The Beer–Lambert law and the gain coefficient

As mentioned in chapter 2, spectroscopy has contributed immensely to the development of modern science, in particular, in ground-breaking experiments in the late 19th and early 20th centuries. One spectroscopic technique deals with the measurement of absorption spectra. For a two-level system, for example with ${E}_{2}\gt {E}_{1}$, the Boltzmann law suggests that at a given temperature, the largest population of molecules lies in the lower energy state (${E}_{1}$). Let us look at the experiment shown in figure 5.1. A broadband light source of various frequencies with a constant intensity ${I}_{0}$ is incident on the sample. The selection of a frequency is achieved using either a prism or a grating spectrograph (known as a monochromator). The sample is considered to be made of infinitesimally thin sheets of thickness ${\rm{\Delta }}x$.

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♣ The term intensity is defined as the power per unit solid angle per unit area of the source (♠ see chapter 10 for details of the solid angle). On the other hand, irradiance is the total power per unit area incident on the body.

The change of intensity $({\rm{\Delta }}I)$ for subsequent sheets of the sample is proportional to their thickness (${\rm{\Delta }}x$) and the incident intensity at the particular sheet ($I\left(x\right)$):

$$\begin{eqnarray*}&&{\rm{\Delta }}I\propto {\rm{\Delta }}{xI}\left(x\right)\end{eqnarray*}$$

or

$$\begin{eqnarray*}&&{\rm{\Delta }}I=-\alpha \,{\rm{\Delta }}{xI}\left(x\right).\end{eqnarray*}$$

Here, $\alpha$ is the proportionality constant known as the absorption coefficient (in cm⁻¹). It is related to the absorption strength of a material. For a sample in liquid form, it is further described as the product of the concentration (C) and the extinction coefficient (${\rm{\epsilon }}$) of the sample. The absorbing medium causes a loss of photons from an incident beam of initial intensity ${I}_{0}$. The detector is used to monitor the transmitted light intensity (I).

♣ In practice, with the help of a beam splitter, two beam paths are generated from the same source for the sample and the reference to measure I and I₀, respectively.

We can rearrange the above equations as

$$\begin{eqnarray}&&\frac{\partial I}{\partial x}=-\alpha I\left(x\right).\end{eqnarray} \tag{ 5.1 }$$

On integrating we obtain the following form of the equation, known as the Beer–Lambert law.

$$\begin{eqnarray}&&\mathrm{ln}\,\left(\frac{I}{{I}_{0}}\right)=-\alpha x\,\mathrm{or}\,I={I}_{0}{e}^{-\alpha x}.\end{eqnarray} \tag{ 5.2 }$$

Here, $\alpha =\epsilon C$ for liquid solutions. The quantity ${\rm{ln}}\left(\frac{{I}_{0}}{I}\right)$ is also known as the optical density (OD) of the sample.

The basic idea is to measure the light transmitted by the sample to find its absorption strength across the frequencies of interest. The effect of the EM wave incident on the medium is schematically shown in figure 5.2. We can see that the amplitude of the incident light wave is reduced after passing through an absorbing medium.

Exercise 5.1.  An optical density of unity is obtained for a sample that has a concentration of 0.1 M when the path length used is 1 mm. What are the other values of thickness and concentration that could be be selected to obtain samples with the same optical density?

Solution: The optical density (OD) is defined as ${\rm{ln}}\left({I}_{0}/I\right)$. For the path length ($x$) and the concentration ($C$) of a sample, we know that its extinction coefficient ($\epsilon$) is a fixed quantity ($\alpha /C=\epsilon$). Equation (5.2) can be rewritten as ${OD}=\epsilon {Cx}$, and we see that to maintain the same value of OD, the parameters $x$ and $C$ have to be adjusted suitably. Therefore, for the given situation, OD = 1 can be obtained, for example, using (i) $C$ = 0.5 M with a sample path length of 0.2 mm, or (ii) ${C}$ = 0.01 M with a path length of 10 mm, and so on.

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The irradiance $I\left(\nu \right)$ of a light source at a photon frequency is the power delivered by the source (or energy crossing in unit time) to the unit area of a body. Therefore, in terms of the energy density of the photons ${\rho }_{\nu }$, the speed of light $c$, and the refractive index $n$ of the medium, it is also written as $I\left(\nu \right)={\rho }_{\nu }\,\frac{c}{n}$. A change in the irradiance, ${\rm{\Delta }}I\left(\nu \right)$, is defined as the total number (M) of photons delivered by a source of an area (${\rm{\Delta }}A$) per unit time (${\rm{\Delta }}t$), as follows:

$$\begin{eqnarray*}&&{\rm{\Delta }}I\left(\nu \right)=\frac{{Mh}\nu }{{\rm{\Delta }}A{\rm{\Delta }}t}.\end{eqnarray*}$$

The energy density (${\rho }_{\nu }$) of M photons for a volume V, is given by

$$\begin{eqnarray*}&&{\rho }_{\nu }=\frac{{Mh}\nu }{V}={Nh}\nu .\end{eqnarray*}$$

Here, ${N}$ is number of photons per unit volume. For sufficiently small ${\rm{\Delta }}x$, we can rewrite the change in irradiance within the light beam as

$$\begin{eqnarray*}&&{\rm{\Delta }}I\left(\nu \right)=\frac{{Mh}\nu }{{\rm{\Delta }}A{\rm{\Delta }}t}\frac{{\rm{\Delta }}x}{{\rm{\Delta }}x}=\frac{{Mh}\nu }{{\rm{\Delta }}V}\frac{{\rm{\Delta }}x}{{\rm{\Delta }}t}={\rm{\Delta }}{Nh}\nu \frac{{\rm{\Delta }}x}{{\rm{\Delta }}t}.\end{eqnarray*}$$

By replacing $\nu$ by ${\nu }_{21}$, which denotes the emission frequency between the two levels, we can express ${\rm{\Delta }}{N}$ as

$$\begin{eqnarray}&&{\rm{\Delta }}N=\frac{{\rm{\Delta }}I\left(\nu \right)}{h{\nu }_{21}}\frac{{\rm{\Delta }}{\rm{t}}}{{\rm{\Delta }}{\rm{x}}}.\end{eqnarray} \tag{ 5.3 }$$

The beam travels through a volume element that has a thickness of ${\rm{\Delta }}x$ and a unit sectional area of ${\rm{\Delta }}A$. As described in chapter 4, after stimulated absorption rate $({B}_{12}\,\rho \,\left(\nu \right)\,{N}_{1}$), the molecules from the higher energy state decay to the lower state with a spontaneous emission rate of (${A}_{01}{N}_{2}$) and a stimulated emission rate of (${B}_{21}\rho \left(\nu \right){N}_{2}$). As shown in figure 5.3, assuming that the spontaneous emission coefficient ${A}_{21}$ is very small compared to that due to stimulated emission (${A}_{21}\ll {B}_{21}$), the net rate of photon loss from the beam per unit volume is given by the following:

$$\begin{eqnarray*}&&{-\frac{{dN}}{{dt}}={B}_{12}\rho \left(\nu \right){N}_{1}-B}_{21}\rho \left(\nu \right){N}_{2}\,{\rm{so\; that}}\end{eqnarray*}$$

$$\begin{eqnarray}&&-\frac{{dN}}{{dt}}=\left({N}_{1}\frac{{B}_{12}}{{B}_{21}}-{N}_{2}\right)\rho \left(\nu \right){B}_{21}.\end{eqnarray} \tag{ 5.4 }$$

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Equation (5.3) can be rewritten as

$$\begin{eqnarray*}&&\frac{\partial N\,}{\partial t}=\frac{\partial I\left(\nu \right)}{\partial x}\frac{1}{h{\nu }_{21}}.\end{eqnarray*}$$

Using equation (5.1), we can rewrite this as

$$\begin{eqnarray*}&&\frac{{\rm{d}}N}{{\rm{d}}t}=-\alpha I\left(x\right)\frac{1}{h{\nu }_{21}}.\end{eqnarray*}$$

Placing the value of irradiance $I\left(\nu \right)$, we get,

$$\begin{eqnarray}&&\frac{{\rm{d}}N}{{\rm{d}}t}=-\alpha {\rho }_{\nu }\frac{c}{n}\frac{1}{h{\nu }_{21}}\end{eqnarray} \tag{ 5.5 }$$

Comparing equations (5.4) and (5.5), and by using $\frac{{B}_{12}}{{B}_{21}}=\frac{{g}_{2}}{{g}_{1}}$, the absorption coefficient (α) can be written as

$$\begin{eqnarray}&&\alpha =h{\nu }_{21}\frac{n}{c}\left({N}_{1}\frac{{g}_{2}}{{g}_{1}}-{N}_{2}\right){B}_{21}.\end{eqnarray} \tag{ 5.6 }$$

Equation 5.6 relates α to the coefficient of stimulated emission, (B₂₁). We already know from equation (4.13) that B₂₁ is related to the spontaneous emission rate coefficient, (A₂₁).Therefore, if we know the decay time of spontaneous emission (${\tau }_{0}$) of the system, the stimulated parameters can be estimated. From this relation, it is obvious that for a higher value of the absorption coefficient, the probability of stimulated emission is higher.

We now know that atoms or molecules in the forms of gases, liquids, or solids give rise to spontaneous/stimulated emission when excited. Any suitable material which is capable of emitting with a high quantum yield, and can therefore be used to generate photons, is known as a gain medium. Examples of popular gain media are He–Ne (gas), dye solutions (liquid), and Nd⁺³:yttrium aluminum garnet (YAG) (solid).

Equation (5.6) gives the absorption coefficient in terms of the populations of the two states involved in the transitions. This equation, when used in combination with the Beer–Lambert law, gives rise to an interesting insight about gain media. We see that the value of $\alpha$ is positive for ${N}_{1}\gt {N}_{2}$ and becomes negative for ${N}_{2}\gt {N}_{1}$. The latter is the situation for population inversion (see the next chapter). For instance, if we replace the negative value of $\,\alpha$ with ${\rm{k}}$ in equation (5.2), we obtain,

$$\begin{eqnarray}&&I={I}_{0}{e}^{{\rm{k}}x}.\end{eqnarray} \tag{ 5.7 }$$

Equation (5.7) indicates that the irradiance increases exponentially with the thickness of the medium for a given value of $k$. Here $k$ is known as the small-gain coefficient, i.e. the gain at a given frequency of the emission spectrum.

Let us now consider figure 5.4 in view of equation (5.7) for negative values of the absorption coefficient ($\alpha =-k).$ By comparing this figure with figure 5.2, we can schematically observe that the transmitted EM wave has an increased amplitude compared to that of the incident wave.

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We know that when the absorption strength (also known as the oscillator strength) of the system is high, the amplitude of the interacting incident light is exponentially reduced by passing through the medium, as shown in figure 5.2. In contrast, for negative values of $\alpha$, the transmitted light gains photons from the gain medium, as shown in figure 5.4. The greater the oscillation strength, the larger the number of photons transmitted by the system. This point shows the importance of a prior estimate of the absorption parameters of a system in order to establish whether it is a suitable laser material for the given wavelength range.

♣ Richard Tolman proposed the idea of negative absorption in 1924.

♣ In terms of the transition dipole moment (♠ see equation (2.8)), the oscillator strength (f) is a dimensionless quantity given by $f=4.319\times {10}^{-9}\int \epsilon \left(\nu \right){\rm{d}}\nu$. Here, $\epsilon \left(\nu \right)$ is the extinction coefficient and the integration is performed over the absorption band.

To obtain a negative value of $\alpha$ (i.e. a positive value of k), we can rewrite equation (5.6) for a particular frequency ${\nu }_{21}$ as follows:

$$\begin{eqnarray}&&k=\frac{n}{c}\left({N}_{2}-{N}_{1}\frac{{g}_{2}}{{g}_{1}}\right){B}_{21}h{\nu }_{21}.\end{eqnarray} \tag{ 5.8 }$$

Note that the difference between the populations (assuming an inverted population between ${E}_{1}$ and ${E}_{2}$) is now given by ${{\rm{\Delta }}N={N}}_{2}-{N}_{1}\frac{{g}_{2}}{{g}_{1}}$. It should be noted that the transmitted amplitude has a net gain of photons and is now given by equation (5.7). This indicates that the amplitude is enhanced exponentially (i.e. amplified!). While k is known as small-gain coefficient, the exponential factor ${e}^{{kx}}$ is termed the overall gain coefficient (${k}_{G}$) or simply the gain.

From equation (5.8), the stimulated emission cross section (${\sigma }_{{\rm{es}}}$) is introduced, as follows:

$$\begin{eqnarray}&&k={\sigma }_{\mathrm{es}}{\rm{\Delta }}N,\end{eqnarray} \tag{ 5.9 }$$

where ${\sigma }_{{\rm{es}}}=\frac{{B}_{21}h{\nu }_{21}}{c}$. It can also be written in terms of ${A}_{21}$ from equation (4.13). Therefore, equation (5.7) can now be rewritten as

$$\begin{eqnarray}&&I={I}_{0}{e}^{{\sigma }_{\mathrm{es}}{\rm{\Delta }}{Nx}}.\end{eqnarray} \tag{ 5.10 }$$

Lasers are also known as optical amplifiers ¹ . We note here that an ideal laser requires a single optical transition between two energy levels. In practice, a correction by a spectral width factor is included in equation (5.8) (♠ see the next chapter).

Exercise 5.2.   The stimulated emission coefficient of a gain medium is of the order of 10⁻¹⁹ cm². Assuming no losses, what is the value of the small-gain coefficient if the population inversion in the two states is equal to 10¹⁹ cm⁻³ ?

Solution: Using equation 5.9, we see that the value of k is 1 cm⁻¹.

The value of the small-gain coefficient $k$ is defined for a stimulated emission transition between two levels. As shown in figure 5.5, we now place the emitter in between two mirrors—a 2D cavity of length L. The reflectivities of the cavity mirrors (1 and 2) are ${R}_{1}$ and ${R}_{2}$, respectively. Let the initial photon density at mirror 1 be ${\phi }_{1}$. As shown below, after the first reflection, the photon density ${\phi }_{2}$ is given by the product of ${\phi }_{1}$ and the reflectivity of the first mirror:

$$\begin{eqnarray}&&{\phi }_{2}={\phi }_{1}\times {R}_{1}.\end{eqnarray} \tag{ 5.11 }$$

After traveling the length L, the photon flux (${\phi }_{3}$) includes the gain of the medium and is given by

$$\begin{eqnarray*}&&{\phi }_{3}=({\phi }_{2}\times {e}^{{kL}})\equiv {\phi }_{1}\times {R}_{1}\times {e}^{{kL}}.\end{eqnarray*}$$

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Similarly, after the second reflection at mirror 2, on reaching mirror 1 (by completing a full round trip), the photon flux (${\phi }_{5})$ is given by

$$\begin{eqnarray*}&&{\phi }_{5}={\phi }_{1}\times {R}_{1}\times {e}^{{kL}}\times {R}_{2}\times {e}^{{kL}}.\end{eqnarray*}$$

For sustained lasing, the initial and final photon densities must at least be identical. We can now define the round-trip gain (G) as the ratio of the initial and final photon densities, as follows:

$$\begin{eqnarray}&&G=\frac{{\phi }_{5}}{{\phi }_{1}}\equiv {R}_{1}{R}_{2}{e}^{2{kL}}.\end{eqnarray} \tag{ 5.12 }$$

Exercise 5.3.  Find the length of the gain medium required to provide an overall gain of three for a small-gain coefficient value (k) of 0.4 cm⁻¹ at a given frequency. If the length of the gain medium is doubled, what is the overall gain?

Solution: In the absence of losses, the (overall) gain coefficient is given by ${k}_{G}={e}^{{kL}}$.

Using the given values, $3={e}^{0.4L}$, we obtain $L$ = 2.75 cm. Furthermore, if the gain length (L) is doubled, the value of ${k}_{G}$ is 9.02.

Losses in the laser cavity arise from several factors. A few of them are: (i) reflection losses at the mirrors, (ii) scattering and absorption due to impurities in the active medium, and (iii) external parameters. The threshold value of the pump rate includes some of these losses. As a consequence, the value of $k$ is reduced. If we denote all such losses by γ, then the net value of the small-gain coefficient ($k^{\prime}$) can be written as

$$\begin{eqnarray*}&&k^{\prime} =k-\gamma .\end{eqnarray*}$$

According to equation (5.12), if the value of $G$ is more than one, the oscillations will grow. However, if $G$ is less than one, the oscillations will die out. Therefore, for the steady state and for continuous-wave operation of the laser, the value of G is taken to be unity. Therefore, equation (5.12) yields

$$\begin{eqnarray}&&{R}_{1}{{R}_{2}e}^{2{k}^{^{\prime} }L}=1\end{eqnarray} \tag{ 5.13 }$$

The situation in which the value of $G$ is unity is known as the condition of gain saturation. At the point of gain saturation, the number of photons extracted from the excited population is insignificant in comparison to the number of the excited molecules. This can be further understood as follows: as shown in figure 5.6, at the commencement of laser operation, the round-trip gain is greater the threshold value (indicated by $G=1$ in the figure). At the onset of the process of stimulated emission, the population inversion value starts to decrease. The existing losses in the cavity further decrease the gain. These processes affect the net value of the gain. As a result, the net round-trip gain may vary, as shown in figure 5.6. When the gain just balances the losses and the value of $G$ remains at unity over a period of time, the power stabilizes.

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For $G=1$, equation (5.13) can be rewritten as $k^{\prime} =\frac{1}{2L}{\rm{ln}}\left(\frac{1}{{R}_{1}{R}_{2}}\right)$. In view of equation (5.9), in terms of population inversion, we have ${\sigma }_{{es}}{\rm{\Delta }}N=k^{\prime}$, so that ${\rm{\Delta }}N=\frac{1}{2L{\sigma }_{{es}}}{\rm{ln}}\left(\frac{1}{{R}_{1}{R}_{2}}\right)$. Here, ${\rm{\Delta }}N$ is known as the critical inversion.

Exercise 5.4.  A He–Ne laser tube is 50 cm long and the reflectivities of the two mirrors are 100% and 99%, respectively. Assuming no losses, what is the value of the small-gain coefficient?

Solution: Assuming no losses, the small-gain coefficient is given by

$k=\frac{1}{2L}{\rm{ln}}\left(\frac{1}{{R}_{1}{R}_{2}}\right)$. For the given values of $L$ = 50 cm, ${R}_{1}$ = 1, and ${R}_{2}$ = 0.99, we obtain the value of $k=1\times {10}^{-4}\,{\rm{cm}}.$

The absorption coefficient (i.e. the parameter $\alpha$) contains vital information about the sample. The absorption strength of the sample (and hence the strength of the transition from the lower to the higher energy level) as well as the population in the upper level immediately after absorption can be estimated from $\alpha$. A number of modern sensing technologies for toxic and explosive materials are based on the measurement of $\alpha$. Therefore, the Beer–Lambert law has tremendous application in science and industry. Notice that the thickness of the sample plays an important role. In practical terms, the thickness should be sufficiently small that the intensity at the first layer (${\rm{\Delta }}x$) of the sample can be considered to be the same as that at the final layer of the sample. For this reason, the value of the optical density (OD) is generally kept at less than one. However, for extremely weakly absorbing systems, the thickness used can be as large as several meters. As mentioned above, α is a function of the concentration (C) and the extinction coefficient ($\epsilon$) of the sample. This means that the concentration or thickness of the sample can (and should) be controlled to achieve an OD value near unity. Therefore, by making use of capillaries several meters long or passing the light through the sample multiple times, it is possible to measure the absorption parameters of extremely dilute samples. The technique of cavity ring-down spectroscopy (CRDS), which is based on absorption coefficient, is used to detect explosives at the ppm level (♠ see question 7).

♣ Cavity ring-down spectroscopy (CRDS) measures the decay of light intensity using multiple passes through the sample.

• 1.  
  (a). What is the significance of the thickness of the sample in Beer–Lambert's law?(b). The optical density of a sample is 0.5 for a path length of 1 cm and a concentration of 1 M. What is the molar extinction coefficient?
• 2.  
  What is a gain medium? In a gain medium, the ratio of the inverted population of the two levels after optical pumping is ${10}^{25}$. At 300 K, what wavelength of light can be emitted by the system?
• 3.  
  Can absorption by a system be related to amplification by the same system?
• 4.  
  If γ represents the total loss of the resonator, how is the threshold gain coefficient modified in terms of the length of the medium (L) and the reflectivity of the resonator mirrors R₁ and R₂?
• 5.  
  For a dye laser at 540 nm, the stimulated emission cross section is ${10}^{-17}$ cm⁻². The population inversion is ${10}^{17}$ cm^−3. Calculate the small-gain coefficient, provided that there are no losses.
• 6.  
  Define the term 'gain saturation'.
• 7.  
  CRDS is a technique based on the phenomenon of absorption. It is used to quantitatively detect atomic or molecular species with high sensitivity [4]. The sample chamber made of a high finesse cavity (♠ see next chapter) is scanned repetitively thousands of times by a laser pulse. The intensity decay of the tunable transmitted laser pulse, i.e. the ring-down decay time ($\tau$) decreases exponentially. The decay time, as shown in the figure, is monitored through one of the mirrors. The rate of decrease is proportional to the reflectivity of the mirrors ($R$), the length of the sample chamber, and the absorption coefficient. The ring-down time $\tau$ in a round trip ${t}_{r}$ is given by $\tau =\frac{{t}_{r}}{2\left(1-R+\alpha L\right)}$. Here, (1 − R) is the reflection loss. Furthermore, a plot of $1/\tau$ vs laser frequency shows that the sample's $\alpha$ is of the order of ${10}^{-8}$. Calculate the value of $\tau$ for a length of 0.5 m for a cavity with no absorption ($\alpha =0).$ What will happen if there are ${10}^{3}$ reflections?