LASING ACTION

When light travels through a material, part of the light energy is absorbed by the atoms in the material. The amount of light absorbed is dependent upon the characteristics of the material and its thickness. Optical components, such as lenses and windows, are made of materials that absorb very little of the light energy in the wavelength region within which they are designed to function. Optical filters are designed to transmit only a particular portion of the light that strikes them. They may attenuate some wavelengths or eliminate some selected wavelengths, and transmit the remaining ones with no change. The absorption of light is a critical process in the optical pumping of solid and liquid lasers.
When a laser beam passes through the active medium of a laser, energy is added to the laser beam through a process called "optical gain."
This module discusses the absorption of light by materials and the gain of a laser medium. The similarities of these two processes are examined, and both are measured experimentally in the laboratory.
ATTENUATION OF LIGHT:
Figure 1 depicts a beam of light traveling through a piece of optical material. Some of the light energy is absorbed by the material, and some is transmitted.

Fig. 1 Attenuation of a light beam
The transmission of the optical material is given by Equation 1.
Equation 1

where: T = Transmission.
E0 = Irradiance of light incident upon the material.
E = Irradiance of light transmitted through the material.

EXAMPLE A: CALCULATION OF THE TRANSMISSION
Given: The light incident upon the material in Figure 1 has an irradiance of 2.5 mW/cm2. The irradiance of the transmitted light is 0.50 mW/cm2.
Find: The transmission.
Solution:
T = 0.20
20% of the light is transmitted.
In some cases, almost no light is absorbed, and the transmission is almost 1.0. In others, there is no transmission at all (T = 0). Reflection and scattering of light together with absorption account for losses in all optical systems, but reflection and scattering are not considered in this module.
THE EXPONENTIAL LAW OF ABSORPTION:
Obviously, any increase in the thickness of the absorbing material will decrease the irradiance of the transmitted light. Figure 2 depicts light traveling through four identical pieces of filter material, each 1 mm thick and each filter absorbing one-half the light incident upon it. Figure 3 is a plot of the transmission of this filter material as a function of total filter thickness. Transmission in this case is based upon incident and transmitted power, rather than upon irradiance. The curve in Figure 3 is called an "exponential curve." It begins at an initial value of 1.0 and approaches zero asymptotically as thickness increases. The degree of transmission for any thickness of a material is given by the exponential law of absorption, as stated in Equation 2.

Fig. 2 Transmission of light through a series of filters

Fig. 3 Transmission as a function of thickness
Equation 2
T = e–kx
where: T = Transmission.
e = The natural logarithm base = 2.718.
k = The absorption coefficient of the material in cm–1.
x = The thickness of the material in cm.
The unit cm–1 in this equation should not be confused with the reciprocal centimeter used to indicate wave number in the previous module. In this case, the absorption coefficient is measured in terms of absorption per centimeter. The absorption coefficient is numerically equal to the reciprocal of the thickness of a specific material that results in a transmission of 1/e (0.368) of the incident light. The units of thickness and absorption coefficient must be reciprocals of one another in order that their product, the exponent of e, will remain a dimensionless quantity.
EXAMPLE B: THE EXPONENTIAL LAW OF ABSORPTION
Given: The absorption coefficient of a material is 5.0 cm-1.
Find: The transmission of pieces of this material having the following thicknesses:
(a) 0.01 cm
(b) 0.10 cm
(c) 1.0 cm
Solution: (a) T = e–kx
(a) T = e–(5.0 cm–1)(0.01 cm)
(a) T = e–0.05
(a) T = 0.951
(b) T = e–(5.0 cm–1)(0.1 cm)
(b) T = e–0.5 (evaluate using ex key on calculator)
(b) T = 0.606
(c) T = e–(5.0 cm–1)(1.0 cm)
(c) T = e–5
(c) T = 0.0067
Equation 2 may be solved for the absorption coefficient through the following steps:
Equation 2


(by taking the reciprocal of both sides of the equation)

(by taking the natural logarithm of both sides of the equation)
Rearrangement of terms yields Equation 3 for the absorption coefficient.


Equation 3

This equation is utilized to solve a problem in Example C.
EXAMPLE C: CALCULATION OF ABSORPTION COEFFICIENT
Given: A 10-mW laser beam strikes a piece of filter material 1.2 cm thick. A beam of 0.42 mW is transmitted.
Find: Absorption coefficient of the material.
Solution:
ABSORPTION AS A FUNCTION OF WAVELENGTH:
Optical filters come in many varieties. Generally, they are used to (1) select one wavelength region over another, (2) select a very narrow wavelength region and exclude all other wavelengths above and below this region, (3) select a broad range of wavelenths which, as a group, are all transmitted with the same intensity.
Optical filters are generally made of colored glass, metals, and thin dielectric films, often arranged in complicated "sandwich" geometries to achieve a special type and degree of filtering of the light incident on them. You will find such names as colored glass filters, thin metallic film filters, hot and cold mirrors, neutral density filters, bandpass filters, narrow band interference filters, and broadband interference filters, if you peruse the section on "Filters," covered, for example, in catalogs of optical fabrication companies such as Melles Griot. You should "walk through" such a section to develop an appreciation for the many types of filters available. Here we shall concentrate on the important types–cutoff filters, narrow bandpass filters, and neutral density filters.
The absorption coefficient (and thus the transmission) of any material is a function of the wavelength of the light striking that material. Until now, discussion in this module has dealt with thickness variations of absorbing materials at a single wavelength. This section assumes a fixed absorber thickness and presents transmission variations as a function of wavelength changes. Three types of optical filters are used as examples.
Figure 4 gives the transmission of two optical filters as a function of wavelength. The blue filter in Figure 4a allows most of the blue light to pass, but absorbs other colors of the visible spectrum, while the red filter (Figure 4b) allows only red light to pass through. This type of filter has a sharp division between high- and low-transmission regions and thus is called a "cutoff filter." Optical components of this type are used to eliminate unwanted wavelengths in many optical systems. Laser safety goggles are an important example of cutoff filters.

Fig. 4 Transmission-versus-wavelength for cutoff filters
Figure 5 depicts the transmission curve of a band pass filter. This type of filter passes a narrow band of wavelengths but blocks all light outside this band. Darkroom safelights contain band pass filters that pass only those light wavelengths to which the film is not sensitive.

Fig. 5 Transmission-versus-wavelength for a "spike" or band-pass filter
Transmission of a neutral density filter is illustrated in Figure 6. This filter is designed to have the same transmission for all wavelengths over a broad range of the spectrum. The gray lenses of most sunglasses are neutral density filters for visible light.

Fig. 6 Transmission-versus-wavelength for a neutral density filter
A term often used to describe the transmission of neutral density filters, laser safety goggles, and other low-transmission filters is optical density. If the optical density of a filter is known, its transmission can be calculated from Equation 4.
Equation 4
T = 10–OD
where: T = Transmission.
OD = Optical density.
Thus, a filter with an optical density of 2.0 has a transmission of 10–2, or one percent. (Table 1 lists some OD values.) Example D illustrates the use of this equation in solving another problem.

TABLE 1. Optical Density Values.
OD Transmission % Transmission
0 1 1.0 100%
1 10–1 0.1 10%
2 10–2 0.01 1%
3 10–3 0.001 0.1%
4 10–4 0.0001 0.01%
5 10–5 0.00001 0.001%
EXAMPLE D: CALCULATION OF TRANSMISSION FROM OPTICAL DENSITY
Given: A neutral density filter has an optical density of 0.6.
Find: Transmission of filter.
Solution: T = 10–OD
T = 10–(0.6)
T = 0.25
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GAIN IN A LASER:
POPULATION INVERSION:
Figure 7 illustrates the normal population distribution of atoms in their various energy levels in a given material at some given temperature.

Fig. 7 Normal population distribution
The lengths of the horizontal lines in this figure represent the number of atoms in each energy state. The majority of the atoms are always in the ground state, and the population of each successively-higher energy state is less than that of any other lower energy state.
The absorption coefficient of this material at a wavelength of light corresponding to the transition from E3 to E2 is proportional to the quantity N2 – N3. A greater difference in population of the two energy states results in a Greater absorption coefficient. If atoms were to be removed from E2 and added to E3 to reduce the population difference, the absorption coefficient would be reduced.
If the populations of the two states are the same, the absorption coefficient is, in effect, zero. Although absorption still occurs, stimulated emission occurs at a similar rate, replacing photons as they are removed by absorption. The two processes balance each other, and the net difference is zero, hence an effective zero absorption coefficient.
In Figure 8, the populations of energy states E2 and E3 have been altered until the quantity N2 – N3 is negative.