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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 the
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Answer #1

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

P=nE

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

E=\frac{6.626\times10^{-34}\times3\times10^8}{514.5\times10^{-9}}\ J=3.86\times10^{-19}\ 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

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