EM Wave Reflection and Refraction: Understanding the Fresnel Formalism

[Apashanka]

For this question I want to clarify that 5k which is the electric field component perpendicular to the incident plane ( the xy plane) will be continuous for reflection and refraction ,
For which none of the options seems correct ,am I right??
The component of magnetic field perpendicular to incident plane will also continuous for reflection and refraction also??
Continous means same in magnitude and direction.
Thank you

 

[Charles Link]

A couple of hints: What is $\vec{k}$ for this wave? (You can see that because the exponent contains $\pm (\vec{k} \cdot \vec{r}-\omega t)$). $\\$ Next question: Is it transverse? i.e. is $\vec{k} \cdot \vec{E} =0$ ? $\\$ I haven't solved it yet to see which answer is correct, but perhaps this will help. $\\$ I think this really belongs in the homework section. I am going to ask the moderators to move it there. In the future, they will want you to post this kind of problem in the homework section and fill out the homework template.

 

[Charles Link]

And I solved it, and I did get one of the answers listed. What is the wave vector $\vec{k}'$ of the reflected wave? That is readily computed. $\\$ Also, at $x=0$, we must have $\vec{E}_{incident}+\vec{E}_{reflected}=0$. This will make the amplitude of the reflected wave equal to the minus of the amplitude of the incident wave, [Edit note: See post 14 below=the previous statement was determined to be incorrect], as you can see once you put in the exponential with $i( \omega t-\vec{k}' \cdot \vec{r} )$, and you set $x=0$ for both the incident and reflected waves.$\\$ With these hints, hopefully you can determine the answer.

 

[Apashanka]

And I solved it, and I did get one of the answers listed. What is the wave vector $\vec{k}'$ of the reflected wave? That is readily computed. $\\$ Also, at $x=0$, we must have $\vec{E}_{incident}+\vec{E}_{reflected}=0$. $\\$ With these hints, hopefully you can determine the answer.

The wave vector for the reflected wave is -3i-4j
I have made a schematic diagram and calculating the magnetic field component perpendicular to the incident plane and also the electric field component perpendicular to the incident plane I got none of the ans to be right.
One schematic diagram I have made

Am I right that first two options will not be correct since the conducting slab is positioned at x=O and the two will represent refracted wave.??

 

[Charles Link]

I don't agree, but let's work it together: I get that the incident $\vec{k}=-3 \hat{i}+4 \hat{j}$, because the exponential is written in the form $i (\omega t-\vec{k} \cdot \vec{r})$. Thereby, I think $\vec{k}'=3 \vec{i}+4 \hat{j}$. $\\$ Please look this over, and determine whether I might have it correct. $\\$ the next step is to write the exponential as $i( \omega t-\vec{k}' \cdot \vec{r})$.

 

[Apashanka]

I don't agree, but let's work it together: I get that the incident $\vec{k}=-3 \hat{i}+4 \hat{j}$, because the exponential is written in the form $i (\omega t-\vec{k} \cdot \vec{r})$. Thereby, I think $\vec{k}'=3 \vec{i}+4 \hat{j}$

Yes that right.
Sorry for the mistake I made above. But what about the perpendicular component of the electric and the magnetic field.

 

[Charles Link]

Yes that right.(typographical error)
Sorry for the mistake but what about the perpendicular component of the electric and the magnetic field.

See post 3, and read my other posts again. You don't need to calculate the magnetic field vector here. You should be able to compute the answer with post 3.

 

[Apashanka]

See post 3, and read my other posts again. You don't need to calculate the magnetic field vector here. You should be able to compute the answer with post 3.

Ohh sir since this is conducting slab so at any x=O the net electric field is 0 ,that why E(incd.)+E(reflected) =0 ,am I right??

 

[Charles Link]

Ohh sir since this is conducting slab so at any x=O the net electric field is 0 ,that why E(incd.)+E(reflected) =0 ,am I right??

$\vec{E}_{incident}=E_1 e^{i(\omega t+3x-4y)}$ and $E_{reflected}=E_2 e^{i(\omega t-3x-4y)}$. Notice even though the $3$ on the $3x$ has the opposite sign, at $x=0$ that term disappears. The result is $E_2=-E_1$.

 

[Apashanka]

Ohh sir since this is conducting slab so at any x=O the net electric field is 0 ,that why E(incd.)+E(reflected) =0 ,am I right??

Sir one more question ,for the incident and reflected em wave from a dielectric ,the components of the electric and magnetic field perpendicular to the plane of incidence they should not suffer any change in magnitude or direction??am I right?

 

[Apashanka]

$\vec{E}_{incident}=E_1 e^{i(\omega t+3x-4y)}$ and $E_{reflected}=E_2 e^{i(\omega t-3x-4y)}$. Notice even though the $3$ on the $3x$ has the opposite sign, at $x=0$ that term disappears. The result is $E_2=-E_1$.

Sir thank you very much for the clarification you have given.
It will help me in clearing the concept.
But sir this is possible since it is a conducting slab,for dielectrics this will not work,am I right??

 

[Charles Link]

For a dielectric, it gets a little complicated. Try reading through this "link": https://ocw.mit.edu/courses/electri...ring-2011/lecture-notes/MIT6_007S11_lec33.pdf

 

[Apashanka]

For a dielectric, it gets a little complicated. Try reading through this "link": https://ocw.mit.edu/courses/electri...ring-2011/lecture-notes/MIT6_007S11_lec33.pdf

Sir once again thanks

 

[Charles Link]

I thought $d$ was the correct answer, but I see the reflected wave that we have is no longer transverse. I'm starting to think the boundary condition we used (from post 3) is incorrect, and we need to look at the parallel (to the surface) components. I need to work through the details. For the components parallel to the surface, we must have $E_{incident}+E_{reflected}=0$, but that does not apply to the perpendicular component. $\\$ Edit: I don't think $c$ is correct either. To satisfy the parallel to the surface boundary condition, we must have $-6 \hat{j} -5 \hat{k}$ in the reflected wave, and to have it be transverse (with $\vec{k}' \cdot \vec{E}=0$ ), the $8 \hat{i}$ needs to stay as $8 \hat{i}$ in the reflected wave.

 

[rude man]

Perhaps the best approach is to do a coordinate transformation such that the plane wave propagates in the (say) z' direction. Then all you'd have to worry about is the sinusoidal variation with position which would be expressed in x' y' z' coordinates instead of the unprimed system.

Then you'd have to reverse-transform back to the unprimed system.

http://homepages.engineering.auckla...sors_05_Coordinate_Transformation_Vectors.pdf

 

[Charles Link]

Perhaps the best approach is to do a coordinate transformation such that the plane wave propagates in the (say) z' direction. Then all you'd have to worry about is the sinusoidal variation with position which would be expressed in x' y' z' coordinates instead of the unprimed system.

Then you'd have to reverse-transform back to the unprimed system.

http://homepages.engineering.auckla...sors_05_Coordinate_Transformation_Vectors.pdf

But then the location of the surface (which is at $x=0)$ becomes a plane of the form $Ax'+By' =0$. It is far simpler to have the equation of the surface be $x=0$.

 

[Charles Link]

See also: https://www.brown.edu/research/labs....labs.mittleman/files/uploads/lecture13_0.pdf and look at the $r_{||}$. Let $n_t=+\infty$, and you see that $r_{||}=-1$. It is important to look at the orientation in their diagram for the sign of the $E_x$ and $E_y$. If I am interpreting it correctly, a $r_{||}=-1$ will have $E_x$ be the same upon reflection, but $E_y$ will get the opposite sign.

 

[Apashanka]

See also: https://www.brown.edu/research/labs....labs.mittleman/files/uploads/lecture13_0.pdf and look at the $r_{||}$. Let $n_t=+\infty$, and you see that $r_{||}=-1$. It is important to look at the orientation in their diagram for the sign of the $E_x$ and $E_y$. If I am interpreting it correctly, a $r_{||}=-1$ will have $E_x$ be the same upon reflection, but $E_y$ will get the opposite sign.

Sir the explanation that

See also: https://www.brown.edu/research/labs....labs.mittleman/files/uploads/lecture13_0.pdf and look at the $r_{||}$. Let $n_t=+\infty$, and you see that $r_{||}=-1$. It is important to look at the orientation in their diagram for the sign of the $E_x$ and $E_y$. If I am interpreting it correctly, a $r_{||}=-1$ will have $E_x$ be the same upon reflection, but $E_y$ will get the opposite sign.

Sir the explanation you have given earlier and what I also thought correct is that the parallel as well as the perpendicular component of electric field on the surface of a conductor has to be 0(of course near the surface it is perpendicular ) otherwise it will lead to charge flow on the surface as well as popping up of charge from the surface,and it will no longer remain conductor
Hence to satisfy this, all the three components of the reflected wave of the electric field has to be negative at the surface (e.g at the origin) propagating in the positive x-y plane.
But taking the continuity of the magnetic field and to satisfy k dot E and k dot B =0, it creates some confusion

 

[Charles Link]

It is necessary for the component of $E$ parallel to the surface to be zero. There is no requirement in this case for the x-component of $E$. $\\$ To work this problem using the Fresnel formalism, the electric field is split into two components of different polarization types: $\\$ 1)The "parallel" component that lies in the plane of incidence and reflection, which in this case is the x-y plane $\\$ 2) The component that is perpendicular to the plane of incidence (which in this case is the $k$ component). $\\$ For the Fresnel formalism, the words "parallel" and "perpendicular" refer to the plane of incidence rather than the plane of the surface. $\\$ It might be helpful to work the simpler case first where the direction of the wave is perpendicular to the surface. In that case, the polarization doesn't matter, and the well know results for the Fresnel coefficients are $\rho=\frac{n_1-n_2}{n_1+n_2}$ and $\tau=\frac{2n_1}{n_1+n_2}$. For this simpler case, note that for a conductive surface with $n_2=+\infty$ that $\rho=-1$ and $\tau=0$. $\\$ (Note: I'm using $r$ and $\rho$ interchangeably. Some books use $r$ for the Fresnel coefficient, while others use $\rho$). $\\$ And for the more difficult problem that you have here, I think it is necessary to look at the orientation of $\vec{E}_r$ specified in the diagram: I think it is incorrect to write $\vec{E}_r=\rho \vec{E}_i$. Instead, the coefficient relates the amplitudes $E_r$ and $E_i$, (with $E_r=\rho E_i$ ), with the direction of the vectors as indicated in the diagram. That's why the sign of the x-component of $E$ is a tricky one in this problem. See if you might agree with me here on this, because this case is one that I have not previously encountered in detail. On the "link" of post 17, on page 11 they mention that "Hecht , (a well-known Optics book=Hecht and Zajac), uses a different sign convention for the reflected field, so that some of the signs are changed,... " . I am going to need to study this in a little more detail, but I am still pretty sure we got the right answer when we worked the problem above.