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

[greswd]

You use a spectrometer whose range includes both wavelengths, or two spectrometers which each cover one of those wavelengths. You hook the spectrometer(s) up to a computer, point the spectrometer(s) at the source, and look for peaks on the intensity vs. wavelength graph that is displayed on the spectrometer software on the computer.

If the range includes both wavelengths, will there be two peaks?

What formula describes this graph? Both versions of Planck's Law only have one peak.

 

[Charles Link]

There is no 923 nm peak in the solar spectrum. None of these two formulas give such a peak.

The ratios between the peak in the $\nu$ spectrum vs. $\lambda$ spectrum are very nearly 5/3. For a 500 nm peak in the $\lambda$ spectrum, this makes for an 830 nm peak in the $\nu$ spectrum. It would be rather unconventional for a physicist to speak in terms of the wave number spectrum and to describe the solar spectrum as peaking at 830 nm.

 

[greswd]

There is no 923 nm peak in the solar spectrum. None of these two formulas give such a peak.

In this function

the peak is at 325 THz, which is 923 nm.

Sorry, can you please answer my earlier question about the mathematical function for the intensity?

 

[greswd]

The ratios between the peak in the $\nu$ spectrum vs. $\lambda$ spectrum are very nearly 5/3. For a 500 nm peak in the $\lambda$ spectrum, this makes for an 830 nm peak in the $\nu$ spectrum. It would be rather unconventional for a physicist to speak in terms of the wave number spectrum and to describe the solar spectrum as peaking at 830 nm.

For 5525 K I got it very close to 923 nm.

By the way, What is the math formula of $V(\lambda)$?

 

[Charles Link]

In this function

the peak is at 325 THz, which is 923 nm.

Sorry, can you please answer my earlier question about the mathematical function for the intensity?

The $V(\lambda)$ for a spectrometer also includes the photodiode response along with the spectrometer throughput which is affected by the wavelength the spectrometer is optimized for (diffraction grating has a blaze angle that will optimize a spread of wavelengths.) These are all of secondary consideration for the question at hand. The OP is interested in the difference between the wavelength and wavenumber blackbody functions. When comparing the $\nu$ and $\lambda$ peaks, please include the temperature that you used. I think you'll find the ratio of 5/3 between them to be fairly accurate.

 

[greswd]

The $V(\lambda)$ for a spectrometer also includes the photodiode response along with the spectrometer throughput which is affected by the wavelength the spectrometer is optimized for (diffraction grating has a blaze angle that will optimize a spread of wavelengths.) These are all of secondary consideration for the question at hand. The OP is interested in the difference between the wavelength and wavenumber blackbody functions. When comparing the $\nu$ and $\lambda$ peaks, please include the temperature that you used. I think you'll find the ratio of 5/3 between them to be fairly accurate.

I used 5525 K, which the OP also used.

 

[Charles Link]

For 5525 K I got it very close to 923 nm.

By the way, What is the math formula of $V(\lambda)$?

For the $\lambda$ peak, you should then get very nearly 923*3/5 nm. Usually the solar spectrum is given to be a little closer to 6000 K. $V(\lambda)$ is the measured spectrometer voltage. It depends on the amplifier gain as well as the photodiode response but will be proportional to the incident irradiance (onto the photodiode) at wavelength $\lambda$, i.e. you can do a spectral run of an unknown source, and then do a spectral run of a blackbody calibration source and the ratio of the measured voltages will be equal to the ratio of the irradiances of each of the two sources at each wavelength.

 

[tade]

For the $\lambda$ peak, you should then get very nearly 923*3/5 nm. Usually the solar spectrum is given to be a little closer to 6000 K. $V(\lambda)$ is the measured spectrometer voltage. It depends on the amplifier gain as well as the photodiode response but will be proportional to the incident irradiance at wavelength $\lambda$, i.e. you can do a spectral run of an unknown source, and then do a spectral run of a blackbody calibration source and the ratio of the measured voltages will be equal to the ratio of the irradiances of each of the two sources at each wavelength.

Wow, so many replies.

5525 K is the value measured by this experiment.

I think $V(\lambda)$ is actually quite crucial to answering my question. What is the general formula for $V(\lambda)$?

 

[Charles Link]

I used 5525 K, which the OP also used.

I think the 524 nm number is then incorrect. Should be closer to 555.

 

[tade]

I think the 524 nm number is then incorrect. Should be closer to 555.

5525 K is the value measured by this experiment.

I think $V(\lambda)$ is actually quite crucial to answering my question. What is the general formula for $V(\lambda)$?

 

[Charles Link]

5525 K is the value measured by this experiment.

I think $V(\lambda)$ is actually quite crucial to answering my question. What is the general formula for $V(\lambda)$?

In general $V(\lambda)= R(\lambda)E(\lambda)$ where $R(\lambda)$ is the complete spectrometer response function and $E(\lambda)$ is the incident spectral irradiance (watts/cm^2/nm). $R(\lambda)$ gets measured in the calibration spectral run using a calibration blackbody at fixed (and known) temperature.

 

[tade]

In general $V(\lambda)= R(\lambda)E(\lambda)$ where $R(\lambda)$ is the spectrometer response function and $E(\lambda)$ is the incident spectral irradiance (watts/cm^2/nm). $R(\lambda)$ gets measured in the calibration spectral run using a calibration blackbody at fixed (and known) temperature.

Can you give me a typical example of $R(\lambda)$ and $E(\lambda)$ for a blackbody?

How do these functions relate to

and

?

 

[Charles Link]

Can you give me a typical example of $R(\lambda)$ and $E(\lambda)$ for a blackbody?

How do these functions relate to

and

?

And I doublechecked your arithmetic=my mistake: Wien's law $\lambda_m *T=2898$ is much more precise than the 5/3 estimate. Experimentally, you need to ensure that the light from the unknown source enters the spectrometer the same way, (i.e. having the same F#), as the light from the blackbody source or all you can get is a relative spectrum. For a blackbody source, spectral brightness (radiance) is designated by $L$. Total radiation per unit area over a hemisphere from the flat source (with a cosine spatial intensity distribution) is given by $M=L*\pi$. To get the irradiance $E$ onto a photodiode for simple geometries (no collection optics) $I=L*A$ where $A$ is the blackbody source area and irradiance $E=I/s^2$ where $s$ is the distance between the blackbody and the photodiode, so that irradiance $E=L*A/s^2$. When there are collection optics included, such as in the spectrometer, along with a diffraction grating, this is all included in $R(\lambda)$ and it would be difficult but normally unnecessary to make any detailed measurements of all the separate factors that go into $R(\lambda)$... Your $B(\lambda, T)$ above is my $L(\lambda, T)$. (precisely).

 

[greswd]

For a blackbody source, spectral brightness (radiance) is designated by $L$. Total radiation per unit area over a hemisphere from the flat source (with a cosine spatial intensity distribution) is given by $M=L*\pi$. To get the irradiance $E$ onto a photodiode for simple geometries (no collection optics) $I=L*A$ where $A$ is the blackbody source area and irradiance $E=I/s^2$ where $s$ is the distance between the blackbody and the photodiode, so that irradiance $E=L*A/s^2$. When there are collection optics included, such as in the spectrometer, along with a diffraction grating, this is all included in $R(\lambda)$ and it would be difficult but normally unnecessary to make any detailed measurements of all the separate factors that go into $R(\lambda)$... Your $B(\lambda, T)$ above is my $L(\lambda, T)$. (precisely).

Then how do we obtain $B_\nu(\nu, T)$?
If we use the same initial functions, how do we get it to peak at a different wavelength?

 

[Charles Link]

Then how do we obtain $B_\nu(\nu, T)$?
If we use the same initial functions, how do we get it to peak at a different wavelength?

Using B's $B_{\lambda}(\lambda)=B_{\nu}(\nu)/ \lambda^2$ . $|d\nu/d\lambda | =1/\lambda^2$ creates this whole dilemma with the $1/\lambda^2$ differences in the two formulas. As a result, when taking derivatives, and setting them equal to zero to find the spectral peak, there is a considerable difference between the location of the peak depending on which format was used. Both functions give identical results for the energy contained between any two wavelengths. They contain precisely the same information. Note that $| B_{\lambda}(\lambda)d \lambda|=|B_{\nu}(\nu) d \nu |$.

 

[greswd]

Your $B(\lambda, T)$ above is my $L(\lambda, T)$.

Any reason as to why it is $B(\lambda, T)$ in this quote of yours and not $B_\nu(\nu, T)$?

 

[Charles Link]

Any reason as to why it is $B(\lambda, T)$ in this quote of yours and not $B_\nu(\nu, T)$?

I do most calculations using the wavelength format. It can readily be converted to wave numbers.

 

[greswd]

There are two functions, $B_\lambda$ and $B_\nu$ , and they are not identical after you apply $c=\nu*\lambda$.

You said:

Your $B(\lambda, T)$ above is my $L(\lambda, T)$. (precisely).

Does this $B(\lambda, T)$ refer to $B_\lambda$ or $B_\nu$?

Why is it one and not the other?

 

[greswd]

Sorry for my insistent questioning but I really want to get to the bottom of this. Thank you.

 

[Charles Link]

Sorry for my insistent questioning but I really want to get to the bottom of this. Thank you.

Normally in spectroscopy, they use $\nu=1/\lambda$, but no, the spectral intensity (density) functions are not given by a simple substitution. For an atmospheric transmission function $\tau(\lambda)$ or detector response function $R(\lambda)$, the simple algebraic substitution $\nu=1/\lambda$ into the functional form does give you the corresponding functional form in the other format (because these are not spectral density functions). For$L(\lambda,T)$ or $L(\lambda)$ or $E(\lambda)$ or $I(\lambda)$ when they are spectral density functions, there is a $1/\lambda^2$ factor that goes along with the conversion. (The E=LA/s^2 formulas I presented above are correct with either format provided you use this same format on both sides of the equation.) Meanwhile, even though we have a slight discrepancy in what we might call the peak of the solar spectrum or peak of a blackbody spectrum , this procedure dos not alter the spectral positions of (narrow) atomic and molecular absorption and emission lines. Hopefully this helps clarify the issue.

 

[greswd]

Your $B(\lambda, T)$ above is my $L(\lambda, T)$. (precisely).

So does this $B(\lambda, T)$ refer to $B_\lambda$ or $B_\nu$?

Why is it one and not the other?

 

[Charles Link]

So does this $B(\lambda, T)$ refer to $B_\lambda$ or $B_\nu$?

Why is it one and not the other?

In a much earlier post above, I used $L(\lambda,T)$ and I have it memorized from many years of spectroscopic measurements and calculations. The $B_{\lambda}(\lambda,T)$ that the OP listed matched it precisely. For a $B_{v}(\nu, T)$, you always need to check the defintion: Is $\nu=1/\lambda$ or is $\nu=c/\lambda$ ? For the wavelength function there is no ambiguity. The one item that is also important here though, is the blackbody function given as $M(\lambda, T)$ or $L(\lambda, T)$ ? The $B$ notation is not standard, and I needed to determine if it was the $L$ form or the $M$ form ($M=L*\pi$). It was indeed the $L$ form.

 

[tade]

In a much earlier post above, I used $L(\lambda,T)$ The $B_{\lambda}(\lambda,T)$ that the OP listed matched it precisely.

Is there a scientific reason as to why the one that matches it precisely is $B_{\lambda}$ and not $B_{\nu}$ ?

I believe that this must be due to the type of apparatus used.

 

[Charles Link]

Is there a scientific reason why the one that matches it precisely is $B_{\lambda}$ and not $B_{\nu}$ ?

I believe that this must be due to the type of apparatus used.

I never memorized a $B_{\nu}(\nu)$ blackbody form. With a diffraction grating based spectrometer, the spectroscopist usually presents spectra in wavelength. There is a FTIR (Fourier Transform spectrometer) with a moving Michelson mirror, and researchers often present those spectra in wavenumber. Both forms are appropriate, although it can be somewhat confusing when first learning the subject. I did check you $B_{\nu}(\nu)$from the other (wavelength) form. You correctly had the $1/\lambda^2$ factor (actually $c/\lambda^2$for your $\nu=c/\lambda$). Your definition uses $\nu=c/\lambda$. It makes a difference whether you graph a spectrum in $\nu=1/\lambda$ or $\nu=c/\lambda$ for your units on the y-axis.

 

[tade]

I never memorized a $B_{\nu}(\nu)$ blackbody form. With a diffraction grating based spectrometer, the spectroscopist usually presents spectra in wavelength.

$B_{\lambda}$ peaks at 524 nm and $B_{\nu}$ peaks at 923 nm.

You say that with a diffraction grating based spectrometer, we can measure $L(\lambda,T)$, which then gives a peak at 524 nm.

Is there any instrument that gives the raw data of $L(\lambda,T)$ which peaks at 923 nm?

 

[Charles Link]

$B_{\lambda}$ peaks at 524 nm and $B_{\nu}$ peaks at 923 nm.

You say that with a diffraction grating based spectrometer, we can measure $L(\lambda,T)$, which then gives a peak at 524 nm.

Is there any instrument that gives the raw data of $L(\lambda,T)$ which peaks at 923 nm?

Once you have $R(\lambda)$ measured with a calibration run, you can measure any blackbody of any temperature that gives off enough energy to measure. When you report your results, for example for a T= 5525 blackbody, if you have units of wavelength on the x-axis it will peak at 524 nm. If you have units of wavenumber on the x-axis, it will peak at a wave number that has corresponding wavelength of 923 nm. The raw data peak is of little significance. There are too many other spectral factors, (detector spectral response, optics, and spectrometer diffraction grating throughput) that affect this.

 

[greswd]

The peak occurs from a result of the function $I_{\lambda}(\lambda)$ or $I_{\nu}(\nu)$. The spectrometer voltage $V(\lambda)$ is likely to also peak somewhere in the vicinity of both of these

So for the Sun which is at 5525 K, at what wavelength does $V(\lambda)$ peak?

 

[Charles Link]

So for the Sun which is at 5525 K, at what wavelength does $V(\lambda)$ peak?

A major factor here is the detector response. If you use a silicon photodiode or any other photodide, they respond to photons so the detector response is typically proportional to $\lambda$ in the region of responsivity. A thermopile type detector (e.g. pyroelectric) will have a flat spectral response. These two detector types will have peaks in the raw voltage spectrum that are quite different with the peak from the silicon photodiode occurring at a longer wavelength. The precise position of this raw voltage peak is really of little significance.

 

[greswd]

A major factor here is the detector response. If you use a silicon photodiode or any other photodide, they respond to photons so the detector response is typically proportional to $\lambda$ in the region of responsivity. A thermopile type detector (e.g. pyroelectric) will have a flat spectral response. These two detector types will have peaks in the raw voltage spectrum that are quite different with the peak from the silicon photodiode occurring at a longer wavelength. The precise position of this raw voltage peak is really of little significance.

With different equipment giving different peaks, how do we end up with the one peak of 524 nm?

 

[Charles Link]

With different equipment giving different peaks, how do we end up with the one peak of 524 nm?

Please read the previous postings far above. That's what the calibration spectral run is used for. Each detector employed needs a separate calibration run.