Physics 60 Interference of Light (8 of 8) Glass Plates

[@kaleidoscope8651]

Mr. van Biezen, I have an immense amount of respect for you. I had my "light bulb" moment exactly at 4:52 of this video :) This is a problem that I've studied numerous times and I almost gave up trying to solve it in the book. Thank you for the videos, the fact that you've put them online with free service to students is very valuable to all of us.

[@MichelvanBiezen]

Glad that these videos are helping students around the world. Keep up the hard work.

[@andrewjustin256]

At what grade did you study this topic?

[@joannaelhaj3571]

Your service to humanity is very well respected and appreciated. Thank you sir.

[@MichelvanBiezen]

Thank you for your kinds words.

[@quornnugget7799]

What do you mean when you say no extra distance has been traveled @2:18, there is a path difference between the two rays so surely there is more distance travelled by the 2nd ray?

[@MichelvanBiezen]

Not at the point where the two slabs touch.

[@quornnugget7799]

@@MichelvanBiezen Ahh I see thank you! I was taught this today in EM and I was so confused!

[@MichelvanBiezen]

Glad we were able to clear it up for you. 🙂

[@andrewjustin256]

Mr. Michel, why didn't we do anything with the fact that the two beams are lamda/2 phase different? I mean why didn't we add lamda over two with 11 lamda? Please inform me!

[@MichelvanBiezen]

Don't forget that when light reflects off a surface that has a higher index of refraction, the light goes through a 180 degree phase shift.

[@nadirgas]

Hello, I can't seem to get why xdist = 1*lambda at the 2nd fringe, and also in general why xdistn = (N-1)*lambda at the Nth fringe

[@MichelvanBiezen]

extra distance = 2 x path across the gap. Then you need to determine what condition you want: constructive or destructive interference. To get a frinch you need destructive interference. Normally for destructive interference you want a phase difference of 1/2 wavelength. But in this case there is a 180 degree phase shift on the bottom boundary, so you need a full wavelength difference for destructive interference.

[@nadirgas]

@@MichelvanBiezen thank you so much!!!

[@melvinjijumathewrajagiri6918]

sir the glass plates has 2 boundaries and light comes from air so why isnt the refraction from the first boundary not shown?

[@MichelvanBiezen]

That is a good question (although I think you meant to write: "why isn't the reflection from the first boundary shown?") Because the refraction from the first boundary is shown. That is done because of simplicity. The ray that is reflected from the second glass plate in part also transmits and refracts into the bottom glass and then in part reflect from the bottom and in part transits through that boundary, etc.

[@melvinjijumathewrajagiri6918]

@@MichelvanBiezen thank you sir

[@sutopamodak6247]

My question is where the interference pattern is making in the air( above the inclined slab ) or in our eyes ?? Because we know that where the two ray meet with each other , the pattern is made

[@MichelvanBiezen]

Notive the position of the "eye" above the top slab. That is where the interference pattern will be observed.

[@ALFPAJARITO]

Dear Michel van Biezen: How big can be "t" in order you still be able to see the interference fringes? This question is related to the other question I made in other video from yoy about how "fat" can by a film in order to be able to see an interference pattern. Thank you in advance.

[@MichelvanBiezen]

You are correct in assuming that theoretically the size of n and t shouldn't matter. But in the real world, other sources of light and interference will creep in. As the separation distance between the plates increases, other light will more easily get in from the sides, and the difference between bright and dim will diminish, making it more difficult to see the fringes. Also the phase differences at angles other than 0 degrees and 180 degrees play a role which will cause the fringes to "smear" when n becomes too large. You can get more understanding by looking at this playlist: PHYSICS 61 DIFFRACTION OF LIGHT kzbin.info/aero/PLX2gX-ftPVXXzgOs4KITgRrBxUluUzCrR

[@ALFPAJARITO]

Dear Michel: Thank you very much for your fast reply. Just one more question: I usualy make measurements with a Fizeau aparatus in orden to check homemade flat mirrors for telescopes, and I always wanted to know why we just see the interference pattern caused for the two reflections you show in your video and why we do not see the total interference between the four reflections that already exist in the system (that is the two reflections you show plus the other two reflections in the upper surface of the upper glass and the bottom surface of the bottom glass)? Finaly, I would like to know if are you planing to make a video about this issue and why there is no interference in the real world when the film is not thin. Best regards

[@santoshmumbai]

Alfpajarito t should be of the order of wavelength of visible light. I.e. max 10 times wavelength. For thick film the two reflected rays originate at far distance and therefore do not have coherence so they do not give pattern. Visit my channel SS Physics. Thanks.

[@abelgideon9413]

In all the formulas in my exam they use 'm' so how does that get into the formulas you use For example there was a question where blue light and red light were going through a thin film interference and the red light was visible(constructive) at the end but the blue light wasn't visible(destructive) Since there was 2 phase shifts I cancelled them out and using the blue light I calculated the thickness with (640nm)/2(1.25) but there right answer ended up being 2(640nm)/2(1.25) where did I go wrong? The givens are wave length of blue light: 640nm wave length of red light: 512nm, n of oil = 1.25 , n of water = 1.33 and the layers where AIR OIL WATER

[@MichelvanBiezen]

m is an integer and can take the values 0, 1, 2, 3, .... etc. If the reflected light has a path length that is equal to an integer multiple of a wave length, the light will be in phase and there will be constructive interference

[@TheBarbarios]

How can we say that at the point of contact of the two glasses and where the hair lies there is destructive interference?

[@MichelvanBiezen]

At the point of contact of the 2 glass plates, the extra distance traveled is ~ 0 and therefore the two waves are 180 degrees out of phase and they experience destructive interference. At the location of the hair, that is a "given" to the problem.

[@TheBarbarios]

I see, thank you very much for your prompt response!

[@Kimi-vn7jz]

Shouldn't we replace the lambda with lambda/n since light goes into different refraction index?

[@MichelvanBiezen]

The extra distance occurs between the plates where there is air (n = 1)

[@Kimi-vn7jz]

@@MichelvanBiezen thankss.

[@IloveNateReuss]

Does that mean you have to know the thickness of the hair is twice the t value you have chosen because the t value changes depending where it is measured... Doesn't that mean you need to know the diameter beforehand? Sorry if this doesn't make sense, I'm just really confused

[@MichelvanBiezen]

No, you don't need to know the thickness of the hair. You can figure it out by counting fringes. We use 2t because the light travels the gap twice, once on the way down and once on the way back up.

[@IloveNateReuss]

Thank you for answering. So you can choose any t and the corresponding fringe?

[@MichelvanBiezen]

I am sorry, I didn't understand this latest question.

[@drhf1214]

for the fringes, are you basically adding the x distance of the fringe to the phase difference of lambda/2? so for fringe 1, x distance = 0 lambda +lambda/2= lambda/2 which is destructive?

[@johnnybatafljeska6368]

Why is Xdistance for the 2nd fringe equal to one lambda?

[@MichelvanBiezen]

+Johnny Batafljeska It is not that it is equal. We set it equal for the specific purpose of figuring out where the fringes and the bright spots are.

[@alperguler4808]

Hello sir, I have something missing in my mind. Obviously, the sunlight is coming from air and entering to the glass in first refraction, as a result of this, shouldn't it be shifted 0.5 lambda ? why do we assume that the light is coming from in glass firstly ?

[@MichelvanBiezen]

There will be a partial reflection and partial transmission at every boundary. Here we just concentrate on the boundaries on either side of the air gap.

[@mihika6730]

I loved this video. Thank you so much. Also, one small doubt (ref: 5:03). If the Refractive indices of the two materials and surrounding were adjusted such that no phase shift occurred for both the rays, and if there were exactly 12 bright fringes (with the 12th bright fringe at the end near the hair strand), then our results would still be the same right? Because we'd consider, for the 1st bright fringe, the x(dist) = 0(lambda) ; for the 2nd bright fringe, the x(dist) = 1(lambda) ; for the 12th bright fringe, the x(dist) = 11(lambda).

[@MichelvanBiezen]

Yes, the extra distance would be 2 x path length distance between the two glass plates and you would set that equal to the integer number of wavelengths

[@MulatieDemis]

Good job Mr.Van Biezen

[@MichelvanBiezen]

Thank you. Glad you liked it.

[@naheeniqra9218]

Hello, are you not including lambda(sub n) because the medium in which thickness is being measured is air?

[@MichelvanBiezen]

That is correct.

[@hamzahameed117]

Why is the phase shift ignored?

[@MichelvanBiezen]

The phase shift is taken into account by letting the first dark fringe be at x = 0

[@noelthomasbejoy3089]

Is lambda the extra path diff bw waves making fringe 2 or the distance bw fringe 1and 2

[@MichelvanBiezen]

The extra path difference is twice the thickness of the gap: xdist = 2t. Then you set that 2t equal to a function of the wavelength to find what you want (a destructive or constructive interference spot). You also need to account for phase shifts at the boundaries.

[@noelthomasbejoy3089]

Thank you.

[@PedroSalazarJulio]

What if we are given the light spots. Can we do the same procedure assuming that on top of the hair there is always a dark fringe? (subtracting then the number of light spots by two)

[@MichelvanBiezen]

Pedro, Sure. There are usually several ways to solve a physics problem In this case it is easier to start with the first dark spot at the point where the two glass plates meet.

[@ricohines9318]

for the second fringe, why the extra distance travelled is 1 lambda?

[@MichelvanBiezen]

+Jayden Li After the second wave travels 1 lamda (1 wavelength), it will again be 180 degrees out of phase with the first wave and thus there will be destructive interference.

[@EliShiff]

Hello Sir, great video, by the way, I have a question. What if you change the value of lambda, does it mean that the diameter of the hair will change also?

[@MichelvanBiezen]

No, just the number of fringes will change.

[@ricohines9318]

Could you tell me what is" in phase" and "out of phase" Thank you sir.

[@MichelvanBiezen]

+Jayden Li The concept of wave interference is explained in this video: Physics - Mechanics: Mechanical Waves (11 of 21) The Interference of Waves (Same Direction)

[@shabarish2727]

when the light gets reflected from the bottom surface it will be refracted again by the glass surface , why is this ignored here ?

[@MichelvanBiezen]

Yes, some of it will continue through the bottom glass slab and reflect at the bottom boundary, and some will go across that boundary and....... The amount of light continuing is rather small (although it does), but we ignore it so we can concentrate on the main phenomenon of interest.

[@shabarish2727]

@@MichelvanBiezen oh ok , i was studying interference and that's why i came to this video and was confused. But i got what you are saying

[@henrybun0503]

Hello Sir, I have a question. If the refractive indices of the slide plate n1, flat plate n2 and the space in between the two plates n3 are changed so that n2>n3>n1, are there any changes to the interference pattern observed?

[@tamana-ezindagi2317]

Respected Sir I want to deposit coating of titania on glass and I want them anti reflective coating ? kindly help me regarding variation in thickness of titania, from 300nm to 500 nm ? kindly upload lectures on refraction and effective medium refraction too

[@drhf1214]

oh i get it. if there is a relative phase shift in a thin film, you would use the 2t=mlambda for DESTRUCTIVE since there is already a pi shift.

[@MichelvanBiezen]

That is correct.

[@adolphmatjukotja7096]

Nicole O , isn't it supposed to be complete lambda(s) n it's half for destructive?

[@drhf1214]

are you asking if its supposed to be (m+.5)lambda for destructive? It would be, yes, if both have pi shifts since there would be NO RELATIVE shift but if only one interface has a pi shift, then the reverse of what we normally do is the equation for constructive interference with one pi shift. Hence, destructive for a thin film with a relative pi shift equation is 2t=mlambda. Hope it made sense

[@adolphmatjukotja7096]

Ooh, I get you, n it physically makes sense, I've an exam at 9am, so it was necessary for me to understand,,, thanks a it really, makes sense

[@pacific_ocean9302]

Thank you for teaching me when my teacher is unable to :)

[@MichelvanBiezen]

Happy to help!

[@neefofway8746]

god bless your soul

[@MichelvanBiezen]

Thanks!

[@aps.007]

Sir u are awesome

[@omurgokcinar4187]

Thank you very much. This is very helpfull for my exam tomorrow

[@TWan-dm2en]

hey man you've saved me! thanks!!

[@aryanmathur636]

Very helpful, thank you!

[@chiaosun3505]

老师的地中海好性感

[@michaelxu8810]

聪明绝顶

[@jonathansum9084]

Let me translate this one in English. He is saying this. Professor's Mediterraneo(the empty spot on the top of the head) is so sexy. In Chinese, Mediterraneo's direct translation is the sea in the ground region, so it also has another meaning which you get it.

[@MrY3110w]

awesome!

[@MrKarpovy]

Nice. But the actual diameter of human hair is about 100 micrometers :) But let's say it was a piece of spider web.