# Absorptance versus absorbance

## Main Question or Discussion Point

I want to know what exactly is the physical significance of absorptance versus absorbance. I have not come across a reference that clearly explains the distinction between the two quantities.

From the definitions:

When light is incident on a material, by finding the fraction that gets reflected and transmitted, we can find the absorptance — the fraction that is absorbed:

$$\underbrace{R}_{\text{reflectance}}+\underbrace{A}_{\text{absorptance}}+\underbrace{T}_{\text{transmittance}}=1$$

Absorbance (also called optical density) follows from the Beer-Lambert law and is written:

$$\text{Absorbance}=\log_{10}\left(1/T\right)$$

For a given sample, I used a spectrophotometer to find both quantities. But the curves appear to give contradicting information about my sample:

Here is the problem:

The absorptance curve above shows that in the red (~680 nm) and in the blue-violet (<450 nm) regions we have approximately the same level of absorption. But the absorbance curve shows that the absorption is much greater in the blue-violet than in the red. These can't both be true.

I believe that these quantities are providing different information about the sample. But what exactly is the difference?

Any explanation would be greatly appreciated.

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DrDu
I can't see no contradiction. Note that the absorbance involves the logarithm and the inverse of transmittance.

HAYAO
Gold Member
You should look closely at the equations.

What is the maximum number that Absorptance can take? (Equation 1)
What is the maximum number that Absorbance can take? (Equation 2)
Why are they different?

A little note that may help you. The variables (R, A, and T) in the first equation are not independent.

You should look closely at the equations.

What is the maximum number that Absorptance can take? (Equation 1)
What is the maximum number that Absorbance can take? (Equation 2)
Why are they different?

A little note that may help you. The variables (R, A, and T) in the first equation are not independent.
Hi HAYAO and DrDu,

I can understand why the curves look different from a mathematical standpoint. I need to know the physical phenomenon described by each curve.

If our aim is to heat a material to cause thermal destruction, which quantity should we be looking at for the selection of wavelength? In various papers, I have seen both absorbance and absorptance curves being taken as the receptiveness of the material to radiation of different wavelengths. This is where they give contradicting information.

Suppose we have two light sources: one is blue and the other one is red. The absorptance curve in my first post says that they would both deposit roughly the same amount of energy into the material. But the absorbance curve says that the material is much more receptive to blue. So, for maximal placement of energy, should we select the red or the blue?

P. S. The maximum value of absorptance is 100%. For absorbance, the range of the function is ##(-\infty, \infty)##, so 100% T (T=1) means 0 absorbance, and as T tends to 0 the absorbance tends to infinity.

HAYAO
Gold Member
I don't agree that the absorptance at 450 nm is the same as the absorptance at 680 nm in the first place. (Nor do I think transmittance at 450 nm is the same as the transmittance at 680 nm.)

And as DrDu said, Absorbance involves logarithm of the inverse of transmittance. From the first graph, the transmittance at 450 nm looks like it is about 0.02 (2%) and 680 nm is about 0.04 (4%). Try putting those numbers into the Absorbance equation (Equation 2). See the difference in the Absorbance values.

roam
DrDu
Absorbtance is the fraction of the intensity of the incident light which really gets absorbed by the substance. So multiply absorbtance by the incident intensity and you will get the energy deposed per second in the substance.

roam
mjc123
Homework Helper
Consider the derivation of the Beer-Lambert law. Light of intensity I passes through a thin section of sample of thickness dx and concentration of absorbing species c. The amount of light absorbed is proportional to the intensity and the amount of absorber per unit area (= concentration x thickness).
dI/dx = -αIcdx where α is a constant we call the extinction coefficient, and is a measure of how strongly the substance absorbs light at that frequency.
Integrating this gives I = I0e-αcd, where d is the sample thickness
or I = I0* 10-εcd, where ε = α/2.303
Absorbance = -log(I/I0) = εcd
Thus absorbance is directly proportional to a quantity that reflects the substance's intrinsic propensity to absorb light. Absorptance is not (neglecting reflectance, it is 1-e-αcd). However, this might not be what you want. For your purposes, you might be more interested in the absorptance of the sample (how much light the sample will actually absorb).

roam
Absorbtance is the fraction of the intensity of the incident light which really gets absorbed by the substance. So multiply absorbtance by the incident intensity and you will get the energy deposed per second in the substance.
Hi DrDu,
Are you sure? Because in some papers they give the deposited energy as the absorption coefficient (not the absorptance) multiplied by the intensity.

Hi mjc123,
But, that extinction coefficient, in some texts is defined as ##\alpha = a + b## where ##a## and ##b## are the absorption and scattering coefficients, respectively. The heat deposition rate is then something like ##Q=a.I##.

This means that absorbance, rather than absorptance, is the quantity that is directly related to heating as a result of irradiation. Isn’t that correct?

Hi HAYAO,
I know that absorptance at 450 is higher than it is at 680, and I agree with the rest of your observations. The actual transmission values are:

$$\begin{array}{c|cc} \lambda \ (nm) & 450 & 680\\ \hline A\ [\%] & 2.7 & 3.7 \end{array}$$

It's at ~300 nm that it becomes 0.02, at which point the decadic logarithmic relationship should produce negative values. But the absorbance is very high in the UV regime.

I understand why the discrepancy exists mathematically. But absorbance still predicts that a larger proportion will be absorbed than it actually is. What are the physical processes that prevent that amount from actually being absorbed by the material?

DrDu
Absorbance only refers to the light which does not get reflected. So a high absorbance does not necessarily mean that much of the light really gets absorbed (rather, it may be reflected before entering the substance). You also have to take in mind that both absorptance as well as absorbance depend on the thickness of the sample. For a thick sample, absorptance will approach 1-R while absorbance goes to infinity. This should make clear that you are more interested in absorptance than absorbance.

roam
HAYAO
Gold Member
Hi HAYAO,
I know that absorptance at 450 is higher than it is at 680, and I agree with the rest of your observations. The actual transmission values are:

$$\begin{array}{c|cc} \lambda \ (nm) & 450 & 680\\ \hline A\ [\%] & 2.7 & 3.7 \end{array}$$

It's at ~300 nm that it becomes 0.02, at which point the decadic logarithmic relationship should produce negative values. But the absorbance is very high in the UV regime.
1) Transmittance is T, not A
2) Transmittance of 0.02 means Absorbance of ~1.7. It will not take negative values. You probably forgot to inverse Transmittance (look carefully at Equation 2).

In a typical absorption spectroscopy, transmittance of ~0.02 is way too low to provide a quantitatively reliable result in terms of the value of Absorbance. Experimental conditions should be set up so that Absorbance is in the range of 0.1 to 0.9, and preferably 0.2 to 0.8. That means transmittance between 80% to 13%.

roam
Absorbance is the log of the ratio of power of incident light on the sample to the power of transmitted light, it doesn't tell you anything about how much light bounces off the sample vs gets absorbed by the sample.

Absorptance is the fraction of light actually absorbed by the sample, determined by accounting for transmission as well as reflectance.

So, one only cares about the light that gets through while the other only cares about the light the sample actually absorbs.

I believe this is the kind of "in words only" explanation you might be interested in (hopefully it's accurate). Everyone else did a much better technical job than I could at differentiating the two for you.

Last edited:
roam
Hi,
As you mentioned absorptance is A=1-R-T. It only shows the attenuation of light by absorption from the material. However Absorbance shows the attenuation of light by different physical processes which can be absorption, scattering and reflection. So I think the reflection of your sample in the blue-violet range is contributed in increasing the attenuation measured by absorbance.