Show that the reflected and transmitted waves are related to the incident wave

[2thumbsGuy]

Homework Statement

Suppose that an electromagnetic wave is moving in the +x-direction, from one medium to another; call them medium 1 and medium 2. The boundary conditions on the electric and magnetic fields are then
E₁= E₂, \frac{1}{μ1}B₁=\frac{1}{μ2}B₂

These equations relate the components of electric and magnetic fields just to the left and just to the right of the interface between the two media, parallel to the surface. (There are no components perpendicular to the surface, so no boundary conditions in this direction.)

As described in class, there will be an incident wave E_I, a reflected wave E_R, and a transmitted wave E_T. Apply the boundary conditions given above and show that the reflected and transmitted waves are related to the incident wave.

Homework Equations

E_R=\frac{(1-β)}{(1+β)}E_I, E_t=\frac{2}{(1+β)} E_I
where
β=\frac{μ_1}{v_1} \frac{μ_2}{v_2}=√\frac{μ_1}{ɛ_2}\frac{μ_2}{ɛ_1}=\frac{μ_1}{μ_2}\frac{n_1}{n_2}
HINT: Recall that B=\frac{1}{c} E,n=\frac{c}{v},c=(\frac{1}{√ɛμ}), and that the speed of the EM wave is different in different media.

The Attempt at a Solution

E_R=((α-β)/(α+β)) E_I and E_T=(2/(α+β)) E_I where α ≡ (cosθ_T)/(cosθ_I ) and β≡μ₁/μ₂ n₁/n₂.
If the E and B fields are parallel to the incident medium and do not have any components vertical to the medium, this means the E and B fields are parallel to the surface (perpendicular to the normal).
α is a function of the angle from the EM wave to the normal, and since the angle between the angle of the incident wave to the normal is 0, α=(cosθ_T)/(cosθ_I )=cos0/cos0=1
In addition, β, which is a constant specific to how the mediums 1 & 2 affect how light travels in them, equals μ₁/μ₂ n₁/n₂ (among other things) So, more completely:
E_R=((cosθ_T)/(cosθ_I )-β)/((cosθ_T)/(cosθ_I )+β) E_I, E_T=2/((cosθ_T)/(cosθ_I )+β) E_I
Then, at the boundary, if:
E_I=E_R+E_T=((cosθ_T)/(cosθ_I)-β)/((cosθ_T)/(cosθ_I)+β) E_I)+2/((cosθ_T)/(cosθ_I)+β) E_I)
Then: E_I=(((cosθ_T)/(cosθ_I)-β)/((cosθ_T)/(cosθ_I)+β)+2/((cosθ_T)/(cosθ_I)+β)) E_I=(((cosθ_T)/(cosθ_I)-β+2)/((cosθ_T)/(cosθ_I )+β))E_I
Since (cosθ_T)/(cosθ_I)=1
E_R+E_T=((1-β+2)/(1+β)) E_I⇒(E_R+E_T)/E_I = (1-β+2)/(1+β)=1

This means that (E_R+E_T)/E_I is proportional, making them equal at the boundary of the mediums.

Please tell me if I'm missing some critical point, all these subtleties are lost on me.

 

[ehild]

As I understand the text of the problem, you need to start from the boundary conditions and derive the relation between E_R, E_T and E_i at normal incidence. You did just the opposite: started out from the general transmission and reflection coefficients and arrived to the boundary conditions.

ehild

 

[2thumbsGuy]

Hmmm, ok. So, then:

E_R=\frac{α-β}{α+β}E_I and E_T=\frac{2}{α+β}E_I where α ≡ \frac{cosθT}{cosθI} and β≡\frac{μ1}{μ2} \frac{n1}{n2}.
If the E and B fields are parallel to the incident medium and do not have any components vertical to the medium, this means the E and B fields are parallel to the surface (perpendicular to the normal).
α is a function of the angle from the EM wave to the normal, and since the angle between the angle of the incident wave to the normal is 0, α=\frac{cosθ_T}{cosθ_I}=\frac{cos(0)}{cos(0)}=1
In addition, β, which is a constant specific to how the mediums 1 & 2 affect how light travels in them, equals \frac{μ1}{μ2} \frac{n1}{n2} (among other things) So, more completely:
ER=\frac{\frac{cosθT}{cosθI}-β}{\frac{cosθT}{cosθI}+β}EI, ET=2/\frac{}{}((cosθT)/(cosθI )+β) EI
Then, at the boundary, if:
EI=ER+ET=\frac{}{}\frac{}{}((cosθT)/(cosθI)-β)/\frac{}{}((cosθT)/(cosθI)+β) EI)+2/((cosθT)/(cosθI)+β) EI)
Then: EI=\frac{}{}\frac{}{}(((cosθT)/(cosθI)-β)/\frac{}{}((cosθT)/(cosθI)+β)+2/((cosθT)/(cosθI)+β)) EI=(((cosθT)/(cosθI)-β+2)/((cosθT)/(cosθI )+β))EI
Since (cosθT)/(cosθI)=1
ER+ET=\frac{}{}((1-β+2)/(1+β)) EI⇒(ER+ET)/EI = \frac{}{}(1-β+2)/(1+β)=1\frac{}{}

 

[2thumbsGuy]

darnit, I hit reply... hang on let me try again. Why can't I delete my own post?

 

[2thumbsGuy]

Here's another way to look at it, according to some of my prof's lectures on waves on a string:

But it doesn't take into account the magnetic field. His lecture handles wavenumber and wave speed independantly, so it doesn't make sense to me how he wants for me to apply E_R=\frac{1-β}{1+β}E_I and E_R=\frac{2}{1+β}E_I

 

[2thumbsGuy]

Should I be adding the magnetic field to the electric one?

 

[ehild]

I suppose you are familiar with Maxwell equations and electromagnetic waves. You know that both E and B are perpendicular to the direction of propagation of the wave and also to each other.

At an interface between two media, that component of the electric field which is parallel with the interface, is equal at both sides.
The same is true for B/μ.
In medium 1 from where the wave is incident, the electromagnetic field is the sum of the incident wave and the reflected one, while there is only a single traveling wave - the transmitted wave - in the second medium.

It is normal incidence as you said already, the whole electric and magnetic fields are parallel with the interface. The boundary conditions mean that

E_i+E_r=E_t,
1/μ₁(B_i+B_r)=B_t/μ₂ .

I think it was shown to you that the amplitude of the electric and magnetic waves are related as B=E/v where is v is the velocity of light in the medium, positive for a traveling wave and negative for the reflected one. Use this relation in the previous formula and solve the system of equations for E_t and E_R in terms of E_i .
Take care. β is incorrectly written in your first post. It has to be a dimensionless parameter: β=(μ₁v₁)/(μ₂v₂).

ehild

 

[ehild]

Should I be adding the magnetic field to the electric one?

No, they are entirely different quantities!


ehild

 

[2thumbsGuy]

OK, that makes more sense. I was hoping I wouldn't have to consider both at the same time, it was simply illogical.

 

[2thumbsGuy]

I think it was shown to you that the amplitude of the electric and magnetic waves are related as B=E/v where is v is the velocity of light in the medium, positive for a traveling wave and negative for the reflected one. Use this relation in the previous formula and solve the system of equations for E_t and E_R in terms of E_i .
Take care. β is incorrectly written in your first post. It has to be a dimensionless parameter: β=(μ₁v₁)/(μ₂v₂).
ehild

Well, I forgot all about B=E/v... so many eqns. OK, so I substituted E/v into 1/μ₁(B_i+B_r)=B_t/μ₂, but it only gets me to E_r = (βE_t)-E_i and E_t = β(E_r+E_i)

That doesn't seem close enough to the solutions in relevant equations. I substituted E_r into the solution for E_t and just came up with E_t = β²E_t, so that's clearly wrong. Did I mis-understand you? Anyway, it's clear that I'm lost. I don't know what steps to take, when I'm done, what to do next, nothing.

 

[ehild]

E_i+E_r=E_t
Using B=E/v transforms the second equation into
(E_i-E_r)/(v₁μ₁)=E_t/(v₂μ₂).
β=(v₁μ₁)/(v₂μ₂),

with that notation:
(E_i-E_r)=βE_t
E_i+E_r=E_t

Add the equations:E_r cancels. Express E_t in terms of E_i.

ehild

 

[2thumbsGuy]

OK, I see how you got that. But why would you add them? What about the first two relevant eqns in #2? These solutions so far do not fall into that format, and if I plug in, say (E_i-E_r)=βE_t it does not resolve.

 

[ehild]

You have got two equations,

E_i+E_r=E_t
E_i-E_r=βE_t.

Two equation, two unknowns. Find E_t and E_r in terms of Ei and β.

It is simple Maths.

If you have a similar system of equations in Maths,

x+y=A
x-y=βA

how will you solve it for x and y in terms of A and β ?

ehild

 

[2thumbsGuy]

Ah.

\frac{1}{μ1}B₁ = \frac{1}{μ2}B₂

\frac{1}{μ1v1}E₁ = \frac{1}{μ2v2}E₂

E₁ = \frac{μ1v1}{μ2v2}E₂

E₁ = βE₂

E_i-E_r = βE_t

Subbing for E_t:

E_i-E_r = β(E_i+E_r)

E_r = E_i-βE_i-βE_r

E_r+βE_r = E_i-βE_i

E_r(1+β) = E_i(1-β)

E_r = \frac{1-β}{1+β}E_i

So simple! Why didn't I see it before?

I can do the other one subbing for E_r.

Thanks for your patience. I was trying to get this done before the end of class but I didn't make it, so I'm hoping for partial credit. Anyway, I still wanted to solve it, so thank you for your help, I really do appreciate it.

 

[ehild]

You are welcome. Next time you will see how to solve a system of equations at once

ehild