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He-Ne lasers are often used in physics demonstrations. They produce light of wavelength 633 nm and a power of 0.500 mW spread over a cylindrical beam 1.00 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) \( E_0 \approx 693 \text{ V/m}, B_0 \approx 2.31 \times 10^{-6} \text{ T} \); (c) \( u \approx 2.12 \times 10^{-6} \text{ J/m}^3 \)

Step by step solution

01

Calculate the Intensity

The intensity of the laser beam is defined as the power spread over the area. The formula for intensity \( I \) is given by: \[ I = \frac{P}{A} \]where \( P = 0.500 \text{ mW} = 0.500 \times 10^{-3} \text{ W} \) is the power and \( A \) is the cross-sectional area of the beam. First, calculate the area \( A \) of the circular beam:\[ A = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{1.00 \times 10^{-3}}{2} \right)^2 \]\[ A = \pi \times (0.5 \times 10^{-3})^2 = \pi \times 0.25 \times 10^{-6} \]\[ A \approx 7.85 \times 10^{-7} \text{ m}^2 \]Now, find the intensity:\[ I = \frac{0.500 \times 10^{-3}}{7.85 \times 10^{-7}} \]\[ I \approx 637 \text{ W/m}^2 \]
02

Find Maximum Electric Field Amplitude

The maximum value of the electric field \( E_0 \) is related to the intensity by the formula: \[ I = \frac{1}{2} c \varepsilon_0 E_0^2 \]Rearranging the formula to solve for \( E_0 \):\[ E_0 = \sqrt{\frac{2I}{c\varepsilon_0}} \]where \( c = 3.00 \times 10^8 \text{ m/s} \) is the speed of light and \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \) is the permittivity of free space.Substitute the known values:\[ E_0 = \sqrt{\frac{2 \times 637}{3.00 \times 10^8 \times 8.85 \times 10^{-12}}} \]\[ E_0 \approx 693 \text{ V/m} \]
03

Find Maximum Magnetic Field Amplitude

The maximum value of the magnetic field \( B_0 \) is related to the electric field by the formula:\[ B_0 = \frac{E_0}{c} \]Using the value of \( E_0 \) from the previous step:\[ B_0 = \frac{693}{3.00 \times 10^8} \]\[ B_0 \approx 2.31 \times 10^{-6} \text{ T} \]
04

Calculate the Average Energy Density

The average energy density \( u \) in the laser beam is given by:\[ u = \frac{I}{c} \]Substitute the intensity calculated in Step 1:\[ u = \frac{637}{3.00 \times 10^8} \]\[ u \approx 2.12 \times 10^{-6} \text{ J/m}^3 \]

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Key Concepts

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

Intensity Calculation
Intensity is an important concept when discussing light sources, as it tells us how much power is being emitted per unit area. For the He-Ne laser, the intensity can be calculated using the formula:
  • \[ I = \frac{P}{A} \]
Here, \( P \) is the power of the laser, and \( A \) is the area over which the power is distributed. In our case, the power \( P \) is given as 0.500 mW, which can be converted to watts for calculation purposes (1 mW = \( 10^{-3} \) W).
The area \( A \) of the cylindrical beam is determined by the formula for the area of a circle \( A = \pi r^2 \), where \( r \) is the radius of the beam. For a beam diameter of 1.00 mm, the radius is 0.5 mm or \( 0.5 \times 10^{-3} \) m. Substituting this into the equation gives:
  • \[ A = \pi \times (0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7} \text{ m}^2 \]
With these values, the intensity \( I \) is calculated as:
  • \[ I = \frac{0.500 \times 10^{-3}}{7.85 \times 10^{-7}} \approx 637 \text{ W/m}^2 \]
Electric Field Amplitude
The electric field amplitude, or maximum electric field \( E_0 \), describes how strong the electric field is in the laser at its peak. It can be calculated using the relationship between intensity and electric field given by the formula:
  • \[ I = \frac{1}{2} c \varepsilon_0 E_0^2 \]
Where \( c \) is the speed of light in a vacuum, approximately \( 3.00 \times 10^8 \text{ m/s} \), and \( \varepsilon_0 \) is the permittivity of free space, about \( 8.85 \times 10^{-12} \text{ F/m} \). By rearranging this equation, we solve for \( E_0 \):
  • \[ E_0 = \sqrt{\frac{2I}{c\varepsilon_0}} \]
Substituting the calculated intensity \( I \) and the constants \( c \) and \( \varepsilon_0 \) gives:
  • \[ E_0 = \sqrt{\frac{2 \times 637}{3.00 \times 10^8 \times 8.85 \times 10^{-12}}} \approx 693 \text{ V/m} \]
Magnetic Field Amplitude
The magnetic field amplitude, \( B_0 \), is another essential parameter of an electromagnetic wave such as a laser. It indicates the strength of the magnetic field at its maximum. The relationship between the electric field and the magnetic field in a wave is:
  • \[ B_0 = \frac{E_0}{c} \]
In this formula, \( c \) is the speed of light. \( E_0 \) from the previous calculation is used here to find \( B_0 \). Given \( E_0 \approx 693 \text{ V/m} \), the magnetic field amplitude \( B_0 \) can be calculated as:
  • \[ B_0 = \frac{693}{3.00 \times 10^8} \approx 2.31 \times 10^{-6} \text{ T} \]
This shows the magnetic field of the laser beam is relatively small due to the high speed of light.
Energy Density
The energy density \( u \) provides insight into how much energy is contained per unit volume in the wave. For electromagnetic waves like laser beams, the energy density is directly proportional to intensity and can be calculated as:
  • \[ u = \frac{I}{c} \]
Here, \( I \) is the intensity of the laser beam and \( c \) is the speed of light. Using the intensity calculated earlier, the energy density is determined by:
  • \[ u = \frac{637}{3.00 \times 10^8} \approx 2.12 \times 10^{-6} \text{ J/m}^3 \]
This formula shows that the energy density is small, which is typical for light beams. The low energy density indicates that, although lasers are very intense, the energy carried per cubic meter is relatively low.

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Most popular questions from this chapter

There are two categories of ultraviolet light. Ultraviolet A (UVA) has a wavelength ranging from 320 nm to 400 nm. It is necessary for the production of vitamin D. UVB, with a wavelength in vacuum between 280 nm and 320 nm, is more dangerous because it is much more likely to cause skin cancer. (a) Find the frequency ranges of UVA and UVB. (b) What are the ranges of the wave numbers for UVA and UVB?

Consider each of the electric- and magnetic-field orientations given next. In each case, what is the direction of propagation of the wave? (a) \(\vec{E}\) in the +\(x\)-direction, \(\vec{B}\) in the +\(y\)-direction; (b) \(\vec{E}\) in the -\(y\)-direction, \(\vec{B}\) in the +\(x\)-direction; (c) \(\vec{E}\) in the +\(z\)-direction, \(\vec{B}\) in the -\(x\)-direction; (d) \(\vec{E}\) in the +\(y\)-direction, \(\vec{B}\) in the -\(z\)-direction.

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.

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 intense light source radiates uniformly in all directions. At a distance of 5.0 m from the source, the radiation pressure on a perfectly absorbing surface is 9.0 \(\times\) 10\(^{-6}\) Pa. What is the total average power output of the source?

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