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Mathewe

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If a laser reads one watt on the LPM, what would 'two' identical lasers read if both dots were focused in the same spot? Would the LPM read 2 watts, or would some form of relativity kick in and the LPM show only 1 watt? What if the dots were spread out and not focused in the same spot?
 



RedCowboy

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Yes the power of the two beams adds up and de-focusing doesn't matter except any losses in your laser, for example if your lasers lens is clipping, however any difference should be minor.

As long as you get your beam onto the sensor area it will register, actually it's usually best to de-focus a bit and spread the beam across your sensor area to prevent damage to the coating, this is more important for more powerful lasers so familiarize yourself with your power meters sensor damage threshold.

Remember safety 1st, wear your laser safety glasses when power testing.
 

Mathewe

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Here's another question for you;
If photons have no mass, what would happen, within the beam(s), if two identical lasers were pointed directly at each other?
What would happen if the two lasers were of differing wavelengths, but of the same power each?

Get ready for a lot of questions, like this. Until I get set up, and receive a package of goodies from DTR's Laser Shop, I have nothing better to do than read, watch videos and dream up "What If" questions like this! :rolleyes:
 

Mathewe

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When 'de-focusing' my laser for testing power... should I only 'slightly' take it out of focus, or do I need to make it 'sloppy, way out of focus' to protect my measuring device?
 

julianthedragon

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If photons have no mass, what would happen, within the beam(s), if two identical lasers were pointed directly at each other?
What would happen if the two lasers were of differing wavelengths, but of the same power each?

Get ready for a lot of questions, like this. Until I get set up, and receive a package of goodies from DTR's Laser Shop, I have nothing better to do than read, watch videos and dream up "What If" questions like this! :rolleyes:

And we have nothing better to do than answer them of course! If you shined 2 lasers directly at each other aperture to aperture, there's one major problem that leaps out, and that is the collimated light from one laser entering the other and damaging the inside. This may or may not be an issue depending on the diode/module/lens setup of the affected laser but I've heard stories of lasers breaking like that. Also keep in mind some of the light will be reflecting off the other lasers' lens and going who knows where

But assuming you get past this, it would look like a lightsaber with two hilts, I guess (let's say you can see the beam). If the lasers were perfectly identical and exactly the same wavelength, I would imagine the inteference of their electromagnetic waves (light) would make a standing wave like this like this. The combined light beams would oscillate between cancelling each other out and summing with each other, but this would be happening so fast that it would average out and probably look normal to the human eye. If you had two identical beams pointed in the same direction instead, you could potentially try to cancel out or double the amplitude of the waves.

This is all hypothesis bc I haven't tested or observed this but it's my best guess.

If the lasers were different wavelengths with perfectly aligned beams of similar width, their colors would mix together just how any color light mixes. You can try this by shining two different color lasers into a white container or box where the light scatters. Blue and red mix to make magenta, green and red to make yellow, etc. RGB lasers might interest you

When 'de-focusing' my laser for testing power... should I only 'slightly' take it out of focus, or do I need to make it 'sloppy, way out of focus' to protect my measuring device?
Less than "way out of focus" is probably best to avoid the clipping RedCowboy mentioned, that's my only 2 cents there
 

Mathewe

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I found this video, today. I find it fascinating and chocked full of potential areas for experimentation and great fun...
 

Mathewe

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Messages
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Due to destructive interference, I would not expect the output to be 2 W. It'd probably be somewhere more than 1.414 W and less than 2 W
Math, please. I'd like to know exactly how you come up with the estimate of 1.414 watts remaining after estimated losses? Just how does one go about calculating the 'approximate' destructive forces as well as the 'estimated' levels of interference? "I have my note-pad and calculator on the ready!"
 
Last edited:

bostjan

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Ok. Well, let's start simple. Say that the two beams are focused to a point (don't ever focus a point on your LPM, though) on the same spot. Say both lasers are identical wavelength and identical power, for convenience. Shine 1W from laser 1 and 1W from laser 2. What is the optical power incident on the spot? If the two light beams strike the spot with exactly the same phase angle, the two beams will add up perfectly to 2W. If the two light beams strike the spot with exactly opposite phase angles (180° difference), the two beams will cancel perfectly to yield 0W.

Next, say that the two beams are focused into spots of identical radius and identically centered, and the two sources are exactly the same distance from the spot from exactly the same opposing angle. For the exact center of the spot, the same principle applies, but for everywhere else within the spot, there is a difference in distance from that point with in the spot to the two lasers. If the two lasers are perfectly in phase, the center of the spot will be twice as bright, and a point half a wavelength away from the center will be completely unilluminated by the lasers, because the beam incident at that point will be phase shifted by 180° between the two beams. A point one which is situated with a difference in distance from the point to each of the lasers of exactly one wavelength will again have perfect constructive interference, because the light travelling from the farther laser completes 360° of an extra wave, placing it back at the same point in the waveform. So, ultimately, you would see a bright spot in the center, then dark rings every odd number of half wavelengths in radius, and bright rings every even number of half wavelengths in radius.

If you move one laser slightly, so that it is farther away from the spot, and, say, refocus it such that the spot size remains constant, the only thing that will change about the situation is that there will be an offset in the "home point" of the fringes from the center of the spot to some ring of radius whatever within the spot. The perceived power of the entire spot would be the root mean square of the sum of the lasers power, or the square root of two, 1.414...W.

If, on the other hand, the two lasers are not exactly the same radius or centered exactly the same or perfectly aligned perfect circles, there will be some area around the spot where both lasers are incident where only one laser is incident. Since there is no interference in that area, the perceived power there would be 2W.

Since I can only assume that the two lasers will never be perfectly aligned spots with perfect overlap, the resultant perceived power must be somewhere between the square root of 2 and 2 times the laser's individual powers.
 

Mathewe

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Oct 3, 2021
Messages
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Points
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Ok. Well, let's start simple. Say that the two beams are focused to a point (don't ever focus a point on your LPM, though) on the same spot. Say both lasers are identical wavelength and identical power, for convenience. Shine 1W from laser 1 and 1W from laser 2. What is the optical power incident on the spot? If the two light beams strike the spot with exactly the same phase angle, the two beams will add up perfectly to 2W. If the two light beams strike the spot with exactly opposite phase angles (180° difference), the two beams will cancel perfectly to yield 0W.

Next, say that the two beams are focused into spots of identical radius and identically centered, and the two sources are exactly the same distance from the spot from exactly the same opposing angle. For the exact center of the spot, the same principle applies, but for everywhere else within the spot, there is a difference in distance from that point with in the spot to the two lasers. If the two lasers are perfectly in phase, the center of the spot will be twice as bright, and a point half a wavelength away from the center will be completely unilluminated by the lasers, because the beam incident at that point will be phase shifted by 180° between the two beams. A point one which is situated with a difference in distance from the point to each of the lasers of exactly one wavelength will again have perfect constructive interference, because the light travelling from the farther laser completes 360° of an extra wave, placing it back at the same point in the waveform. So, ultimately, you would see a bright spot in the center, then dark rings every odd number of half wavelengths in radius, and bright rings every even number of half wavelengths in radius.

If you move one laser slightly, so that it is farther away from the spot, and, say, refocus it such that the spot size remains constant, the only thing that will change about the situation is that there will be an offset in the "home point" of the fringes from the center of the spot to some ring of radius whatever within the spot. The perceived power of the entire spot would be the root mean square of the sum of the lasers power, or the square root of two, 1.414...W.

If, on the other hand, the two lasers are not exactly the same radius or centered exactly the same or perfectly aligned perfect circles, there will be some area around the spot where both lasers are incident where only one laser is incident. Since there is no interference in that area, the perceived power there would be 2W.

Since I can only assume that the two lasers will never be perfectly aligned spots with perfect overlap, the resultant perceived power must be somewhere between the square root of 2 and 2 times the laser's individual powers.
"Where's that button for 'DOUBLE'-LIKE'!" :giggle:
 




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