# Absorption coefficient and Linear Optical Susceptibility

PhysicsTruth
Homework Statement:
For a complex refractive index ##n^*=n+ik##, establish the relationship between the absorption coefficient and linear optical susceptibility. Take ##(n+ik)^2 = \epsilon = 1 + \chi##
Relevant Equations:
##(n+ik)^2 = \epsilon = 1+ \chi##
##I=I_0 e^{-\alpha z}##
##\alpha = \frac{4\pi k}{\lambda}##
##\alpha## is considered to be the absorption coefficient for a beam of light of maximum intensity ##I_0##. It's related to the complex part of the refractive index as we have shown above. Now, I have a doubt. Should I solve for ##k## from the quadratic equation in terms of the linear optical susceptibility ##\chi## directly, or should I assume a complex form of ##\chi## and separate the real and imaginary terms and then proceed?

## Answers and Replies

Homework Helper
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Homework Statement:: For a complex refractive index ##n^*=n+ik##, establish the relationship between the absorption coefficient and linear optical susceptibility. Take ##(n+ik)^2 = \epsilon = 1 + \chi##
I've (extremely) limited knowledge of this topic. But, since no one else has answered yet, see if this helps...

First note that:
##(n+ik)^2 = \epsilon = 1 + \chi##
should be:
##(n+ik)^2 = \epsilon_r = 1 + \chi##

The equation tells you that susceptibility, ##\chi##, and relative permittivity, ##\epsilon_r##, are being treated as complex quantities.

... or should I assume a complex form of ##\chi## and separate the real and imaginary terms and then proceed?
That sound like the way to go. It only requires simple algebra to express the real and imaginary parts of ##\chi## in terms of ##\alpha## (along with ##n## and ##\lambda##).

vanhees71 and PhysicsTruth
PhysicsTruth
Yeah, I've done that thankfully. Thanks for the heads up!

vanhees71 and Steve4Physics