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?

 

[Steve4Physics]

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 $\lambda$).

 

[PhysicsTruth]

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