How to convert from nm to cm-1
Anthony Cook
cm? For example 1550 nm equal how many inverse cm?
Next, how do I convert linewidth in nm to linewidth in inverse cm?
For example a linewidth of 10 MHz equals how many inverse cm? Or, a
linewidth of 0.5 picometers = how many inverse cm? Consider a laser
source at 1550 nm, for example.
Anthony Cook
Petter Hagberg
Hi Anthony,
I am mildly surprised that NASA has problems with freshman wave-physics,
hope this is not a joke..?
Inverse cm is the cgs unit for the wavenumber k, and as you know,
k=2*pi/lambda. Now 1550 nm equals
1.55*10^-4 cm so the wavenumber k would be 2*pi/1.55^10-4 = 40536.7/cm.
When you consider linewidths the problem is a bit more intriguing (say
second year physics...:-)
lets call the wavelength L and the frequency f: c=f*L where c is the
speed of light. Substituting k for L you get
c=f*2*pi/k so k=2*pi*f/c (=omega*c, omega is the angular frequency
usually used). When considering linewidths we have an uncertainty df in
the frequency. Differentiating the relation between k and f yields dk
=2*pi*df/c. In your example with df = 10 MHz,dk=0.2094 /cm. If you have
the linewidth in wavelength instead (dL) we have (again) k=2*pi/L.
Differentiating this we get dk=(-)2*pi*dL/L^2. You can omit the minus
sign, dk is really an absolute value.With a linewidth dL of 0.5 pm
=5*10^-11 cm at a wavelength of 1.55*10^-4 cm we get dk = 0.013/cm.
Best regards,
Petter Hagberg
Mahendra Prabhu
> cm? For example 1550 nm equal how many inverse cm?
The absolute wavelength can be converted to inverse cm by the simple
equation:
k (cm^-1)=1/lambda (units in cm)
eg. 1550nm = 1550e-7cm
k (cm^-1) = 1/1550e-7
= 6.4516e3
> Next, how do I convert linewidth in nm to linewidth in inverse cm?
> For example a linewidth of 10 MHz equals how many inverse cm? Or, a
> linewidth of 0.5 picometers = how many inverse cm? Consider a laser
> source at 1550 nm, for example.
To calculate the frequency shift (in cm-1 region) accurately, one needs
to calculate it only in frequency regime (cm-1 regime).
k1 = 1/lambda1;
k2= 1/lamda2;
del_lambda = 0.5e-12
del_k = k1-k2
k1-k2 = 1/lambda1 - l/lambda2
or
del_k = del_lambda/(lambda1 * lambda2)
In your case as your del_lambda is 0.5e-12
and lambda is 1550e-9
del_k = 0.5e-10/(1550e-7)^2 <--- here i have assumed lambda1 is
approx equal to lambda2
del_k = 0.0021
Hope this answers your question
--
Mahendra Prabhu
Research Student
Institute for Laser Science
University of Electrocommunications
Chofugaoka 1-5-1 Chofu-shi,
Tokyo 1828585
---------------------------------
Phone: +81-424-435714 (O)
+81-424-891958 (R)
070-6566-4610 (Mobile)
Fax: +81-424-858960
home: http://prabhu.peaceful.nu
Petter Hagberg
You are right, I was thinking of the circular wavenumber. It is probably
more appropriate to use the ordinary inverse wavelength instead of the
circular wavenumber. Omit the factor 2*pi in my reply and (I hope) we get
the same thing.
/Petter
Pieter Kuiper
<pra...@ils.uec.ac.jp> wrote:
> The absolute wavelength can be converted to inverse cm by the simple
> equation:
>
> k (cm^-1)=1/lambda (units in cm)
The definition k = 2*pi/lambda is much more usual.
The factor comes from the way one writes travelling waves:
u(x,t) = sin(k*x - omega*t)
omega = 2*pi*f = 2*pi/T
--
Pieter...@itn.hh.se http://www.hh.se/staff/piku/
Anthony Cook
time in a somewhat hectic schedule. While we are all scientists, we
all have areas of expertise. In the areas in which we do not spend
most of our effort, many times we do what is most efficient to find
the answers we need. Yes, I could have spent some time digging in the
proper places to find the answer to the questions I posed, or I could
draw upon the knowledge of the members of a newsgroup for a
time-efficient answer. Perhaps I, or someone else, can save you some
time in the future when you need a quick answer to a question outside
the area of your concentration.
Anthony
Petter Hagberg
No intent to offend, it just felt a little funny to see NASA in the
organization line...I just took the opportunity to make a little joke. But
there was a serious undertone in it: in my mind any scientist working in the
fields of physics/technology/space research/lasers/etc should know this or be
able to figure it out (e. g. by looking at the units...). I think that the
problem here is that you did not recognize the unit (cm^-1). Its an awkward
relic of the cgs system that still is used widely in spectroscopy, mainly I
think for historical and conventional reasons. In my mind, eV or nm or
possibly Hz are clearer in for example a term scheme of an atom. I guess that
you just wondered why the hell one would use cm^-1 as an energy unit (which
it is used as..) and I agree with you on that one. Also, since I am working
in the field of laser science, this seems like the "basics of the basics" for
me, while I am probably completely ignorant of the fundamental cornerstones
of your field...but you easily get home blind... :-)
Best regards,
Petter
Anthony Cook
No offense taken, just a little note to let you know where I was
coming from. By looking at the units, my first inclination was to
convert 1550 nm into cm and then take the inverse, but I would have
been off by 2*pi. That's the kind of info I was looking for. I know
I could have found the answer if I dug in my Physics book, but chose
the lazy route and asked here. :-)
And then for the linewidth it gets more involved, so I took the
opportunity to ask both the simple and the more complex questions. I
convert all the time from nm to MHz when dealing with linewidths (and
this can be tricky if you don't know the correct way to do it; and the
answer isn't immediately available in a lot of books), but we have
never dealt directly in inverse cm for wavelength or energy.
Thanks for your inputs. I'm on the road now to working in inverse cm.
Andy Alexander
: The definition k = 2*pi/lambda is much more usual.
If he's talking about spectroscopic notation (which I am assuming he is),
then simply,
nu_bar (cm-1) = 10^7 / lambda (nm)
so, e.g., 9398 (cm-1) = 10^7 / 1064 (nm)
The 10^7 is just nanometers -> centimeters conversion. No 2*Pi.
cheers,
andy
User702495
>cm?
Remember 100.
100 inverse centimeters is 100 microns.
50 inverse centimeters is 200 microns.
Etc.
Regards, Nelson Wallace
PS. 100 microns is 100,000 nm
Nelson Wallace's Optics Site:
http://members.aol.com/user702495/index.htm
Petter Hagberg
> Petter,
> No offense taken, just a little note to let you know where I was
> coming from. By looking at the units, my first inclination was to
> convert 1550 nm into cm and then take the inverse, but I would have
> been off by 2*pi. That's the kind of info I was looking for. I know
> I could have found the answer if I dug in my Physics book, but chose
> the lazy route and asked here. :-)
>
> And then for the linewidth it gets more involved, so I took the
> opportunity to ask both the simple and the more complex questions. I
> convert all the time from nm to MHz when dealing with linewidths (and
> this can be tricky if you don't know the correct way to do it; and the
> answer isn't immediately available in a lot of books), but we have
> never dealt directly in inverse cm for wavelength or energy.
>
> Thanks for your inputs. I'm on the road now to working in inverse cm.
>
>
Anthony,
You were on the right track from the beginning. The factor 2*pi is necessary when
you compute the circular wavenumber or the wave vector k. The inverse wavelength
is commonly used in laser science and then you omit the factor 2*pi. Check your
literature and see which definition they use.
/Petter
Nikos Michalopoulos
1cm is (10 to the -7th power) nm , example:1550nm=1550*10(-7)cm
1cm is (10 to the -10th power) pm
about how you can convert f Hz to L m the equation is: L=c/F , c: speed of light.
if the means of transmission is the air or else you have to use another equation and I don't remember it now :)
Nikos Mihalopoulos
Anthony Cook wrote:
Could anyone tell me how to convert absolute wavelength into inverse
cm? For example 1550 nm equal how many inverse cm?
Next, how do I convert linewidth in nm to linewidth in inverse cm?
For example a linewidth of 10 MHz equals how many inverse cm? Or, a
linewidth of 0.5 picometers = how many inverse cm? Consider a laser
source at 1550 nm, for example.
Anthony Cook
Prof Harvey Rutt
Pieter Kuiper wrote:
> In article <375CDE05...@ils.uec.ac.jp>, Mahendra Prabhu
> <pra...@ils.uec.ac.jp> wrote:
>
> > The absolute wavelength can be converted to inverse cm by the simple
> > equation:
> >
> > k (cm^-1)=1/lambda (units in cm)
>
> The definition k = 2*pi/lambda is much more usual.
I would agree this is the usual definition of 'k' in wave theory. But that
is not the conventional 'wavenumber', it would probably be called circular
wavenumber or similar, a rare term.
But in normal spectroscopy it is certainly NOT usual to include the factor
of 2 pi.
There are thousands of infrared & visible instruments calibrated in
wavenumbers, & vast numbers of papers & books giving spectroscopic
constants & tables, & I've never seen one of those include the 2 pi. It is
just one over wavelength in cm.
Incidentally, it should be the vacuum wavelength strictly; (not in air) at
high resolution that matters.
Harvey Rutt
tschramNOSPAM
> you compute the circular wavenumber or the wave vector k. The inverse wavelength is
> commonly used in laser science and then you omit the factor 2*pi.
the same is valid for the use of FTIR-spectroscopy for chemical characterisations.
This is probaly also the reason for the still frequent use of this rather odd unit.
You won't be able to convince a chemist to use another unit, since he would have to
translate almost whole his spectral library that is used as fingerprints for the
studied molecules.
AES
Angstroms (which is 1 micron) is, fortuitously but fortunately, also
(exactly) 10,000 wavenumbers.
So, just work inversely from there -- e.g. 5000 Angstroms (middle of the
visible) is 20,000 wavenumbers; or, 10.6 microns (CO2 laser) is a little
under 1000 wavenumbers.
Combine that with the rule-of-thumb approximation that energy in eV =
1.24/wavelength in microns, and you're pretty much set: 5000 Angstroms
is also 2.48 â 2.5 eV.
Also useful to remember that 1 wavenumber = 30 Ghz (almost exactly; to the
degree that c = 3); so 5000 Angstroms is also 60,000 GHz = 6 * 10^14 Hz.
Allan Colquhoun
Slight slip of the finger I think, visible light is ~600,000 GHz ie 600 THz,
though
6*10^14 Hz is correct.
I noticed the following symmetrical pair and always wondered why no one else
used it.
At a wavelength of 547 nm the frequency is 547 THz.
I know the exact number is 547.53... but it is the concept that is important
and
then the ability to scale. Has anyone used this? Are there any references?
Allan Colquhoun