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)

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

Read more about the propagation of EM waves

Ionizing Radiation

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

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