The discussion centers on calculating the imaginary part of the index of refraction (ni) for an electromagnetic (EM) wave with a wavelength of 1 mm, which is attenuated to 10% of its original intensity after traveling 10 cm in a material. The complex refractive index is expressed as N = n + iκ, where κ represents the imaginary part. The relationship between intensity and the electric field is established, showing that intensity decreases exponentially with distance, leading to the determination of the imaginary part of the refractive index.
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What have you learned about EM wave propagation in absorbing media?June_cosmo said: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...
Thanks! That is helpful!ehild said: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.