Published on *METEO 300: Fundamentals of Atmospheric Science* (https://www.e-education.psu.edu/meteo300)

Gases, liquids, and solids can all absorb radiation. Since we are most interested in gases, we will use a gas to develop the equation that we need for absorption.

Suppose we have a volume of uniformly distributed gas that absorbs. Let's suppose that there is a beam of radiation that passes into this medium. Let's narrow this beam down so that all the radiation is traveling in essentially the same direction, so now we are interested in the irradiance that is traveling in a narrow range of angles, which is called the radiance, which is denoted by *I* and has units of W m^{–2} steradian^{–1} nm^{–1}. Refer back to Lesson 6.5 [1] to refresh your memory about what steradians are.

In an infinitesimal slab of that medium, a certain fraction *dI* of the radiation with radiance *I* (W m^{–2}) is absorbed. *dI* is equal to the absorption cross section, *σ* (m^{2}), of a single molecule or particle multiplied by the number density, *n* (# m^{–3}) of absorbers, the length of the light-path through the slab, *ds* (m), and the amount of radiance *I* itself :

$$dI\left(\lambda \right)\text{}=\text{}-I\left(\lambda \right)\sigma \left(\lambda \right)n\text{}ds$$

Note that both *I* and *σ* are functions of wavelength. If there is a beam of radiation with a certain wavelength, then the chance of absorbing the radiation depends on the cross-sectional area of each molecule or particle, which quantifies the efficiency with which the molecule absorbs the radiation with that particular wavelength. The absorption cross sections are largest if the energy of the radiation's wavelength matches the transition energy of an electron, atom, or molecule.

Integrating this expression over the total thickness of the volume, we get the **Beer–Lambert Law**:

$$I\left(\lambda \right)={I}_{o}\left(\lambda \right)\mathrm{exp}\left[-\sigma \left(\lambda \right)ns\right]$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaamysamaabmaapaqaa8qacqaH7oaBaiaawIcacaGLPaaacqGH9aqpcaWGjbWdamaaBaaaleaapeGaam4BaaWdaeqaaOWdbmaabmaapaqaa8qacqaH7oaBaiaawIcacaGLPaaaciGGLbGaaiiEaiaacchadaWadaqaaiabgkHiTiabeo8aZnaabmaapaqaa8qacqaH7oaBaiaawIcacaGLPaaacaWGUbGaam4CaaGaay5waiaaw2faaaaa@4DF0@

where *I _{o}* is the radiance at the front edge of the path and

When *σns *= 1,* I*(*λ*)*/I _{o}*(

For a uniform gas, the optical depth equals *σns* and is often given by the symbol *τ* (tau). If the optical depth gets large, then very little radiation gets through.

Credit: Wikipedia [3]

Ozone has an absorption cross section of 1.0 x 10^{–19} cm^{2} at a wavelength of 310 nm. If the average ozone concentration in the ozone layer is 2 x 10^{12} molecules cm^{–3} and the thickness of the ozone layer is 25 km, what is the fraction of sunlight that gets through when the sun is directly overhead?

ANSWER:

$$\frac{I\left(\lambda \right)}{{I}_{o}\left(\lambda \right)}=\mathrm{exp}\left(\mathrm{\u20131}\times {10}^{-19}{\text{cm}}^{2}\cdot 2\times {10}^{12}{\text{cm}}^{-3}\cdot 25\times {10}^{5}\text{cm}\right)=0.6$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@6A4A@

At 310 nm, 60% of the solar UV gets through to Earth’s surface. Yet at 290 nm, where *σ* = 1.09 x 10^{–18} cm^{2}, only 0.4% of the solar UV gets through.

Once we know the absorption coefficients of all of the absorbing molecules between the sun and the air volume of interest and the number densities of the absorbers, we can determine how much radiation will be absorbed at every wavelength.

The angle of the Sun with respect to the zenith (directly overhead) is called the solar zenith angle (SZA) and is 0^{o} overhead and 90^{o} on the horizon. The SZA affects the total path through absorbers, and thus is important. We often make an assumption that the atmosphere is a parallel plane (see figure below).

The result is that we can write $ds=\frac{1}{\mathrm{cos}\left(SZA\right)}dz=\mathrm{sec}\left(SZA\right)dz$ in the differential form of Beer's Law:

$$\frac{dI\left(\lambda \right)}{I\left(\lambda \right)}=-\sigma \left(\lambda \right)n\text{sec}\left(\text{SZA}\right)dz$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeWaaSaaa8aabaWdbiaadsgacaWGjbWaaeWaa8aabaWdbiabeU7aSbGaayjkaiaawMcaaaWdaeaapeGaamysamaabmaapaqaa8qacqaH7oaBaiaawIcacaGLPaaaaaGaeyypa0JaeyOeI0Iaeq4Wdm3aaeWaa8aabaWdbiabeU7aSbGaayjkaiaawMcaaiaad6gacaqGGaGaae4CaiaabwgacaqGJbWaaeWaa8aabaWdbiaabofacaqGAbGaaeyqaaGaayjkaiaawMcaaiaadsgacaWG6baaaa@5168@

which gives:

$$I\left(\lambda ,z\right)={I}_{0}\left(\lambda \right)\mathrm{exp}\left[-\text{sec}\left(\text{SZA}\right){{\displaystyle \int}}^{\text{}}\sigma \left(\lambda \right)n\left(z\right)dz\right]$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaamysamaabmaapaqaa8qacqaH7oaBcaGGSaGaamOEaaGaayjkaiaawMcaaiabg2da9iaadMeapaWaaSbaaSqaa8qacaaIWaaapaqabaGcpeWaaeWaa8aabaWdbiabeU7aSbGaayjkaiaawMcaaiGacwgacaGG4bGaaiiCamaadmaabaGaeyOeI0Iaae4CaiaabwgacaqGJbWaaeWaa8aabaWdbiaabofacaqGAbGaaeyqaaGaayjkaiaawMcaa8aadaqfGaqabSqabeaacaaMb8oaneaapeGaey4kIipaaOGaeq4Wdm3aaeWaa8aabaWdbiabeU7aSbGaayjkaiaawMcaaiaad6gadaqadaWdaeaapeGaamOEaaGaayjkaiaawMcaaiaadsgacaWG6baacaGLBbGaayzxaaaaaa@5DB3@

Often, the integral is called the optical thickness, or optical path, where *s*_{1} is one point along the path and *s*_{2} is another. The exponential of the optical thickness, factoring in the sec(SZA), equals exp[–sec(SZA) *τ*(*s*_{1},*s*_{2})] and is called the transmittance, often designated with the symbol *t*, where *τ* is the optical depth.

$$\tau ({s}_{1},{s}_{2})\equiv {\displaystyle \underset{{s}_{1}}{\overset{{s}_{2}}{\int}}\sigma (s)\cdot n(s)\cdot ds}$$

For matter that absorbs and does not scatter, the absorptivity, *a = 1 – t*. Note that in this case the emissivity

Usually, but not always, *σ* is a function of altitude because it often is a function of pressure and temperature, which vary for different altitudes.

Here is a video (2:03) further explaining Beer's Law:

**Links**

[1] https://www.e-education.psu.edu/meteo300/node/684

[2] mailto:MathType@MTEF

[3] http://en.wikipedia.org/wiki/Beer%E2%80%93Lambert_law