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

1. Dec 5, 2011

### 2thumbsGuy

1. The problem statement, all variables and given/known data
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 EI, a reflected wave ER, and a transmitted wave ET. Apply the boundary conditions given above and show that the reflected and transmitted waves are related to the incident wave.

2. Relevant equations
ER=$\frac{(1-β)}{(1+β)}$EI, Et=$\frac{2}{(1+β)}$ EI
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.

3. The attempt at a solution
ER=((α-β)/(α+β)) EI and ET=(2/(α+β)) EI where α ≡ (cosθT)/(cosθI ) and β≡μ12 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 μ12 n1/n2 (among other things) So, more completely:
ER=((cosθT)/(cosθI )-β)/((cosθT)/(cosθI )+β) EI, ET=2/((cosθT)/(cosθI )+β) EI
Then, at the boundary, if:
EI=ER+ET=((cosθT)/(cosθI)-β)/((cosθT)/(cosθI)+β) EI)+2/((cosθT)/(cosθI)+β) EI)
Then: EI=(((cosθT)/(cosθI)-β)/((cosθT)/(cosθI)+β)+2/((cosθT)/(cosθI)+β)) EI=(((cosθT)/(cosθI)-β+2)/((cosθT)/(cosθI )+β))EI
Since (cosθT)/(cosθI)=1
ER+ET=((1-β+2)/(1+β)) EI⇒(ER+ET)/EI = (1-β+2)/(1+β)=1

This means that (ER+ET)/EI 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.

2. Dec 6, 2011

### ehild

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

ehild

3. Dec 6, 2011

### 2thumbsGuy

Hmmm, ok. So, then:

ER=$\frac{α-β}{α+β}$EI and ET=$\frac{2}{α+β}$EI 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{}{}

4. Dec 6, 2011

### 2thumbsGuy

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

5. Dec 6, 2011

### 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 ER=$\frac{1-β}{1+β}$EI and ER=$\frac{2}{1+β}$EI

6. Dec 6, 2011

### 2thumbsGuy

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

7. Dec 6, 2011

### 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 travelling 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

Ei+Er=Et,
1/μ1(Bi+Br)=Bt2 .

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 travelling wave and negative for the reflected one. Use this relation in the previous formula and solve the system of equations for Et and ER in terms of Ei .
Take care. β is incorrectly written in your first post. It has to be a dimensionless parameter: β=(μ1v1)/(μ2v2).

ehild

8. Dec 6, 2011

### ehild

No, they are entirely different quantities!

ehild

9. Dec 6, 2011

### 2thumbsGuy

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

10. Dec 6, 2011

### 2thumbsGuy

Well, I forgot all about B=E/v... so many eqns. OK, so I substituted E/v into 1/μ1(Bi+Br)=Bt2, but it only gets me to Er = (βEt)-Ei and Et = β(Er+Ei)

That doesn't seem close enough to the solutions in relevant equations. I substituted Er into the solution for Et and just came up with Et = β2Et, 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.

11. Dec 6, 2011

### ehild

Ei+Er=Et
Using B=E/v transforms the second equation into
(Ei-Er)/(v1μ1)=Et/(v2μ2).
β=(v1μ1)/(v2μ2),

with that notation:
(Ei-Er)=βEt
Ei+Er=Et

Add the equations:Er cancels. Express Et in terms of Ei.

ehild

Last edited: Dec 6, 2011
12. Dec 6, 2011

### 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 (Ei-Er)=βEt it does not resolve.

13. Dec 6, 2011

### ehild

You have got two equations,

Ei+Er=Et
Ei-Er=βEt.

Two equation, two unknowns. Find Et and Er 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

14. Dec 8, 2011

### 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

Ei-Er = βEt

Subbing for Et:

Ei-Er = β(Ei+Er)

Er = Ei-βEi-βEr

Er+βEr = Ei-βEi

Er(1+β) = Ei(1-β)

Er = $\frac{1-β}{1+β}$Ei

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

I can do the other one subbing for Er.

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.

15. Dec 8, 2011

### ehild

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

ehild