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Q: Mapping Table from IR Wavelength to Temperature of source ( Answered,   3 Comments )
Subject: Mapping Table from IR Wavelength to Temperature of source
Category: Science
Asked by: dadadee-ga
List Price: $50.00
Posted: 23 May 2003 08:57 PDT
Expires: 22 Jun 2003 08:57 PDT
Question ID: 207741

I'm looking for a table of values mapping the wavelengths of Infra-red
radiation emission to specific temperature values (of a given source)
between 35.0 to 40.0 degrees celsius (0.1 deg celsius resolution). My
search has yielded an approximate value of 10 microns for temperatures
in this range but I require the specific mappings.

An example of how the table might look like:

IR Wavelength (microns)       Temperature
-----------------------       -----------
8.0                           35.0
8.1                           35.1
8.2                           35.2

These are fake figures.

I will also need to know if there are manufacturers which produce IR
filters of the various IR wavelengths (to cut out radiation
corresponding to certain temperatures) listed in the table.


Best wishes,

Request for Question Clarification by mathtalk-ga on 24 May 2003 21:04 PDT
Hi, dadadee-ga:

Racecar-ga has correctly sketched out for you the existence of a
spectral "peak" in the black-body radiation curve which depends on
absolute temperature.  Such a peak is rounded, rather than sharp,
however, which adds to my confusion about what you require in
connection with manufacturers of filters "to cut out radiation
corresponding to certain temperatures".  Can you point out an example
of the sort of filter you have in mind?

regards, mathtalk-ga
Subject: Re: Mapping Table from IR Wavelength to Temperature of source
Answered By: welte-ga on 25 May 2003 14:18 PDT
Hi racecar-ga,  Thanks for your question.

The general formula for the peak wavelength of blackbody radiation
emitted by a warm body is as follows:

Lambda [microns] = 2900 / T [deg K],

where Lambda is the wavelength of the light in microns, and T is the
temperature in Kelvin.  To convert from Kelvin to Celcius, Kelvin =
Celsius + 273.15, so the equation becomes
Lambda[microns] = 2900 / (T [celsius] + 273.15).

I generated an Excel spreadsheet to chug through the numbers and
create the following table:

Temperature   Peak Wavelength
(Celsius)          (Microns)
30.0                9.56622
30.1                9.56307
30.2                9.55991
30.3                9.55676
30.4                9.55362
30.5                9.55047
30.6                9.54733
30.7                9.54418
30.8                9.54104
30.9                9.53790
31.0                9.53477
31.1                9.53164
31.2                9.52850
31.3                9.52537
31.4                9.52225
31.5                9.51912
31.6                9.51600
31.7                9.51288
31.8                9.50976
31.9                9.50664
32.0                9.50352
32.1                9.50041
32.2                9.49730
32.3                9.49419
32.4                9.49108
32.5                9.48798
32.6                9.48487
32.7                9.48177
32.8                9.47867
32.9                9.47558
33.0                9.47248
33.1                9.46939
33.2                9.46630
33.3                9.46321
33.4                9.46012
33.5                9.45704
33.6                9.45395
33.7                9.45087
33.8                9.44779
33.9                9.44472
34.0                9.44164
34.1                9.43857
34.2                9.43550
34.3                9.43243
34.4                9.42936
34.5                9.42630
34.6                9.42323
34.7                9.42017
34.8                9.41711
34.9                9.41406
35.0                9.41100
35.1                9.40795
35.2                9.40490
35.3                9.40185
35.4                9.39880
35.5                9.39576
35.6                9.39271
35.7                9.38967
35.8                9.38663
35.9                9.38359
36.0                9.38056
36.1                9.37753
36.2                9.37449
36.3                9.37147
36.4                9.36844
36.5                9.36541
36.6                9.36239
36.7                9.35937
36.8                9.35635
36.9                9.35333
37.0                9.35031
37.1                9.34730
37.2                9.34429
37.3                9.34128
37.4                9.33827
37.5                9.33526
37.6                9.33226
37.7                9.32926
37.8                9.32626
37.9                9.32326
38.0                9.32026
38.1                9.31727
38.2                9.31428
38.3                9.31129
38.4                9.30830
38.5                9.30531
38.6                9.30233
38.7                9.29934
38.8                9.29636
38.9                9.29338
39.0                9.29041
39.1                9.28743
39.2                9.28446
39.3                9.28149
39.4                9.27852
39.5                9.27555
39.6                9.27258
39.7                9.26962
39.8                9.26666
39.9                9.26370
40.0                9.26074

There are several ways to go about filtering out these wavelengths,
depending on your specific application and how narrow a band around
the peak you will need to filter.  Without details of your project, I
will only be able to make some general suggestions in this regard. 
Feel free to ask for clarification.

If you're looking a static (non-changing) object, then possibly the
most cost-effective method of putting such a project together (if
you're thinking of small quantities) would be to use so-called "wedge
filters."  These filters have a different peak transmission at
different positions along the filter (i.e., at one end, they may only
pass 9.0 microns, at the other end, they may only pass 9.5 microns). 
You could position a single filter in the path of the incoming light
and simply vary the position (e.g. using a stepping motor) to change
the center of the filtration.  The variation in the peak is linear
along the position of the filter, making the use of a stepping motor
much easier.  After looking at several companies, the best bet for
such a filter at these wavelengths seems to be Barr Associates.  
They're both in the US and UK.  Here's a link to their wedge filter
page (UK):

Barr Associates can make a narrow band pass filter to your
specifications.  They've worked in the astronomy / astrophysics arena,
so they can make fairly narrow and complicated spectra band filters. 
They have some of their IR filter profiles online at this site:
The animated figure on the above page demonstrates the principle of
the IR wedge filter.  The figure only goes out to 7microns, however
Barr Associates is one of the few companies that goes to longer
wavelengths in filter design.  They state that they can design filters
out to 35 microns.

You would need to move the filter across the light path, buiding up an
image (e.g. on a CCD camera or film) then block the light path while
the filter moves through the range of wavelengths you don't want, then
uncover the aperature as the filter moves to wavelengths you want to
keep.  This may be too slow or complicated to implement, depending on
your project.

The US home page of Barr Associates is here:

Another simpler, but more expensive, option would be to buy notch
filters (filters that pass all wavelengths except a narrow band). 
These could be placed in the light path and would filter out the
unwanted wavelengths without moving anything.  This solution may work
better for more dynamic objects, where you can't take the time to move
a wedge filter across the entire light aperature.  Barr Associates
also makes custom notch filters.  They also make filter arrays, so you
could mount many filters on a wheel and spin it to the needed

Finally, Barr can make custom Edge filters, which have transmission
spectra that look like a step-function (on for a range of wavelengths,
off for wavelengths above or below a cutoff).  Here's a link for

Transmission spectra for samples of all these types of filters are
available by clicking on one of the filter selections on each given
page in the lower right corner.

Another possible supplier is NDC Infrared, although they state that
their narrow band pass filter upper limit is 5000 nanometers (5

If your project is of an academic nature, you may be able to
collaborate with an academic lab.  One such lab is the Infrared
Multilayer Lab at University of Reading (UK), Department of
Here's a link to their filter page:
and a link to an example of their custom notch filter work:

A good dicsussion of blackbody radiation can be found at:

I hope this was helpful.  Please feel free to request clarification or
provide more details on your project if you require other filter
suggestions.  Also, let me know if you require more digits in the peak
wavelength table above, or a wider range of temperatures.

Subject: Re: Mapping Table from IR Wavelength to Temperature of source
From: racecar-ga on 23 May 2003 11:23 PDT
A warm body emits a wide range of wavelengths.  If you're interested
in the wavelength at which power peaks, I think you need to specify
whether power per unit frequency (wavenumber) or power per unit
wavelength.  I think they don't peak at the same place.  In any case,
the Planck radiation law will tell you all you want to know.  I
plugged 30 C and 40 C in and got that the power per unit wavelength
peaks at 9.57 microns and 9.27 microns respectively (note that higher
temperature correspond to shorter wavelengths, unlike in your table).
Subject: Re: Mapping Table from IR Wavelength to Temperature of source
From: dadadee-ga on 24 May 2003 20:08 PDT
Thanks for the information. As mentioned, I would need the 
complete table for wavelength against temperatures in the 
range of 35 degrees celsius to 40 degrees celsius at 0.1 
degree intervals. I would also need a list of manufacturers 
who are able to produce the IR filters for the corresponding 
temperatures before I can accept the submission as an answer. 

Thank you once more.

PS: Please feel free to clarify any doubts that you may have.

Subject: Re: Mapping Table from IR Wavelength to Temperature of source
From: leoj-ga on 18 Aug 2003 14:15 PDT
Wow, $50 for this?  At least he should have gotten it in generic form:

peak wavelength = 0.000293 m/Temperature (in Kelvin)

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