According to this Herschel discovers infrared light Herschel measured the “temperature” of the colours increasing from violet through to red and beyond. Now I know it’s not the only possible representation of the spectrum, but most often you’ll see the diagram where wavelength is on the x-axis and irradiance on the y, and sunlight peaks somewhere in the middle of the visible spectrum, and based solely on a naive interpretation of that there’s something funny with Herschel’s results. Would I get the same result today with more sophisticated detectors, and if so, what’s the explanation?
Well, the temperature of Herschel’s thermometers would depend on how strong the light is but also on how well the light was absorbed (as oppposed to being reflected) by the thermometers. If the thermometers absorbed much more efficiently in the infrared, then the infrared light would heat them more, even if the intensity of another color was a bit more. Additionally, even if the thermometer was perfectly black in all wavelengths, it’s possible the table it is sitting on could be better at absorbing infrared, so heats up more in the infrared areas, heating the thermometer there.
That’s my guess, at least until CalMeacham corrects me.
i’d guess that the sunlight peak in the visible spectrum is there because the visible spectrum is going to be where our eyes can get the most light
So when you’re talking about the different amounts of light in different wavelengths, you’re talking about something like the graphic given here in Wikipedia? This diagram shows both the light released by the sun (the yellow curve) and how much actually makes it to the Earth’s surface (the red curve), which is what Herschel would have been measuring.
Looking at that diagram, the biggest change we see in the sun’s visible spectrum at the Earth’s surface is less light at the blue/violent end of things, which is what Herschel measured as being cooler than the redder end of the spectrum. Perhaps the drop-off in intensity at the blue end partially explains his observation of cooler temperatures there.
The rest of the colors differ much less in intensity at the surface of the Earth… but the intensity does drop off a lot as we get into infra-red, where Herschel got the highest temperature of all.
That makes me think it’s not the intensity of sun’s light producing the increasing temperature of the thermometer. It’s either the material of his refracting lens (which might absorb some light in the bluer range) or the material of the thermometer (which might absorb more light in the redder range). Either of those separately, or both combined, would produce the results he saw.
I chose not to go into it in the OP, since I only have a vague recollection of the details, but the peak in the visible is IIRC an “artifact” of using wavelength as the x-axis. If you use frequency instead, you get a different peak, but I’ve had trouble finding it before and was uncertain if it even explained the phenomenon, so I hoped someone more knowledgable would take the bait.
I learned this reading about allegations that your guess is a myth, and that the important issue as regards to the development of vision is which wavelengths best penetrate water, but as I just mentioned, the science and calculations were a bit too much for me at the time.
You could be right, but emboldened but the lack of authorative answers so far I tried my hands at researching the details of various peak wavelengths myself, and according to the Wikipedia article on Wien’s displacement law, the peak in terms of power per unit optical frequency is in the near infrared.
Here’s the relevant source: Some paradoxes, errors, and resolutions concerning the spectral optimization of human vision - American Journal of Physics
Here’s a webpage that will display, if for example you input temperature, the peak wavelength for several different plots. I was surprised at how much the effect was. The Sun peaks at yellow/green as a function of wavelength, but in infra-red as a function of frequency.
One of the webpages in the OP’s cite discusses the reasons for the differences, and as others have noted, there vare many contributing factors:
http://home.znet.com/schester/calculations/herschel/index.html
You do have to realize that Herschel’s experiment was the first to demonstrate even the existence of infrared light. Although the Wikipedia page on infrared claims that the existence had been suggested prior to this, I doubt if that suggestion was well-known. Herschel’s finding that the temperature of a thermometer held below the red actually went up was a revelation. To expect him to produce a spectrum of sunlight was premature, at best. Although Newton had split sunlight with his prism, and recombined it to prove the rays were constituents of light, and had even associated lengths with each of the colors, he had stubbornly refused to consider light as a wave phenomenon. The first person to make a table of colors versus wavelength was Thomas Young, about four years after Herschel’s experiment. What we think of as an obvious idea – plotting the strength of colors of light versus their wavelength – wasn’t even a concept when Herschel did his experiment.
Another thing – Herschel discovered infrared light by putting his thermometer below red. Why didn’t he discover ultraviolet at the other end, but putting thermometer beyond the violet end, and seeing a similar temperature rise?
I suspect he did try that, and didn’t see a result. But scientists are notorious for not reporting non-results (except in famous cases like Michelson-Morley). Why didn’t he see a result? Most glasses block ultraviolet light. (both the prism and the thermometer bulb were made of glass) It’s another of those factors that contribute to the differences in response that he saw.
When Ritter discovered ultraviolet light the year after Herschel (something he did inspired by Herschel’s work), he did it by detecting the rays in a completely different fashion – he placed a piece of paper soaked in silver chloride solution beyond the violet end. There was no glass thermometer bulb intervening (I don’t know what his prism was made of). Silver chloride was known to darken when exposed to light, so he was taking a chance that any unseen rays would, too. He was right, luckily. As a result, Ritter called the new rays “Chemical rays”. “Ultraviolet” came later.
septimus writes:
This was the topic of my very first Light Touch article for Optics and Photonics News. I entitled it “Yes, I see, it’s Obvious”, about my observation that the peak of a curve plotted in wavelength will NOT be the same value as a peak plotted in inverse wavelength. I got a big argument from a professor, until I went through the mathematics to prove it. I thought the observation would be worth a brief article. But he responded that, oh. no, it was “obvious”. (After I’d spent half an hour arguing with him about it). The article I eventually wrote brought more mail than any other one I did subsequently, with people vociferously denouncing me and other just as vigorously supporting me. (including two former heads of the OSA, one on each side of the issue). One of my supporters brought up precisely this example of the peak of the graph of transmitted sunlight, if plotted both ways.
Aside: Once, when teaching an astronomy class, I told the students about both of those experiments. The test included a question “Name one way that light other than visible light can be detected”. Most students said “a thermometer” or “photographic film”, but one clever student pointed out that you could detect UV by getting a suntan from it.
OK, I’m not getting it. Can you point us dummies to the math about this somewhere?
I didn’t call anyone a 'dummy", and it might not be obvious. I’m certainly not going to try to write the math in this thread, but I’ll explain:
In a case where the integrated area under a curve is conserved – that is, where the area has to be the same, no matter how you express and plot the curve – the same must hold over pieces of the curve as well. In the case of absorption spectra plotted as optical density vs. wavelength over some range this is the case – you can figure out the number of absorbing centers from that area. And it shouldn’t make a difference whether I plot it as wavelength or as the inverse of the wavelength, the area between the two ends (provided they correspond to identical wavelengths) has to be the same.
Think of this mental model of the graph – take two glass plates, separated by a slight distance. Put in water-tight dividers that go from the front to the back wall. Space them equally, and say that they correspond to 300 nnanometers, 310 nanometers, 320 nanometers, etc. Fill each of the rectangular volumes with colored water corresponding to the average value of your spectrum over that range. If you look at this “living graph” from the side, it looks like a somewhat blocky graph of the spectrum.
Now change to wavenumber, proportional to 1/wavelength. If you want to keep the walls at positions that correspond to 1/300 nm and 1/310 nm and 1/320 nm, but you sill want the plot to be linear, you have to move the dividers. The widths of the chambers are no longer equal at 10 nm, but are going to vary in width. As you do so, the height in the individual chambers will vary, as well, and won’t be the same height relative to each other that they were. The chamber that used to have the highest vlevel of liquid (if your dividers were originally spaced close enough together to give the resolution required) now doesn’t any more – it’s shifted slightly to one side, because THAT chamber wasn’t made as wide by the variable change.
This will always be the case, unless your spectrum consists of sharp lines with no width. It becomes most noticeable as your peak width approaches the absolute value of the wavelength.
This won’t work if the value of the graph isn’t conserved – if you plot as percent transmission, for instance. (Although that doesn’t mean the peak doesn’t shift – it doesn’t shift in the same way, though, and the argument is more complex) But I first noticed this when working with curves where the area was conserved, and which had a linear wavelength scale on the bottom and a nonlinear wavenumber scale on the top (or vice-versa). I saw this in Russian papers a lot, for some reason. And I eventually realized that the peak wavelength in a plot linear in wavelength would not correspond to the peak wavenumber on a linear wavenumber plot, although the peak on a single graph with a linear and non-linear scale disguised this fact.
You can do all the above in mathematical form, but, believe me, I don’t want to try to code it in here.
I was just two clicks away! Didn’t notice the link though. I’ll blame bad web design.
Good to have the correct answer to replace my half-studied assumptions.
Nope, you certainly didn’t: that was me calling myself a dummy, in what was meant as semi-humorous self-deprecation. I did not mean to imply that you were anything other than respectful (as you always are) to others here.
Anyway,I think I get the gist of it, now, so thanks for the explanation.