Final Exams and Video Playlists: www.video-tutor.net/
@DudeWhoSaysDeez4 жыл бұрын
You have a video for every topic that I am ever stuck on. Thank you 🙏🏻
@nknm8 ай бұрын
Thank you so much! These videos are *literally* a lifesaver. Every time I turn to this channel it has exactly what I'm looking for.
@anjani36524 жыл бұрын
You are saving my whole degree ✌
@minimarsed2 жыл бұрын
english isnt even my native language but this is what im resorting to because i have no clue what my teacher is saying, thank you for this man
@katlegokokoana34766 жыл бұрын
God bless you bro... you just made prepared for the physics test.
@mojo61126 жыл бұрын
So how'd you do?
@Daniel-nm6pf2 ай бұрын
@@mojo6112 6 years, and we never knew how he did
@mojo61122 ай бұрын
@@Daniel-nm6pf hhhh
@bennettgrado23822 ай бұрын
Everyone asking about the dark fringes being a half, it’s because when you calculate the slits, for single slits we divide the width of the split by 2, so for the dark fringes it’s d/2 sin(theta) = 1/2 +)lambda, you multiply the 2 over making d sin(theta) =2 lambda for the first DARK fringe, it’s confusing because the double slits is d sin(theta) =mlamda for the contructive
@__ICT__1288 Жыл бұрын
U taught this topic better than my actual teacher, thanks
@pikachu51883 жыл бұрын
At 9:31 The angle (Theta1) should be in degrees and not meters.
@thai272 жыл бұрын
yes boy
@mojo61126 жыл бұрын
10:38 that's physics for ya.
@eshi20004 жыл бұрын
you should make a video on thin film diffraction !
@manuelsojan90937 жыл бұрын
At 9:33 why is the angle in meters??
@kamfei47746 жыл бұрын
You are right, why is that angle in meter? hahaha maybe it's some careless mistake up there...
@RaneMatthew6 жыл бұрын
You're correct, it should have been in degrees. The value itself though, is correct :)
@joeramirez90425 жыл бұрын
Mark Wahlberg gonna help m get this A
@ManzilSafarAurBaateinOfficial3 жыл бұрын
Please answer 🙏 Consider a slit of width a producing a diffraction pattern, if 'm' is either positive or negative integer than diffraction minima occur under the conditions 1- M-lamda = a sin 2- 2M lambda = a sin 3- m lambda = 2 a sin 4- (2m+1) lambda = 2 a sin
@amnaumar78032 жыл бұрын
for dark fringe, shouldn't m be a decimal value like 0.5, 1.5, 2.5 etc?
@vladimirsheydorov3776 Жыл бұрын
This what I am confused about, in some book examples I saw m being an integer for destructive interference and in some for constructive??? So annoying...
@vladimirsheydorov3776 Жыл бұрын
Never mind, for single slit destructive interference m is integer and for multi-slit constructive interference m is integer
@TwistedKrizZ33 Жыл бұрын
Thank you so much
@Medhansh07Ай бұрын
hey concept is bit wrong as a very small area is under study (fringes in large region will not be visible properly) the angle remains small ,
@daliaalshawi99075 жыл бұрын
Thank you you are amazing
@kuttygundalchannel54554 жыл бұрын
Thank you
@piglink106 жыл бұрын
0:43 For double slit, wont the bright fringes all have the same intensity? (Same amplitude). The one which you drew isnt it for single slit ?
@raajkumar80394 жыл бұрын
Yup
@randychong87704 жыл бұрын
Thanks
@ramaghalayini82823 жыл бұрын
Thanks ! This helped me a lot🤩
@sudukikellington35946 жыл бұрын
Thank u. You are amazing!
@TheLover67316 жыл бұрын
I thought since the angle is bigger than 10 degrees then we cannot use the formula y1= Ltan(theta) because it may be used for small angles. Therefore, wouldn’t we use the formula y=L(m+0.5)lambda/d to find y1? If so, then the distance of the central maximum would be equal to 1.43m.
@casbox26672 жыл бұрын
I think n+0.5 is what I've learned in class as well
@xXishibashiX6 жыл бұрын
This video gives me but a "whiff" of hope
@alainfabrice5094 Жыл бұрын
I almost left without liking the video 😅👍
@mashanatanzon557 Жыл бұрын
On question one, shouldn;t the answer be 0.952m? We know that theta = lambda/slit width. And that lambda is = opposite over adjacent, making (680x10^-9)/(2x10^-6)=(W/2)/1.4. Resulting in width = 0.952m. Thank you for clarifying this!
@estherguberman6710 Жыл бұрын
I got the same answer
@Benjicmm10 ай бұрын
10:15 theta = 0.122 meters? shouldn't it be in degrees?
@schenzur5 жыл бұрын
9:33 why aren't we multiplying theta by 2
@fauzansyauqi28154 жыл бұрын
why must?
@jackwilson47104 жыл бұрын
He didn't need to because he halfed the width to get y1
@BlazinblaiserVods8 ай бұрын
Why do we do this is Radians and not Degrees when finding the angle?
@aaryansrivastava78353 жыл бұрын
thank you :)
@nirelleksenasitchoa76526 ай бұрын
When you say to use the second formula if theta is really small, what range would you suggest for theta? (anything less the 5 degrees?)
@tomainterpunkcijamioc5 ай бұрын
Any angle that satisfies sinx≈tanx (three/four decimals)
@luckyprosperity64084 жыл бұрын
thanks!
@isaitalvarado57272 жыл бұрын
what does theta need to be smaller than in order to be small enough to use the (y)(d)=(L)(m)(lamda)?
@RelevantDad Жыл бұрын
Because of how that equation is derived. It is created by assuming that sin(theta) equals tan(theta), which is a close approximation for very small angles.
@samb.7395 Жыл бұрын
@9:35 it should be degrees not meters
@nemuirostorageroom Жыл бұрын
spent 4 hours watching my prof's lecture, and I feel like I have a brain fog. Literally 1 minute into the video (the part where he says the amplitude of the bright fringe decreases progressively), I went "OHHHHHHHH".
@MegaTiffanym3 жыл бұрын
Why didn't you convert the 1.6 cm to meters to be consistent with the rest of the units?
@ethanhope99353 жыл бұрын
why does destructive interference happen on single slit?
@blackart25393 жыл бұрын
Sir how to relate wavelength and intensity in single slit
@husnaadibah59243 жыл бұрын
I guess it's something related to interference?
@SciD19 күн бұрын
No. A single-slit should not produce a wave interference pattern, yet it does. It's basic reflection off both sides of the slit, according to the Law of Reflection. The Huygens Principle is an ad hoc mathematical contraption to force waves where they do not exist.
@thai272 жыл бұрын
In example 2, why m is 1 sir
@SnapOfLife-b1z4 ай бұрын
For a double slit, the intensity of the dark and bright shown on the screen will be the same no🤔
@olympichigh90142 жыл бұрын
resolution questions?
@richardrivaldo69414 жыл бұрын
hello how about phasor and I-y diagram for 4 slits interference?
@derekliu49175 жыл бұрын
9:40 your theta should be in degrees
@oscarobioha5955 жыл бұрын
ahh thank u!. I was scrolling to look for someone who caught it
@lilianawong48195 жыл бұрын
can we assume that sin(theta) is = to y/L? For the first problem I paused the video and solved for y first and got a completely different answer. Isolating y I got: y= (m*lambda*L)/d = 476m ? Or can I only assume that with double slits? please anyone respond T.T
@boofmonkey54553 жыл бұрын
maybe
@matthewkoob7600 Жыл бұрын
what a god
@souvikdeyiipcmbi14963 жыл бұрын
man my teachers say that intensity remains constant for ydse but u are saying different plx help
@nabiha57723 жыл бұрын
intensity does not remain constant in a single slit diffraction, that's experimentally proven that the bright fringe in the centre is the brightest and the farther you go fringes/light get dimmer. You can have a rough observation on that in a dark room with partially closed door and bright light outside
@lojfiojo47255 жыл бұрын
Isn’t theta 22?
@macavmoudbhawanxutra8766 жыл бұрын
Can you solve jee advance paper?
@p.n1453 Жыл бұрын
I want explanation the greatest nation
@rushvi16116 жыл бұрын
i thought as it's a minumun it would be sinø = ((mlamda)/2)/d
@oscarobioha5955 жыл бұрын
single slit is different
@faizan4929Ай бұрын
Please change the background of your videos. Black looks absurd to eye😢
@zubairahmed17186 ай бұрын
I just don't understand a single topic idk why
@deanshetlar63614 жыл бұрын
try this new 2 edge diffraction setup, over 100 nodes can be seen: kzbin.info/www/bejne/fnenpGtrmKqcgas
@redeyejay95746 жыл бұрын
someone help me my grandpa is abusing me and threw me off chair