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
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
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...
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!
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.
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".
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.
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
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
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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
God bless you bro... you just made prepared for the physics test.
So how'd you do?
@@mojo6112 6 years, and we never knew how he did
U taught this topic better than my actual teacher, thanks
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
you should make a video on thin film diffraction !
Mark Wahlberg gonna help m get this A
Thank you so much
10:38 that's physics for ya.
At 9:31 The angle (Theta1) should be in degrees and not meters.
yes boy
This video gives me but a "whiff" of hope
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
Thanks
Thank you
I almost left without liking the video 😅👍
Thanks ! This helped me a lot🤩
Thank you you are amazing
thank you :)
At 9:33 why is the angle in meters??
You are right, why is that angle in meter? hahaha maybe it's some careless mistake up there...
You're correct, it should have been in degrees. The value itself though, is correct :)
for dark fringe, shouldn't m be a decimal value like 0.5, 1.5, 2.5 etc?
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...
Never mind, for single slit destructive interference m is integer and for multi-slit constructive interference m is integer
Thank u. You are amazing!
thanks!
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!
I got the same answer
what a god
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.
I think n+0.5 is what I've learned in class as well
Why do we do this is Radians and not Degrees when finding the angle?
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".
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?)
Any angle that satisfies sinx≈tanx (three/four decimals)
9:33 why aren't we multiplying theta by 2
why must?
He didn't need to because he halfed the width to get y1
In example 2, why m is 1 sir
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 ?
Yup
why does destructive interference happen on single slit?
For a double slit, the intensity of the dark and bright shown on the screen will be the same no🤔
10:15 theta = 0.122 meters?
shouldn't it be in degrees?
Sir how to relate wavelength and intensity in single slit
I guess it's something related to interference?
Why didn't you convert the 1.6 cm to meters to be consistent with the rest of the units?
@9:35 it should be degrees not meters
resolution questions?
I want explanation the greatest nation
what does theta need to be smaller than in order to be small enough to use the (y)(d)=(L)(m)(lamda)?
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.
hello how about phasor and I-y diagram for 4 slits interference?
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
maybe
Can you solve jee advance paper?
man my teachers say that intensity remains constant for ydse but u are saying different plx help
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
9:40 your theta should be in degrees
ahh thank u!.
I was scrolling to look for someone who caught it
Isn’t theta 22?
I just don't understand a single topic idk why
i thought as it's a minumun it would be sinø = ((mlamda)/2)/d
single slit is different
whatta g
someone help me my grandpa is abusing me and threw me off chair
try this new 2 edge diffraction setup, over 100 nodes can be seen: th-cam.com/video/HFDs85cEiKI/w-d-xo.html
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Penty
Thanks
You are awesome thank you