What is the physical significance of absorbance? Is it not possible to directly decipher the amount of radiant energy absorbed by the sample from transmittance? Why is that negative log of Transmittance is called absorbance?

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    $\begingroup$ I think you should post one question per one post. $\endgroup$ – Mockingbird Jun 22 '17 at 10:28
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    $\begingroup$ The are all closely related, stemming from the main question. $\endgroup$ – Girish Kumar Kaitholil Jun 22 '17 at 11:36

Absorbance is useful because it is additive. That is, it's absorbance which is used in Beer's law:

$$A = \epsilon \cdot c \cdot l$$

While you can certainly make a version of this law which uses transmittance instead of absorbance, the math becomes much less straightforward.

The explanation of why absorbance is additive falls out of the derivation of Beer's law. The official derivation relies on calculus, but you can get a sense of things by thinking of the solution not as a single thing, but as a succession of "slabs":

Take a single thin slab of a colored material. This absorbs a certain amount of light at a given frequency. But it doesn't absorb a fixed amount of light. Instead, each photon that passes through has a certain probability of encountering an absorbant molecule and being absorbed. The more light, the more absorbed. For a given slab under typical conditions, we can define the percent transmittance (%T), which is the fraction of light that goes through that material.

But what happens if you start varying the width of the material. Or equivalently, if you stack several of such slabs together. The amount of light going through one slab would be %T. The amount going through two slabs would be %T times %T. (For example, if the slab had a %T of 50%, then the first slab would absorb half the light. The second slab would only "see" half the original light, and would then absorb half of that light. So the amount being transmitted out the back of the second slab would be only 0.5*0.5 = 25% of the total amount.) Three slabs gives you %T * %T * %T. Generally, for N slabs, you'd get only $(\%T)^N$ of the light coming through. (Note the necessity of converting from percentages to fractions to do the calculation here.)

Okay, but how about varying concentration? Well, since the chance that a photon is absorbed is proportional to the chance it interacts with an absorbing molecule, then if you double the concentration, you double the number of interactions that could happen. If you work out the math, it turns out that (in dilute solutions) doubling the concentration is equivalent to just doubling the path length - you effectively have two "slabs", just compressed into the space of a single slab. The effect is the same as increasing the path length. So for a solution at twice the concentration, you's have $(\%T)^2$ the transmittance.

Note the use of exponents. This is all because when you combine two things which absorb, you have to multiply their percent transmittance. Exponents are hard to work with. So instead, we do a trick which can convert mulexponentstiplication into addition: take a logarithm. If we work with logarithms of the transmittance, all of those multiplications are now additions, and instead of using exponents for path length and concentration, we multiply by each. Taking the negative log is just another trick to make things work out reasonably: it's more convenient to have the measurement move in the same direction as concentration and path length (such that more absorbance means a larger concentration), so we stick a negative sign on the result of the logarithm to make things easier.

That's basically it. You certainly could do all your calculations with transmittances, but the equations and reasoning gets more complicated. If you use absorbances, you get the nice Beer-Lambert law, which states the concentration and path length are proportional to absorbance.


Beer's law relates transmitted intensity to concentration $[c]$ of the solution at each wavelength $\lambda$ as $$ I_{trans_\lambda}=I_0e^{-\epsilon_\lambda [c]L}$$ where L is the sample path-length and $\epsilon$ the extinction coefficient of the molecule at wavelength $\lambda$ for an incident intensity $I_0$. (The extinction coefficient measures how strongly the molecule absorbs at each wavelength). Thus taking logs produces $$ \ln(I_{trans_\lambda}/I_0)=-\epsilon_\lambda [C]L$$ The product $\epsilon [C]L$ is also called the optical density which can be called the absorbance A. Thus Beers law can be written as $$ I_{trans_\lambda}=I_0e^{-A_\lambda}$$

It is clearly matter of convenience to use absorbance rather than transmittance as this is directly proportional to concentration. (In ir spectroscopy however, transmittance is still commonly used.)

[ Note that it is usual in practice to define the Beer Lamber law with a power of 10 and not e (as above) thus $\log_{10}$ is used rather than $\ln \equiv \log_e$]


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