Geometric Optics for EM wave

EM wave propagation

An EM wave goes through the athmosphere of Earth (considered inhomogeneous) having an index of refraction of $n(z)$. This index of refraction depends on altitude $z$ as:

$n(z) =1+a*exp(-\frac {z}{H})$ where $a =3*10^{-4}$ and $H = 9 km$ is the altitude.

Please find a system the  equations for the position and wave vector evolution as $x$, $y$, $z$, $k_x$, $k_y$ and $k_z$. (Start from the definition of the speed of the EM waves then consider the definition of the wave vector.)

Let the speed of the EM wave in vacuum be $C$. Then along x, y and z axes the speed of the wave will be

x:   $Vx = C$,   $X =X0 + C*t$

y:  $Vy = C$,  $Y = Y0+C*t$

z:  $Vz = C/n(z)$,  $dZ = [C/n(z)] *dt$

$dz =C/(1+a*exp(-z/H)) *dt$

$(1+a*exp(-z/H))*dz = C*dt$

by integrating (from $Z0$ to $Z$) we get the equation

$z-a*H*exp(-z/H) + [- z0 + a*H*exp(-z0/H) ] = C*t$

$z-a*H*exp(-z/H) = z0-a*H*exp(-z0/H) + C*t$

We write the equations for the wave vector $k = 2*pi/\lambda = 2*pi/(V*T)$

$Kx = 2*\pi/(C*T) = constant$ (since C is constant)

$Ky= 2*\pi/(C*T) = constant$ (since C is constant)

$Kz(z) =2*\pi / (C*T/n)= n(z) *2*pi/(C*T)$

Some radiation is used on products of meat to lower the levels of microbial pathogens. For meat this is maximum 4.5 kGy. If electrons of 1.6 MeV are used for 5 kg of beef

what is the number of electrons to attain the allowable limit?

The equivalence between Gy (Gray) and J/s is written as

$1 Gy (Gray) = 1 J /1 kg$

Therefore if D is dose of radiation (in Gy) we have

$D = Et/m$ where $Et$ is the total energy.

If $E$ is the energy of an electron then we write the numer of electrons as

$N =Et/E = m*D/E = 5*4.5*10^3/(1.6*10^6*1.6*10^{-19}) =8.789*10^{16}$ electrons

($1 eV =1.6*10^{-19} J$)