Thin Film Interference part 2

look at 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 I don't know anything L hear that light can travel through and then it's also going to have a certain wavelength and the oil and the wavelength and 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 shoot it could even have a different wavelength in 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 get back to my eye so I'm not going to worry about this part how do we find which first of all which one do we use in this condition do we use the wavelength in the first medium a in the air or to 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 the light that reflected 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 he if it emerges also right at that point say this wave over here this wave cycle so 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 that'll be destructive we want to know how much this cycle got progressed by this wave traveling through here and so we need to know the wavelength in here that's the wavelength that actually matters 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 oyel order the thin-film how do we find this wavelength well it's not too hard the most straightforward conceptual way to think about it is that a 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 reflects no matter what it does 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 divide both sides by the wavelength and I get that this is equal to the speed of the wave over the wavelength so I can replace that over here the a so frequency and material a is just speed of light an a because of this formula over wavelength of the light in region a you think of a is air it doesn't have to be air but it could be was in this case frequency of the light in B which as our oil would be the speed of light and region B divided by the wavelength in region B okay so now we can just solve for wavelength and region B and we get that wavelength in our thin film will equal I multiply both sides by wavelength and 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 light in the air that you can look up because that everyone knows you can look it up it's online and speed the light in the oil I just told you what that was if you're given these speeds take the ratio speed of light oil divided by the speed of light in the air multiplied by the wavelength in that err 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 sometimes they give you the index of refraction and you're like ah 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 be 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 eight divided by the speed and 8 + NB would be three times ten to the eighth over the speed and B so how would this change well shoot I could just solve this for VA if I wanted to I get V eight I'm running out of room here excuse me equals C over n a and VB what would VB equal VB would equal the speed of light in a vacuum divided by index of refraction and be all right now to imagine just plugging into here I'm going to plug in VB which is this for VB now 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 V equals well look C over NB is what I get C over n B divided by C over n a from this condition times lambda a I can simplify that the C's cancel and then one over n B divided by one over na just gives me n a 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 do the oil speed divided by the outside speed multiplied by the wavelength in the air but if you're given the indexes of refraction you 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 wavelengths got to go down so I just look over here and I just make my ratio if they give me ends shoot I just make sure my ratio of ends gives me a smaller number a number less than one that I multiply by to get my lambda in the thin film and if I get that they gave me speeds shoot I just take my speeds I took 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 to 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 Oh can't even look at that anymore all right let's do this so we know what do we know we know from 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 so let's just write this we know velocity in this region is a the thin-film region will call be then we know that V in B divided by V in a times wavelength in region a which is oftentimes the air this would give you constructive or if you are 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 two times T should be well half integers so we could do M plus 1/2 times this same thing wavelength in the thin film which again is VB over VA times lambda a or same thing half integers times wavelength but if you had index of a fraction you'd want to use na over N be 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 half 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 way it 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 if this region a is air well the index of refraction and air is just the 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 1 over NB x lend a or in other words M times lambda a over NB would give you constructive and M plus 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 I mean 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 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 pie 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 reflected 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 confuses people hopefully you do well with that these are the ways you deal with it