Optics | Energy Density, Irradiance

In summary: The equation I'm using is E=Iradiance*c*ε0/2Yes, you can use that equation to calculate the irradiance. Just plug in the values you calculated for energy and the speed of light in the medium (which you correctly calculated to be 2*108 m/s). You should end up with an irradiance of 1.57*1011 W/m2, which is slightly higher than the irradiance in free space. In summary, the pulse length, energy density, and irradiance for a laser pulse with duration 1 ns, diameter 1 cm, and total energy 1 mJ were calculated for both free space and a medium with refractive index n=1
  • #1
heycoa
75
0

Homework Statement


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.


Homework Equations


Irradiance = c*ε0/2 * E02
Energy Density = .5*ε0*E2 + .5*B20
Frequency=c/n
Wave number k=n*ω/c

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.

Please help me understand this. All your help and time is greatly appreciated!
Thank you
 
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  • #2
Hello heycoa,


heycoa said:

Homework Statement


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.


Homework Equations


Irradiance = c*ε0/2 * E02
Energy Density = .5*ε0*E2 + .5*B20
Frequency=c/n
Wave number k=n*ω/c

The Attempt at a Solution


I do not know how "energy density" and "total energy" differ and/or are related.

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.

heycoa said:
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.

Please help me understand this. All your help and time is greatly appreciated!
Thank you


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
heycoa said:

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.

Essentially, if you know the energy and divide it by the volume, you get energy density. And to get energy (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

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 transferred per second per squared metre, http://en.wikipedia.org/wiki/Intensity_(physics )).

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.
 
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  • #4
tthtlc said:
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

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
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
heycoa said:
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.

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).

heycoa said:
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?


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.

heycoa said:
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?

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).
 

1. What is energy density in optics?

Energy density in optics refers to the amount of energy per unit volume in a given area. It is typically measured in joules per cubic meter or watt-hours per cubic meter.

2. How is energy density related to irradiance?

Energy density and irradiance are closely related as they both measure the amount of energy present in a given area. However, irradiance specifically refers to the amount of energy per unit area, while energy density takes into account the volume of the area.

3. What is the formula for calculating energy density in optics?

The formula for calculating energy density in optics is: Energy Density = Energy / Volume. This means that the energy density is equal to the total amount of energy divided by the volume of the area in which it is present.

4. How does energy density affect the behavior of light?

Energy density plays a significant role in the behavior of light as it determines the intensity or brightness of the light. Higher energy density leads to a brighter light, while lower energy density results in a dimmer light.

5. What factors can affect energy density in optics?

The main factors that can affect energy density in optics include the intensity of the light source, the distance from the source, and any obstructions or absorption of light in the area. Additionally, the type of material the light is passing through can also impact energy density.

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