A small helium-neon laser produces a light beam with an average power of 3.9 mW

Question

A small helium-neon laser produces a light beam with an average power of 3.9 mW and a diameter of 2.8 mm. (Assume the wavelength of light is 632.8 nm.)

(a) How many photons per second are emitted by the laser?
................... photons/s

(b) What is the amplitude of the electric field of the light wave? Compare this result with the electric field at a distance of 1 m from an incandescent light bulb that emits 100 W of visible light.

Explanation / Answer

Ephoton = h c / lambda

= (6.626 * 10-34 * 3 * 108) / (632.8 * 10-9) = 3.14 * 10-19 J

P = E / delta T = 3.9 * 10-3 W

Pavg / Ephoton = 3.9 * 10-3 / 3.14 * 10-19

= 1.24 * 1016 photons/s

-------------------------------------------------

I = P / A = 3.9 * 10-3 / pi r2 = 3.9 * 10-3 / (pi (2.8 mm/2)2) = 633.37 W/m2

E = sqrt [2 I / c E] = sqrt [2 * 633.37 / 3 * 108 * 8.85 * 10-12]

= 690.74 V/m

I = 100 / 4 pi * 12 = 7.96 W/m2

E = sqrt [2 I / c E] = sqrt [2 * 7.96 / 3 * 108 * 8.85 * 10-12]

= 77.44 V/m

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