A small helium-neon laser emits red visible light with a...

A small helium-neon laser emits red visible light with a power of 3.20 $\mathrm{mW}$ in a beam that has a diameter of 2.50 $\mathrm{mm}$ . (a) What are the amplitudes of the electric and magnetic fields of the light? (b) What are the average energy densities associated with the electric field and with the magnetic field? (c) What is the total energy contained in a $1.00-\mathrm{m}$ length of the beam?

Transcript

00:01 Hello students in this question we have helium neon laser helium neon laser that emits red red visible light with a power p equals to 3 .2 millivot okay in a beam having diameter d it is equals to the 2 .5 mm okay so for the part a we have to find the amplitude of electric field and magnetic field of the light so now we can write that the laser beam intensity i will be given by power p divided by area that is pi by four multiplied by the diameter d square okay so it can be written as four multiplied by this four will become in the numerator and power is 3 .2 milli watt that is 10 to the power minus 3 watt divided by pi and diameter is 2 .5 mmm that is 10 to the power minus 3 meter and this whole square so intensity will be equals to the 652 watt per meter square we can write that the intensity i is also equals to the 1 by 2 epsilon not multiplied by c multiplied by e0 square where e0 is amplitude of electric field so e not from here will be 2 i divided by c epsilon not and this under root okay now we can substitute all the values okay so two multiplied by intensity is 652 watt per meter is square and divided by c which is speed of light that is 3 into 10 to the power 8 meter per second but glad by epsilon knot is 8 .8 .mere by 10 to the power minus 12 per me per meter okay so after solving we obtain that amplitude of the electric field e0 is 70 1 volt per meter okay this is the answer for the amplitude of electric field and magnetic field amplitude will be equals to the e0 divided by the c so 701 volt per meter divided by c which is 3 multiplied by 10 to the power 8 meter per second so amplitude of the magnetic field b not will be equals to the 2 .34 multiplicity by 10 to the power minus 6 tesla.

02:32 So these are the answers for the part a of the problem.

02:37 Now moving to next part b in which we have to determine the average energy density, you average energy density associated with the electric field as well as the magnetic field...

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