Question

He-Ne lasers are often used in physics demonstrations. They produce light of wavelength 633 nm and a power of 0.750 mW spread over a cylindrical beam 1.35 mm in diameter (although these quantities can vary).
Part A

What is the intensity of this laser beam?
I=___________W/m^2

Part B

What is the maximum value of the electric field?
Emax=__________V/m

Part C

What is the maximum value of the magnetic field?
Bmax=__________uT

Part D

What is the average energy density in the laser beam?
Uav=____________J/m^3

Answer 1

A) $524.0 W/m^2$

B) 0.531 V/m

C) $1.77\cdot 10^{-3} \mu T$

D) $2.5\cdot 10^{-12} J/m^3$

Explanation:

A)

The intensity of an electromagnetic wave is given by

$I=\frac{P}{A}$

where

I is the intensity

P is the power of the wave

A is the area over which the radiation is spread

In this problem:

$P=0.750 mW = 0.750\cdot 10^{-3}W$ is the power of the light emitted by the laser

This power spread over a cylindrical beam, so the area over which it spreads is the area of the base of the cylinder, which is the area of a circle:

$A=\pi (\frac{d}{2})^2$

where

$d=1.35 mm = 1.35\cdot 10^{-3} m$ is the diameter of the beam

Therefore, the intensity of the laser beam is:

$I=\frac{P}{\pi (\frac{d}{2})^2}=\frac{0.750\cdot 10^{-3}}{\pi (\frac{1.35\cdot 10^{-3}}{2})^2}=524.0 W/m^2$

B)

The relationship between power of an electromagnetic wave and maximum value of the electric field is the following:

$P=\epsilon_0 E^2 c$

where

P is the power

$\epsilon_0$ is the vacuum permittivity

$c$ is the speed of light

E is the maximum value of the electric field

In this problem we have:

$P=0.750 mW = 0.750\cdot 10^{-3}W$ is the power of the light emitted by the laser

$\epsilon_0 =8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

$c=3.0\cdot 10^8 m/s$ is the speed of light

Solving for E, we find the value of the electric field:

$E=\sqrt{\frac{P}{\epsilon_0 c}}=\sqrt{\frac{0.750\cdot 10^{-3}}{(8.85\cdot 10^{-12})(3.0\cdot 10^8)}}=0.531 V/m$

C)

For an electromagnetic wave, the relationship between amplitude of the electric field and amplitude of the magnetic field is:

$E=cB$

where

E is the maximum value of the electric field

c is the speed of light

B is the maximum value of the magnetic field

In this problem we have:

E = 0.531 V/m is the maximum value of the electric field

$c=3.0\cdot 10^8 m/s$ is the speed of light

Solving the equation for B, we find the maximum value of the magnetic field:

$B=\frac{E}{c}=\frac{0.531}{3.0\cdot 10^8}=1.77\cdot 10^{-9} T = 1.77\cdot 10^{-3} \mu T$

D)

The energy density of an electromagnetic wave is the amount of energy per unit volume of the wave.

The average energy density of an electromagnetic wave is related to the amplitude of the electric field by the equation:

$w=\epsilon_0 E^2$

where

w is the energy density

$\epsilon_0$ is the vacuum permittivity

E is the maximum value of the electric field

Here we have for this laser beam:

$\epsilon_0 =8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

E = 0.531 V/m is the maximum value of the electric field

Substituting, we find the average energy density of the laser beam:

$w=(8.85\cdot 10^{-12})(0.531)^2=2.5\cdot 10^{-12} J/m^3$