Thin-film interference?

[SaintJude]

Thin films are on the AAMC syllabus but not in Kaplan (?! kaplan talks about everything, it even discusses spaghetti for goodness sakes..and it's gonna leave out thin films??) and of course not even in EK...I mean let's not even go there.

What are we supposed to know about thin-film interference. Where is a MILSKI when you need one??

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[MedPR]

Thin films are on the AAMC syllabus but not in Kaplan (?! kaplan talks about everything, it even discusses spaghetti for goodness sakes..and it's gonna leave out thin films??) and of course not even in EK...I mean let's not even go there.

What are we supposed to know about thin-film interference. Where is a MILSKI when you need one??

It's like double refraction. You have three different media with different indexes of refraction. It follows the same rules as a simple problem like light bending when it goes from air into water.

The thing to be careful of is that the light undergoes a 180deg phase shift whenever it reflects off of a media that has a higher index of refraction. I think I have this right, but someone will correct me if I'm wrong so let's see.

Say you have air (n=1), water (n=1.3), and glass (n=1.5) all stacked on one another. When the light hits the air-water barrier, some will reflect and some will refract into the water. The light that reflects will undergo a 180deg phase change. When the refracted light hits the water-glass barrier, some will again reflect and some will refract. The light that reflects will undergo a 180deg phase change and re-enter the air at the air-water barrier. This light ray will constructively interfere with the ray that originally reflected at the air-water barrier since both rays underwent a 180deg phase change. If the glass had a lower index of refraction than water, however, the light reflected at the water-glass barrier would not have undergone a 180deg phase change and once it passed back up into the air, it would destructively interfere with the ray originally reflected at the air-water interface.

 

[milski]

Well, MedPR already covered the basics. As for how much you'll need to know for the test - no idea. You'll probably get by just knowing that the interference depends on the wavelength and the thickness, so for different thickness of the layer you'll see different colors.

Also, when we talk about three medias here, you really need two boundaries between different medias. In other words A-B-A with a thin layer of B will also exhibit this effect. A well know example of that are soap bubbles.

 

[SaintJude]

Thank you both of you!!

I'm attaching a picture that illustrates what MedPr so clearly explained.

So this picture shows a thin film like that created with a soap bubble (n =1.4). The thin film represents the thickness of the soap bubble boundary that creates a thin separation b/w air. The top interface and the bottom (the inside of the bubble filled with air) is then of an index of n=1.

When a wavelength of light initially hits the bubble, some of it will reflect back. As MedPr noted, b/c ray 1 is moving into a higher index, the reflected ray (Ray 2) exhibits a phase change while the refracted light ray (ray 3) will continue on to hit the bottom interface. Again, some of the light will reflect. This second reflected ray of light (Ray no.4) will not exhibit a phase change, b/c the bottom interface has a lower index of refraction than the thin film. So ray 3 is in phase with ray 4, which will then refract back into the top interface (ray 5). So ray 1 and ray 5 are in phase, since refracted light never inverts.

EDIT: Actually, one can say there will thus be a net phase inversion of ½lambda. Since ray 1 is in out of phase with ray 2, but ray 5 is in phase with ray 1. Whether constructive or destructive will then depend on thickness, I think...

 

[milski]

I may be misreading what you're saying but some of your conclusions are incorrect. You cannot conclude if rays 2 and 5 will be in phase or not just based on the type of reflection from the second interface. The phase difference will also depend on the additional distance traveled by the ray inside the thin film. For the same media you will get both constructive and destructive interference (and whatever is in between) for different wavelengths.

Let's consider the case when the reflection does not change the phase. In that case the shortest additional distance that ray 5 needs to travel to have constructive interference with ray 2 is λ (the wavelength of the light). That means that if the thickness of the film is λ/2, you'll get constructive interference for λ. You can also add nλ to the path of ray 5 without changing anything, so you'll get constructive interference for all thicknesses which are (n+1)λ/2. In a similar way, you'll get destructive interference when the thickness is λ/4+nλ/2.

The case where there is a phase change during the reflection can be derived in the same way, except you get an additional π of phase change, so you'll end up with λ/4+nλ/2 for constructive and (n+1)λ/2 for destructive interference.

I don't know if you should know the formulas or not but you do need to know that there is a combination of thickness, wavelength and type of media which gives you constructive/destructive interference. Changing any of the three will change the interference results.

 

[milski]

EDIT: Actually, one can say there will thus be a net phase inversion of ½lambda. Since ray 1 is in out of phase with ray 2, but ray 5 is in phase with ray 1. Whether constructive or destructive will then depend on thickness, I think...

Yes. 👍

 

[SaintJude]

Thanks, milski for verifying. Whoever has you as a classmate at med school (or student TA!)= lucky.