Laser Power Needed for Beam Visibility by Wavelength

[phosphoressential]

Hi All!
First real post. Real happy to be here.

This is something everyone thinks about at some point when dealing with what power you need to achieve a visual effect with a certain wavelength of light. What I would like to see is a graph of the milliwattage needed at any wavelength to achieve a "visible beam" in reasonably low lighting. I've never seen one, so I decided to go ahead and bang one out and get the idea going.

I would love any and all help to approve/improve/disprove any of the work i've done. Thanks!

First, my current results:

(Almost) Full Visible Spectrum


Zoomed for Resolution Blue to Yellow


The main two concepts I used were that our eyes are only so sensitive to a certain color, and only a certain amount of a certain color will escape the beam for us to perceive and see the beam.

I started with a plot of the human eye's sensitivity to different wavelengths (blue/right axis). Data was generated from http://www.cvrl.org/lumindex.htm, which purportedly gives the "CIE 'physiologically-relevant' luminous efficiency functions consistent with the Stockman & Sharpe cone fundamentals."

I then plotted the amount of light that will scatter from the beam when it is traveling through air (red/right axis). The true proportion of light that gets to our eyes from the beam is very small, so I normalized it to the visible spectrum in order to more easily graph it.

We then combine these effects by multiplication and normalize across the spectrum, and we get the purple (right axis) plot. This is the visibility of a light beam going through air (at a certain viewing angle) at different wavelengths. This shows us how different colors relate to each other at any certain power level. We now need to figure out what we mean by a "visible beam," and set a reference point. With the knowledge I have and some mild research, I decided to go with a standard 5mW 532nm green laser pointer. They are very common, considered safe for the general public with minimal precautions. They can also be seen reasonably well in a reasonably low lighting environment from most viewing angles.

To apply the sensitivity plot to the reference point and calculate the final plot, I took the reciprocal of the sensitivity, normalized it to the value at 530nm (close enough to 532), and multiplied it by 5, the power of our reference beam at 532. This produces the green plot (left axis), which supposedly shows the laser power needed for a beam to be as visible as a 5mW 532nm beam in the same conditions at any visible wavelength. From the limited research I've done, the numbers seem pretty reasonable. What do you think?


That's all I can think to say right now. What are other significant factors I should include? As far as I can think, LIV curves are inconsequential because we're talking about optical power. Feedback please!

Cheers! Happy lasing!
Nigel

 

[Encap]

There is no such chart of or graph of "beam visibility" for real reasons ---whatever you come up with is wrong as you do not comprehend fundamentally why a "beam " is visible at all. You are mixing apples and oranges -- leaving out the fundamental reality.
It is not a matter of output power or wavelength.
A "laser beam" is invisible in a vacuum regardless of wavelength or power,

I will try to steering you towards reality of laser beam visibility rather than imagination about same --here goes.

Beam visability to an observer is complicated real world phenomenon of light, eye sensitivity, and reflection from particle/aerosols in the air the details of which are not intuitively obvious.

At sea-level, one cubic inch (1 inch x 1 inch x 1 inch) (16.39 cm3) of "air" contains approximately 400 billion billion (4*1020) air molecules, each moving at about 1600 km/hr (1000 miles/hr), and colliding with other molecules and anything else they come into contact with about 5 billion times per second. This is the reason for "air pressure". The amount of particles in that air that can reflect a portion of a laser beam's light back to your eye determines if you can see it or not.

It all depends upon atmospheric conditions--a beam you can see extremely well in fog or area with high concentration of particulate matter in the air can be almost invisible in clean clear air

Laser beam visibility is highly dependent on ever changing atmospheric conditions and aerosols in the air.
You never actually see the laser beam --what you see is the reflections from particles in the air.

"In a vacuum, the laser beam itself would be invisible - regardless of power or color. As a laser beam passes through Earth's Atmosphere some of the photons encounter large airborne particles which reflect some of the light back to an observer. This only creates intermittent tiny bright flashes of light or "knots" in the beam - it is not why we can see the beam itself.

It is extremely small airborne particles called aerosols having a diameter significantly less than the wavelength of the light that causes the beam to become visible.

The effect of minute particles scattering light is called Rayleigh scattering and it's most noticeable effect is to turn the daytime sky blue. Rayleigh scattering causes photons to be scattered in a roughly spherical manner around these particles. Some of the light is scattered forward (in the direction of the beam), a lesser amount is scattered to the sides and about the same amount that is scattered forward is scattered backwards towards the light source. This backwards scattering is why the beam is more visible to people standing near the astronomer using it, than people standing some distance to the side. The more of these minute particles there are in the atmosphere, the mo re Rayleigh scattering there is."
From :RASC Calgary Centre - The Atmosphere, Astronomy and Green Lasers http://calgary.rasc.ca/atmosphere.htm

Really you need to understand aerosols in air --is very interesting actually--you need to know what they are locally and density of same to be able to figure laser beam visibility possibilities of one place compared to another place on the earths surface. Example: the most aerosol-laden air in the United States today pales in comparison to Asia. So laser beams are more visible there, generally speaking. Depending on the season and weather conditions, surges of aerosols can make their way into the atmosphere almost anywhere on Earth

See NASA Earth Observatory page for an excellent explanation of aerosols in air with lots of pictures and charts of the planetary distribution of aerosols : Aerosols: Tiny Particles, Big Impact : Feature Articles

 

[paul1598419]

^^^^^THIS^^^^^^^

Nice job, Encap. I can't think of anything to add to this. It is true that a laser in outer space is invisible even at a green wavelength and at many thousands of watts. That is because space is a vacuum and has no particulates or aerosols.

 

[phosphoressential]

There is no such chart of or graph of "beam visibility" for real reasons ---whatever you come up with is wrong as you do not comprehend fundamentally why a "beam " is visible at all. You are mixing apples and oranges -- leaving out the fundamental reality.
It is not a matter of output power or wavelength.
A "laser beam" is invisible in a vacuum regardless of wavelength or power,

I will try to steering you towards reality of laser beam visibility rather than imagination about same --here goes.

Beam visability to an observer is complicated real world phenomenon of light, eye sensitivity, and reflection from particle/aerosols in the air the details of which are not intuitively obvious.

At sea-level, one cubic inch (1 inch x 1 inch x 1 inch) (16.39 cm3) of "air" contains approximately 400 billion billion (4*1020) air molecules, each moving at about 1600 km/hr (1000 miles/hr), and colliding with other molecules and anything else they come into contact with about 5 billion times per second. This is the reason for "air pressure". The amount of particles in that air that can reflect a portion of a laser beam's light back to your eye determines if you can see it or not.

It all depends upon atmospheric conditions--a beam you can see extremely well in fog or area with high concentration of particulate matter in the air can be almost invisible in clean clear air

Laser beam visibility is highly dependent on ever changing atmospheric conditions and aerosols in the air.
You never actually see the laser beam --what you see is the reflections from particles in the air.

"In a vacuum, the laser beam itself would be invisible - regardless of power or color. As a laser beam passes through Earth's Atmosphere some of the photons encounter large airborne particles which reflect some of the light back to an observer. This only creates intermittent tiny bright flashes of light or "knots" in the beam - it is not why we can see the beam itself.

It is extremely small airborne particles called aerosols having a diameter significantly less than the wavelength of the light that causes the beam to become visible.

The effect of minute particles scattering light is called Rayleigh scattering and it's most noticeable effect is to turn the daytime sky blue. Rayleigh scattering causes photons to be scattered in a roughly spherical manner around these particles. Some of the light is scattered forward (in the direction of the beam), a lesser amount is scattered to the sides and about the same amount that is scattered forward is scattered backwards towards the light source. This backwards scattering is why the beam is more visible to people standing near the astronomer using it, than people standing some distance to the side. The more of these minute particles there are in the atmosphere, the mo re Rayleigh scattering there is."
From :RASC Calgary Centre - The Atmosphere, Astronomy and Green Lasers http://calgary.rasc.ca/atmosphere.htm

Really you need to understand aerosols in air --is very interesting actually--you need to know what they are locally and density of same to be able to figure laser beam visibility possibilities of one place compared to another place on the earths surface. Example: the most aerosol-laden air in the United States today pales in comparison to Asia. So laser beams are more visible there, generally speaking. Depending on the season and weather conditions, surges of aerosols can make their way into the atmosphere almost anywhere on Earth

See NASA Earth Observatory page for an excellent explanation of aerosols in air with lots of pictures and charts of the planetary distribution of aerosols : Aerosols: Tiny Particles, Big Impact : Feature Articles

Hi Encap!

I am, in fact, well aware of what you speak.

I then plotted the amount of light that will scatter from the beam when it is traveling through air (red/right axis).

As you can see by my graph, the red line is labeled the Rayleigh Scattering plot. This is what I described as the "amount of light scattering from the beam." The Rayleigh distribution and the visual spectrum sensitivity were combined with the reference of 5mW 532nm to create the power plot. Since my graph is referenced to however a 532nm 5mW beam of light looks in a given atmosphere, it is independent of atmospheric composition. If the conditions were great, you'd see the 532nm beam better, and you'd also see it's equivalents better if they were at the power shown by the graph for a certain wavelength.

In my spreadsheet I now have the refractive index, n, in the Rayleigh scattering function as a variable and as I change it, the scattering proportions change, but once it is normalized across the spectrum by dividing by the max value, the curve is the same. This proves that the relationship between scattering different wavelengths is independent of atmospheric conditions. The conditions at any moment can of course be changing constantly, but if an atmospheric condition is chosen, then the relationship between power, wavelength, and visibility in my graphs should be true.

Nigel

 

[CurtisOliver]

Good post Encap. I have nothing to add apart from this. You haven’t accounted for luminous efficacy properly. 532nm is not a peak in both photopic and scotopic conditions. Considering you are talking about dark lighting conditions then you are going to have to consider scotopic data instead. This gives a luminosity peak at around 507nm. Rayleigh scattering and the scotopic shift would favour the shorter wavelengths.

Edit: On a positive note. Good effort. It’s better to try and get it wrong than it is not to try at all.

 

[phosphoressential]

Good post Encap. I have nothing to add apart from this. You haven’t accounted for luminous efficacy properly. 532nm is not a peak in both photopic and scotopic conditions. Considering you are talking about dark lighting conditions then you are going to have to consider scotopic data instead. This gives a luminosity peak at around 507nm. Rayleigh scattering and the scotopic shift would favour the shorter wavelengths.

Edit: On a positive note. Good effort. It’s better to try and get it wrong than it is not to try at all.

Good point. I was trying to model low light, so I should've been using the scotopic response curve since the beginning. No hurt in having both! Here are the updated charts including scotopic eye sensitivity as darker colors. I'll remind again, Rayleigh scattering has been at the base of my model the whole time. It is the red line, which is favoring low wavelengths as you said.

 

[CurtisOliver]

What wavelength are you using for your peak of your photopic curve? (Required Power)
Because the peak is at 555nm for photopic vision. Also if incorporating photopic data a lot more power will be needed for a 'visible' beam even if facing towards the emitter.

 

[hakzaw1]

OP
I really enjoyed reading your 'intro' in the WELCOME/NEWCOMERS section.. I liked the part where you mentioned where you live.... we have some great members there. And lots here have the same hobbies as you!!!

;-(

 

[phosphoressential]

What wavelength are you using for your peak of your photopic curve? (Required Power)
Because the peak is at 555nm for photopic vision. Also if incorporating photopic data a lot more power will be needed for a 'visible' beam even if facing towards the emitter.

Hi CurtisOliver!
My photopic curve peaks at 555, and scotopic at 505 nm. Both the photopic and scotopic curves I am using are from the International Commission on Illumination (CIE), and are kindly hosted here by University College London.



hakzaw1, hey I'm getting there! I needed impetus to participate at all and this was it. I will be sure to formally introduce myself. Maybe that's why everyone seems so quick to judge lol

 

[hakzaw1]

well.. it is a bit rude to ask anyone for help and NOT introduce yourself first... and not just on this forum.. everywhere you go.. its considered to be a polite way to start out--agree?..


btw some here skipped the intro only to make it long after joining our friendly forum . better late than never-- be sure to use your location in the TITLE of your 'intro' like : Hello-- new member here from texas''.. etc
I have been here almost 10 years... & I would never 'steer' you wrong.. trust me (& us)

 

[CurtisOliver]

Hi CurtisOliver!
My photopic curve peaks at 555, and scotopic at 505 nm. Both the photopic and scotopic curves I am using are from the International Commission on Illumination (CIE), and are kindly hosted here by University College London.

View attachment 62399

hakzaw1, hey I'm getting there! I needed impetus to participate at all and this was it. I will be sure to formally introduce myself. Maybe that's why everyone seems so quick to judge lol

Your light blue line is correct and standard. It is your green photopic line that is favouring the wrong wavelength peak. This is also notable in your table.

 

[phosphoressential]

Your light blue line is correct and standard. It is your green photopic line that is favouring the wrong wavelength peak. This is also notable in your table.

CurtisOliver, I'm not quite understanding. I thought you said that it should be at 555 nm. My numbers from the CIE agree with that.