Index of Refraction: Calculate ni for 10 cm Attenuation

[June_cosmo]

Homework Statement

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

Homework Equations

3. The Attempt at a Solution [/B]
so n_i=\sqrt{\frac{\epsilon\mu}{\epsilon_0\mu_0}}
but I still don't get the imagery part...

 

[ehild]

Homework Statement

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

Homework Equations

3. The Attempt at a Solution [/B]
so n_i=\sqrt{\frac{\epsilon\mu}{\epsilon_0\mu_0}}
but I still don't get the imagery part...

What have you learned 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.

 

[June_cosmo]

What have you learned 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.

Thanks! That is helpful!