question about photon densities

[Fonduman]

I was just wondering, if 2 lasers of the same power rating but of different wavelengths were being used, wouldnt the laser of shorter wavelength have a lower photon density in the beam? could this contribute to visibility or lack of? I guess it cant be too big an effect seeing as red isnt very visible.

 

[Meatball]

I was just wondering, if 2 lasers of the same power rating but of different wavelengths were being used, wouldnt the laser of shorter wavelength have a lower photon density in the beam? could this contribute to visibility or lack of? I guess it cant be too big an effect seeing as red isnt very visible.

I have not heard of photon density as you put it.

What you are referring to are two different things.

A violet laser, and a red laser are placed next to each other. The red laser is pointed at an LPM and the reading is 5.000mws. Next, the violet laser is measured to be 5.000mws.

Each laser is capable of doing the same amount of "work" as the other. The difference, is that each violet photon carries with it, more energy than any red photon.

So yes, 5mws is 5mws. The violet laser has fewer photons, or energy "packets" than the red laser has.

Visibility is something completely different. Our eyes pick up the color spectrum in a rough sort of bell curve from violet to red.

5mws of green in a sense, will appear MUCH brighter than 5mws of violet, red, or blue etc.

Hope this makes sense for ya..

 

[Moptsp]

I have not heard of photon density as you put it.

What you are referring to are two different things.

A violet laser, and a red laser are placed next to each other. The red laser is pointed at an LPM and the reading is 5.000mws. Next, the violet laser is measured to be 5.000mws.

Each laser is capable of doing the same amount of "work" as the other. The difference, is that each violet photon carries with it, more energy than any red photon.

So yes, 5mws is 5mws. The violet laser has fewer photons, or energy "packets" than the red laser has.

Visibility is something completely different. Our eyes pick up the color spectrum in a rough sort of bell curve from violet to red.

5mws of green in a sense, will appear MUCH brighter than 5mws of violet, red, or blue etc.

Hope this makes sense for ya..

Is that backwards? Shorter wavelength means higher frequency to keep speed constant. So I thought this would mean that violet is of higher energy than red. I could be wrong, so please correct me if so, Meatball.

Like gamma rays are on the far side from violet in the short wavelengths and red is in the higher wavelengths like radio waves.

Though I still think I'm mixing up a concept.

*ps: First time I have posted here in months! xD*

EDIT:
Oh, I just re-read your post. So if the photon is of higher energy, there are less of them? If so, how does this ratio work?

 

[RA_pierce]

I think what he is saying is that a short wavelength laser has fewer photons per milliWatt than a long wavelength laser, because each short wavelength photon carries more energy.

Is that backwards? Shorter wavelength means higher frequency to keep speed constant. So I thought this would mean that violet is of higher energy than red. I could be wrong, so please correct me if so, Meatball.

Like gamma rays are on the far side from violet in the short wavelengths and red is in the higher wavelengths like radio waves.

Though I still think I'm mixing up a concept.

*ps: First time I have posted here in months! xD*

EDIT:
Oh, I just re-read your post. So if the photon is of higher energy, there are less of them? If so, how does this ratio work?

 

[Moptsp]

I think what he is saying is that a short wavelength laser has fewer photons per milliWatt than a long wavelength laser, because each short wavelength photon carries more energy.

Thank you, RA_pierce! That makes sense!

 

[Fonduman]

yeah I understand all that. by photon density I meant photons per unit volume. I was just wondering if less photons in a higher energy beam will affect visibility at all. actually, isnt the curve skewed a bit to the right? this would explain it possibly.

 

[Meatball]

The bell curve? It is skewed to the right for daytime visibility, and it is skewed to the left for nighttime visibility. For normal visibility situations, a slight skew is not going to matter very much since visibility is relative to each and every person that views the beam/dot.

Having less photons in a higher energy beam will not directly affect visibility at all. You must remember that photons are just another way to think about light. I usually think of light as a wave, before I think of it as a particle carrying a defined amount of energy.

500mws of green is MUCH more visible, than 500mw of red. Which beam has less photons?

 

[Fonduman]

EDIT:
Oh, I just re-read your post. So if the photon is of higher energy, there are less of them? If so, how does this ratio work?

in terms of ratios, E=hf, f=v/wavelength
so E=hv/wavelength, or E=k/wavelength with k=hv
means photon energy is inversely proportional to wavelength. 400nm photons have twice as much energy as 800nm ones therefore?
and the number of photons fired out (over a course of a second for example) will be, for 500mw for example, 500(Joules)/photon energy (in joules)
ofc joules can be converted.
for one sec
power/E=(power*wavelength)/k=wavelength*C (C=power/k)
so photons per sec fired is directly proportional to wavelength i think.
not an overly difficult calculation overall, might look a little complicated in this format for some but easy to think through.
correct me if I've messed something up lol.

 

[Fonduman]

500mws of green is MUCH more visible, than 500mw of red. Which beam have less photons?

I never said I thought it was what caused all visibility differences, just that it might be something that affects it to a degree. thanks for considering anyway

 

[Moptsp]

in terms of ratios, E=hf, f=v/wavelength
so E=hv/wavelength, or E=k/wavelength with k=hv
means photon energy is inversely proportional to wavelength. 400nm photons have twice as much energy as 800nm ones therefore?
and the number of photons fired out (over a course of a second for example) will be, for 500mw for example, 500(Joules)/photon energy (in joules)
ofc joules can be converted.
for one sec
power/E=(power*wavelength)/k=wavelength*C (C=power/k)
so photons per sec fired is directly proportional to wavelength i think.
not an overly difficult calculation overall, might look a little complicated in this format for some but easy to think through.
correct me if I've messed something up lol.

Ah. Ok, thanks for the info.

 

[Cyparagon]

The bell curve?

 

[chipdouglas]

WiN

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