Young's double slit equation | Light waves | Physics | Khan Academy
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- Опубликовано: 6 июл 2014
- Let's derive a formula that relates all the variables in Young's double slit experiment. Created by David SantoPietro.
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I had such a difficult time learning this in class for some reason. I really appreciate this. I don't know why but watching this made perfect sense to me. Maybe I'm just more of a visual learner. My instructor teaches concepts sometimes without illustrating. It makes it hard for me to visualize.
kenpachi zaraki Can’t help but read this in Zaraki’s voice :)
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I was stupid lost my login password. I appreciate any tips you can offer me
thank u very much sir...i cant focused in my online class and my test is in three days and thankfully you save my lifeee
I'm a physics student and i'm planning to get BIG marks in order to become a teacher, when i do that i will use the best method to teach my students, these and alot of other videos made it very clear how my method would be .. thanks i hope i will be as helpful as you guys :)
Best of luck!
U a teacher yet?
@@carpetman9191 Almost, I became top of class.. now our names were presumably send to the government.. I'm waiting for the next season to come and see if I became a university teacher :) my dream might come true soon
@@Hunar1997 congradulations! I'm sure you will do great.
@@carpetman9191 Thanks :D
awesome explaination :) !!
Every faculty must teach like this one !!
Explained way better than any textbook! Thanks so much
hats of to u guys
u guys r just amazin'
u make every thing so easy for us .
love de khan academy
2:04 why the paths will be the same length? Aren't they cathetus and hypotenuse of right triangle, and can't be same length?
Well sir i understand it as if the distance between the slits is very small as compared to the distance from the screen . say micro meters compared to a meter or so . For small angles the length becomes same .I hope i am right in my explanation.
saharsh pruthi yes but for the interference pattern of light this approximation must be accurate enough for the wave length of light, roughly 550 nano meters, so im affraid i am also confused as to why this works. He did not explain this assumtion at all
Joe 1234 I'm also kinda confused about that, because if one angle is 90 an the 2 opossite sides are the same, that means the angles are the same and then we have a triangle with 2 90degree angles (not posible)
yes well it doesnt hold to Pythagoras' theorem, however i believe that there is a convincing case that exists for the use of the small angle approximation here (which settles it) but it seems no one on youtube is willing to share it
Actually after that length as they are very close
You could consider them parallel
So only that small part is your delta x
I worked out a formula for Δx without using the sine approximation...
I used the law of cosines a few times. If "d" is the distance between the centers of the slits, "L" is the perpendicular distance between the slit system and the back wall, and "y" is the height to the first order point, then:
Δx = ( 2dy + d^2 ) / ( 2 * sqrt((y + d/2)^2 + L^2) )
The sine approximation (Δx = dsinθ) is good when "d" is very small compared to "y" (and to a lesser extent "L").
What screen recording software do you use? I would like to know a good one that allows me to pause recording.
Thank you for this great explanation this helped me so much with school i am from Turkey by the way
The most important part of the video (just before and after 2:30) was quite poorly explained. Please consider making another video for students that demonstrates parallelism of lines, why the path lengths would be considered equal and why you placed the second theta where you had.
it made my concepts crystal clear....
thank you soo much .. I'm in my senior year in high school and this year will control what college I will enter and that's after 3 months hhhh .. you helped alot thank you soo much again ♥
Wow amazing!!! You explain wonderfully
Now this is a really good explanation for someone who knows nothing about Young's experiment and has to get it prepared for tomorrow's exam !!!👍
Wow what a good teaching 👏
very well explained thank you
Thanks for uploading that video! It's really helpful: )
please explain ...why both of the angles are same
exactly what my question was!
I don't get it too. How did he equate the angle between the blue lines and the angle between the white lines?
pause the video at 6:36 for observation
in both triangles,
there are 90° angles--------> (1)
further, for constructive we know that :
Δx =n λ
so Δx (in red) and λ (in white, right beside the text "1st")
can be considered to have a constant ratio
Δx
----- = constant
λ
hence by triangle congruence, both of those angles are same
by triangle similarity
It's not drawn to scale, but in reality the two slits are really close together. This makes the rays from both slits essentially parallel and to the middle white line. If the angles are that small, both thetas are almost identical. you can try drawing the triangles yourself www.geogebra.org/geometry
Awsome video! Keep making more!
thanks for teaching well through online video. :)
Finally got it👐 thank u💜
Arouynd 2:15, I notice a lot of people are pausing. I think he draws the right-angle symbol in the wrong place and it should be at the center, white line. Otherwise the argument doesn't seem to work. But I'm not a physicist and have not checked the argument all the way to the end.
Nice explanation thank you
Good explanation
I love you...
Best teacher on RUclips
sorry for this silly question, but why do we number "m" the way we do, such as 1, 1.5, 2, 2.5, etc? What does the number mean? Is it similar to how electrons are quantized?
Very much detailed and nice explanation 👍🏻
How did he equate the Q angles in the two triangles
Can someone please help me, I don't understand why when a question asks for the third order minima we take m=2, but when it asks for third order maxima we take m=3? Is the nth order not the same as order number=n for destructive?
AMAZING!
Hello. I just watch this video, is very good! thanks. But I want you check at 2:16. I have a question on this. If you take the right angle on where it is marked the two light blue paths marked are not equal. you most take the right angle on the white line so the both light blue paths have the same lenght... am I right?
i noticed it too bro! you're right, they screwed up here!
in the double slit experiment, the difference between distance between the slits and distance between each slit and the screen is significantly high, which makes the two lines approximately equal, but if you use the method in another application, where the distances are comparable, you can't make this approximation, as it would lead to a fatal error in calculations.
someone explain me the relation of lambda on screen ...like how the crust and trough meet the screen
what is the effect of light intensity ion the fringe spacing??
THANK YOU SO MUCH I LOVE YOU
Thank you!
1:58 I have a problem. How are those path lengths equal? I mean that's a right angled triangle in which hypotenuse is always greater, isn't it?
I suspect the right-angle is supposed to be on the white line, and it's supposed to be an isoceles, not right-angled triangle. I haven't watched it all though.
Is there a possibility of making a real photon experiment where there are not enough photons to make a solid line?
Are there ways of seeing photons through a prism and make the different wavelength go trough different slits?
How can the hypotenuse in a right triangle from the path lengths be equal to the opposite??
what grade or stufy level yall in, rn 2021
Im confused at 2:10. Since if as he claims the 2 blue lengths are equal, then this forms an iscoles triangle and so the base angles are also equal. However, since the lower base angle is 90 degrees, this would also make the upper angle 90 degrees and you would not have a triangle but just a line. Consequently, the only thing that makes sense here is if this is just acttually an approximation. If this is the case then it would explain why he said the distance to the screen most be a lot further than the slit difference, as the approximation would get better then. If this is what he means here, why did he not clarify that for large comparable distance this is a good approximation.
Isn't the equation for this a=D(lambda)/d, where a is the fringe spacing and D is the distance from the screen to slit and d the distance between the 2 slits?
AmaZing
can anyone explain in 2:17 why that 2 lentgh will be equal if it is right angle? and why the angle is the same is theta?
is it true the output on the wall will change depending if the waves are being observed or not? IE: if the output is not being observed it would appear as if there was only 2 intense points with equal intensity?
2:37 how did you know that the angles are the same?
Thank you😁
Still not convinced by the hand wavey justification given in all explanations as to why it's ok to assume the two paths minus the path difference form a right triangle. They don't they obviously form an isosceles triangle as the whole point is that they are the same length!
in 6:18, how to measure the angle???
Thanks 😍
1:56? why is that line drawn and how is it perpendicular and how is deltaX's length below that line. How is that for sure?? looks like an assumption . can someone explain to me
+Yunik Maharjan Yes, It is an assumption. We can use this assumption because this normal to deltax has a very smal size, and we know that as the normal approaches zero, the two "paths" will equal.
How are double slit test results changed when using polarized light either aligned with the slits or perpendicular to them, as well as with helical? Thx
Amazing!
How is sin equal to base /hypotenuse ?
just saved my life! thank you
what is m for?
maximum or minimum. It depends on what value they hold at a certain location on the screen. If it's a half-number, its a minimum (dark spot) and if its a whole number, then its a max (bright spot)
I thought the formula was
wave length = (distance between slits * distance between 2 bright fringes) / Distance from laser to board
I hate to be that guy, but just in case any one else is confused, I gotta make a small correction. At 2:20, he says that the two sides highlighted in turquoise would be the same length, but that's impossible since one of the turquoise sides is the hypotenuse of a right triangle while the other side is not the hypotenuse.
It’s in the far limit it’s true
Thanks, great video.
2:34 - Why are those two angles the same? Am I missing something obvious here?
Thomas Bradley They are approximately the same if you assume the distance is large enough between the screen and the slits. Similarly the blue lines are only approximately equal.
Thomas Bradley Okay so i think he made an errror or assumed that they were the same length due to the small difference of lengths, but as far as I know, the orange line should be perpendicular to the white line that makes theta, not the blue line.
By triangle similarity
+Tee Ponsukcharoen could you please explain
They are similar triangles
I didnt understand why the two angles are the same and what it has to do with the second plane being far away . someone please help .
Can someone please explain why the angle opposite to the Δx side in the right triangle is equal to theta? Theta originally being the angle between the line cutting through the central maximum and the line going to the int. fringe.
It's been 3 years, did you already find out why?
It's been 5 years did u get??
how are the angles the same at 2:34, don't say that the distance between the screen and the source is much larger than the distance between the slits... I know that but how does that allow you to assume that the angles are the same
www.geogebra.org/geometry. Try drawing the same triangles here, because the distance between the two slits are sooo small(drawing a reallly tiny triangle). you will see that the theta in the orange triangle is almost equal to the theta in the white triangle
How can the length be same..There are two right angles,If u make one right angle
You can also replace Sin for y/H where y is position on the screen from the center of it and H (Hypotenuse) is distance to the screen from the center of the origin. If you move it around (and do some approximations Like making H = Distance from center to center) you can get the positions on the screen based on the wavelength, d and D.
y = (D/d) n Lambda
I know this formula is correct but Y/D is tan theta
Kabir Manrai tan theta is approximated to be equal to sin theta when the distance between the slits (d) is very small compared to the distance from the slits to the screen.
lpenney49 oh yeah, thanks
super
3:33 those 2 blue sides of the right angle triangle definitely do not have the same lenght, or am I just tripping?
Why do we take the paths of the waves to be parallel? That's the only way the two thetas are equal.
why are those two angles the same?
i guess the angle is gonna be very small....so can't we just assume sin Θ = Θ (in radians)
Super
how have you taken the 2 path lengths to be equal? then the other side of the 90° would also be a 90° and this would have to be equal to the opposite angle sonce the two path lengths are equal...then it would not form a ∆ at all
Exactly. That's the same thing that I've been thinking.
It doesn't really make sense.
I still don't understand why the screen has to be very far from the slit :(
Thanks though, you cleared up many of my misconceptions
check part one perhaps when he starts drawing each waves or you can look closely at the drawing here and see that if the screen was close the waves wouldnt interfere.Late reply but i love the question lol
1:58
How the hell did they make the holes so small and so close back then?
the remainder of the pads form a straight angle triangle. They cannot be equal, based on Pitagora theorem. Why did you say they are equal? in this drawing delta x is not the pad difference! Is larger than the pads difference
Why are angles the same ?
that angle theta was so random what happened there!
2:00 he says "if this is right angle right here," then the two purple paths are equal. Not at all! They are a cathetus and a hypotenuse of a triangle.
where L >>>>>>>> d they are very close to equal.
For very very very very small thetas , they are assumed to be almost equal
5:55 summary.
*Green Green Green Green Green!*
what if the wavelengths are different?
Michael Jayasuriya Then you wont get regual fringes, if any at all. They wont interfere properly
is sin theta changeable or constant here?
You are wrong at drawing where the right angle is supposed to be in 2:00. It is not necessarily equal with how you presented it. The right angle is supposed to lie on the white line above the central line. You did not show your "basic trigonometry" in 2:14 or so. That's maybe why the errors stockpiled. Please reupdate your video.
My optics teacher has a so special dialect that I cant't understand him.
i love you
umm... slit spacing are small, the screen is far away... small... far away...
Wow i didn’t think about using angle to measure wavelength
We are taught to use this fomula: Δx = λD/a
D is the distance between the slits and the screen in meter
a is the distance between two slits in millimeter
Δx is measured in centimeter
And i should have the wavelength in micrometer
Thanks for the knowledge
Is ø pronounced theta or tita
Ajay Priyadharshan its called as phi
Phi
The cursor really makes it too difficult for me to understand about what he's teaching..can barely see it
Lies, lies, all lies. You must make the the cross line at right angles to the *centerline*. This make the difference in distances tan θ, not sin θ. You then have to approximate tan θ as sin θ for small angles. You get your answer, but it is only an estimate based on the closeness of tan and sin for small angles.
This formula is a disgrace, you can absolutely get the answer without assuming the hypotenuse and the adjacent being the same. I don't get it, this is physics, how can we just assume something like this. In my IB textbooks, they don't even explain how to get the formula, and this video is very ambiguous concerning the assumptions there.
I mean, here's the thing. You can argue that the difference between the hypotenuse and the adjacent is small, but we are taking wavelength into consideration here, which is in the nanometer level, the error would be big compared to the wavelength.
I hate physics
I’ve been saying since 2013 that it seems extremely obvious that all electrons and photons are in orbit with a dark matter particle.
Based on Fermilab and other recent findings, I now think electrons are made of an electron neutrino entangled in orbit with a dark matter muon neutrino, explaining their erratic orbits around nuclei and explaining superposition and uncertainty.
I think photons are a pair of electrons entangled in orbit together in apparent an axial or helical polarizable wave-like movement depending on the direction of their rotation as they travel. This explains the double slit experiment for example and the speed of the rotation explains electromagnetic wavelengths (and visible colors, etc).
What do you think of this Mudfossil University video?:
‘’Light Duality Solved and Seen’’, Sep 29, 2022