you have tree components: transmitted light, absorbed light and scattered (reflected) light.

http://spie.org/samples/PM229.pdf

http://ase.tufts.edu/biomedical/research/fantini/publications/NIRS_experimental/4_Comp_Series_Photochem_Photobiol_4_211_2004.pdf

try google, it is basic spectroscopy

Not really. Transmitted light differs from the incident one by a degree of sample's absorption ( or extinction=absorption+scattering in a more general case) plus the part that has been reflected. Hence, measuring just the reflectance tells you nothing about what has been absorbed and in the same way, measuring the transmittance tells you only how much light has passed through the sample hiding details about reflection and absorption.

Absorbance and transmittance are related on the other hand. See the technical note attached. 

I guess the answer probably depends on what type of spectroscopy you are referring to.

In general, transmittance is the ratio between the intensity of light passing through some material and the intensity of incident light.

Tr = I_tr/I_0

On the other hand, reflectance is the ratio between the intensity of light reflected off a surface and the intensity of incident light.

Rf = I_rf/I_0

If we assume a scenario based on transmissive spectroscopy, the intensity of light that passes through a sample (I) will be equal to the incident intensity (I_0), minus the contributions of reflective and absorptive processes. Assuming there are no absorptive processes, then all intensity losses are due to reflection, so the transmittance will be exactly the inverse of reflectance (but, again, only if there is no radiation absorption by the sample).

So, in this case...

I_tr = I_0 * (1-reflectance) * (1-absorptance)

I_tr/I0 = transmittance = (1-reflectance) * (1-absorptance)

which, if there is no absorptance (and only in this case), reduces to...

transmittance = 1-reflectance

If, on the other hand, we assume a scenario based on reflective spectroscopy (e.g. ATR spectroscopy), then what is being measured is radiation that is being reflected off the sample, so in these cases it probably doesn't make as much sense to talk in terms of "transmittance", but simply in terms of "reflectance".

All of this assumes, of course, no losses due to light diffusion, which is probably wishful thinking.

Thanks Tomé .  That is what I was looking for.  Barring the numerous factors, I was aware of the simple conversion of Absorption to %T, and while it seemed that reflected light was the difference or left-over of the transmitted light, (possibly an inverted relationship) , I could not find any conversion calculation.  Thanks again.

Serge,  Thanks for the tip and the technical note.  Much appreciated.

What information do you want to obtain out of your reflectance data?

Since the light will to some extend interact with the reflecting surface, it might be possible to obtain chemical information. However the intensity of your signal will be lower in comparison to what you can obtain out of a well prepared transmission experiment.

If the reflecting surface is quite smooth, the light will mainly reflect specular. As a result your obtained spectrum (If you would record one), might show derivative shaped bands. Using the Kramers-Kronig theorem, you can convert this reflected data to absorbance (or transmittance) data.

If the surface is more rough, the light will be reflected diffuse. In this case the Kubelka-Munk theorem can help you to improve the interpretation of your reflected data.

Agree with Serge Grabtchak , he has given nice correct answer.

I am using a Perkin Elmer Instrument ,and its software (I can give the details) has a conversion componet of the spectra between the two.....That is from %R it can give an ''inverse' of this spectra by ploting log(1/R(^-1)) v/s Wavelength...Can anyone help to answer this question on this basis ?

My substrate is silicon so i cnt neglect absorbance to make T = 1 - R cn any1 help how to calculate absorption coefficient from Reflectance data

a = log (1/r)

 Kebelka-Monk Function is well known method for this purpose. if you calculate F(r) then you can convert R% to T or Abs. but it is just estimation. actually T+R+A=1.

see also

http://www.labcognition.com/onlinehelp/en/convert_y-axis_unit.htm

 I have simple question. If absorption is zero. How can I convert reflection coefficient in dB to transmission coefficient?

Hi.... I have attached a file for your clarification. Please see reference 1. 

https://www.perkinelmer.com/lab-solutions/resources/docs/APP_Thin-films.pdf

Daryl:

I see this is an old question, but I don't feel it's been properly answered in this forum yet. So let me add a few comments.

Reflection (%Reflectance, etc.) is an interface, or surface phenomenon. While Transmittance (%Transmittance, etc.) is a bulk phenomenon.

For example, anti-reflection coatings are often applied to lenses to increase the amount of light that is coupled from the air, across the air/glass interface, and into the bulk of the lens. Then, on the other side of the lens, another similar coating again increases the light coupled across that interface. If the coating is efficient, only a fraction of a percent or so is reflected at each interface (for bare glass, normally about 4% would be reflected at each interface). Hence, with an AR coating, the majority of losses are those that occur in the bulk. If the lens is made of high quality optical glass, and the wavelength is in the UV/VIS/NIR region, bulk losses are very small. For example, N-BK7 (Schott glass) has a bulk absorption coefficient of about 0.0024 cm-1.

When you think about reflectance in this way, converting between transmittance and reflectance is rather simple. The transmittance (through the coating/interface) is simply T = 1-R.

Another way to view this is via the Fresnel Equations (see: https://en.wikipedia.org/wiki/Fresnel_equations) that describe the reflection of light at an interface based on the index of refraction of the two materials at the interface, the angle of incidence, and the polarization of the light with respect to the plane of incidence. Under certain circumstances the light could be coupled through the interface and into the bulk with zero reflection, even without an anti-reflection coating. Conversely, for other conditions, light could be totally reflected with zero coupled across the interface. Again, this is a description of a surface phenomenon.

On the other hand, once coupled across the interface and into the bulk, transmittance and absorbance now have a new, simple relationship. By definition, T = 10-A (-log(T) = A). You may already know of the famous Beer-Lambert law (from freshman chemistry) that relates molar absorption coefficient, concentration and path length: A = epsilon * c * l. Not all materials obey the Beer-Lambert law, but most (optically linear) materials do.

So, if you're talking about some kind of optical component, like a lens or a window, and you're using it in its design wavelength band, then you may be able to ignore bulk losses and use the %reflection figure to calculate %transmittance. Just remember there's an interface at both sides: one on the incident side and another on the exigent (cool word!) side. So Transmittance becomes: T = (1-R)2. If you like, it's easy to include bulk losses, just subtract them: T = (1-R)2 - alpha*thickness. Where alpha is the absorption coefficient.

Obviously, these examples are highly simplified. And I haven't even begun to talk about what happens with metals, or non-dielectrics in general. If you really want to get into it, Maxwell's equations and a bit of quantum mechanics enter the discussion. Not to mention physics, chemistry, materials science .... But I'm pretty sure that's not part of your question.

Hi

May have a look attached file

In general, reflection, transmission and absorption depend on the wavelength of the affected radiation. Thus, these three processes can either be quantified for monochromatic radiation (in this case, the adjective “spectral” is added to the respective quantity) or for a certain kind of polychromatic radiation. For the latter, the spectral distribution of the incident radiation has to be specified. In addition, reflectance, transmittance and absorptance might also depend on polarization and geometric distribution of the incident radiation, which therefore also have to be specified.The reflectance r is defined by the ratio of reflected radiant power to incident radiant power. For a certain area element dA of the reflecting surface, the (differential) incident radiant power is given by the surface’s irradiance Ee, multiplied with the size of the surface element, thus

dFe,incident = Ee dA

And also the reflection of light, R, normally incident onto a plane surface is described by the Fresnel equation (see attached file)

Wow, even after such a long time, there is still no comprehensive answer to this question! First of all, it is somewhat restricting to say that reflectance is an interface or surface phenomenom while transmittance refers to the bulk. Such a view is influenced by the Bouguer-Beer-Lambert law which actually assumes that there are no interfaces or by the assumption that the sample is semiinfinite, meaning that there is no second (plane-parallel) interface, from which part of the transmitted light could also be reflected back, or that the sample is so thick and strongly absorbing that light does not reach the second interface (or, again, the first).

Also, it is not clear if the the question refers to specular or diffuse reflectance. Note that for the latter, there are only highly approximate solutions, whereas for the former, a conversion from reflectance to transmittance is always possible, but might be not an easy and straightforward task.

For this conversion, you should first determine what is the correct optical model to represent your sample. In a transflection experiment, e.g. you usually have a semiinfinite incidence medium, a thin layer, and a semiinfinite exit medium (actually a higly absorbing/reflecting medium like a metal). In this case, you usually have coherent mixing of the light amplitudes. A thick, but weakly absorbing sample might be represented by a incident medium / layer / exit medium if its interfaces are plane-parallel but it could be so thick that incoherent superposition of the waves could be assumed.

Once you have chosen the correct optical model, you have to determine the optical constants, i.e. the dielectric function (or the dielectric function tensor, if your sample is anisotropic). If you have the dielectric function then you can calculate for whatever optical model the transmittance. Of course, also for the original sample for which you know the reflectance.

To do that, in the general case 2x2 or 4x4 matrix formalism are employed (see e.g. Article Generalized matrix method for calculation of internal light ...

Article Modelling IR spectra of polycrystalline materials in the lar...

Thanks Thomas. I haven't done anything with this project since asking the question, but you just revived my interest.

I'm going to end this thread. Thanks again.

A=1-R-T

can we get absorbance from reflectance through this formula??

(1-R)2/ 2R*2.30= Absorbance

Iqra Khalid There are a lot of different kinds of absorbances... (specular) reflectance absorbance, transmittance absorbance and, what you mention would be, (diffuse) reflectance absorbance. I would advise against using any of them, but the last one is probably the worst (approximation)...

Dear Thomas Mayerhöfer I have results of the DRS test for my samples, I wanna convert these R-Wavelength data to Abs-Wavelength, Would you please help me to how can I do it?

According to the Kubelka-Munk theory, K should be identical to the absorption coefficient. Accordingly you would have to evaluate the product 4*Pi*K*d/lamda where d is the length of the lightpath to obtain absorbance.

Can some one explain me why blue shift in UV absorbance was observed after cobalt doping ???

When you think about reflectance in this way, converting between transmittance and reflectance is rather simple. The transmittance (through the coating/interface) is simply T = 1-R. ... Conversely, for other conditions, light could be totally reflected with zero coupled across the interface.

Here are very detailed explanations of the connections between the two quantities, certainly also valid for other spectral ranges:

Preprint Wave optics in Infrared Spectroscopy

Gul Bint Ghulam dear, as I know that you have a doping to ZnO, the Fermi energy level rises to near the conduction band edge, as a result the band gap value increases, which is defined as the blue shift. I hope it is ok.