Question

1. (3 points) Consider a beam of light of wavelength 514.5 nm emitted by a laser. The diameter of the beam is 2.00 mm and there are 1.30 X 1018 photons emitted per second. a) Find the peak electric field for the electromagnetic wave that constitutes the beam. b) Find the peak magnetic field for the electromagnetic wave that constitutes the beam. (Hint: Recall the average intensity of electromagnetic waves. Also remember that intensity is power per unit area.)

Answer 1

If n photons are emitted per second then power of the laser is

п.

E is the energy of each photon

Each photon has a wavelength given by

$\lambda=514.5\ nm=514.5\times10^{-9}\ m$

So, energy of each photon is

$E=\frac{hc}{\lambda}$

h is the Planck's constant and c is the velocity of light

Putting the values

6.626 x 1034 x 3 x 10 E = 514.5 x 10-9 J 3.86 x 1019 J

So, power of the laser is

$P=1.3\times10^{18}\times3.86\times10^{-19}\ W=0.502\ W$

Diameter of the beam is 2 mm

So, radius of the beam is

$r=\frac{2}{2}\ mm=1\ mm=0.001\ m$

So, intensity o the laser beam is

$I=\frac{P}{\pi r^2}=\frac{0.502}{\pi\times0.001^2}\ W/m^2=1.6\times10^5\ W/m^2$

a) We know average intensity corresponding to an electromagnetic wave of peak electric field is given by

$I=\frac{1}{2}\epsilon_0cE^2$

$\epsilon_0$ is the permittivity of free space

Putting the values,

$1.6\times10^5=\frac{1}{2}\times8.85\times10^{-12}\times3\times10^8\times E^2$

$or,\ E=\sqrt{\frac{2\times1.6\times10^5}{8.85\times10^{-12}\times3\times10^8}}\ N/C=10978.5\ N/C$

This is the peak electric field.

b) If corresponding to the above electric field E we have a peak magnetic filed B then we must have from the property of EM wave that

$c=\frac{E}{B}$

$or,\ B=\frac{E}{c}=\frac{10978.5}{3\times10^8}\ T=3.66\times10^{-5}\ T=36.6\ \mu T$