Question

A small He-Ne laser produces a 632.8 nm beam with an average power of 3.5 mW and diameter of 2.4mm

a) How many photons per second are emitted by the laser?
b) What is the amplitude of the electric field inside the beam? (Hint: You should consult your EP book, the internet, or equation 3.10 for this problem)
c) How does this electric field compare with one produced by a 100W light bulb at a distance of 1 meter?

Answer 1

a)

Energy of each photon where is the wavelength

E = (4.14 x 10^-15)/(632.8 x 10^-9) = 6.542 x 10^-9 J = 6.542 nJ

Power = energy per unit time (second)

Number of photons emitted per second = power/Energy per photon = 3.5 mW/6.542 nJ = 535004.585 photons/second

b)

The total area of cross section of the beam = 3.14 x (2.4mm/2)^2 = 4.5216 x 10^-6 m^2

in 1 second, the wave forms a cylinder of length 3 x 10^8 m.

Volume of this cylinder dV = 4.5216 x 10^-6 x 3 x 10^8 m^3 = 1356.48 m^3

Energy contained in this cylinder U = (1/2) x x (E^2) x dV ------(1)

where E is the amplitude of the Electric field

From the power of the wave, we know that the total energy contained in this cylinder = 3.5 mJ

Substituting this in equation 1,

E^2 = 3.5 x (10^-3) x 2 /( x 1356.48) = 0.58309 x 10^-6 -----(2)

E = 0.76 x(10^-3) (N/C)

c)

Energy of a 100 W bulb is distributed over a whole sphere of area 4 x x r^2.

The power available per unit area for a sphere of 1 m = 100/((4 x3.14 x (1^2))

=7.95 W/(m^2)

Assuming this energy to be a laser with a diameter of 2.4 mm, with the same diameter, but starting 1 m away from the bulb, the power contained in this laser would be

= 7.95 x 4.5216 x 10^-6 = 35946.72 x 10^-3 W

Substituting this value in (2) and then taking square root, we know that the Electric field from the bulb is approx, 100 times stronger than the laser.