Homework Help: Optics | Energy Density, Irradiance

1. Feb 1, 2013

heycoa

1. The problem statement, all variables and given/known data
Suppose a laser produces a pulse of light with duration 1 ns, a diameter of 1 cm, and a total energy of 1 mJ. In free space, the pulse length is this 30 cm, the energy density is 42 J/m3, and the irradiance is 1.3*1010 W/m2. Calculate the length, energy density, and irradiance if the pulse with the same duration, diameter, and total energy is instead in a medium with refractive index n = 1.5. Compare the electric fields for the two cases.

2. Relevant equations
Energy Density = .5*ε0*E2 + .5*B20
Frequency=c/n
Wave number k=n*ω/c
3. The attempt at a solution
I do not know how "energy density" and "total energy" differ and/or are related. I also am having a difficult time distingushing the separation between free space and a medium with a refractive index. But perhaps my biggest struggle is how to put to use the length, pulse duration, and diameter. I guess this is a geometrical problem however, I just cannot understand how to utilize the measurements.

Thank you

2. Feb 1, 2013

cepheid

Staff Emeritus
Hello heycoa,

As the name suggests, energy density is the amount of energy per unit volume. If you consider the pulse to be a cylinder of length 30 cm and diameter 1 cm, and you calculate the volume of this cylinder, and then you divide the total pulse energy (1 mJ) by this volume, you'll get an energy density of 42 joules per cubic metre (they already did the calculation for you). Interesting random tidbit: pressure and energy density have the same units.

In a medium, light travels more slowly than it does in vacuum. If you think about it, what determined the pulse length was the pulse duration (how long the laser was on for) and the speed of light. The distance between the two ends of the pulse was just the distance that light could travel in that amount of time. In the medium, light travels more slowly, which means that it travels a shorter distance in the same amount of time, and the pulse is shorter.

3. Feb 1, 2013

tthtlc

Essentially, if you know the energy and divide it by the volume, you get energy density. And to get enery (in Joules), it is jut the power (in Watts) multiply by the duration (in secs).

And for free space vs medium with refractive index, the energy dissipation comes in the form of Lambert Beer law, which relate the change in light intensity as it passed from one medium to another medium. It is a classical approximation with lots of prerequisite conditions:

http://en.wikipedia.org/wiki/Beer–Lambert_law

Energy of light, using the photon calculation, for example:

http://www.chemteam.info/Electrons/calc-energy-freq-wavelength.html [Broken]

can be estimated if the frequency/wavelength is known. But this is only for a photon, not intensity. (by definition, intensity is rate of energy transfered per second per squared metre, http://en.wikipedia.org/wiki/Intensity_(physics [Broken])).

As it passed from one medium to another medium, since the change in intensity can be related to refractive index (via Lambert Law), u can calculate the new rate of energy transferred, and thus the "energy density" (which energy tranferred per volume).

Hope you can continue the rest of the calculation.

Last edited by a moderator: May 6, 2017
4. Feb 2, 2013

cepheid

Staff Emeritus
It's much simpler than all this. The original post says to assume the pulse has the same energy and duration as the original one.

5. Feb 2, 2013

heycoa

So I believe I've done it correctly. I calculated the speed of light in the index of 1.5 to be 2*108 m/s. That times a duration of 1 ns = .2 m (which makes sense as it is shorter than that of free space), and the volume of our "light cylinder" is pi*(.0052)*.2 = 1.57*10-5 m3. From the info already given, I calculated the volume of the light cylinder in free space to be 2.36*10-5 m3, and that volume times the given energy density of 42J/m3 gives a total energy of 9.9*10-4 J. Taking that total energy and dividing by our light volume in a medium I get an energy density of 63 J/m3, which is greater than free space but that makes sense as the cylinder is shorter.

That is most of the problem done, however I need help on calculating the irradiance. Do I just use the equation I mentioned in my original post and use the calculated total energy (9.9*10-4 J) squared for E and solve for irradiance? Or is that incorrect, do I use my energy density in the medium?

Also, the question asks me to compare the electric fields for the two cases and I am not sure what the difference between the two is. Is it simply that in free space the electric field propagates further? Or is there a quantitative result that I can answer with?

6. Feb 2, 2013

cepheid

Staff Emeritus
Well this is correct, although you didn't have to recalculate the energy of the pulse, it's already given in the problem as 1 mJ, and all you did was reproduce that result (with some rounding error).

Remember that irradiance is power per unit cross-sectional area (of the beam). So you can get the irradiance directly from the energy density by considering the geometry of the situation. This cylinder is sliding past you, the observer, and it has a certain energy per unit volume. A certain volume passes you per unit time, which means that a certain amount of energy passes across the perpendicular area of the beam per unit time.

I think it wants you to compare the quantitative values of the two fields. Remember that the irradiance is the magnitude of the time-averaged Poynting vector, and that this depends on E in a certain way. I imagine that you would have to figure out the permittivity ε of the medium (as opposed to for free space, ε0).