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

[Charles Link]

Just to make things very clear, I am talking about an IDEAL spectrometer.

Where does 2898 come from?

Wien's law for a blackbody is $\lambda_{max}T=2898$ microns deg Kelvin (see also post #85-if you take the derivative of the Planck wavelength function and set it equal to zero, you can derive Wien's law). Meanwhile one other item is the points in a diffraction grating spectral run are normally equally spaced in wavelength, so that with a flat detector response, your raw data would look very much like the processed wavelength spectrum. The optical layout (insides) of a diffraction grating spectrometer is fairly standard. There are a couple of minor variations, but basically the standard version is an Ebert spectrometer. Another similar variety if I remember correctly is called a Czerny-Turner. The optics basically send a collimated beam (parallel rays=plane wave) incident on the grating by having the entrance slit of the spectrometer at the focal point of the first parabolic or spherical mirror and then the "far field pattern" off of the grating is created in the focal plane of a second focusing mirror=parallel rays at the same angle $\theta$ onto the mirror focus at the same position $x=f \cdot \theta$ in the focal plane etc. The exit slit of the spectrometer is placed in the focal plane of the second focussing optics. Normally both slits are chosen to have the same slit width. There is often a trade-off between resolution by going to a very narrow slit versus the amount of energy that reaches the detector. Wide slit=plenty of energy, but lower resolution. ... editing...Additional item worth mentioning is the grating is rotated during a spectral run to get the spectrum. The textbook equation applies $m\lambda=d (\sin(\theta_i)+\sin(\theta_r) )$ for the location of a given wavelength. (The spectrum is most often observed with m=1, but the higher orders can be used.) Most often the grating is of the reflective variety (the lines on the grating can be thought of as "Huygens mirrors" (as opposed to Huygens sources from the slits of a transmission grating.) The appropriate gear mechanism is usually designed into the rotation of the grating so that the grating steps through equal wavelength increments as it steps through the spectrum.

 

[tade]

Remember Charles, we're sticking to ideal spectrometers.

Wien's law for a blackbody is $\lambda_{max}T=2898$ microns deg Kelvin

For $B_\lambda$ it is 2898 microns, approximately 3000. For $B_\nu$, it is 5099 microns, approximately 5000. This is where you get 3/5, as mentioned earlier.

For T=6000 K, this would be near $\lambda=500$ nm. ((2898/6000)*1000)nm to be more precise.

I asked you about an ideal spectrometer. Why would the measurement be consistent with 2898 microns and not 5099 microns?

 

[Charles Link]

Remember Charles, we're sticking to ideal spectrometers.
For $B_\lambda$ it is 2898 microns, approximately 3000. For $B_\nu$, it is 5099 microns, approximately 5000. This is where you get 3/5, as mentioned earlier.I asked you about an ideal spectrometer. Why would the measurement be consistent with 2898 microns and not 5099 microns?

The energy that comes though the slit is proportional to $L_{\lambda}(\lambda,T)\Delta \lambda =L_{\nu}(\nu,T)\Delta \nu$ where the $\Delta$ 's are the wavelengths or wavenumbers covered by the slit. If $\Delta \lambda$ remains constant throughout the spectral run, the (raw) spectrum will look like $L_{\lambda}(\lambda,T)$. If $\Delta \lambda$ is constant, you can compute the variation of $\Delta \nu=\Delta \lambda/\lambda^2$ . Notice we can really use either side of the first equation above to do our computation of the energy that comes through. Alternatively, if $\Delta \nu$ remained constant throughout the run, then $\Delta \lambda$ would not. The observed spectrum would then be $L_{\lambda}(\lambda,T)\lambda^2$. Note though, even in an ideal spectrometer, $\Delta \lambda$ will not be precisely constant throughout the spectral run. The equation $\Delta \lambda =(d)(\Delta x)/f$ is an approximation and not exact. That's why a calibration spectrum is usually necessary for accurate results.

 

[vanhees71]

drvrm, vanhees71, The Bill, any suggestions?

I've still no clue, what this discussion is about. The Planck spectrum as an energy distribution per frequency looks different from the same distribution expressed in terms of wave length with the relation $\lambda=c/\omega$. This is so because a distribution transforms as a distribution under coordinate transformations. In this case of only one independent variable ($\nu$ or $\lambda$) it's simply the chain rule
$$\frac{\mathrm{d} f}{\mathrm{d} \nu}=\frac{\mathrm{d} f}{\mathrm{d} \lambda} \frac{\mathrm{d} \lambda}{\mathrm{d} \nu}.$$
Now when you calculate the value $\nu_{\text{max}}$, where the frequency distribution peaks and compare it to the $\nu_{\text{max}}'=c/\lambd_{\text{max}}$ where $\lambda_{\text{max}}$ is the wavelength where the wave-length distribution peaks, are slightly different. This is not very surprising since the distribution functions look different depending on which variable, $\nu$ or $\lambda$, you relate it to.

To measure the spectrum you simply measure the intensity of the black-body radiation and either plot it against $\nu$ or $\lambda$ respectively. There is no principle problem in determining either of the two "maxima" of the corresponding functions. There is not too much physics in this, except that it characterizes roughly the relevant order of magnitude of frequencies/wave lengths which are dominating the spectrum of a black body. There's also Wien's displacement law about the location of $\lambda_{\text{max}}$ as function of $T$:

https://en.wikipedia.org/wiki/Wien's_displacement_law

 

[greswd]

The Planck spectrum as an energy distribution per frequency looks different from the same distribution expressed in terms of wave length with the relation $\lambda=c/\omega$. This is so because a distribution transforms as a distribution under coordinate transformations.
Now when you calculate the value $\nu_{\text{max}}$, where the frequency distribution peaks and compare it to the $\nu_{\text{max}}'=c/\lambd_{\text{max}}$ where $\lambda_{\text{max}}$ is the wavelength where the wave-length distribution peaks, are slightly different. This is not very surprising since the distribution functions look different depending on which variable, $\nu$ or $\lambda$, you relate it to.

The maths is pretty clear cut, I think I understand the OP's concern clearly, and its not the math.

I think he is confused about the physical aspects of the spectrum, which is why he's asking about experimental results.

Physically, at which wavelength is there the greatest number of photons? In reality, there can only be one wavelength at which this occurs, not two.

And let's assume that the detector is ideal, so that we don't have to deal with all the optical defects and corrections etc.

 

[greswd]

If $\Delta \lambda$ remains constant throughout the spectral run, the (raw) spectrum will look like $L_{\lambda}(\lambda,T)$.
Alternatively, if $\Delta \nu$ remained constant throughout the run, then $\Delta \lambda$ would not. The observed spectrum would then be $L_{\lambda}(\lambda,T)\lambda^2$.

Why would the value of the peak wavelength change depending on whether $\Delta \lambda$ or $\Delta \nu$ was constant?

Doesn't the spectrum only peak at one wavelength, therefore making it physically impossible to measure two different peak wavelengths?

 

[Charles Link]

Why would the value of the peak wavelength change depending on whether $\Delta \lambda$ or $\Delta \nu$ was constant?

Doesn't the spectrum only peak at one wavelength, therefore making it physically impossible to measure two different peak wavelengths?

You need to study the posts above in more detail. I think if you do, you will conclude they are correct, including #93. The OP has asked some very good questions-similar ones that I have had from other students who were first learning the details of spectroscopy and diffraction grating spectrometers. The OP appears to have figured out a good portion of it. It does take work to figure out the details, and the subject does contain quite a number of details.

 

[tade]

You need to study the posts above in more detail. I think if you do, you will conclude they are correct, including #93. The OP has asked some very good questions-similar ones that I have had from other students who were first learning the details of spectroscopy and diffraction grating spectrometers. The OP appears to have figured out a good portion of it. It does take work to figure out the details, and the subject does contain quite a number of details.

But this quote above is not an attempt to answer greswd's question in #96. This quote does not follow.

He understands what I'm trying to get at.

 

[greswd]

You need to study the posts above in more detail. I think if you do, you will conclude they are correct, including #93. The OP has asked some very good questions-similar ones that I have had from other students who were first learning the details of spectroscopy and diffraction grating spectrometers. The OP appears to have figured out a good portion of it. It does take work to figure out the details, and the subject does contain quite a number of details.

He's right, I feel that you're not moving in the direction of trying to answer the question.

 

[Charles Link]

I answered the questions the best I can. In asking about the raw spectrum from a spectrometer, it was necessary to include some finer details about a typical spectrometer. One additional item with these spectrometers is there is an overlap of the orders (different m's and different wavelengths) so that a couple of long pass order sorting filters typically need to be employed in a spectral run. In any case, if you managed to pick up on a couple of the concepts, I think it was a very worthwhile discussion. Even the blackbody Planck function contains plenty of detail that can take a lot of time to digest.

 

[greswd]

I answered the questions the best I can. In any case, if you managed to pick up on a couple of the concepts, I think it was a very worthwhile discussion.

I don't know, I think some parts have been an unnecessarily repetitive discussion.Should I take it that you don't know how to answer the question? (#96).

Not that there's anything wrong with that. I still thank you for your effort, help and time.