I measured reflectance and transparence of copper oxide thin films and when I calculated refractive index in range 400-600 nm I have no results because coefficient absorption so high.
mathematiquelly under square root negative result we can’t calculated. but physically what does mean?
N= (1+R)/(1-R) + ((4R /(1-R)^2 +k^2 ))^0.5
is any films have the same phenomen?
Dear Ben Messaoud Ouarda can you give me a reference from where you formula for n comes from? Also, can you provide your reflectance and transmittance spectrum to see if there is a problem with one or the other?
Dear Ben Messaoud Ouarda , the equation that you are using is probably derived for a single interface between two materials. It will not take into account interference effects that occur in thin films (it means that the apparent oscillation of refractive index on your graph may not be real).
Thomas Mayerhöfer is right, it would help to see the reflectance and transmittance spectra.
hello Thomas Mayerhöfer, those our spectrum of transmittance and reflectance
and the formula of refrective index is below this is what i found in literature.
N= (1+R)/(1-R) + ((4R /(1-R)^2 -k^2 ))^0.5
It looks like the transmittance is very low below 600 nm. From the formula you used I suspect you used the Beer-Lambert approximation to calculate from T via -log10T the absorption index k, right? This must fail, not only because the high instrumental errors below 600 nm, but also because you also have to consider the high and varying reflectance in this spectral region. In addition, as Daniel Poitras pointed out, you have to factor in interference effects. Can you tell us more like how thick is your layer and what type of substrate did you use? How does its transmittance and reflectance spectra look like?
i used a glass as substrate and my layer is so thick about 194 nm
the formula of calculate k and ansorption coefficient alpha
k = αλ/4π
α = 1/ d ln [ (1 − R) 2 /2T + ((1 − R)2 /2T + R2 )1∕2 ]
If you have a thin layer, then above formula will not work because of, as already mentioned, interference effects. This is easy to see, because alpha is in this case definitely not a linear function of the thickness.
P.S.: What kind of glass did you use as substrate?
Glass microscope slides
i repeat this sample at 0.075 mol/l twice and i have the same problem while the other sample at different molarity (0.025,0.05and 0.1 mol/l) i don't have any problem.
There are a few ways to obtain the transmittance and reflectance equations in cases with more than one interface, one of them is the matrix approach [as described in https://en.wikipedia.org/wiki/Transfer-matrix_method_(optics) ]. However, your will need to fit your R and T spectra with the resulting equation in order to extract n and k. The calculation could be simpler if your film was thicker, with many interference maxima and minima in its spectra, so that you could be able to use an envelop method and extract n and k more directly; the following paper describes the method, and is freely available on researchgate [ Article Cumhuriyet Üniversitesi Fen Fakültesi The optical constants ...
].
If you like, show us the transmittance and reflectance spectra for the samples that work. Its counterintuitive, but if interference effects are at play, absorbance can be higher for thinner samples...
In addition to the sources Daniel Poitras mentioned, have a look at my lecture notes under Preprint Wave optics in Infrared Spectroscopy
(the corresponding equations certainly also work in the UV/Vis). Good luck with your project!- Badges
- Science method
- Similar topics
- Methodology
- Crystallography
- X-ray Diffraction