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I Different Blackbody Intensity Peaks, how do we measure it?

  1. Apr 21, 2016 #1
    For the Blackbody Spectrum, there are two versions of the formula, one for wavelength and the other one for frequency:

    b57bebb5337f0de6333ec9bc85688c08.png 46a1efc581519117de92da6afb5a8e78.png

    The peak intensities for both occur at different wavelengths (or frequencies).

    How do scientists measure the spectral radiance of blackbodies?

    Are there TWO types of equipment, one for [tex]B_\lambda[/tex] and the other for [tex]B_\nu[/tex], such that each device yields a peak at a different frequency?
  2. jcsd
  3. Apr 21, 2016 #2
    as frequency and wavelengths are related by c= frequency x wavelength ,where c is the velocity of light , one may not need two types of experiments to measure intensity variation with either wavelength or frequency.
    just an example of measurement technique;
    For the experiment, a tungsten filament light bulb was used as the emitting source.
    The filament, in conjunction with an apparatus that could record the relative intensity of light at various frequencies or wavelength, . The apparatus used contained a tungsten filament light as the black body emitter, a diffraction grating to separate the various wavelengths of light and a photodiode to detect relative intensities of incident light.
  4. Apr 21, 2016 #3
    But then how would we obtain the two different peak values?
  5. Apr 21, 2016 #4
    does any experimental curve give different peak value for a wavelength and corresponding frequency?
    pl. quote results.
    or you want to say that c=nx wavelength does not hold experimentally?
  6. Apr 21, 2016 #5
    your formula may be incorrect or depicting some other parameter-not the intensities

    some facts;

    The amount of radiation emitted in a given frequency range should be proportional to the number of modes in that range. The best of classical physics suggested that all modes had an equal chance of being produced, and that the number of modes went up proportional to the square of the frequency.

    But the predicted continual increase in radiated energy with frequency (dubbed the "ultraviolet catastrophe") did not happen. Nature knew better.
  7. Apr 23, 2016 #6
    You can do the math and see for yourself.

    From wikipedia of plancks law: "Evidently, the location of the peak of the spectral distribution for Planck's law depends on the choice of spectral variable"

    Also, read the section "Peaks" under "Properties"
  8. Apr 25, 2016 #7


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    By maximizing the corresponding distribution functions. Where do you see a problem with that? The peak values are just rough characterizations of the distributions without much further physical significance!
  9. Apr 26, 2016 #8
    What do you think these formula represent?
    They don't have peaks anyway.
    Do you understand what are these?
  10. Apr 26, 2016 #9

    Charles Link

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    Any spectral density intensity function has a slightly different shape when displayed in wavelength space than it does in frequency space. These two are related by ##|I_{\lambda}(\lambda)d\lambda|=|I_{\nu}(\nu)d\nu| ##. where ## \nu=1/\lambda ## so that ## |d\nu/d\lambda |=1/\lambda^2 ##. It is straightforward to convert from one to the other, and most any spectroscopist has little trouble converting their experimental results to which ever form they need to compare to. You are correct that the "peaks" for the blackbody curve, (as well as any other smooth spectral function), do occur at different wavelengths for the two graphs. The energy found between any two wavelengths is the same for both graphs, and the graphs by either method are entirely consistent with each other. You may still be asking how this is done experimentally. The blackbody is often used as a calibration source, and the spectrum is measured at 100 or more spectral points. A spectrum is run for an unknown source, and then a calibration spectral run is performed over the same set of wavelengths using a calibration blackbody source (at a fixed temperature). Let ## V(\lambda) ## be the detector voltage at wavelength ## \lambda ##. Assuming the detector responds linearly to incident energy, (which is often the case), ## V_s(\lambda)/V_{bb}(\lambda)=I_s(\lambda)/I_{bb}(\lambda) ##. (other geometric factors, area, distance, etc., can also be included in this formula). Depending on whether you display the results in wavelength or frequency, the appropriate blackbody spectral function ## I_{bb}(\lambda) ## (or ## I_{bb}(\nu) ##) is inserted into the equation. It is straightforward for anyone to take a displayed graph of ## I_s(\lambda) ## or ## I_s(\nu) ## and convert it to the other form. One additional item is your blackbody function is not entirely correct. This may be an early proposed version or an approximation of some kind. In wavelength form, it correctly reads ## L_{\lambda}(\lambda, T)=2hc^2/((\lambda^5)(exp((hc/(\lambda k T))-1))## and ##I_{\lambda}(\lambda,T)=L_{\lambda}(\lambda,T)A_{bb} ## where ##A_{bb} ## is the area. (Your blackbody formulas are using ##\nu=c/\lambda ## which is ok, but the formulas are not completely correct. In addition, the usual convention in spectroscopy is that ## \nu=1/\lambda ##.)
    Last edited: Apr 26, 2016
  11. Apr 27, 2016 #10


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    They are distribution functions of energy per frequency or wave length, and of course they have peaks. In the Wikipedia article you find these curves!
  12. Apr 27, 2016 #11

    Charles Link

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    As I mentioned in post #9, the formulas the OP has are incorrect, and upon closer inspection, neither of them even has a peak. The OP still asks a very good question, but would do well to start with correct blackbody formulas.
  13. Apr 27, 2016 #12
    The question was for the first poster. :)

    Not as they are given in the OP. These have no peak.
    They may be the high temperature approximation of the complete functions.
  14. Apr 27, 2016 #13

    Charles Link

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    Hopefully the OP returns to see all of our inputs. :-)
  15. Apr 29, 2016 #14
    I thought the peak values can be physically measured?
  16. Apr 29, 2016 #15

    You are right. I accidentally uploaded the Rayleigh-Jeans Law instead of Planck's Law. I meant to use Planck's Law.

    Now it is too late to edit it though.
  17. Apr 29, 2016 #16
    Why would we need different instruments? That's like saying you'd need different rulers to measure items that are 5cm long from the one used to measure items that are 7cm long. As long as the peak of the spectrum is within the wavelength range of the spectrometer you are using, you can measure the peak. For example, the commonly used 3B student spectrometers can measure an intensity curve from 360nm to 800nm, which includes the entire visual spectrum along with a little ultraviolet and a good chunk of short infrared. That allows one to measure the peaks in the sun, common lightbulbs, and many other sources.
  18. Apr 29, 2016 #17
    Thanks Charles, but I'm still confused. What do 'bb' and 's' stand for?

    http://qdl.scs-inc.us/2ndParty/Images/Charles/Sun/SolarSpectrum2_wbg.png [Broken]
    This is the solar spectrum.

    For 5525 K, and following Planck's Law 3ffb6801557271888f2563643e4dfd5d.png , it peaks at about 524 nm.

    For 5525 K, and following 982b68ac28df611584f05c7d56a26bb3.png , it peaks at about 325 THz, or 923 nm.

    What kind of instruments allow us to obtain 524 nm And/Or 923 nm?
    Last edited by a moderator: May 7, 2017
  19. Apr 29, 2016 #18

    drvrm, vanhees71, The Bill, any suggestions?
  20. Apr 29, 2016 #19

    Charles Link

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    The "s" subscript stands for the unknown source whose spectrum is being measured, and "bb" is the calibration blackbody. Its temperature can be measured with a thermocouple and kept constant. Calibration blackbodies are commercially available. A typical temperature used for them is 1000 C (1273 K), and the absolute spectral intensity output is typically within +/- 1% of the Planck blackbody function. Meanwhile silicon photodiodes (also available commercially) can be used for measurements wavelengths from 400 nm to 1000 nm. Their response is not the same at all wavelengths and similarly a (diffraction grating based) spectrometer does not have the same throughput for all wavelengths, thereby the need for a complete calibration blackbody spectral run. One other interesting thing about the blackbody function is that the function integrates precisely from 0 to infinity to give the radiation per unit area ## M=\sigma T^4 ##. (Note ## M=L*\pi ## )
    Last edited by a moderator: May 7, 2017
  21. Apr 29, 2016 #20
    Let's say that we are measuring the solar spectrum.

    We want to obtain the 524 nm and the 923 nm peaks; what is the difference in the experimental procedures for both scenarios?
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