Effect on visibility of thin films due to interference

[brochesspro]

Homework Statement

Two thin films of the same material but different thickness are separated by air. Monochromatic light is incident on the first film. When viewed normally from point A, the second film appears dark. On normal incidence, from point B:
a. the first film will appear bright
b. the first film will appear dark
c. the second film will appear bright
d. the second film will appear dark

Relevant Equations

Path difference due to thin film of thickness t and of refractive index μ is 2μt.

So, what I do not understand 1st and foremost is, when they say "see the film", does the light have to be reflected from the film or can I consider refraction of light through thin film also a case using which image can be formed? To be honest, I have solved this question before, but the problem is that I do not remember how to solve it anymore. I do not even know where to start. So please help me solve this problem again. Also, the answer is D.

 

[brochesspro]

So I tried assuming the case where only reflection of light rays produces image.

Sorry for the clumsy ray diagram. But I satisfied the condition that film 2 should be dark for A. In 1 case, hard boundary occurs at film 1 and in the other case, film 2. But then how do I check for B? And is there any way that I can minimize the no. of reflections needed? Since I do not really remember doing this much last time.

 

[kuruman]

This is a problem in logic. The films have different thicknesses which means that each film can produce either constructive of destructive interference upon reflection in which case the transmission through each will be, respectively, destructive or constructive. We assume no in-between interference.

Does this help?

 

[brochesspro]

We assume no in-between interference.

What do you mean by this?

So, what I do not understand 1st and foremost is, when they say "see the film", does the light have to be reflected from the film or can I consider refraction of light through thin film also a case using which image can be formed?

Could you answer this part of my doubt as well? eg. If the light rays pass through both the films without reflection and reach A, does A see film 1 or 2 or both? Or does he even see an image? I think he can not see either film. Am I right?

 

[kuruman]

It's not a matter of images. You don't "see" light if light does not reach point A. Conversely, if light reaches point A, you "see" it. Here, we are told that light does not reach point A. It's like placing an opaque screen at the location of Film 2.

 

[hutchphd]

To the OP: I, too, am totally confused by the question. If possible please do not waste your precious time on it. Otherwise please discuss it with whomever is assigning it to you.

 

[brochesspro]

To the OP: I, too, am totally confused by the question. If possible please do not waste your precious time on it. Otherwise please discuss it with whomever is assigning it to you.

Here is the the solution given in the book.

I have understood absolutely nothing of this solution. Please help me understand it if it made sense to you. And sure, if no one here solves it, I will figure it out and explain it here.

It's not a matter of images. You don't "see" light if light does not reach point A. Conversely, if light reaches point A, you "see" it. Here, we are told that light does not reach point A. It's like placing an opaque screen at the location of Film 2.

No, we might not see the film even if the light rays reaches our eyes if they differ by a phase of π. This is due to thin film interference. One famous example is Newton's Rings.

 

[haruspex]

I have understood absolutely nothing of this solution. Please help me understand it if it made sense to you. And sure, if no one here solves it, I will figure it out and explain it here.

What makes it tough is the typos.
The diagram labels the right hand thickness $t_1$ instead of $t_2$.
The text following it has A and B swapped over.
$t_3$ should be $t_2$,

That said, I find the conclusion difficult to believe. It implies the right hand film is absorbing much of the light. I need to think about it more.

No, we might not see the film even if the light rays reaches our eyes if they differ by a phase of π. This is due to thin film interference. One famous example is Newton's Rings.

You are making a distinction that does not exist. Waves getting through but cancelling is equivalent to their not getting through.

 

[hutchphd]

Do you really have to solve this?? None of it makes any sense, and trying to solve a badly worded problem is an exercise in how to solve a badly presented problem, not physics. Why bother?

 

[brochesspro]

What makes it tough is the typos.
The diagram labels the right hand thickness $t_1$ instead of $t_2$.
The text following it has A and B swapped over.
$t_3$ should be $t_2$,

Exactly what I thought.

Do you really have to solve this?? None of it makes any sense, and trying to solve a badly worded problem is an exercise in how to solve a badly presented problem, not physics.

Could you tell me what exactly about this problem is badly presented or alternatively, what should the question have been? Since I did understand it, but am just not sure how to solve it. Though I do get your point.
I'll try finding out a solution. I will see you tomorrow.

 

[hutchphd]

I really don't like the term "film looks dark" If I turn out the light the film looks dark. I would restate the problem but I don't really understand what they are trying to present to you. I believe it is a bad attempt at some clever problem Clearly the subject is (antireflctive or totally reflective) thin films. The thing you need to not forget is the π phase change at the reflection off of the higher index medium. This effect true for all waves (not only light) and so it is good to understand this. Soap bubble problems are my favorite. If you play with soap bubbles you will in fact notice a point where the skin gets invisible because it gets so thin that it no longer reflects any order.

 

[kuruman]

No, we might not see the film even if the light rays reaches our eyes if they differ by a phase of π. This is due to thin film interference. One famous example is Newton's Rings.

We never see the film; we see the light that passes through it. Ignore Film 2 for the time being and look at Film 1. Imagine one observer to the right of the film at B and another observer to the left of th film at, say, C. If observer B sees light (constructive interference), observer C will see dark. That's because of the phase shift by π upon reflection between air and dielectric on the left side of Film 1. No such phase shift occurs for the light that is refracted straight through and the light that is reflected an even number of times within the film. We have to assume that we have constructive interference if we combine at B what emerges from Film 1. Without this assumption, we have the trivial result that no light passes through Film 1 to reach Film 2.

Now let's look at Film 2. We are told that it is dark, which means that no light makes it through, i.e. destructive interference at "system II" at A where the refracted and double reflected rays are combined. Thus, the phase difference between rays that reach II is an odd multiple of π. The question is what happens at B when the reflected rays from that film are combined at "system III"?

Well, I think the diagram shown in the solution is flawed because it shows the interference of reflected rays from the right interface of Film 2, all of which differ by even multiples of π. It shoud show instead the interference between the multiply reflected rays at the right interface and the reflected rays at the left interface which differ by odd multiples of π. That's because of the additional phase shift by π at the left interface of Film 2. This means constructive interference at "system III".

To summarize, I think that, relative to the phase of the monochromatic source,

1. the phase of the light at "system I" is an even multiple of π
2. the phase of the light at "system II" is an odd multiple of π
3. the phase of the light at "system III" is an even multiple of π.

So now how does one answer the question? In the region between the films we have transmitted rays from Film 1 traveling to the right and reflected rays from Film 2 traveling to the left. Their phases differ by even multiples of π which means constructive interference. If I plant myself in the region between films and look either way, I will see monochromatic light coming from each filter.

If this is a multiple choice question, the correct answer is "None of the above, both films are bright." If this is a multiple answer question the correct answer is "a and c". I know I may be stretching this a bit, but it's the only way I see to make sense of this question.

 

[haruspex]

I find the conclusion difficult to believe. It implies the right hand film is absorbing much of the light. I need to think about it more.

… indeed, it should be obvious that, as long as light is coming through the first film, the second cannot look dark both sides. All of the light intensity reaching it is either transmitted or reflected.
The solution author appears clueless. What book is this?

 

[kuruman]

The solution author appears clueless. What book is this?

The solution author appears clueless. What book is this?
I am not sure that this is out of a textbook. It looks like that this is a problem that someone put up on the web (some sort of "Olympiad" perhaps?) and then propagated by bots and/or humans. Looks like we got infected affected too. See

• https://www.doubtnut.com/qna/644106842
• https://www.toppr.com/ask/question/two-thin-films-of-the-same-material-but-different-thickness/
• https://www.sarthaks.com/1737035/fi...ted-monochormatic-incident-first-viewed-norma
• https://www.vedantu.com/question-an...lass-12-physics-cbse-6080e89fb167b0459abe010c

Of particular interest is the last link that provides a rationale for answer (d). Here is the logic

1. Because no light reaches point A, there must be destructive interference at either Film 1 or Film 2.
2. If this destructive interference occurs at Film 1, then Film 2 will receive no light and therefore appear dark from point B.
3. If this destructive interference occurs at Film 2, then Film 2 will still appear dark from point B.

Argument 2 is fine. Argument 3 is specious. Film 2 is not a black hole.

It also looks like the author of the problem is not the same as the author of the solution. The structure of the multiple choices indicates that the author of the problem had in mind that only one of the answers is correct. The author of the solution says that the answer of what Film 1 appears from B cannot be answered because "$\dots$ (the) relation between $t_1$ an $t_2$ has to be known."

If the author of the problem also authored the solution, this brings back the old problem of multiple choice questions, if the answer is "True" under some circumstances and "False" under others, is there a default answer? Yes, I think the default answer is "badly constructed problem."

 

[brochesspro]

I really don't like the term "film looks dark" If I turn out the light the film looks dark. I would restate the problem but I don't really understand what they are trying to present to you. I believe it is a bad attempt at some clever problem Clearly the subject is (antireflective or totally reflective) thin films. The thing you need to not forget is the π phase change at the reflection off of the higher index medium. This effect true for all waves (not only light) and so it is good to understand this. Soap bubble problems are my favorite. If you play with soap bubbles you will in fact notice a point where the skin gets invisible because it gets so thin that it no longer reflects any order.

I think I get what you mean, but I would personally rather believe that the problem is correct but I am not able to solve it, atleast as of now. I will keep your point about phase change of waves in mind. Also, could you give an example of a soap bubble problem? Thank you.

I request that we all use this diagram as convention from now. The previous one is just stupid. By the way, I think I might have a solution.

From the POV of A, the path difference is 2μt₂. Since there is no phase change of π anywhere, the conditions for maxima and minima do not change. So, since A does not see 2nd film, The path difference is an odd multiple of λ/2. Hence, 2μt₂ = (n-½)λ.

From the POV of B₁ again, the path difference is 2μt₂. But here, since a phase change of π occurs, the conditions for maxima and minima interchange. So, now, the condition for maxima is for the path difference to be an odd multiple of λ/2. So, from the above conclusion 2μt₂ = (n-½)λ, B₁ is able to see the 2nd film clearly.

As for B₂, since we no information about t₁, we can not conclude anything about the path difference and hence the 1st film can be either seen or unseen for B₂.

What do you guys think about this, I might not be getting the correct answer, but this is the closest I have gotten to solve this problem.

The solution author appears clueless. What book is this?

Well, it is from this book(probably an older edition of it.)
https://www.amazon.in/Optics-Modern...p_n_condition-type:8609962031&s=books&sr=1-10

Looks like we got infected affected too.

That time I accidentally started a pandemic.

It also looks like the author of the problem is not the same as the author of the solution. The structure of the multiple choices indicates that the author of the problem had in mind that only one of the answers is correct. The author of the solution says that the answer of what Film 1 appears from B cannot be answered because "… (the) relation between t1 an t2 has to be known."

If the author of the problem also authored the solution, this brings back the old problem of multiple choice questions, if the answer is "True" under some circumstances and "False" under others, is there a default answer? Yes, I think the default answer is "badly constructed problem."

I agree with you, that is usually the case with these books, since they need to provide a load of problems for students to practice, the authors instead usually pick up questions from various sources and just provide solutions for these problems themselves, sometimes not even that, there are very few exceptions to this(if any).

 

[kuruman]

I request that we all use this diagram as convention from now. The previous one is just stupid. By the way, I think I might have a solution.

Actually, the diagram is post #15 is not appropriate for analyzing the situation correctly. The picture below shows the five regions in space where there are EM waves. In regions I and V you have, respectively, the incident and transmitted waves traveling to the right. In the other three regions you have waves traveling to the left (red) and to the right (black). This becomes a problem of matching boundary conditions at the four interfaces sorting out the wave amplitude in each region. The solution is facilitated with the transfer matrix method. That said, I am sure the author of the problem did not intend it to be this complicated. I am mentioning this because you need to understand that, in region III between films, there is only one wave traveling to the right and only on traveling to the left.

 

[haruspex]

Actually, the diagram is post #15 is not appropriate for analyzing the situation correctly. The picture below shows the five regions in space where there are EM waves. In regions I and V you have, respectively, the incident and transmitted waves traveling to the right. In the other three regions you have waves traveling to the left (red) and to the right (black). This becomes a problem of matching boundary conditions at the four interfaces sorting out the wave amplitude in each region. The solution is facilitated with the transfer matrix method. That said, I am sure the author of the problem did not intend it to be this complicated. I am mentioning this because you need to understand that, in region III between films, there is only one wave traveling to the right and only on traveling to the left.

View attachment 337271

It is not given to be a laser, and the separation of the two films is not given as small, so isn't there a problem applying the transfer matrix method across region III?

 

[kuruman]

It is not given to be a laser, and the separation of the two films is not given as small, so isn't there a problem applying the transfer matrix method across region III?

Sure, but the separation is not given as large either. If the separation is large one can find the transmitted intensity through the first film and then use as the incident intensity on the second. Or one can match boundary conditions at the interfaces just as one would do with a double-barrier problem in quantum mechanics. My point in post #16 was not to advocate for a full blown treatment but to show to the OP that the situation is more complicated than shown in post #15.

I think OP's analysis is fine. The simple truth is that if Film 2 looks dark from A, it must look bright from B unless Film 1 looks dark from B in which case both films will look dark from B.

 

[hutchphd]

I think I get what you mean, but I would personally rather believe that the problem is correct but I am not able to solve it, atleast as of now.

What I mean is that your everyone's time would be far better spent if you get some "soap bubble" solution (dish detergent and glycerin typically ) and look at some soap bubbles in sunlight. Formulate your own questions and if you cannot answer them, ask for help here. This is a far more useful exercise than answering a very contrived question badly asked. I cannot help you with this question for reasons already explained. I will be happy to attempt answers to other questions, well and carefully stated.

 

[brochesspro]

Sure, but the separation is not given as large either. If the separation is large one can find the transmitted intensity through the first film and then use as the incident intensity on the second. Or one can match boundary conditions at the interfaces just as one would do with a double-barrier problem in quantum mechanics.

Well, we are dealing with films whose thickness is of the order of the wavelength of visible light, and I do not know what the transfer matrix method is, as it is not in our syllabus, so I think it must be safe to assume that the distance between the films is much larger than either of their thickness.

This is a far more useful exercise than answering a very contrived question badly asked. I cannot help you with this question for reasons already explained. I will be happy to attempt answers to other questions, well and carefully stated.

It is fine, I understand what you mean. And thank you for all your help till now.

I am mentioning this because you need to understand that, in region III between films, there is only one wave traveling to the right and only on traveling to the left.

I do not understand this part, and does it make my analysis wrong by any chance?

I think OP's analysis is fine. The simple truth is that if Film 2 looks dark from A, it must look bright from B unless Film 1 looks dark from B in which case both films will look dark from B.

Could you please elaborate?

 

[haruspex]

I do not know what the transfer matrix method is

https://en.wikipedia.org/wiki/Transfer-matrix_method_(optics)
I don’t expect you to understand that text (I wish I could) but it demonstrates that the book solution is simplistic. The topic is too advanced for the author.

 

[brochesspro]

https://en.wikipedia.org/wiki/Transfer-matrix_method_(optics)
I don’t expect you to understand that text (I wish I could) but it demonstrates that the book solution is simplistic. The topic is too advanced for the author.

I do not think this phenomenon matters in this problem, since it is not a part of our syllabus.

 

[haruspex]

I do not think this phenomenon matters in this problem, since it is not a part of our syllabus.

It isn't some obscure phenomenon, it is how light behaves. There is no benefit in teaching invalid methods that lead to wrong answers.

 

[brochesspro]

It isn't some obscure phenomenon, it is how light behaves. There is no benefit in teaching invalid methods that lead to wrong answers.

So if you are implying that this problem can not be solved without using this method, then this problem is out of our syllabus. And hence I shall give up on it, so could you please confirm this? And if so, could you please send the complete solution if it is different from the ones above, since there is no way I am getting this problem on my own if it uses a concept out of our syllabus.

 

[haruspex]

So if you are implying that this problem can not be solved without using this method, then this problem is out of our syllabus. And hence I shall give up on it, so could you please confirm this? And if so, could you please send the complete solution if it is different from the ones above, since there is no way I am getting this problem on my own if it uses a concept out of our syllabus.

I am saying it needs a valid method. Whether there is a simpler approach than the quite general solution that I linked, I do not know. What we can say is the method in the book cannot be valid since it leads to the impossible result that the second filter absorbs all the light landing on it.

 

[brochesspro]

I am saying it needs a valid method. Whether there is a simpler approach than the quite general solution that I linked, I do not know. What we can say is the method in the book cannot be valid since it leads to the impossible result that the second filter absorbs all the light landing on it.

Then what about my method? Is there a problem with it? I am asking this because I think I am very close to the correct answer(maybe even the correct solution.)

 

[haruspex]

Then what about my method? Is there a problem with it? I am asking this because I think I am very close to the correct answer(maybe even the correct solution.)

The analysis in post #15 has two problems, in my view.
First, it arbitrarily stops after two rays headed towards the viewer. In reality, there is an infinite sequence of internal reflections, leading to an infinite sequence of paths to the viewer.
Secondly, it fails to consider the relative strengths of the paths. A reflection in going from $n_1$ to $n_2$ may be a different fraction of the intensity than the reverse case; the second rightwards ray leaving the film will be weaker than the first.

The matrix transfer method deals with these complexities by treating it all as a single waveform.

At least your analysis does not lead to an obvious fallacy, so may well be a reasonable approximation to what happens in some cases. That does not mean it will be approximately right in all cases.
A simpler and safer argument is to note that, assuming it absorbs nothing, all the light entering the second film has to come out some time. Since we are told it looks dark from the right it must look bright from the left.

 

[brochesspro]

Secondly, it fails to consider the relative strengths of the paths. A reflection in going from n1 to n2 may be a different fraction of the intensity than the reverse case; the second rightwards ray leaving the film will be weaker than the first.

Why does that matter, the path difference being a suitable multiple of λ or λ/2 should be enough, right?

The matrix transfer method deals with these complexities by treating it all as a single waveform.

Is there any way to do this problem without using this method? Using just the basics of interference and such?

First, it arbitrarily stops after two rays headed towards the viewer. In reality, there is an infinite sequence of internal reflections, leading to an infinite sequence of paths to the viewer.

Will it work if we assume that the intensity of wave decreases the more distance it travels, so by the time the light from other possible interactions strikes the retina, it loses intensity. So in my opinion 1 interaction should be enough if we follow this assumption.

 

[haruspex]

Why does that matter, the path difference being a suitable multiple of λ or λ/2 should be enough, right?

Is there any way to do this problem without using this method? Using just the basics of interference and such?

Will it work if we assume that the intensity of wave decreases the more distance it travels, so by the time the light from other possible interactions strikes the retina, it loses intensity. So in my opinion 1 interaction should be enough if we follow this assumption.

As far as I know, it is feasible that the first leftwards reflection, being an external reflection, could be weaker than the third one, which has undergone three internal reflections.

 

[brochesspro]

As far as I know, it is feasible that the first leftwards reflection, being an external reflection, could be weaker than the third one, which has undergone three internal reflections.

Sorry, I didn't get you.

 

[haruspex]

Sorry, I didn't get you.

Suppose on hitting the film from the left only 5% is reflected. The other 95% reaches the far side of the film, where 75% is reflected. (Is this possible? I don't know.) So then 75% of that is reflected from the LHS, the other 25% emerging. The reflected portion returns to the RHS, 75% of that being reflected and 25% emerging as the third ray directed leftwards.
We have, in the first three rays going leftwards from the film:
5% in the first
95%*75%*25%=15.7% in the second
95%*75%*75%*75%25%=9% in the third

 

[brochesspro]

Suppose on hitting the film from the left only 5% is reflected. The other 95% reaches the far side of the film, where 75% is reflected. (Is this possible? I don't know.) So then 75% of that is reflected from the LHS, the other 25% emerging. The reflected portion returns to the RHS, 75% of that being reflected and 25% emerging as the third ray directed leftwards.
We have, in the first three rays going leftwards from the film:
5% in the first
95%*75%*25%=15.7% in the second
95%*75%*75%*75%25%=9% in the third

So this supports my assumption, right?

 

[haruspex]

So this supports my assumption, right?

That was an explanation of how the third ray to the left could be more significant than the first one, so an analysis that only considers the first two is incomplete.

 

[brochesspro]

That was an explanation of how the third ray to the left could be more significant than the first one, so an analysis that only considers the first two is incomplete.

So in short, no, right?
If so, I think I will stop thinking about this problem.

 

[haruspex]

So in short, no, right?
If so, I think I will stop thinking about this problem.

Good.