Powers of Laser Beams
Most recent answer: 04/04/2011
Q:
I know that red, green and blue lasers have different properties and that blue and green are the strongest. I'm thinking about testing color (wavelengths) and how it affects the "potency" of the beam (how strong it is, how far it shows). I dont't want to waste hundreds of dollars on a 500mW laser but at the same time I want the difference in strength to be measurable. My questions are:
1. At what point (in mW)do the different laser's strengths become easily measurable?
2. What is he best way to measure the power (for distance I'll just compare with landmarks, but for strength should I shine it on a thermometer? any suggestions?)?
3. How much would all the lasers cost? (must all be same mW (I couldn't find any red lasers past 5mW until I hit 500 or some other rediculous number mW))
Please respond ASAP. It would also be nice to include a website where all the lasers could be found.
- Sam O (age 13)
Ashland, Or, US:
- Sam O (age 13)
Ashland, Or, US:
A:
The key thing here is to figure out what it is you're actually trying to measure. Words like "potency" and "strength" don't have very specific meanings.
If what you're trying to find is how much energy the beam delivers per second (say how much it can heat up some little target) then all you need to know is the power. 100 mW means 100 milliJoules/second, etc. The color doesn't matter.
If what you're interested in is how visible the beam is, then your question is really more about your eyes than about the laser. How much light energy does it take for you to see at different frequencies? That involves both the background light and your eyes. In the simplest case, maybe all the background light is absent. Then all that counts is the dark-adapted sensitivity of your eyes to different colors. You can read about that: , .
Notice that you'll get different results for the comparative "potency" of different colors if you happen to have some form of color-blindness, or are interested in the visibility for some other animal, or want to know the effectiveness for driving photosynthesis, etc.
Mike W.
If what you're trying to find is how much energy the beam delivers per second (say how much it can heat up some little target) then all you need to know is the power. 100 mW means 100 milliJoules/second, etc. The color doesn't matter.
If what you're interested in is how visible the beam is, then your question is really more about your eyes than about the laser. How much light energy does it take for you to see at different frequencies? That involves both the background light and your eyes. In the simplest case, maybe all the background light is absent. Then all that counts is the dark-adapted sensitivity of your eyes to different colors. You can read about that: , .
Notice that you'll get different results for the comparative "potency" of different colors if you happen to have some form of color-blindness, or are interested in the visibility for some other animal, or want to know the effectiveness for driving photosynthesis, etc.
Mike W.
(published on 04/04/2011)
Follow-Up #1: different color lasers
Q:
So is there any significant difference in laser color? I'm talking about wavelengths (Red lasers wavelengths average at about 671 nanometers, green at 532, blue at 473 and violet beams at only 405) and whether their size has any affect. Also, what I'm planning to test is:
1. Energy delivered (That small target idea sounds nice, so should I point it to a thermometer or if not what?)
2. Distance (like I said, I would use landmarks)
3. Visability (Since I want to make this as accurate as possible, are there any ways I can measure this electronically?)
Thanks
-Sam O
- Sam O (age 13)
Ashland, Oregon, US
- Sam O (age 13)
Ashland, Oregon, US
A:
I'm still not sure exactly what the issues are here, but will try to respond to what I think you're asking.
1. The test to see how much a little target warms up is just a way to compare the powers of the different lasers. If the manufacturers have correctly labeled their output powers, it will not provide new information. If you wish to do this as a check, it might work to take a small test tube filled with a mixture of all sorts of food dyes, to make sure it does a good job of absorbing all wavelengths. You then need a very small thermometer. Perhaps a little diode thermometer measured with a little digital voltmeter would work at low cost.
2. All the lasers will send light an infinite distance, if nothing gets in the way. Some have beams which spread out more (have more divergence) and you can measure that by measuring the spot size as the target is set farther away. You can also try using lenses to reduce the spot size. You'll find that some lasers can be focused better on distant spots than others.
3. "Visibility" depends on who is viewing. If you were to build an automated system, you'd still have to calibrate it by comparing to how visible you thought the beam was. You could try to see how far away you could put a white sheet of paper and still see the laser spot. The answer will depend on things like how much other light is around, whether you have any color-blindness, etc.
Mike W.
1. The test to see how much a little target warms up is just a way to compare the powers of the different lasers. If the manufacturers have correctly labeled their output powers, it will not provide new information. If you wish to do this as a check, it might work to take a small test tube filled with a mixture of all sorts of food dyes, to make sure it does a good job of absorbing all wavelengths. You then need a very small thermometer. Perhaps a little diode thermometer measured with a little digital voltmeter would work at low cost.
2. All the lasers will send light an infinite distance, if nothing gets in the way. Some have beams which spread out more (have more divergence) and you can measure that by measuring the spot size as the target is set farther away. You can also try using lenses to reduce the spot size. You'll find that some lasers can be focused better on distant spots than others.
3. "Visibility" depends on who is viewing. If you were to build an automated system, you'd still have to calibrate it by comparing to how visible you thought the beam was. You could try to see how far away you could put a white sheet of paper and still see the laser spot. The answer will depend on things like how much other light is around, whether you have any color-blindness, etc.
Mike W.
(published on 04/10/2011)
Follow-Up #2: laser power
Q:
So is Mw the measurement of overall power, or the power used? What I'm trying to do is compare lasers with the same initial power output but with different wavelengths. IE, Two lasers are given the same amount of power, like, a violet and a red. I'm trying to find out if there is a real difference on which one delivers the most "output" (like heating a target).
Thanks,
Samuel O
P.S. I'm interviewing a professor on light andoptics this Wednesday, so that might help.
- Sam O (age 13)
Ashland, Oregon, US
- Sam O (age 13)
Ashland, Oregon, US
A:
I think you mean mW (milliwatt) not MW (megawatt). I don't believe young people are allowed to have megawatt lasers. At any rate, the standard power rating given (e.g. 2 mW) is for the output optical power. If the laser plugs into the wall, you should be able to calculate it's input power from the current rating given.
So the standard mW power rating describes exactly the same physical quantity that goes into the ability of the laser to heat up an absorbing target.
Mike W.
So the standard mW power rating describes exactly the same physical quantity that goes into the ability of the laser to heat up an absorbing target.
Mike W.
(published on 04/13/2011)
Follow-Up #3: laser output power
Q:
So a 30mW red, green, blue and violet laser would all have the same output? If so, then should I divide the input power (Watts, right?) by the mWs?
- Sam O (age 13)
Ashland, Oregon, US
- Sam O (age 13)
Ashland, Oregon, US
A:
Those 30mW lasers would have the same output power, although not, of course, the same output since they have different color light.
You ask if you should divide the input power, typically some Watts, by the output power. I'm puzzled by that question. Without some idea of what question you are trying to answer, there's no way to say what you should do to answer it.
Mike W.
You ask if you should divide the input power, typically some Watts, by the output power. I'm puzzled by that question. Without some idea of what question you are trying to answer, there's no way to say what you should do to answer it.
Mike W.
(published on 04/19/2011)