Example 6.3Consider an epidermis skin tissue in which the absorption
coefficient is 300 cm−^1 at 308 nm and 35 cm−^1 at 633 nm. Using Eq. (6.3),
what are the absorption lengths for each of these wavelengths at which I(x)/
I 0 = 1/e = 0.368?
Solution: (a) From Eq. (6.3) at 308 nm the intensity ratio isIðxÞ
I 0¼expðÞ¼lax exp½ð 300 Þx¼ 0 : 368Therefore300 cm^1x¼ln 0: 368 ¼ 0 : 9997which yields x = 3.33× 10 −^3 cm = 0.033 mm.
(b) Similarly, at 633 nm35 cm^1x¼ln 0: 368 ¼ 0 : 9997which yields x = 2.86× 10 −^2 cm = 0.286 mm.6.2.2 Absorption in Biological Tissues
In general, absorption in biological tissues is due largely to water molecules and
macromolecules such as proteins and chromophores [ 7 , 22 – 25 ]. By definition, a
chromophoreis the part of an organic molecule that absorbs certain wavelengths
and transmits or reflects other wavelengths. The reflection and transmission process
causes the molecule or compound to have a specific color. As Fig.6.8shows, water
is the main absorbing substance in the infrared region. Because water is the prin-
cipal constituent of most tissues, its absorption characteristic is of importance in
light-tissue interactions.
In the UV and visible regions the important absorbing substances include the
aromatic amino acids (such as tryptophan, tyrosine, and phenylalanine), proteins,
melanins, DNA, and porphyrins (which include hemoglobin, myoglobin, vitamin
B12, and cytochrome). The aromatic amino acids exhibit large absorption peaks in the
200 – 230-nm spectrum and have local peaks in the 250–280 nm range. The absorption
coefficients of three of these amino acids at their local peak absorption wavelengths
are listed in Table6.1. Proteins typically have a peak absorption value at about
280 nm, which depends on the concentrations of amino acids, such as tryptophan.
Two important absorbing substances in tissue are the chromophore melanin and
the protein hemoglobin.Melaninis the basic chromophore that gives human skin,
6.2 Absorption 157