Talk:Beer–Lambert law

Howzat Again?

"...relates the attenuation of light to the properties of the material through which one substance like light, neutrons or host rarefied gases is traveling"

Is light now a substance?

It's not clear what the defining characteristics are of the class "substance like light, neutrons or host rarefied gases". Would whey (traveling through cheesecloth) count?

So if you have neutrons passing through a material, light is attenuated? What light is affected?

I realize that it's difficult to explain highly technical subjects to fourth-graders, but it would make this highly important page MUCH more useful if was written clearly, instead of simply tossing keywords on the page, without bothering to ensure the sentences make sense.12.193.238.99 (talk) 23:08, 20 September 2014 (UTC)

I agree with the concerns expressed that the introduction had become needlessly convoluted. Concepts were introduced that were never developed further. Advanced concepts were included that were confusing. I have edited the introduction with the goal of highlighting the key elements. A note to the commenter, all users are welcome to edit Wikipedia, and your insights were valuable, so I suggest you follow through and give editing a try. ronningt (talk) 00:39, 22 September 2014 (UTC)

Comments

• Equations:"The frequency scale in molecular spectroscopy is often in cm-1, wherefore the lineshape function is expressed in units of 1/cm-1, which can look funny but is strictly correct." Correct me if I am wrong, but doesn't 1/cm^-1 just equal cm? Kbw1988 (talk) 12:43, 15 May 2008 (UTC)
• Fellow annonymous editor, indeed a lot of people (myself included) are used to your notation for writing the equation. However, if you have not realzied, all those other equations on the article call absorbance α and optical path &l; - not only the equations, that neat diagram which would have to be re-edited if we would change the notation. Not to mention that anyone with minimal common sense will read the legend underneath the equations and immediately identify α with ε within ten seconds, even if they find it a bit odd at first. So, we shall take the easy way on this and avoid an endless discussion on mere names... I better stop here.--Duplode 03:36, 14 April 2006 (UTC)

Dr Bob, I'm not sure where you are from...but google beer lambert and mostly everyone in our World (Earth) uses A = ε b c.

Dr Bob...the reason we introduce other variables is because not everyone is like dr bob and some of us use different variables. : )

• No, you did not introduce other variables; you introduced other names for variables. If you call absorptivity α, ε or whatever you wish it's still absorptivity. Therefore there is no need of introducing more letters and causing additional confusion when there is a clear legend under the equations.--Duplode 07:50, 8 April 2006 (UTC)

Yes, but wouldn't it be better if we used standard IUPAC nomenclature "A = εcl" instead?

Does this make sense: "If concentration c is expressed in moles per unit volume, α is a molar absorptivity usually given in units of mol cm^-2?" Shouldn't it say in units of cm² mol^-1? Srnec 04:30, 16 February 2006 (UTC)

Dimensional analysis agrees with you, but perhaps there is some other error.--Srleffler 05:24, 16 February 2006 (UTC)

• That was just a simple error, it should be mol^-1 cm^-2. I've fixed it and also recast the law in terms of log10, since that seems to be the most common way the law is expressed (my lab spectrophotometers all work that way). This also makes it compatible with related articles in WP (e.g. optical density), and inconsitancy is bad. Unfortunately, in general optics it's also pretty common to see it in the exponential form, so some confusion is possible. --Bob Mellish 20:43, 16 March 2006 (UTC)

• It would be totally cool to define 'N' in the second equation. 208.252.219.2 (talk) 00:34, 25 February 2010 (UTC)

• In the derivation section, how can the fraction of absorbed photons be negative? Either the language or the sign of the equation for differential absorption is incorrect. 208.252.219.2 (talk) 00:38, 25 February 2010 (UTC)

  Defined. Materialscientist (talk) 00:39, 25 February 2010 (UTC)

Picture

Am I the only one who feels the picture is unsuitable? The rhodamine is fluorescing - the light that is diminishing with greater pathlength is not the incident beam but light arising from the absorption of the incident beam. that's adding a whole new layer of confusion - could someone post a similar picture with a non-fluorescing material? Or a picture of a pane of glass from the from and side, showing how the increased pathlength leads to increased absorption resulting in the glass looking green from the side — Preceding unsigned comment added by 86.46.54.236 (talk) 16:30, 6 July 2012 (UTC)

This picture should be indeed removed. The figure illustrates fluorescence and not absorption! Trelam (talk) 19:48, 11 February 2015 (UTC)

Transmittance

I think the article should more explicity explain the relationship beween absorbance and transmittance, or point to an article that does cover this. It is sort of in the article already (Ii/Io), but that wouldn't be clear to the uninitiated. ike9898 17:43, 14 March 2006 (UTC)

• Another observance about this page. With the italic font being used, the characters representing intensity (I) and distance (l) both look confusingly similar.

Derivation added

I've added a simple derivation of the law, with an explanation of the problem at high concentrations. I've used the notation in the previous section, although I do not like it - for reasons that others have already mentioned. It seems we're married to the notation used in the figure, and this is unfortunate. ike9898 - I added a line defining transmittance after seeing your comment.

This is actually my first contribution, and my first attempt at LaTex. The fonts don't look quite right, and I would welcome anyone attempting to clean it up.

Axewiki 12:51, 13 September 2006 (UTC)

I fixed some of the Tex, and removed the last paragraph. The law works even for high concentrations, because the thickness of the slab is dz which is infinitesimally small. For a slab of finite thickness, the chance that there will be an overlap decreases as the thickness of the slab decreases, until it is zero when the slab is infinitesimally thick. PAR 03:00, 14 September 2006 (UTC)

I am not familiar with the common notation for this law, but the article should be consistent. In the first part, base 10 and l is used:

$I_{1}/I_{0}=10^{\alpha lc}\,$

while below, base e and $\ell$ is used

$I_{1}/I_{0}=e^{\alpha \ell c}$

Thanks PAR - your Tex is much better. I also agree with your comment about consistency between base 10 and natural logs - I have fixed that. However, I must insist that the Beer Lambert law breaks down at high concentrations, and it does so for the reason I gave. Even the author of the first section on "Equations" mentions this point. In your 'talk' paragraph above, you mention that the slab becomes "infinitesimally thick". I think you meant "thin", but no matter, the breakdown has nothing to do with the calculus or the thickness of the slab, only with the likelihood that one molecule along the path of a photon eclipses another, making the second molecule "invisible" to whatever detector is measuring the transmitted light. This likelihood of such an eclipse increases with increasing concentration. Breakdown of the Beer-Lambert law at high concentrations is a universally recognized phenomenon that I routinely use as a teaching demonstration for students in my lab. Please don't take offense that I reinserted the final paragraph in this section.

Axewiki 16:05, 14 September 2006 (UTC)

This looks a lot like the few other derivations I was able to find on the web.... neffk (talk) 06:23, 21 November 2009 (UTC)

I left a nice clean derivation on this website in 2006, and came back now only to find it replaced by a messy cluttered derivation, wholly unsuitable for my students. Symbols are unnecessarily complex, difficult to read, and in at least one case, undefined. Rather discourages efforts to edit ... Axewiki (talk) 03:19, 13 July 2015 (UTC)

Beer-Lambert Lambert and Beer laws

Beer-Lambert law is a conection of Beer law and Lambert law, so in the very first line is untrue information. Both are now historical, but...

the Lambert law states for that absorption is proportional to the light path length,

the Beer law states for that absorption is proportional to the concentration of absorbing species in the material.

So if you have time to fix it, please do it. Probably it will need a minor changes in biographical articles about Beer and Labbert. --82.139.42.165 23:16, 25 September 2006 (UTC)

I wonder what Bouguer did. In all Soviet textbooks the law is called "Bouguer-Lambert-Beer Law", whereas american textbooks don't even mention Bouguer at all. Was he a communist? :) Cubbi 21:44, 17 March 2007 (UTC)

As these were three distinct individuals en-dashes (–) should be used, rather than hyphens (-)

—DIV (128.250.80.15 (talk) 09:34, 22 April 2008 (UTC))

P.S. Hyphens shouldn't be used in mathematical notation either! Some of the exponents in the article have them. —DIV (128.250.80.15 (talk) 09:41, 22 April 2008 (UTC))

History

Precise references are needed to back up the history section, particularly with regards the role of Bouguer. Ga2re2t 17:00, 25 October 2007 (UTC)

Relationship of k to alpha

I found the equation $\alpha ={\frac {4\pi k}{\lambda }}$ (4th equation from this article) to be confusing

I reasoned, since alpha has various sets of units (see paragraph below) so must k, yet the way k is derived in the article Extinction coefficient indicates that k has only one set of units, and if the equation above is correct alpha must have only one set of units, which comes back to my point that I am confused.

So I wanted to ask/plead that someone who knows whats going on, amend either the paragraph below quoted from the article or the equations, such that the whole thing is less confusing. Or at least so that the units for the equation $\alpha ={\frac {4\pi k}{\lambda }}$ are clear.

"The units of c and α depend on the way that the concentration of the absorber is being expressed. If the material is a liquid, it is usual to express the absorber concentration c as a mole fraction i.e. a dimensionless fraction. The units of α are thus reciprocal length (e.g. cm⁻¹). In the case of a gas, c may be expressed as a density (units of reciprocal length cubed, e.g. cm⁻³), in which case α is an absorption cross-section and has units of length squared (e.g. cm²). If concentration c is expressed in moles per unit volume, α is a molar absorptivity (usually given the symbol ε) in units of mol⁻¹ cm⁻² or sometimes L mol⁻¹ cm⁻¹." — Preceding unsigned comment added by 139.67.205.181 (talk)

I think it is ok to have multiple units for $\alpha$ and $c$ but only one unit for $k$. The reason being that the if you use the equation $\alpha ={\frac {4\pi k}{\lambda }}$ to relate $\alpha$ and $k$, then \alpha will be in units of cm^-1 (or L^-1) and correspondingly the concentration $c$ will have to be expressed in mole fraction. However, for a given substance, the concentration can be expressed in other units by using appropriate constants. for example, it is common to refer to both frequency ( $f$ ) and angular frequency ( $\omega$ ) as frequency, even though they have different units (Hz) and (rad/s) respectively. The two can be readily interchanged by noting that $\omega =2\pi f$. Anyhow, I will try to fix the statement soon, so that it is not confusing. Thanks for the comment. -Myth (Talk) 05:30, 9 February 2007 (UTC)

Something wrong with alpha

Something is wrong with the first section (2007-03-22). There's a missing ln(10) somewhere. This is obvious, since A is defined in terms of log10, but the rest of the constants are non-log quantities. I'm not sure where it goes, though. —The preceding unsigned comment was added by 192.249.47.9 (talk) 19:04, 22 March 2007 (UTC).

Destructive edit

A user named "Anarotram" deleted a section of the derivation I composed and left the message "I deleted the last comment that appears entirely wrong. The requirement on the slab thickness dz is not involved in the integration step. I deleted the last comment that appears entirely wrong. The requirement on the slab thickness dz is not involved in the integration step.". There is no record of this user making any other edits to Wikipedia pages.

This comment does not explain what Anarotram believes is wrong. The integration is clearly over dz, so the comment is puzzling. Also, the sentence beginning with "The requirement ..." is fractured English, so I'm not even sure what it means.

The effects of error in this assumption is a basic spectroscopic phenomenon that I have students prove to themselves when they first start working with a spectrophotometer. The phenomenon is real, and has a very simple basis. If Anarotrim has another explanation it should be entered into the discussion before a widely accepted explanation is merely deleted. Axewiki 21:13, 18 August 2007 (UTC)

There is an error in the passage in question: "Implicit in the integration step is an extension of this assumption, namely that one particle cannot obscure another particle in any other slab." If this were true then the extinction of light through an absorptive medium would be linear with path length rather than logarithmic.

For example, suppose that a given layer of the medium allows 50% of the incident light to pass through it (i.e. the suspended particles obscure half of the area). The Beer-Lambert law says that two such layers would allow 0.5² = 25% of the incident light to pass through. This is because only half of the particles in the second layer are in a location where additional light can be blocked. However, if we add the constraint that none of the particles in the second layer may coincide with the particles in the first then they must fill the remaining 50% of the area. The total transmission would be 0% rather than 25%.

I believe that the problematic assumption in the Beer-Lambert law is that the absorbance and incident light intensity are uniform across each layer, dz. The result of this assumption is that the Beer-Lambert law requires an infinite path length to completely absorb the incident light. Hawryluka 22:32, 22 August 2007 (UTC)

Hawryluka - your example actually proves my point. One of your layers passes 0.5¹ of the light. Two layers passes 0.5². Three layers, 0.5³, etc. The exponent is the number of layers, or path length. The logarithm of the fraction of light that passes is proportional to the path length. This is easily proven in common experience with a pair of 50% transmittance filters: look through 2 of them and you see 25% of the original light intensity, not zero. In theory, you do indeed need an infinite number of them to completely absorb light. The constraint you mention does not occur in common forms of absorption spectroscopy, where the Beer-Lambert law is essential to an understanding of what's happening. I appreciate the discussion, rather than a destructive edit.Axewiki 23:28, 23 August 2007 (UTC)

Yes, thanks for the discussion. What I am trying to say is that the exponential extinguishing of the light occurs precisely because particles in one layer can eclipse particles in other layers. The constraint I mentioned (that no eclipsing is permitted) is taken from the paragraph in the article beginning "It is instructive to consider...", which cannot be correct. Hawryluka 17:25, 24 August 2007 (UTC)

Ahhh - I see the problem! I think my text was technically correct, because I was pointing out that the Beer-Lambert law makes this assumption implicitly, yet it cannot be absolutely correct, and it becomes increasingly incorrect at higher concentrations. However, my text also appears to have been confusing so I have re-written it. Better? Axewiki 20:02, 24 August 2007 (UTC)

The revised version is less confusing. Thanks! Hawryluka 23:32, 25 August 2007 (UTC)

I also don't agree that the deviation from the Beer law at high concentrations occurs due to "eclipsing". Here's an exert from Skoog Fundamentals of Analytical Chemistry

Real Limitations to Beer's Law

Beer's law describes the absorption behavior only of dilute solutions and in this sense is a limiting law. At concentrations exceeding about 0.01 M. the average distances between ions or molecules of the absorbing species are diminished to the point where each particle affects the charge distribution, and thus the extent of absorption. of its neighbors. Because the extent of interaction depends on concentration. the occurrence of this phenomenon causes deviations frorn the linear relationship between absorbance and concentration. A similar effect sometimes occurs in dilute solutions of absorbers that contain high concentrations of other species, particularly electrolytes. When ions are very close to one another. the molar absorptivity of the analyte can be altered because of electrostatic interactions, and this can lead to departures from Beer's law.

So it's because of electrostatic interactions, not eclipsing. I agree with Axewiki that eclipsing is the reason the Beer Lambert law is exponential and not linear to transmittance.129.100.101.57 (talk)

Molar (decadic) absorption coefficient

IUPAC recommends that ε be called the molar (decadic) absorption coefficient. See [1], [2], and especially [3].
— DIV (128.250.204.118 08:20, 7 September 2007 (UTC))

Absorptivity redirect

I am under the impression that absorptivity should not redirect to this page.

Absorptivity, as it pertains to heat transfer and infrared radiation/optics, is a simple ratio of the irradiation absorbed over the total irradiation. Where, the total irradiation is the radiative flux incident to a surface, both from direction emmission and reflection from all other surfaces in a system. Additionally this unit is dimenssionless and is also given by the greek letter alpha.

The strict definition of absorptivity in infrared radiation is completely analagous to reflectivity and the emissivity.

Furthermore, several wikipedia pages reference and link to this page on absorptivity however they are obviously refering to the dimensionless absorptivity and not to the absorbance that this article discusses.

I could and probably will fix this inconsistency, by the end of the year hopefully, with the addition of a start up article on absorbtivity.

I would appreciate feed back on the discussion page or my talk page. The Lamb of God (talk) 16:32, 17 June 2009 (UTC)

Absorptivity is equal to emissivity, right? So maybe it should redirect there.... (Of course, that page would require some additional text explaining it.) --Steve (talk) 22:00, 17 June 2009 (UTC)

Absorptivity is only equal to the emissivity in an isothermal inclosure, that is, an object in thermal equillibrium. So, in regards to black bodies, the absorptivity is always equal to the emissivity and both have a value of 1.

This is given in Kirchoff's Law. And as you have probably discovered the absorptivity refered to in Kirchoff's Law and the emissivity articles link to this article.

I am working on expanding the Kirchoff's Law page also, (which is where the issue started in the first place.) So, in conjuction I will see what I can do.

In the mean time I do not know what absorptivity should link too???

I may simply create a 1 paragraph article in the short term and direct it there or follow your suggestion and do the 1 paragraph in emissivity.The Lamb of God (talk) 14:47, 18 June 2009 (UTC)

Naperian / Decadic

Please use the terms Naperian and decadic to distinguish between natural and common logarithms. This is what IUPAC does. IUPAC generally tries to reconcile differing notations between different fields. I've personally never seen $\alpha$ used in a decadic system. Also, Naperian vs decadic doesn't really have anything to do with the analyte's state of matter. The solid phase isn't even mentioned! Naperian and decadic definitions are just a matter of different fields having different traditions. 128.146.32.223 (talk) 01:54, 2 September 2009 (UTC)

Or please use the terms natural and decadic to distinguish between Naperian and common logarithms? The meaning of a Naperian (Napierian) logarithm is ambiguous in this context. The meaning of a decadic logarithm is unambiguous.

I usually defer to IUPAC, but on this they seem to be off-kilter; furthermore, "natural log." is by far the more common phrase in my experience.

—DIV (138.194.12.32 (talk) 04:43, 7 October 2009 (UTC))

"Please use the terms Naperian and decadic to distinguish between natural and common logarithms." These two comments look like an example of absurdly preferring gobbldegook over plain language. If you mean common logs just say that, and the same for natural logs. There is nothing wrong with simply saying what you mean... —Preceding unsigned comment added by 82.32.49.157 (talk) 11:13, 2 February 2011 (UTC)

The naperian/decadic adjectives can be applied to nouns such as "absorbance", "absorption coefficient", and "optical density". There is nothing wrong with saying "natural logarithm" or "common logarithm", but "natural absorption coefficient" and "common absorption coefficient" are not terms that are in use AFAIK. This is where the IUPAC definitions are most useful. These adjectives are certainly not gobbledygook! They allow a technical writer to clarify which base is in play with a single word, which would normally require a sentence or more. 99.11.197.75 (talk) 04:25, 17 July 2012 (UTC)

Verbosity

This article is ridiculously verbose. There are way too many examples, at most one example is needed. In particular the "Chemical Analysis" and "Beer–Lambert law in the atmosphere" sections could simply be deleted. The discussion about different units is unimportant to a general audience. (As a matter of fact, it can be highly confusing to specialists.)

I completely disagree —Preceding unsigned comment added by 128.120.188.230 (talk) 04:19, 4 December 2009 (UTC)

Is aborbance a word?

I've tried googling for it (define: aborbance) and I've tried m-w.com--neither one defines it. If I search for aborbance, I find lots of references to it (over a 1000, iirc), but my guess they are all careless misspellings.

I was going to correct the spelling, but I have just a little bit of doubt. (If someone does correct it, search the page, I think there is more than one occurrence.)

Rhkramer (talk) 11:40, 2 November 2010 (UTC)

Prerequisites

I have edited the Prerequisites paragraph, presenting the 2 conditions which were originally stated under item (2) as separate requirement (now 2 & 3, the remainder being renumbered up to 6). I have no argument with these requirements. But, when they are stated explicitly, which is a good thing to do, it is clear that the illustration (which involves fluorescence) cannot be a good example of the law in action, since it transgresses condition (3) "no scattering" and also condition (6) "no influence on medium". For the sake of consistency, either a better illustration should be found, or the reasons for ignoring these transgressions (change in wavelength - the B-L law can be applied to the original green light only) should be given. Dr Andrew Smith —Preceding unsigned comment added by 82.32.49.157 (talk) 11:07, 2 February 2011 (UTC)

I think scattering should be discussed better than just saying "no scattering is allowed". The atmosphere example incorporates scattering, for example! Scattering is a legitimate contributer to the attenuation coefficient, even if it makes the absorption harder to measure. I put some discussion of this point in Attenuation coefficient#Attenuation versus absorption...maybe something like that would be appropriate here too.

Anyway the statement in the prerequisites section is a bit extreme, I think a reader will read that and say, "OK well since everything scatters a little bit I guess the Beer-Lambert law never applies in the real world. The real assumption is that light is absorbed faster than it is scattered, assuming you are measuring the absorption length.

I don't see why fluorescence is ruled out by #6. As long as most of the dye is in the ground state most of the time, then the green laser light is indeed a "non-invasive probe". Certainly some text could be added that the fluorescence is just functioning as a marker of how much laser light there is. Also, why do you think there's scattering of green light? --Steve (talk) 23:12, 2 February 2011 (UTC)

Water it down a bit maybe?

I'm a third year pharmacy student, and I've done my first year of basic chemistry, and even I find it difficult to follow this page. I can't imagine how this looks to person with a non-science background. I of all people understand the need to explain things properly, but this page seriously needs to be simplified somehow. I've not seen anything resembling "A = ε b c", nor do I think I have seen a mentione of ε1%. Seriously, please do something about this Chairman Xi (talk) 05:31, 10 March 2012 (UTC)

Shadowing & law breakdown.

It is implicit in the law that shadowing must occur. This is because the incident intensity on the infinitesimally thin slab is dependent on the output intensity of the slab before. If no shadowing occurred, there would be no need to perform calculus and the entire slab would be considered as one. Putting this falsehood back in the article is lying to people. If you think better wording should be used, then try to word it better, but leave in the fact that the shadowing is required for the derivation of the law to hold true (which actually means the law breaks down at low concentration/pathlength/extinction coefficient) not high).Black.jeff (talk) 20:30, 17 October 2013 (UTC)

Validity of atmospheric section questioned

I would like to point out that an editor of the French article has deleted the section on the atmosphere as "totalement faux" (totally false). When I asked for further explanation on the talk page, that editor and one other explained that the equation given does not apply to diffusive media such as the atmosphere. See discussion in French at fr:Discussion:Loi de Beer-Lambert#Suppression de section Loi de Beer-Lambert dans l'atmosphère.

This raises the question of whether the section should be removed from the English article also. I don't know enough to judge whether the objections are valid. Perhaps those editors who are more expert in optical theory could read the French discussion and comment. Dirac66 (talk) 00:59, 31 December 2016 (UTC)

Difficult article

I ended up watching some youtube videos explaining the law, which is a LOT simpler than the whole article there.

I have no problem with anyone explaining things in a complicated way per se, but wikipedia should also teach people; and from this point of view, the article is worded and structured in a WAY too complicated manner. Not everyone is a maths genius - the article was clearly written by physicists mostly. 2A02:8388:1603:CB00:3AD5:47FF:FE18:CC7F (talk) 11:57, 10 December 2017 (UTC)