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This version reflects the comments of the core participants as reviewed and incorporated in accordance with CORD's FIPSE-supported Curriculum Morphing Project. MODULE 1-6
(1) 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. (2) 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." (3) 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. |
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| where: | T = Transmission. E0 = Irradiance of light incident upon the material. E = Irradiance of light transmitted through the material. |
(6) Example A illustrates the use of this equation in solving a problem.
| 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. |
(7) 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.
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| View a tutorial on Absorption and Filters (Optics Guide-Melles Griot - Filters) After viewing, close the window to return to this lesson. |
(8) 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
T = ekx
| where: | T = Transmission. e = The natural logarithm base = 2.718. k = The absorption coefficient of the material in cm1. x = The thickness of the material in cm. |
(9) The unit cm1 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.
(10) Example B demonstrates the use of Equation 2 in solving problems.You should be able to evaluate terms like ekx easily with the help of a scientific calculator, using the ex key.
| 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 = ekx (a) T = e(5.0 cm1)(0.01 cm) (a) T = e0.05 (a) T = 0.951 (b) T = e(5.0 cm1)(0.1 cm) (c) T = e(5.0 cm1)(1.0 cm) |
(11) 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
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(12) This equation is utilized to solve a problem in Example C.
(12a) Equations like T=ekx (Eq 2) and
k=
l n
(Eq 3) involve exponential functions and logarithmic
functions. If you are not comfortable with handling these, the following quick review may
help you.
When a function is expressed as the base e raised to a certain power, either negative or positive, it can be evaluated quickly with your scientific calculator. For example, evaluate e5.7.
When a function is expressed as a natural logarithm, l n x, for example, the scientific calculator again makes the evaluation easily. Note in the derivation from Eq 2 to Eq 3, taking the l n of ekx simply results in the exponent to which e had been raised. Thus:
l n(e2) = 2
l n(e2.1) = 2.1
l n(ekx) = kx
Now evaluate a term involving the "l n" function: l n 5.7
| 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: | |
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View a tutorial on Absorption and Filters (Optics Guide-Melles Griot - Filters) Suppliers of Optical Filters: Suppliers of Laser Safety Eyewear After viewing, close the applet screens to return to this lesson. |
(12b) 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 typescutoff filters, narrow bandpass filters, and neutral density filters.
(13)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.
(14) 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
(15) Figure 5 depicts the transmission curve of a bandpass filter. This type of filter passes a narrow band of wavelengths but blocks all light outside this band. Darkroom safelights contain bandpass 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
(16) 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
(17) 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 = 10OD
| where: | T = Transmission. OD = Optical density. |
Thus, a filter with an optical density of 2.0 has a transmission of 102, 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 | 101 | 0.1 | 10% |
| 2 | 102 | 0.01 | 1% |
| 3 | 103 | 0.001 | 0.1% |
| 4 | 104 | 0.0001 | 0.01% |
| 5 | 105 | 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 = 10OD T = 10(0.6) T = 0.25 |
| Applet demonstrating Normal Population Distribution and Population Inversion After viewing, close the applet window to return to this program. |
18) 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
(19) 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.
(20) 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.
(21) 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.
(22) In Figure 8, the populations of energy states E2 and E3 have been altered until the quantity N2 N3 is negative.

Fig. 8 Population inversion between states E2 and E3
(23) The population inversion between E2 and E3 now requires that the value of the absorption coefficient also must become negative. This condition is called a "population inversion" because the populations of the two atomic states have been inverted with respect to each other. When a population inversion exists, the absorption coefficient in the equation T = ekx is negative; and the transmission is then necessarily greater than one. Under these conditions, stimulated emission occurs at a rate greater than the rate of absorption, and the light passing through the material is amplified.
(24) The gain of such an optical amplifier is dependent upon the quantity N3 N2; therefore, greater population inversion produces greater gain. To increase the gain of a laser, some method must be utilized to increase of the upper energy state of the lasing transition. Since the lasing process itself will transfer atoms to the terminal state of the losing transition, some process must be available to reduce the population of this lower energy state. If atoms are allowed to remain in this terminal state, the population of the state will increase to the point at which a population inversion no longer exists, and the losing activity of the medium will cease.
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(25) The amplifier gain of a laser medium is given by Equation 5.
Equation 5
Ga = e ax
| where: | Ga = Amplifier gain. a = Gain coefficient. x = Length of active medium. |
The gain coefficient usually is expressed in cm-l, with the active medium length in cm; but sometimes the gain is in m-1, and the active medium length is in m. In solving problems, one must ensure that the unit of gain coefficient is the reciprocal of the unit of length.
(26) Example E illustrates the use of Equation 5 in solving a problem.
| EXAMPLE E: AMPLIFIER GAIN OF A LASER | |
| Given: | A ruby rod is 10 cm in length and has a gain coefficient of 0.23 cm-1. |
| Find: | The gain of the amplifier. |
| Solution: | Ga = eax Ga = e(0.23 cm-1)(10 cm) Ga = e2.3 Ga = 10 |
(27) Equation 5 is of the same form as Equation 2 (the exponential law of absorption) with a replacing k and Ga replacing T. If these quantities are substituted into Equation 3 for the absorption coefficient, the result is Equation 6 for the gain coefficient.
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(28) Example F illustrates a problem solved by use of Equation 6.
| EXAMPLE F: CALCULATING GAIN COEFFICIENT | |
| Given: | A HeNe laser tube 50 cm long has a gain of 1.08. |
| Find: | The gain coefficient. |
| Solution: | ![]() |
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(29) Module 15, "Absorption and Emission of Light," discussed briefly the width of spectral lines and the factors leading to broadening of a line. The same factors apply to the wavelength range over which the active medium provides gain in a laser. Figure 9 exhibits the general shape of the gainversuswavelength curve for most lasers. In all cases, the active medium provides gain, not at a single wavelength, but over a narrow range of wavelengths. The effect of the gain profile upon laser output will be discussed further in Module 17, "Optical Cavities and Modes of Oscillation."

Fig. 9 Gain as a function of wavelength
| Applet demonstrating Energy Transfer in Gas Lasers After viewing, close the applet window to return to this program. |
(30) The excitation mechanism in most gas lasers is a direct-current discharge through the gas (the active medium). Gas atoms gain energy through collisions with energetic free electrons produced in the gas ionization process. In ion lasers, the atoms are pumped upward in energy through several intermediate energy states by multiple collisions. In lasers, such as the HeNe and CO2 types, that contain gas mixtures, the electron collisions excite one type of atom or molecule to an upper energy state. This excited atom or molecule then collides with an atom or molecule of the losing gas, and energy is transferred through this collision.
(31) The depopulation of the lower losing level E2 in gas lasers may occur through spontaneous emission of photons or through additional collisions, either with atoms or with the walls of the container.
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(32) Most solid lasers (and liquid lasers) employ optical pumping. Light from a flashlamp is absorbed by atoms in the active medium, raising them to "level" E4 (Figure 10). In solid and liquid lasers the level E4 is not a single level; rather it consists of a series of energy levels, or pumping bands as depicted in Figure 10. The fast decay transitions from E4 to E3 often are radiationless transitions that contribute to heating the laser active medium.

Fig. 10 Energy-level diagram for Nd:YAG
(33) In order for the optical energy from the flashlamp to be absorbed efficiently, its wavelength must correspond to a transition from the ground state to an available pumping band of the active medium. For Nd:YAG crystal, the strong pumping band has a wavelength range of 790820 nm, and the weak pumping band has a range of 730760 nm. The lamps used with these lasers, therefore, must produce their maximum output at these wavelengths for efficient laser operation. This matching process is called "spectral matching" of the pump source and active medium.
(34) The transition from E2 to E1 usually is a radiationless transition. Both the population of energy level E2 and the rate of decay from that level depend upon the temperature of the active medium. Water cooling normally is employed in solid lasers to remove waste heat and to depopulate the lower lasing level by reduction of the ambient heat level.
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(35) In a threelevel laser (Figure 11), the lower losing level is the ground state. This type of laser can have a population inversion only if more than half the atoms in the ground state are pumped to higher energy states. In the fourlevel laser, such intense pumping is unnecessary because a high groundstate population does not affect the population inversion between the upper and lower laser levels. The ruby laser is the only important three-level laser.

Fig. 11 Three-level laser
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(36) Figure 12 is the energylevel diagram of a fourlevel laser model. The lasing transition occurs between E3 and E2. Energy level E3 is called the upper lasing level and is a metastable state, with a lifetime of 106 seconds or longer.

Fig. 12 Four-level laser
(37) Atoms remain in this metastable state for a relatively long time, increasing both the population inversion and the probability of stimulated emission. Level E2 has a short atomic lifetime, which means that atoms leave this energy state quickly and return to the ground state.
(38) An upward transition to the metastable state is just as unlikely as a downward transition from it; therefore, the pumping of atoms directly from the ground state to the upper lasing level would be impractical. An alternative energy level, E4, is a short-lived atomic state above E3, to which atoms may be pumped more easily. Since short energy transitions are more likely to occur than long energy transitions, most of the atoms pumped to atomic state E4 can be counted on to decay rapidly to atomic state E3, the metastable state.

a. Transmission of the filter.
b. Absorption coefficient.
c. Transmission of a piece of the same material 2.5 mm thick
for HeNe laser light.
General Comment from Glenn Oliver
a. Neutral density.
b. Bandpass.
c. Cutoff.
a. 0.6
b. 2.5
c. 4.0
a. Normal population distribution.
b. Population inversion.
a. Gain of the tube.
b. Power of the beam leaving the tube.
a. Fourlevel.
b. Threelevel.
a.Pumping in gas lasers.
b.Pumping in solid lasers.
c.Depopulation of the lower losing level in fourlevel gas lasers.
d.Depopulation of the lower losing level in
fourlevel solid lasers.

HeNe laser
Optical power meter
Set of gelatin or plastic color filters (6 each: blue, green, red)
1meter optical bench
2 pin carriers
Micrometer
Filter holders
Beam expander
Lab jack
General Comment from Neil Miller
HeNe laser
Optical power meter
Alignment table with 3point adjustment
Meter stick
1meter optical bench
Brewster window HeNe laser tube
Power supply for tube
Pin carrier
Tube holder

In this experiment, the student will measure the transmission of three filter materials as a function of filter thickness and calculate the absorption coefficient of each material.

Fig. 13 Experimental setup for absorption measurements
In this experiment, the student will measure the small signal gain of a HeNe laser tube and will calculate the gain coefficient.

Fig. 14 Experimental setup for gain measurement
KEEP YOUR HANDS AWAY FROM THE HIGHVOLTAGE TERMINALS WHEN THE POWER SUPPLY IS ACTIVATED. Turn the power supply switch to the ON position and verify operation of the laser tube. Then turn off the power.

DATA TABLE 1. EXPONENTIAL LAW OF ABSORPTION
| Equipment | Manufacturer/Model No. | ||
| Laser Optical Power Meter |
|||
| Laser Power Output, PO _______________ mW Average Filter Thickness _______________ mm |
|||
| Number of Filters | Transmitted Power, Pt | ||
| Blue | Green | Red | |
| 1 | |||
| 2 | |||
| 3 | |||
| 4 | |||
| 5 | |||
| 6 | |||
DATA TABLE 2.
| Color of Filter | Absorption Coefficient k (n), cm1, at n = 4.74 × 1014 Hz; l = 632.8 nm |
| Blue | |
| Green | |
| Red |
DATA TABLE 3. GAIN IN A LASER
| Active length: x = __________ cm Power with Brewster window tube off: P1 = _______________ mW Power with Brewster window tube on: P2 = _______________ mW Gain: Ga = P2/P1 = _______________ Gain coefficient: a = _______________REFERENCE |

Anderson, John D. Gasdypamic Lasers: An Introduction. From the Series Quantum Electronics Principles and Applications, edited by Yohhan Pao. New York: Academic Press, 1976.
Barnes, Frank S. ed. Laser Theory. New York: The Institute of Electrical and Electronics Engineers, Inc. (IEEE Press), 1972.
Driscoll, Walter G. and Vaughn, William, eds. Handbook of Optics. New York: McGrawHill Book Company, 1978.
Lengyel, Bela A. Lasers. 2nd ed. New York: WileyInterscience, 1971.
O'Shea, Donald C.; Called, W. Russell; and Rhodes, William T. Introduction to Lasers and Their Applications. Reading, MA: AddisonWesley Publishing Co., 1978.
Pressley, Robert J., ed. CRC Handbook of Lasers, with Selected Data on Optical Technology. Cleveland, OH: The Chemical Rubber Company, 1971.
Shortley, George and Williams, Dudley. Elements of Physics. 5th ad. Englewood Cliffs, NJ: PrenticeHall, Inc., 1971.
Siegman, A.E. Introduction to Lasers and Masers. New York: McGrawHill Book Company, 1971.
Verdeyen, Joseph T. Laser Electronics. Englewood Cliffs, NJ: PrenticeHall, Inc., 1981.
Yariv, Ammon. Introduction to Optical Electronics. New York: Holt, Rinehart and Winston, Inc., 1971.
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