# Refraction index

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1. Jan 20, 2016

### June_cosmo

1. The problem statement, all variables and given/known data
Within a certain material, an EM wave with = 1 mm is attenuated to
10% of its original intensity after propagating 10 cm. Determine the imaginary part of the index
of refraction ni

2. Relevant equations
3. The attempt at a solution

so $$n_i=\sqrt{\frac{\epsilon\mu}{\epsilon_0\mu_0}}$$
but I still don't get the imagery part...

2. Jan 21, 2016

### ehild

What have you learnt about EM wave propagation in absorbing media?
IN complex notation, the electric field moving in the + x direction and having angular frequency ω is described with the function $E = E_0 e^{i(\frac{2π}{λ_0}Nx-ωt)}$, where λ0 is the vacuum wavelength and N is the refractive index.
The complex refractive index has both real and imaginary parts, N=n+iκ. Replacing it into the wave formula, you get
$$E = E_0 e^{i(\frac{2π}{λ_0}(n+iκ)x-ωt)}=E_0 e^{-\frac{2π}{λ_0}κx+i(\frac{2π}{λ_0}nx-ωt)}=\left (E_0 e^{-\frac{2π}{λ_0}κx} \right) e^{i(\frac{2π}{λ_0}nx-ωt)}$$,
which is a wave with exponentially decreasing amplitude.
The intensity is proportional to the square of the magnitude of the electric field I ∝|E|2. From here, you get the change of intensity with the distance and the imaginary part of the refractive index.

3. Jan 24, 2016