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
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.
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.
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)
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.
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