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## Physics library

### Unit 14: Lesson 2

Interference of electromagnetic waves- Constructive and Destructive interference
- Young's double slit introduction
- Young's double slit equation
- Young's double slit problem solving
- Diffraction grating
- Single slit interference
- More on single slit interference
- Thin Film Interference part 1
- Thin Film Interference part 2

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# Thin Film Interference part 2

Let's work out a few details on how thin film interference works. Created by David SantoPietro.

## Want to join the conversation?

- At11:00it starts to explain m but I am still at a loss as to what m signifies. How would we know what to plug into "m" when reading a word problem?(29 votes)
- You don't really need m here, it is just that if the thickness of the thin film you get is ANY integer times the wavelength, then the interference will be constructive, and if the thickness of the thin film is any integer + a half times the wavelength, then the interference will be destructive. Of course again this will be the other way around if you have the case of one of the two waves being phi shifted. If you are reading a word problem, you would be given the thickness of the film. You are not able to find it out by multiplying the wavelength by some "m"-value, unless it is stated in the word problem for example that the thickness of the film is two times the wavelength, in which case you wouldn't even need to know the thickness because you know how thick it is in relation to the wavelength, and that is all you need to know to find out whether the interference will be destructive of constructive.(11 votes)

- What if the rays are not perpendicular to a surface? Won't there be more to the path difference ?(16 votes)
- Excellent question, and yes you are right, the path difference does get larger. That's why if you look at thin film interference head on and then look at it from the side, you will see the interference pattern shift in a way that is equivalent to the film getting thicker.(21 votes)

- In my physics book for thin film interference, constructive interference says the equation is 2t = (m + .5) lamda(n) why do you have them opposite??(4 votes)
- Hi, in the video, David said that if light goes form a medium that have large v into a different medium have a small v then the light wave will have a pi shift, and so the equation for the const and dest will be switched. Maybe this is the case in your textbook.(3 votes)

- Its not clear to me what "m" stands for here. It usually means the order of the maxima, but how does that apply here?(4 votes)
- Consider the equation 2t = m/\ ( /\ = wavelength) ,which is applicable for constructive interference if there no or n/\ times change in the path difference. Now by rearranging the equation, we get "t = m/\ / 2" . So, as it is clearly seen now , "m" is merely a parameter which determines whether constructive or destructive interference will take place for the thickness "t".(2 votes)

- I think I get it all apart from the fact that you don't give us when there is a pi shift. Is it when the third medium is optically denser than the second one?(2 votes)
- A pi shift is when you have at least 2 substances and the 2nd medium is slower, or as you put it, denser.(3 votes)

- At about9:30, why are you multiplying by λA instead of λB? I thought we were supposed to multiply by the wavelength within the film?(2 votes)
- If you have lambdaB then by all means, use that. The formula then is 2t=m*lambdaB. However, usually you are only given the wavelength in air. The (vB/vA)lambdaA is how you convert lambdaA into lambdaB.(3 votes)

- Why isn't a definite pattern observed in oil thin film interference (as seen is double slit interference)?(4 votes)
- I bet you can figure that out for yourself if you think about it a little bit. Compare what's happening to the light rays in double slit to what's happening in thin film.(0 votes)

- can someone explain me how the pi shift is determined? like if the ray travels from slow-fast,no pi shift or otherwise(1 vote)
- If the ray is traveling towards a medium in which it would usually travel slower in and reflects, there will be a pi shift. See "Thin Film Interference Part One".(3 votes)

- how do you determine which value of m to use??(2 votes)
- Does this concept have any part of explanation in the formation of a rainbow?(2 votes)
- No. But here is a link that does: http://www.physicsclassroom.com/class/refrn/Lesson-4/Rainbow-Formation(1 vote)

## Video transcript

This wavelength, this
just says wavelength, but there's going to
be a certain wavelength in the air or this material one. I'll call it lambda A. It could be air. It could be anything out here
that light can travel through. And then it's also going to have a certain wavelength in the oil. And the wavelength in this
oil is going to be different. I'll call this wavelength B. It's going to be different
from the wavelength in this first material if
the speed is different. It could even have a different
wavelength than the water, but we don't worry about the
wave traveling down here. So let's not confuse ourselves. Some of the light will pass through here, but it's not necessarily
going to get back to my I. So I'm not going to worry about this part. Which one do we use in this condition? Do we use the wavelength in the
first medium, A, in the air? Or do we use the wavelength in the oil? Well, we use the wavelength in the oil. Most definitely, we use
always this wavelength here. Both of these are always the
wavelength in the thin film. So, the wavelength the
light had in the thin film because that was the
portion of the path where the light traveled an extra distance. So that's the part that will matter. We want to know how much further
up, if this first light ray when it reflected was right here. We want to know where's this other one going to be at when it emerges. Because if it emerges
also right at that point, say this wave cycle, say it also emerges exactly at that point here. Well, they'll be constructive. But if it emerges over here,
that's only 180 degrees, that's 180 degrees out of phase. Then it would be destructive. We want to know how much
this cycle got progressed by this wave traveling through here. So we need to know the wavelength in here. That's the wavelength
that actually matters. The wavelength in the thin film. Often times you're not given the wavelength in the thin film. You're given the wavelength in the air, or whatever this material is. So if you know this
wavelength, how do you get the wavelength in the
oil, or the thin film? How do we find this wavelength? It's not too hard. The most straight forward
conceptual way to think about it is that frequency doesn't change. The frequency in material
A is going to equal the frequency of the light
when it enters material B. Frequency is determined by the source. If it's the sun up here
that emitted that light ray, or the laser, that's
what's determining the frequency of this particular light ray. And that stays the same,
whether it reflects whether it refracts,
no matter what it does. The frequency stays the same. That's a useful thing to know. And it's useful in this case because... Well, how do we relate
this to wavelengths? We know the speed of a wave equals wavelength times the frequency. And so if I wanted to
solve this for frequency, I'd divide both sides
by the wavelength and I get that this is equal to the speed of the wave over the wavelength. I can replace that over here. VA, so frequency and
material A, is just speed of the light in A,
because of this formula. Over, wavelength of the light in region A. You can think of A as air. It doesn't have to be
air, but it could be. It was in this case. Frequency of the light
in B, which is our oil, would be the speed of light in region B, divided by the wavelength in region B. Okay, so now we can just
solve for wavelength in region B, and we get that wavelength in our thin film will equal... I multiply it both sides
by wavelength in B, and then I multiply these
out, and what you'll end up getting is velocity of
the light, speed of the light in region B, divided by speed
of the light in region A. That factor, times the
wavelength of the light in A. So this is one way to determine. If you're given the
speeds like I did here... Remember I gave you the speed
of the light in the air. That you can look up because everyone knows you can look it up. It's online. And speed of the light in the oil, I just told you what that was. If you're given these
speeds, take the ratio, speed of the light in oil divided by the speed of light in the air. Multiply by the wavelength
in the air, this first medium and you'll get the wavelength in the second medium, which is the oil. This would be what you'd plug
into these formulas up here. What if you're not... Sometimes you're not even given the speed. What if you're given
the index of refraction? And you're like, "Ugh,
index of refraction, shoot". Well if they gave you the
N in region A, and instead gave you the N in region B, instead of giving you the speeds, remember index of refraction is defined to B. Speed of light in a vacuum divided by the speed of light in that material. So, NA, index of refraction
of region A is just three times ten to the eighth,
divided by the speed in A. And NB would be three times ten to the eight over the speed in B. How would this change? I could just solve this
for VA if I wanted to. I'd get VA... I'm running out of room here, excuse me. Equals C over NA and VB would
equal the speed of light in a vacuum divided by
index of refraction in B. Now I'm actually just plug it into here. I'm going to plug in VB,
which is this, for VB. I'm going to plug in VA,
which is this, for VA. I'd get a new condition. What would that new condition be? That new condition would
say that wavelength in B equals C over NB, is what I get. C over NB divided by C over NA from this condition, times lambda A. I can simplify that. The Cs cancel and then one
over NB divided by one over NA, just gives me NA over NB
times lambda in the air. So here's another one. Here's another condition. Another way to find it. This would also equal the wavelength in the oil, or the thin film. So if you're given the speeds, you can take the ratio of the speeds. You'd do the oil speed
divided by the outside speed. Multiply by the wavelength in the air. But you're given the
indexes of refraction, you'd take the outside index of refraction divided by the inside index
of refraction times lambda A. You're thinking, "Oh my god, how am "I going to remember all this?" Here's how I remember it. I know if I go from air to oil,
light's going to slow down. And if light slows down,
frequency stays the same. So if light slows down,
wavelength's got to go down. So I just look over here
and I just make my ratio. If they give me Ns, I just make sure my ratio of Ns gives me a smaller number. A number less than one that I multiply by to get my lambda in the thin film. And if they gave me speeds,
I just take my speeds. I take my ratio of my speeds so that I get a smaller lambda in the thin film. Just got to be careful. Make sure you're really slowing down. They might give you a
problem where for some reason this wasn't air, some other material. It was going faster through the oil than it would be through here. Then you'd want to make this ratio more than one when you multiply. But, if in doubt you
can always fall back on, the two frequencies are
equal, and use that. And that's the three things you got to worry about for thin film. These are the conditions. Make sure you pay attention
to whether there's a pi shift and make sure you always use
wavelength in the thin film. So if we wanted to... This looks messy. I'm sorry about this. This looks horrible. We could turn this all into
one super duper equation. Let's do that. Let's get... I can't even look at that anymore. Alright, let's do this. So we know for constructive
we should have 2T equals integer, so M is
zero, one, two, three, times the wavelength in
the oil or the thin film. But, I don't want to
solve for that every time. We already did it. Let's just write this. We know velocity in this region is A. The thin film region, we'll call B. Then we know that V in B divided
by V in A times wavelength in region A, which is often times the air, this would give you constructive. Or, if you're unlucky and
you got index of refraction then you could use
index of refraction in A divided by index of refraction
in B, times the wavelength in the A region, and that would be what you would need for constructive. We could do the same
thing for destructive. 2 times T should be, 1/2
integers, so we could do M plus a 1/2 times this
same thing, wavelength in the thin film, which again
is VB over VA times lambda A. Or, same thing, 1/2
integers times wavelength. But if you had index of refraction you'd want to use NA over NB, lambda A. This would give you destructive. Only other thing to worry about is if there's a relative pi shift. You would flip flop these conditions. These 1/2 integers would
give you constructive and the integers would
give you destructive. If one wave, and only one
wave, gets a pi shift. If one wave gets a pi shift
then you swap these conditions. This still looks a
little bit intimidating. I'm sorry. That looks intimidating. So sometimes it gets even better. If this region A is air... The index of refraction
in air is just one. So that cleans things up a little bit. This just becomes one and
this up here just becomes one. So you end up with M times
one over NB times lambda A. Or in other words, M times lambda A over NB would give you constructive. And M plus a 1/2 times lambda A over NB would give you
destructive for every M equals zero, one, two, three, and so on. Zero, can you really have zero? Yeah you could, kind of. If M equals zero, that would
say the thickness is zero. Doesn't that mean you have no thin film? Well, sort of, but if this
thickness is very small compared to the wavelength
of the light, it's as if the thickness is zero and
that will give you either constructive or destructive
depending on the pi shift. And sometimes you can do that. Anti-reflective coating
is often just a coating that's so thin, completely thin, compared to the wavelength of
light, which just makes all the light reflect destructively. Because in that case you do get a pi shift and these integers give
you destructive points. So that's thin film interference. Sometimes it confuses people. Hopefully, you do well with it. These are the ways you deal with it.