Fundamentals of Materials Science and Engineering: An Integrated Approach, 3e

GTBL042-19 GTBL042-Callister-v2 September 13, 2007 13:59

Revised Pages

770 • Chapter 19 / Optical Properties

(a) (b)

Conduction

band

Bandgap

Valenceband

Photon

emitted,

 1 = ΔE^1

Photon

absorbed

Impurity

level

Energy ΔE 2

ΔE 1

ΔE 2

ΔE 1

ΔE

(c)

h

Photon

emitted,

 2 = ΔEhh^2

Photon

emitted,

Phonon

generated

having

energy ΔE 1

 2 = ΔE^2

Figure 19.6 (a) Photon absorption via a valence band-conduction band electron excitation

for a material that has an impurity level lying within the band gap. (b) Emission of two

photons involving electron decay first into an impurity state, and finally to the ground state.

(c) Generation of both a phonon and a photon as an excited electron falls first into an

impurity level and finally back to its ground state.

The intensity of the net absorbed radiation is dependent on the character of the

medium as well as the path length within. The intensity of transmitted or nonabsorbed

radiationI′Tcontinuously decreases with distancexthat the light traverses:

Intensity of

nonabsorbed

radiation—

dependence on

absorption

coefficient and

distance light

traverses through

absorbing medium

IT′=I 0 ′e−βx (19.18)

whereI′ 0 is the intensity of the nonreflected incident radiation andβ, theabsorption

coefficient(in mm−^1 ), is characteristic of the particular material; furthermore,βvaries

with wavelength of the incident radiation. The distance parameterxis measured

from the incident surface into the material. Materials that have largeβvalues are

considered to be highly absorptive.

EXAMPLE PROBLEM 19.1

Computation of the Absorption Coefficient for Glass

The fraction of nonreflected light that is transmitted through a 200 mm thickness

of glass is 0.98. Calculate the absorption coefficient of this material.

Solution

This problem calls for us to solve forβin Equation 19.18. We first of all rearrange

this expression as

IT′

I 0 ′

=e−βx

Now taking logarithms of both sides of the above equation leads to

ln

(

IT′

I′ 0

)

=−βx