IB Physics: Thin Film Interference

[@yan_man23]

Did you mean 0 or 2 inversions at 18:21? Thanks for the videos by the way.

[@donerphysics]

Thanks...should be 0 or 2 inversions.

[@aravsriagarwal4056]

@@donerphysics Thanks for the clarification!

[@greengoku10]

Hey, I just wanted to thank you for these videos. You're a brilliant teacher and make my love for physics even greater. I sincerely hope your viewership increases exponentially, because more people need these videos. So thanks again

[@MB-zo2xw]

you're such a god, you explain things that IB expects you to know but are never mentioned in their books

[@ananyabeura7224]

Thank you so much for this. i don't know why the IB expects you to know this without even explaining or mentioning it in the text books

[@donerphysics]

It is thin (some pun intended.)

[@raghavcherukuru4936]

Legend, this helped me a lot.

[@noenicolas2815]

Exactly what I needed. Perfect explanation

[@donerphysics]

Glad it helped!

[@VALR1able]

This video was very helpful. Thank you very much.

[@donerphysics]

Glad it was helpful!

[@neet5080]

Great explanation. I was so amazed.

[@donerphysics]

Nice to hear!

[@abhishekpillai5207]

Hi! At the question in 19:59, how come the second ray isn't inverted when coming back out from the methyl iodide film to the glass?

[@donerphysics]

There is no phase change because the index of refraction of the second material is larger than that of the final material.

[@ediewhittington3255]

Thank you! Really well explained!

[@donerphysics]

Glad it was helpful!

[@longtranhai2761]

The plus a 0.5 for 1 inversion is not included in the formula. Do we just add it ourselves if there is one inversion and in that case will it not be +lambda/2? Thank you for the videos.

[@niya.s.]

You probably don't care anymore but no you don't need to add the 1/2 for 1 inversion cus it's already included in the formula. In the case of the destructive interference the 1/2 cancels out and then multiplies by lambda on both sides, which is why the IB formula says 2dn=m*lambda. And for constructive interference, the IB formula subtracts by 1/2 on both sides then multiplies by lambda on both sides which is why 2dn = (m + 1/2)*lambda. And, Mr. Doner said to swap the equations for 0 or 2 inversions which makes sense since it'd be the equations if 1/2 wasn't add to 2dn/lambda.

[@ashishsehrawat9091]

You are awesome sir

[@donerphysics]

Thank you. Please spread the word about the channel, become a subscriber or a member.

[@spifuntastic621]

20:34, didn't you say we have to convert the wavelength value according to the medium? So shouldn't we convert the wavelength from 500nm to that in methyl iodide?

[@donerphysics]

That is what the 1.76 does.

[@sachinprabhuram5765]

I did a question on N19 that says that when there are two inversions like in 13:24 there is a path difference of 2d. How? Question is Q10 b) (ii) in N19 for reference

[@yewweeyak3226]

Shouldn't you account for the refractive index of glass as well?

[@donerphysics]

Neither of the interfering rays travel through the glass so no.

[@vanessawertheim]

Thank you so much for this helpful video!

[@donerphysics]

Glad it was helpful!

[@ezu8501]

YOURE A GOD

[@zhenghongchen8365]

Mr. Doner, in question 1( 21:00), which answer is the correct answer or all the answers are the correct answer? Please further clarify a little bit 😂 thanks

[@donerphysics]

At the thicknesses indicated the reflected rays will be in phase.

[@jasonfong7394]

Hi, sorry to bother you. I was wondering if you could break down the optically dense --> rare --> dense scenario for thin film interference. Thanks!

[@donerphysics]

most common is air-film-air where we get a single phase inversion. the other situation you will see is air- more dense-even more dense, where there are two phase inversions which is equivalent to none at all.

[@ezu8501]

Hey, can you explain what the m represents? do you have any videos explaining?

[@donerphysics]

It is explained thoroughly in previous videos.. We get constructive interference when the the path difference is an integer number of wavelengths. m represents that integer.

[@andreshiguematu9207]

Souldn't you divide the index of refraction of the film by the index of refraction of the surroundings in order to appropriately convert lamba from the film to lamba in the surroundings?

[@donerphysics]

Yes, we must generally use the relative index of refraction. The example was the common case in which the first medium is air for which n=1.

[@sidninja9518]

23:38 would the answer be 189nm because rearranging the equation gives us t=(1/2lambda)/2n? I’m confused how the denominator becomes 4n

[@donerphysics]

There is a 2 on the LHS as well as the RHS. The one the LHS represents a half wavelength of path difference that is necessary for destructive interference. The one on the RHS represents twice the thickness since the reflected beams travel through the thickness twice.

[@letseat1098]

Sir, what are the factors affecting the thin film and how to protect the thin film?

[@donerphysics]

What thin film are you referring to?

[@justalittlebitoflove6520]

In the first example, I considered the air as the first medium, and thus concluded two inversions. Why is air not considered in this case?

[@donerphysics]

Good question. Most problems involve air. There are two ways to know. One would be by having experiences with this type of application (which is unlikely) , and the other would be to recognize that no thickness is given for the glass and therefore we can not treat it as a thin film, just as we do not treat the air as a thin film.

[@aviratthakor8974]

I am totally lost. Why at 10:35 do we need to multiply by the index of refraction of the film?

[@aviratthakor8974]

And why would there be constructive or destructive interference if there are 0 or 2 inversions (18:09)? Wouldn't the reflected rays be in the same phase?

[@donerphysics]

The wavelength changes whenever the medium changes. The higher the index of refraction the shorter the wavelength. The path difference is created when the light travels through the film, so we must use the wavelength within the film.

[@donerphysics]

@@aviratthakor8974 Good. O or 2 inversions has the same effect. To get the rays out of phase we need the light that travels through the film to travel an extra half wavelength or one and one half wavelengths etc.

[@taliyaloz8878]

Great content! Thank you!!

[@engila9295]

Hello! Thank you for the video but I have 2 questions: 1. What does the m stand for in the equation? 2. Why are the interference equations still applied for 0 or 2 inversions? I thought interference patterns only occured for 1 inversion.

[@donerphysics]

1. m can be any number 0,1,2,3 etc., so m is a counter. Waves produce destructive interference when they are out of phase by half of a wavelength, or 1 and a half wavelengths etc. 2. No, we get stable interference patterns when two waves of the same frequency pass through each other.

[@engila9295]

Chris Doner but why was m=0 used when thin film’s thickness was assumed to be minimal? How would the phase difference and thickness denote the number for m?

[@donerphysics]

At what time in the video are you referring to?

[@engila9295]

Chris Doner around 20:10

[@donerphysics]

@@engila9295 20:10 is the summary. Is there a specific case that you are referring to? I have already answered your question in general.

[@tauhid9983]

@22.55 Hey Mr Doner, could you please explain why there are 2 inversions

[@donerphysics]

Any time the index of refraction increases we get an inversion. The index increases twice.

[@tauhid9983]

@@donerphysics So if for example the index of refraction of the film is equal to the index of refraction of the medium through which light come out of the film, would we get one inversion?

[@donerphysics]

@@tauhid9983 If they truly have the same index, they are the same material and there is no reflection.

[@tauhid9983]

@@donerphysics therefore no inversion... hence if there's reflection between mediums, like you've marked the 2 point of reflection on the video, there will be 2 inversions

[@tauhid9983]

Hell Mr Donner, could you explan whether or not, with respect to thin films, will be an inversion for an air wedge

[@donerphysics]

The (somewhat) relevant reflections are glass to air (no inversion) and air to glass (an inverstion), however, for air wedges the inversion isn't important since you are usually just interested in the idea that everytime a new fringe is produced, the path difference (2t) has increased by one wavelength.

[@tauhid9983]

@@donerphysics So in IB question... we can take any equation 2t=nλ/2 or 2t=nλ..... to find for example the thickness of the film