For convenience sake let’s say you have 2 identical lasers, one is blue and one is red. And you shine it on lead (so none of the light leaks through) until the lead doesn’t heat up anymore. Would the temperature change at all between the different color lasers. It doesn’t have to be red or blue, it could be microwave or x ray, just different colors is nessisary.
This question has lots of different answers depending on exactly what you want to know, I’ll ask it in a few different ways and provide answers to them
Does the color of a photon change its energy content?
Yes, color is the result of how energetic a photon is, red is low energy compared to blue, so a single blue photon contains more energy than a single red photon.
Does the color influence how much energy is transferred to an object?
Yes, the color of an object is the result of white light hitting it and it reflecting back that color. For example leaves are green because they reflect green light while absorbing most others. In a way the color of an object is the color that the object rejects. So if you have a blue object it will reflect the blue laser more than the red one, so the red laser will heat it more because it’s being absorbed.
Which color heats lead the most?
That is a very interesting question, to know this you need to look at the absorption spectrum of the material, i.e a graph showing you which wavelengths are more absorbed by the material, so if I’m reading correctly the spectral lines in the wiki page for lead https://en.m.wikipedia.org/wiki/Lead it seems a cyan/greenish laser would be most effective in heating it (or I might be reading it completely opposite and that is the least effective color for heating lead)
Does a blue laser produces more energy than a red laser?
It depends, lasers emit photons, and while a single blue photon contains more energy than a single red photon, thousands of red photons contain more energy than a single blue photon. So it depends on how many photons each laser emits, if it’s the same amount then yes blue lasers will output more energy, but that’s not a given. In fact while I’m not intimately familiar with the physics of lasers, if we asume a perfect energy conversion from electricity to photons if two lasers use the same energy input they should have the same energy output, which would mean less photons for the blue laser.
if we asume a perfect energy conversion from electricity to photons
That’s not really necessary. Lasers are usually rated by the output power already.
Yes, but the input power moat likely varies, and that’s because the energy conversion is not perfect, depending on the mechanics some energy might be dissipated as sound or most likely heat. But since I don’t know the specifics I can’t account for it, and it’s possible that red lasers require more energy than blue lasers because of a quirk in the way they’re generated currently. If we ignore that and imagine a magic box that can convert 100% of the energy given into specific color photons, for any given input it will generate more photons when configured to red than to blue, because a single blue photon contains more energy than a single red photon, therefore you need more red photons to get to the same level of energy.
If you’re measuring energy input by rating on the box yes, but instead of assuming that is your measurement data, they assumed perfect conversion.
It has the same end result descriptively but I’d argue that perfect conversion paints an easier to understand theoretical image, vs a more rigid practical image
doesn’t have to be red or blue, it could be microwave or x ray
Technically those wouldn’t be LASER (light amplification by stimulated emission of radiation.) but MASER or XASER.
I’m probably the only asshole left that cares about this though.
So where do you stand on infrared lasers? Light is already a junk term for the EM spectrum that we can see, otherwise holding no specific importance.
Oh I have a question! Do all frequencies on the EM spectrum emit photons? Like, when gamma rays or X rays or microwaves hit something are photons bouncing off/being absorbed and we just can’t see them?
All EM waves are photons.
When X or gamma ray photons interact with matter, e.g. by Compton scattering, i.e. hitting electrons of an atom. Thereby the photons loose discrete amounts of energy, leading to an increase of their wavelengths, and the electrons are then lifted on corresponding higher energy levels. When the electrons ‘fall’ back onto their base levels, additional photons are emitted.
Microwave photons, however, have to low energy for this kind of interaction. They e.g. induce vibrations and oscillations of molecules which is perceived as a temperature increase.
Ooh neat! Thank you, very well described for a layman such as myself
You’re welcome.
Anyway, a maser or a xaser would create completely different responses on the material. There’s a good reason we use different names.
Eh, I think it’s helpful to point that out. If someone hit a dead end researching lasers, not making it out of the visible spectrum, that could explain why. Maybe they missed the line stating where laser ends, maybe the article assumed the reader would know that already.
You’re telling me only visible light counts as light to lasers?
explains all the CO2 infrared lasers out there.
The EM waves of near spectra, UV and IR, are commonly often referred to as ‘light’ as well. Wikipedia even states:
In physics, the term “light” may refer more broadly to electromagnetic radiation of any wavelength, whether visible or not. In this sense, gamma rays, X-rays, microwaves and radio waves are also light.
Which fits to my perception of physicists, where in astronomy every element starting at Lithium is referred to as ‘metal’.
It’s different wavelengths. It’s our eyes that fail to actually be able to see in certain ones.
Red light will generally carry less energy than blue light, so if you want a red and blue laser with the same power output, you need more intensity on the red laser.
Also the colour affects how some materials absorb the energy. A material that is good at reflecting red light won’t heat up as quickly from a red laser, for example.
it could be microwave or x ray, just different colors is nessisary.
Colour just means different wave lenghts of light. So x-ray is a different “colour” from microwave.
@toboggonablaze is essentially correct, but let me try explain it in a slightly different way.
Lasers do a bunch of things to basically shoot a stream of photons at something. There’s basically two ways you can affect how much energy comes out of a laser, you can make the stream denser (more photons per second) - called intensity, or you can increase the energy in each photon.
The weird part about photon energy is that higher energy photons are of a different “color”, where red is lower than green, is lower than blue, is lower than gamma rays, etc.
So changing the color of a laser already means you’ve changed how much energy it can output.
Then there’s another part of your question: how lead gets heated up. Different materials respond differently to different types/wavelengths of light, an example you might be familiar with is that glass panes let through visible light, but not the heat from the sun, or that water also is see through, but can easily be microwaved (by microwaves - low frequency light).
Basically, a material can be more or less “translucent” in certain frequencies. I’d like to look lead up for you, but Google isn’t cooperating today. But basically, there are frequencies that lead will be more and less susceptible to.
That’s probably not what you meant with the question, but if that’s the application you want to use the laser for, you might want to take it into consideration.
So, in summary: color is energy, intensity is energy, you can change both independently, so your question doesn’t quite make sense.
Also, different targets will heat differently, also not making it a fair comparison.
Basically, a material can be more or less “translucent” in certain frequencies. I’d like to look lead up for you, but Google isn’t cooperating today. But basically, there are frequencies that lead will be more and less susceptible to.
For macroscopic objects, there really isn’t a single answer. Something as generic as “a plate of lead(oxide)” can be all over the spectrum depending on texture, exact composition, oxidation levels, etc etc. there’s a reason why lab-grade filters and mirrors cost so much money, it’s hard to get a narrow frequency range.
It also rapidly changes as the material heats up, melts, breaks down, reacts with the air, etc.
When you say “they’re identical” what properties between the two lasers do you want to force to match?
Okay, so I don’t know much about lasers at all so I don’t know why making them identical but changing the color doesn’t work. I just want the same amount of light to be produced in the same intensity but different colors
Then, as I understand, both lasers produce beams of the same intensity (photons per area) and with the same cross section area, i.e. the number of emitted photons per unit of time is identical, the blue laser delivers more power than the red one as tobogganablaze already wrote, since blue light (high frequency, short wavelength) photons have higher energy than red light photons (low frequency, long wavelength). And the rest is up to the absorption properties of the material hit by the laser.
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The power needed to power the laser, if identical, is what determines the temperature here, not the color.
A thousand watt laser is going to make the lead melt while a 5 watt laser won’t.
Another limiting factor is how fast the lead/target material can dissipate the heat. A 10kg block of lead is going to stay a lot cooler than a 5 gram piece of lead.
The colour of the laser has no influence on the final temperature, at all.
The photons are just transferring energy from a to b. And whether they can do it fast or slow doesn’t really matter. Both blue and red photons are subject to the law of conservation of energy. The amount of energy you put in is the amount of energy you get out.
In reality there is probably a difference in how efficient a blue and red laser diode are, so there are losses. And if one is slower than the other, the target material is losing energy to the surrounding air. And that’s why you’ll get different temperatures.
In an ideal setting with exactly identical lasers, where the only difference is the colour, the end result is also exactly the same.
You misspelled color
Both “color” and “colour” are valid spellings.
cooler
Im in the privileged position where I can say that I was educated to pronounce and write in British English. And while I’m not a native speaker, I’m confident that my knowledge of the English language surpasses most of native speakers in the US, if not 99%. My spell checker is biased for us English.
Now if you excuse me I have to go back to my flat, with my lorry having chips and candy floss in the boot. Do you even elevator bro?
I’m by no means a laserologist but my layperson understanding is that laser colors are achieved through different means of generating the laser - using various gases etc. So a red or green laser have different properties. Now there may be a way to achieve colors optically and the light itself being consistent but from the bit I recall green ones took quite a bit more power to generate than red so I think would be hotter.
….or I’m WAYYYYY off. 🥸😜
