Fyneman exercise I ch31-3

Q.

Noting that the electric field of a light wave traveling throgh a medium of refractive index $n$ is $E=E_0e^{i\omega(t-nz/c)}$,
a) show that, if $n=n' -in''$,

$E=E_0 e^{-n''\omega z/c} e^{i\omega(t-n'z/c)}$
b) use the equation $n-1=\frac{Nq^2}{2\epsilon_0 m}\frac{1}{\omega_0^2-\omega^2+i\gamma\omega}$ to find the rate at which the intensity of a beam of radiation whose frequency is exactly equal to the natural frequency $\omega_0$ of an atom is attenuated.

A.

a)

Substitute $n=n' - in''$,

$E=E_0 e^{i\omega(t-n'z/c+in''z/c)}=E_0 e^{i\omega(t-n'z/c)} e^{i\cdot i n''z/c}=E_0 e^{i\omega(t-n'z/c)}e^{-n''z/c}$

b)

$\omega_0=\omega$,

$n-1=\frac{Nq^2}{2\epsilon_0 m}\frac{1}{i\gamma\omega}=-i\frac{Nq^2}{2\epsilon_0 m}\frac{1}{\gamma\omega}$

$\therefore n''=\frac{Nq^2}{2\epsilon_0 m}\frac{1}{\gamma\omega}$

$I=I_0 \left|e^{i\omega(t-n'z/c)}e^{-n''z/c}\right|^2=I_0 \left| e^{i\omega(t-n'z/c)} \right|^2 \left|e^{-n''z/c}\right|^2$
, where $\left| e^{i\omega(t-n'z/c)} \right|^2=1$ and $\left|e^{-n''z/c}\right|^2=e^{\frac{Nq^2}{\epsilon_0 m}\frac{1}{\gamma\omega}\frac{z}{c}}$

$\therefore I=I_0 e^{-\frac{Nq^2}{\epsilon_0 m}\frac{1}{\gamma\omega}\frac{z}{c}}$

So, attenuation rate of intensity $I$ is

$e^{-\frac{Nq^2}{\epsilon_0 m}\frac{1}{\gamma\omega}\frac{z}{c}}$