In complex refractive index on a material, $n=n'+ ik$, the imaginary part $k$ is physical meaning, as it shows absorption in the material but it is an imaginary. How we measure an imaginary values in physics with imaginary numbers?

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    $\begingroup$ Do you know how to describe oscillation in terms of $e^{i\omega t}$? $\endgroup$ – Floris Apr 16 '14 at 4:31
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    $\begingroup$ Mathematically I know. But I do not understand the meaning of imaginary parameters like k. $\endgroup$ – eglim Apr 16 '14 at 4:46

You are getting confused by the fact that we use an imaginary number (a mathematical construct) to describe a physical phenomenon.

But it is just that - a mathematical trick. You can perfectly well describe a attenuating wave with

$$e^{ikx}e^{-k'x}$$ Which can be made more compact by combining the two k's into $$e^{i(k+ik')x}$$

If you are OK with the former notation (and it should be easy to see how to measure the distance over which the amplitude of the wave decays - that is a very "real" quantity) then you should accept the second notation as just a mathematical way to rearrange things for convenient manipulation.

I hope that clears things up a bit.


The previous answer is good enough but I want to add my point of view.
The fact that an eye and el. devices ultimately do not register the amplitude of the radiation but its intensity (energy), the intensity we get by squaring the real and imaginary units thus they are no longer imaginary.
That is, the imaginary part $k$ is an imaginary unit only in intermediate calculations and not in the final formula.
Incidentally, physically $k$ shows how radiation is reduced for a specific wavelength depending on the path length.
For instance:
$\tau$ - transmittance (internal) in lay with thickness $l$ $[m]$
$\alpha = -ln(\tau)/l$ $[1/m]$ - absorption coefficient (in general case it is lg)
$\lambda$ $[m]$ - wavelength
$k=\lambda \alpha/(4\pi)$ - extinction ratio


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