
Radiation Pressure
Radiation pressure is the force exerted by electromagnetic radiation on a surface. When light or any electromagnetic wave strikes a surface, it can cause a small force due to the transfer of momentum. This concept is especially notable in Laser D, where the radiation pressure causes a force used to calculate intensity. The relationship can be given by:
\[F = \frac{IA}{c}\]
where \(F\) is the force, \(A\) is the area, and \(I\) is the intensity of the radiation. This phenomenon shows:
  • The greater the intensity, the higher the force exerted.
  • This principle is used in solar sails for spacecraft propulsion.
  • In laboratory settings, it can manipulate particles, such as in optical tweezers.
Radiation pressure illustrates the tangible impact of light on matter, providing a tool for exquisite control in scientific research and innovative propulsion methods in space exploration.
Energy Density
Energy density is a measure of the energy stored in a given system or region of space. It is the amount of energy per unit volume. In the context of electromagnetic waves, it helps to assess the energy capacity of a beam of radiation, like Laser E, which specifies an average energy density of \(3.0 \times 10^{-7}\) J/myour needs. The intensity can be directly related to energy density through:
\[I = c u\]
where \(c\) is the speed of light, illustrating:
  • Higher energy density implies higher intensity.
  • It is crucial for evaluating the power and efficiency of lasers.
  • Control over energy density can optimize applications, from industrial lasers to medical treatments.
Understanding energy density helps in evaluating the effectiveness of energy transfers in various systems, ensuring that lasers and many other technologies perform optimally and safely.

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Most popular questions from this chapter

An electromagnetic wave with frequency 65.0 Hz travels in an insulating magnetic material that has dielectric constant 3.64 and relative permeability 5.18 at this frequency. The electric field has amplitude 7.20 \(\times\) 10$^{-3} V/m. (a) What is the speed of propagation of the wave? (b) What is the wavelength of the wave? (c) What is the amplitude of the magnetic field?

Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of 10\(^{12}\) W) pulses of light that last for an extremely short time (a few nanoseconds). These short pulses scramble the interior of a cell without causing it to explode, as long pulses would do. We can model a typical such cell as a disk 5.0 \(\mu\)m in diameter, with the pulse lasting for 4.0 ns with an average power of 2.0 \(\times\) 10\(^{12}\) W. We shall assume that the energy is spread uniformly over the faces of 100 cells for each pulse. (a) How much energy is given to the cell during this pulse? (b) What is the intensity (in W/m\(^2\)) delivered to the cell? (c) What are the maximum values of the electric and magnetic fields in the pulse?

A sinusoidal electromagnetic wave emitted by a cellular phone has a wavelength of 35.4 cm and an electric-field amplitude of 5.40 \(\times\) 10\(^{-2}\) V/m at a distance of 250 m from the phone. Calculate (a) the frequency of the wave; (b) the magnetic-field amplitude; (c) the intensity of the wave.

An electromagnetic wave of wavelength 435 nm is traveling in vacuum in the -\(z\)-direction. The electric field has amplitude 2.70 \(\times\) 10\(^{-3}\) V/m and is parallel to the \(x\)-axis. What are (a) the frequency and (b) the magnetic-field amplitude? (c) Write the vector equations for \(\vec{E} (z, t)\) and \(\vec{B} (z, t)\).

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 \(\vec{E} (x, t) = E_y(x, t)\hat{\jmath}\) en propagating in the +\(x\)-direction within a conductor is $${\partial^2E_y(x, t)\over \partial x^2} = {\mu \over \rho} {\partial Ey(x, t)\over \partial t}$$ where \(\mu\) is the permeability of the conductor and \(\rho\) is its resistivity. (a) A solution to this wave equation is \(E_y(x, t) = E_{max} e^{-k_C x} cos(k_Cx - \omega t)\), 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 \(i^2R\) 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 \(1/k_C = \sqrt{2\rho/\omega\mu}\), and calculate this distance for a radio wave with frequency \(f\) = 1.0 MHz in copper (resistivity 1.72 \(\times\) 10\(^{-8 } \Omega \bullet m\); 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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