An instructor’s laser pointer produces a beam of light with a circular cross section of diameter 0.900 mm and a total power output of 4.50 mW. The beam size stays nearly constant when the instructor uses the laser pointer in the classroom. What is the amplitude E 0 of the electric field of the laser’s light?

 = 2309 V/M

Explanation:

The average power emitted by the electromagnetic wave is:

where:

S = intensity of the electromagnetic wave

A = cross sectional area of the beam

Replacing A with  and making S the subject of the formula; we have:

Intensity of electromagnetic wave in terms of electric field is;

where is the maximum value of the electric field

Making   the subject of the formula; we have:

 =

 = 2309 V/M

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