Problem 29 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

(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 1: 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:\[ I = \frac{0.500 \times 10^{-3}}{7.85 \times 10^{-7}} \]\[ I \approx 637 \text{ W/m}^2 \]

Step 2: 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:\[ E_0 = \sqrt{\frac{2 \times 637}{3.00 \times 10^8 \times 8.85 \times 10^{-12}}} \]\[ E_0 \approx 693 \text{ V/m} \]

Step 3: 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:\[ B_0 = \frac{693}{3.00 \times 10^8} \]\[ B_0 \approx 2.31 \times 10^{-6} \text{ T} \]

Step 4: 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:\[ u = \frac{637}{3.00 \times 10^8} \]\[ u \approx 2.12 \times 10^{-6} \text{ J/m}^3 \]

Key Concepts

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.