Question

28, Mo EO, o=0 o=0 X 45° Hi o k

medium1 : ε_0, μ₀

medium2 : 2ε_0, μ₀

Condition : 1GHz TM electromagnetic wave, H_i0 = 1A/m

What is poynting vector of field that exists in medium2?

I think that we can't calculate the ponyting vector of field exists in medium 2 in the total reflection case.

Answer 1

Answer :

Given data :

Medium 1 : ε0, μ0

medium 2 : 2ε0, μ0

angle $\theta = 45\degree$

So Know that,

​​Poynting vector $\vec{S} = \vec{E}\times \vec{H}$    and

$S = EH \sin\theta$

where $E = \mu_o c H$

So,
S = MocHº sin
   = $S =\frac{\mu_0c(1)^2}{\sqrt2}$ = $S =\frac{\mu_0c}{\sqrt2}$    .

It is a case of total reflection,

because in this case,

refractive index of medium 2 = √2 n

refractive index of medium 1 = n

by applying snell's law,

we can find the value of angle of refraction throgh medium 1 = 90° which is a total reflection case.