EM Wave Reflection and Transmission Between 3 Materials

In summary, the conversation discusses a problem involving a wave incident from air on an anti-reflective coating, which generates two new expressions for the reflected and transmitted waves through the coating. These expressions can be related to the waves in air and in the coating using Fresnel reflection and transmission coefficients. The problem of the reflected total wave can be solved using an infinite series or by using a collective total wave amplitude for left and right going waves at the interfaces within the coating. The conversation also discusses the calculation of amplitudes for the waves leaving the coating and the use of infinite series in the calculation.
  • #1
Mr_Allod
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Homework Statement
Electromagnetic radiation is incident normally on a material whose index of refraction is ##n##. Show that the reflected wave can be eliminated by covering the material with a layer of a second material whose index of refraction is ##n^{\frac 1 2}## and whose thickness is one-quarter of a wavelength.
Relevant Equations
##\tilde {\vec E} (z,t) = \tilde E_{0} e^{i(k_1z- \omega t)} \hat x##
##\tilde {\vec B} (z,t) = \frac 1 v \tilde E_{0} e^{i(k_1z- \omega t)} \hat y##
Boundary Conditions:
##\epsilon_1 E^{\perp}_1= \epsilon_2 E^{\perp}_2##
##B^{\perp}_1= B^{\perp}_2##
##\vec E^{\parallel}_1= \vec E^{\parallel}_2##
##\frac 1 {\mu} \vec B^{\parallel}_1=\frac 1 {\mu} \vec B^{\parallel}_2##
Hello there. I set up the problem like this, I have a wave incident from air on the anti-reflective coating consisting of:
##\tilde {\vec E_I} (z,t) = \tilde E_{0_I} e^{i(k_1z- \omega t)} \hat x##
##\tilde {\vec B_I} (z,t) = \frac 1 v \tilde E_{0_I} e^{i(k_1z- \omega t)} \hat y##

This wave gets both reflected and transmitted through the coating generating two new expressions:
##\tilde {\vec E_{R_1}} (z,t) = \tilde E_{0_{R_1}} e^{i(-k_1z- \omega t)} \hat x##
##\tilde {\vec E_{T_1}} (z,t) = \tilde E_{0_{T_1}} e^{i(k_2z- \omega t)} \hat x## (I will only write out the E-field expressions to keep things shorter)

I can now relate the expressions for air and coating with: ## \tilde {\vec E_{R_1}} + \tilde {\vec E_{T_1}} = \tilde {\vec E_I}## .

##\tilde {\vec E_{T_1}}## is then incident on the material of index n, once again both transmitting and reflecting:
##\tilde {\vec E_{R_2}} (z,t) = \tilde E_{0_{R_2}} e^{i(-k_2z- \omega t)} \hat x##
##\tilde {\vec E_{T_2}} (z,t) = \tilde E_{0_{T_2}} e^{i(k_3z- \omega t)} \hat x##

The expressions between coating and material can be related with: ## \tilde {\vec E_{R_2}} +\tilde {\vec E_{T_2}} =\tilde {\vec E_{T_1}}##

##\tilde {\vec E_{R_2}}## then travels back towards the interface between the anti-reflective coating and air, and its here that I start to get confused. Do I need to account for another transmission and reflection as the wave is leaving the anti-reflective coating? And if so how would this affect the expressions I already for the interfaces between other media? Because as I'm going now it seems I would be indefinitely accounting for reflections and transmissions. Eventually I believe I should be able to show that ##\tilde {\vec E_{R_1}}## and another wave leaving the reflective coating (##{\vec E_{T_3}}## perhaps?) are perfectly out of phase and with equal amplitudes therefore causing destructive interference. But I must be missing some key steps in between so I'd appreciate it if someone could give me a hand.
 
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  • #2
This one is most easily worked using Fresnel reflection and transmission coefficients:
## \rho=\frac{n_1-n_2}{n_1+n_2} ## and ## \tau=\frac{2n_1}{n_1+n_2} ##.
The problem of the reflected ## E_{reflected \, total} ## can be done either as an infinite series of multiple reflections plus transmissions, or solved by using a collective total wave amplitude for left going and right going waves at the two interfaces inside the dielectric coating. Suggest you just keep the phase term for optical path distance, and leave out the ## i \omega t ## dependence. It's been a few years since I did this particular calculation, but it's a fair amount of algebra. Note that intensity ## I ## is proportional to ## n |E|^2 ##.
 
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  • #3
Mr_Allod said:
Do I need to account for another transmission and reflection as the wave is leaving the anti-reflective coating? And if so how would this affect the expressions I already for the interfaces between other media?

Yes you do but it will just reiterate the calculation you have just done, providing successive terms in an infinite expansion. You can then recognize it as a power series and solve for a closed form. The expressions for the interfaces are correct but need subsequent reflections added in.
This is a very useful and interesting calculation.
 
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  • #4
Charles Link said:
This one is most easily worked using Fresnel reflection and transmission coefficients:
## \rho=\frac{n_1-n_2}{n_1+n_2} ## and ## \tau=\frac{2n_1}{n_1+n_2} ##.
The problem of the reflected ## E_{reflected \, total} ## can be done either as an infinite series of multiple reflections plus transmissions, or solved by using a collective total wave amplitude for left going and right going waves at the two interfaces inside the dielectric coating. Suggest you just keep the phase term for optical path distance, and leave out the ## i \omega t ## dependence. It's been a few years since I did this particular calculation, but it's a fair amount of algebra. Note that intensity ## I ## is proportional to ## n |E|^2 ##.

I did see that I could relate ##\tilde E_{0_I}##, ##\tilde E_{0_R}## and ##\tilde E_{0_T}## using:
$$\tilde E_{0_R}=|\frac{n_1-n_2}{n_1+n_2}|\tilde E_{0_I}$$
$$\tilde E_{0_T} = (\frac{2n_1}{n_1+n_2}) \tilde E_{0_I}$$
I was planning on using these to calculate the amplitudes once I got a set of expressions for the two waves leaving the coating in the -z direction. I kind of lost sight of that after getting stuck in the transmissions and reflections.

Unfortunately I am not as comfortable using infinite series as I should be, so if at all possible I would like to avoid them. Do you think you could please expand a little on what you mean by a collective total wave amplitude?
 
  • #5
I wish I could draw a picture, etc, but let me try to describe it in detail. I just did the complete calculation again last night and it worked. Consider going left to right: Call ## E_{incident}=E_1 ##, and the total wave at the first interface that is going to the left ## E_4 ##. Just to the right of the first interface, call the wave collection going to the right ## E_2 ## , and the leftward one ## E_3 ##. Now go to the second interface. Call the rightward going one just inside the interface ## E_5 ##, and the leftward one ## E_6 ##. Finally, for completeness, call the one exiting the system after the second interface ## E_7 ##. The different amplitudes are linearly related by the Fresnel coefficients and/or phase factors that can readily be written down. I will show you how to do this in the next post.
 
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  • #6
Notice ## E_4=\rho_o E_1+\tau_{21}E_3 ## where ## \rho_o=(1-n_c)/(1+n_c) ## and ##\tau_{21}=2n_c/(1+n_c) ##.
Meanwhile ## E_2 ## and ## E_5 ## are related by a phase factor ## e^{i \pi/2} ##. Similarly for other cases. Omit ## E_7 ## for now. There are 5 unknowns ## E_2 ## through ## E_6 ##. You need 5 equations. It is then readily solved. I can supply them for you, if you get stuck, but let's see if you can write them all out.
additional comment: You don't need absolute values on the ## \rho's ##. The sign is important because the ## \pi ## phase factor off of a more dense medium is included inside them in the form of a minus sign, i.e. ## e^{i \pi}=-1 ##. Thereby ## \rho=(n_1-n_2)/n_1+n_2) ##.
 
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  • #7
and a follow-on: The relation connecting ## E_3 ## and ## E_6 ## is simply a phase factor. The one connecting ## E_5 ## and ## E_6 ## is also fairly straightforward. The last relation you need is writing ## E_2 ## in terms of ##E_1 ## and ## E_3 ##. You may not need these extra hints, but you might find them useful.
 
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  • #8
Thank you very much, this is incredibly helpful. I will try to work my way through the analysis and solve the equations. Hopefully I'll be able to sort my way through it!
 
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  • #9
Just a follow-up on this one: It is worth commenting on the infinite series approach, where a geometric series is formed as a portion of the wave keeps reflecting off both interfaces and a part of each of these components winds up transmitting across the left interface and adding to the total reflected amplitude.

For just two interfaces, the infinite series method is also readily workable, but for more than two interfaces, the multiple reflections get very hard to manage mathematically, and the alternative approach that is used in the above posts is much more straightforward.
 
  • #10
Charles Link said:
Just a follow-up on this one: It is worth commenting on the infinite series approach, where a geometric series is formed as a portion of the wave keeps reflecting off both interfaces and a part of each of these components winds up transmitting across the left interface and adding to the total reflected amplitude.

For just two interfaces, the infinite series method is also readily workable, but for more than two interfaces, the multiple reflections get very hard to manage mathematically, and the alternative approach that is used in the above posts is much more straightforward.

I understand your point, as a matter of fact since you kindly explained the collective wave amplitude method I have since also tried the infinite series and got a similar answer, although for some reason I calculated a coating thickness of ##\frac {\lambda}{4n_1}## (with ##n_1## being the refractive index of the coating) as opposed to the ##\frac \lambda 4## given in the question which makes me think I made a mistake somewhere. Although I did get the refractive index be equal to ##\sqrt n## as given in the question.
 
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  • #11
Very good. The thickness of the coating is ## \frac{\lambda_{nc}}{4} =\frac{\lambda_o}{4n_c} ##. You computed it correctly. :)
 
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  • #12
Charles Link said:
Very good. The thickness of the coating is ## \frac{\lambda_n}{4} =\frac{\lambda_o}{4n} ##. You computed it correctly. :)
Oh I see, so it was referring to the wavelength in the material, not the original wavelength of the wave! That makes so much more sense, thank you for pointing that out!
 
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  • #13
Yes, it's the wavelength in the coating. It has to be the right thickness so that ## 2d ## in the coating causes a phase change of ## \pi ##.
 

Related to EM Wave Reflection and Transmission Between 3 Materials

1. What is EM wave reflection and transmission?

EM wave reflection and transmission refer to the behavior of electromagnetic waves as they encounter different materials. Reflection occurs when the wave bounces off the surface of a material, while transmission occurs when the wave passes through the material.

2. How do different materials affect EM wave reflection and transmission?

The properties of a material, such as its density and conductivity, determine how it will interact with electromagnetic waves. Materials with high conductivity, such as metals, tend to reflect EM waves, while materials with low conductivity, such as glass, tend to transmit them.

3. What is the role of the angle of incidence in EM wave reflection and transmission?

The angle at which an EM wave strikes a material's surface, known as the angle of incidence, affects the amount of reflection and transmission that occurs. The greater the angle of incidence, the more likely the wave is to be reflected rather than transmitted.

4. How does the wavelength of an EM wave affect its reflection and transmission?

The wavelength of an EM wave also plays a role in its reflection and transmission. Generally, shorter wavelengths are more likely to be reflected than longer wavelengths. This is why visible light, which has a shorter wavelength, is reflected by mirrors, while radio waves, which have longer wavelengths, can pass through walls.

5. Can EM waves be completely reflected or transmitted?

Yes, it is possible for EM waves to be completely reflected or transmitted, depending on the properties of the materials involved. For example, a metal surface can reflect all incident EM waves, while a vacuum will allow all EM waves to pass through without any reflection or transmission.

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