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Area of ​​Expertise - Optics, spectroscopy

The extinction is a term derived from the Latin extinctio "extinction" for the weakening of the intensity of waves, especially of radiation when passing through matter. The extinction is based on the physical processes of scattering and absorption.

During absorption, part of the incident energy is converted into another form of energy through interaction with matter; with (elastic) scattering the direction of propagation changes (Tyndall effect), with inelastic scattering the deflection is associated with a loss of energy.

The intensity I.(x) increases, according to Lambert-Beer law, exponentially with the distance covered in the medium x away:

I.(x)=I.0eαcxI.0=original intensityα=natural, molar extinction coefficientc=concentration

The extinction coefficient characterizes the strength of the interaction of the radiation with the medium. It is a material constant that depends on the wavelength of the incident radiation.

The wavelength dependence of the extinction is used in spectroscopy.

See also: absorption, transmission, Lambert-Beer law, elastic scattering process

Learning units in which the term is discussed


Absorbance, optical density, Absorbance, Measure for the opacity of a sample: E. = log I.0/I.D. = log 1 /D., whereby I.0 Intensity of the incident radiation, I.D. Intensity of the transmitted radiation, D. Permeability I.D./I.0, Lambert-Beer law.

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Extinction (optics)

In physics in particular, the definition of extinction using the natural logarithm can also be found. In astronomy, the extinction is given in size classes. The physical quantity diabatie takes into account the exponential increase in extinction with the layer thickness of the irradiated material by forming a further logarithm.

The processes of absorption, scattering, diffraction and reflection are generally involved in attenuation. In analytical applications, see Lambert-Beer law, scattering and diffraction are often insignificant and the reflection losses are taken into account by empty or parallel measurements in I 0 < displaystyle I_ <0>>. Then, instead of absorbance, (decadal) absorbance is used. absorbance ), Absorptivity or decadic absorbance (standard-compliant designation). [1] A criticism is that the reduction to the intensity does not take into account the wave properties of light. Therefore, the extinction is generally non-linearly dependent on the path length and the concentration. [2]

The substance-specific strength of the attenuation related to the path length is indicated with the extinction coefficient and the absorption coefficient. If the wave nature of the light is taken into account, the result is that the coefficients are not material-specific, but depend on boundary conditions such as the surrounding medium, shape and heterogeneity of the material, etc. [2]

The extinction of solutions is called the optical density (abbreviation OD). If a specific substance is examined, its concentration in the solution can be determined using its extinction coefficient. OD measurements are generally used to determine the concentration of DNA solutions, protein solutions and to control the cell density in suspension cultures. [3]

Exercise collection physics: extinction through optical filters

When calculating the reflection losses, it must be taken into account that the optical filter absorbs light, i.e. has a complex refractive index. However, only the real part of the refractive index is known.

Calculation of the imaginary part of the refractive index

We get the absorption coefficient by using the formula 0.73 = e - a ⋅ d R < displaystyle 0.73 = e ^ <- a cdot d_>> reshape to a < displaystyle a>:

Why can the imaginary refractive index be neglected?

The degree of reflection with complex refractive indices is the same according to Fresnel's formulas

How big are the reflection losses on the two filter surfaces?

The reflectance is (the imaginary part is neglected):

The optical filter must therefore have a thickness of 3.8 m m < displaystyle 3.8 , mathrm > have.

This new transmission level should now be 50% = 0.5 < displaystyle 50 \% = 0.5>. So

The concentration of the dye particles must therefore be increased by a factor of 2.2 < displaystyle 2.2>.

Similar questions

I am writing an exam on Tuesday about, among other things, complex chemistry. I understand what absorption and transmission are all about, I know how to calculate them with the help of the Lambert-Beer law and and.

But I don't understand the exact meaning of the extension. Somehow the 2-lg of the transmission or something like that and you can calculate with it more easily, I was told.

Unfortunately, I do not understand it, it is a total burden on me.

What exactly is it with the extinction / extinction coefficient for you?

Many thanks in advance.

Hello people :) I'm doing a physics test tomorrow (optics, 7th grade) and unfortunately I don't know exactly what absorption is :( A friend said that: absorption is when light is reflected and it creates a shadow. That's the answer correct? If not can someone explain that to me?: 3 Thank you in advance :)

Dear Community, For a physics thesis I have to be able to explain these three terms, absorption, reflection and scattering. But I don't know what they mean for all of them and I can't explain them at all. Can someone help me please who has a clue? Thanks very much!

Hi, can someone give me examples of absorption .. I only have mirrors

i'm training to be a chemical technician 1st Lhrjr. Can someone explain EXACTLY what adsorption and what is absorption? But please not with 10,000 technical terms, I just can't get along -.-. Would be very nice, thank you & lt3

Hello, can you do it like that ?:

A AND A = A (this is idempotence)



In the Internet, however, it is written everywhere that at the beginning of the absorption there is no A AND (A OR (A AND B)) but A AND (A OR B) and then the distributivity law should be used. Can someone explain that to me?

I wanted to ask what examples of absorption exist in everyday life. I don't think I have to say more: D

Thank you in advance!

Wavelength = 600nm and extinction = 0.123, how can I now calculate the extinction coefficient and the concentration?

So how do I get from the wavelength to the extinction coefficient?

The light can be thought of as electromagnetic waves or photons. Light can be absorbed, emitted or reflected by substances. During absorption, an electron is brought to a higher level by the photon and when the electron falls back again, the photon is emitted in a scattered manner, other light components are reflected and this sum of this absorption and reflection then results in the color. This is my understanding of light and color. But what actually happens during reflection? Is there an absorption with & quot; directed & quot; emission? What happens when light goes through glass? There is only interaction of the photon with the electrons or some kind of "directional absorption and emission". Thank you for answering the question.

I came across something that I can't understand with the best will in the world. If you place two polarization filters one behind the other, one perpendicular to the direction of transmission of the other (90 °), the light is completely absorbed.

If I now have 3 filters that are each offset by 45 ° to each other, there is no longer complete absorption.

As I understand it, it doesn't matter whether I put another filter between the two 90 ° offset filters that filters out more light that is in the wrong direction, because the light is neither rotated by it, nor is a new light source added.

For this reason, I would be very happy to receive a detailed explanation.

3 answers

If you want to carry out concentration measurements with a photometer, you first have to create a calibration curve (calibration curve can only be said officially). So you specify certain concentrations and could measure the associated transmission. You would get a curved calibration curve.

In scientific test evaluations, however, it is nice when you get straight lines. This is often possible by taking the logarithm. That is why the logarithm plays such a big role.

According to Lambert-Beer's law, the extinction is directly proportional to the concentration of the solution investigated. The irradiated layer thickness d and the molar extinction coefficient are constant. However, the extinction is the negative logarithm of the transmission.

If you now plot E over c, you get a straight line and have an ideal calibration curve to investigate unknown concentrations of these (not other!) Solutions.

Transmission and extinction (extinction) are likely to be the flipside of the same coin.

The absorption of light makes an object dark or opaque to specific wavelengths of light. Glass and water are opaque to ultraviolet light, but transparent to visible light (above about 400 nm).

When a material absorbs light of certain wavelengths in the spectrum, they are logically not reflected at the same time as the other wavelengths, and only the latter arrives in the eye of the beholder.


  • A red tomato mainly reflects the portion of the red light, while green and blue are absorbed. The pigment lycopene (E 160d), a carotenoid that absorbs light with a wavelength of 450nm to 550nm, is responsible.
  • The leaves of green plants use a pigment called chlorophyll for photosynthesis, which absorbs the blue and red wavelengths of the spectrum and reflects green.
  • The heme group in hemoglobin is, for example, a chromophore that absorbs light in the range between 520 and 600 nm (i.e. blue to green) and therefore appears red.
  • Substances in a transparent medium that absorb a certain wavelength or a certain wavelength band are called filters.

Note: Color is not a physical property of light (even if the wavelength is one), but a perception.

Color strength

  • The wavelength of the absorbed light waves determines the hue and the coloristic purity of the colorant.
  • The degree of light absorbed determines the color strength.

In the sub-category of soluble dyes (see colorants), the color strength only depends on the chemical purity (and of course the concentration in the medium).

In the pigment subcategory, the color strength depends in particular on the average diameter and the distribution density of the pigment in the medium.

  • If, on the one hand, the pigments are relatively large and, at the same time, are not so common in the medium, some of the light rays do not hit any pigment.
  • If the pigments are so small that they become more or less transparent for the light, the desired effect of absorption occurs only to a lesser extent.

Both ratios lead to a more or less significant loss of color strength. A pigment can only optimally develop its characteristic optical properties if both the average diameter and the distribution in the medium (dispersion) are adjusted accordingly during production.

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