M from Assam

Nanum thala

I am Also Bro💖

🥰🥰🥰same

Me too

Hey are u preparing for JEE or NEET?

interestedvideos.com/physics-youtube-channels/

yes very easy - Take a piece of paper and a razor blade. Make 2 cuts with the razor blade in the paper close together (less than 1 mm apart). Hold the 2 cuts in front of one eye and look at a bright light - you will see interference,

Thank u very much sir.

@@lecturesbywalterlewin.they9259 That was amazing!!! wow!!

here they are: kzfaq.info/get/bejne/osBjlbOGsNWZqpc.html kzfaq.info/get/bejne/h69hZrZhmd2clIE.html

I explain that in my 8.03 lectures (Fresnel Eqs).

It's my pleasure

Glad to hear that

refraction

I cover all this in my lectures. I cannot add to that.

The sign of the reflection coefficient indicates if the change of medium flips the wave ( - ) or not ( + ). The first change does (r = -0.2), the second doesn't (r = +0.2)

Most welcome

Put the chalk in the direction your hands moves while drawing with the angle of 135°

I explain this in detail in my lecture

use google - also google Lasers

sun light

you can choose one or the other.

I do not solve problems of viewers, I teach Physics. I cover the physics in my lectures - or use google.

Thank you for suggestion sir .

good question - if N photons reach the double slit per sec a major fraction of N would not emerge from the slit but would either be reflected or absorbed by the slit.

@@lecturesbywalterlewin.they9259 thank you sir

who called it "equal thickness"? The thickness of colorful oil is not the same everywhere. If that were the case, there would not be so many different colors.

You're welcome 😊

there is supposed to be ONLY 1 at the beginning at that one can be by-passed,

use google - light is both (photons and waves)

Plz tell plz plz Plz Plz Plz

use google

@@lecturesbywalterlewin.they9259 Google doesn't has answer(appropriate)

google: "The sign convention for lenses is similar to that for mirrors. Again, take the side of the lens where the object is to be the positive side. ... Similarly, a converging lens always has a positive f, and a diverging lens has a negative f. The signs associated with magnification also work the same way for lenses and mirrors."

@@lecturesbywalterlewin.they9259 but professor we already used them while deriving the equation. Why should we use them again?

+Shing Lau how many minutes in the video?

+Lectures by Walter Lewin. They will make you ♥ Physics. Thanks for reply! It was about 47:20

+Shing Lau n1-n2 0 minus =>phase change pi. + => no phase change.

+Lectures by Walter Lewin. They will make you ♥ Physics. Thanks for replying! that's exactly what confusing me, I don't understand why. The proof seems lie in the boundary conditions of EM waves. ha probably I should not have skipped the lessons before in this course.

how many minutes into my lecture?

@@lecturesbywalterlewin.they9259 41:35

This is lecture 20. I covered this in detail in Lecture 18. kzfaq.info/get/bejne/laphrZmll-CVZmw.html

Lectures by Walter Lewin. They will make you ♥ Physics. Thanks a lot!!

That's a good question. Let's make the wavelength much larger than the width of each slit. Do it with radio waves. No radio waves will go through any slit which is only 0.1 mm wide.

@@lecturesbywalterlewin.they9259 Tnq so much proffesor for give me your valuable time... Sir may you prefer some best books for physics from which you were teaching regularly?? Please sir..

yes, there will be interference, however, you may only be able to see it if the film is very thin.

thank you :)

Blue because it has a shorter wavelength. The distance of the secondary Maxima from the central is proportional to λ. (d sin θ =nλ)

@@potato733 But if we look b/w central maxima and secondary maxima there is a First Minima. So due to smaller wavelength violet would have its Minima and if its a minima for violet, violet would not appear and hence shouldn't it be red?

White

@@bhend1 we will see 🔮Violet colour near to central maxima.... bcz it has shorter wavelength....☺

@@potato733 No..... We see Blue in scattering phenomena...😊

I cover interference in my 8.02 and 8.03 lectures.There are also many on the web. Try the one by Feynman kzfaq.info/get/bejne/aNN5npaop7Col3k.html

Lectures by Walter Lewin. They will make you ♥ Physics. Sir that Richard feynman lectures are really great... But is there any website/page that covers detailed topics of graduation physics? ...

use google

watch the lecture again

use google

+adam lands Interference happens at lambda = 2 * n * t / m for normal incident light where n is the index of refraction, t is the thickness and m is an integer from 1 up. This means the difference between two adjacent lambda's that show interference (i.e. lambda where m=1 and m=2) is largest with low m and goes to 0 as m goes to infinity. For a relatively thick film (with n=1.5) of 1 cm lambda_1 (lambda where m=1) is 3 cm. lambda_2 is 1.5 cm and so on. For visible light, which has wavelengths of 400 nm to 800 nm, m is of the order of 60000. Since the difference in wavelength between lambda_60000 and lambda_60001 is so small there is no noticeable difference between these two wavelengths and you will not see interference as "all" wavelength in the range 400 nm to 800 nm overlap. .

Lectures by Walter Lewin. They will make you ♥ Physics. Thank you professor!

how many minutes into the lecture?

they probably define a fringe as an area ring (with certain width) which has only ONE color. That ring has then ONLY thickness. Different colors, different thickness.

but sir even for one color as we go from central dark ring diffrence b/w nth and n-1 th ring get so close then how can we say fringe of equal thickness even for one color thanks for reply and giving your precious time

encyclopedia2.thefreedictionary.com/Fringes+of+Equal+Thickness

From the description of the respective lectures 👍

they are below the video (after you open the video) as PDF files. Alternatively open my Playlists "Assignments, Exams, Solutions and Lecture Notes." There is one for each of my 3 MIT course: 8.01, 8.02 and 8.03

use google

Lectures by Walter Lewin. They will make you ♥ Physics. You're a better teacher than Google 😁

So can I mate...The Dutch 'g' is an astonishing thing Edit:To be fair, the name 'huygen' itself is astonishingly beautiful too

take a simple double slit or silngle slit diffraction pattern. Look at the curves of intensity vs angle. The constructive interference intensity peaks are high - that's where the energy of the dark areas are.

@@lecturesbywalterlewin.they9259 Thank you for your reply. In the single slit and double slit cases, there are constructive interference as well as destructive interference. But in the case of thin films cases, if the films are very uniform, then all the lights can be destructive interference without any constructive interference. That means the light intensity is 0 (or very low)all over the area. In that case, where does the energy go?

do you have walter lewin sir number

The basic idea of interference is that the waves are IN PHASE for constructive and 180 degrees out of phase for destructive interference.

telescopes lenses lose 4% of the light intensity. Use google.

did you overlook that v is diffrent in other medium? you can;t just square 0.8. Watch lecture 18 at 57:34

to first order, within certain limits which depend on the ratio of the slit width and the wavelength of the light, the intensity of the central max will scale linearly with the slit width.

@@lecturesbywalterlewin.they9259 Thank you sir

+Saurav Suman :) :) :)

If you determine through which slit a photon goes, then there will be no interference. This is the GREAT mystery of QM

professor i am not determining through which slit photon goes I am considering that if there is a phase difference between the light that reaches two slits then where will central maximum shift. Is it above the central line or below it?

your thinking is wrong. You think Newtonian that is not allowed

Hi Pushpendra, If I may, I will take a stab at your question here, with a more lengthy answer. You ask how the central area (wave) of light in the interference effect, and (I'm guessing) the whole pattern, is affected if you know there is a phase difference between the photons going through the top and bottom slits. What you need to know is that the wave function of light is determined by a probability distribution of different frequencies at which the photon can be found when observed. This means there is an uncertainty as to exactly what the frequency is. Fourier analysis describes how this is equivalent in the case of position, and how uncertainty in frequency is linked to uncertainty in position. When position is certain, frequency is less certain and vice-versa. Now regarding a phase difference among photons, if we take for our double-slit experiment a light composed of a single frequency, and if we consider in this light there are phase differences between photons, the frequency of the light's wave function will be unchanged (unless destructive interference completely kills the wave function, which kills the experiment entirely). This means having different phases of light through the slits maintains the same uncertainty of position we already had in the video (because the uncertainty of the frequency of the light remains unchanged). This uncertainty of position is what creates the wavepacket-like pattern you see in red on Mr. Lewin's screen. Indeed this represents the probability distribution of the position of the photons passing through the slits. For more on the uncertainty principle, see MIT's OCW 8.04 course, available on KZfaq. I hope this helped. :- )

I guess I understand what you are trying to ask @pushpendra , given that the wave reaching the second slit (say the bottom one) already has an additional phase difference as compared to the first one, then the wave from the first one will have to cover an extra path to compensate for its lag thus shifting the central fringe (i.e. zero path diff fringe) and hence the entire pattern towards the second slit i.e. downwards !

yes

glad to hear from you, Will try to get more knowledge from u as much as possible. Thanks a lot😊