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## Physics library

### Course: Physics library > Unit 14

Lesson 2: Interference of electromagnetic waves- Constructive and Destructive interference
- Young's double slit introduction
- Young's double slit equation
- Young's double slit problem solving
- Diffraction grating
- Single slit interference
- More on single slit interference
- Thin Film Interference part 1
- Thin Film Interference part 2

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# Young's double slit problem solving

An example problem involving a double slit experiment. Created by David SantoPietro.

## Want to join the conversation?

- Why wasn't d (the distance between the two slits) calculated as the distance from the center of one opening to the center of the other opening, 1500nm instead of 1300nm? This seems more intuitive, to me.(9 votes)
- The question is stating that the 200 nm slits are spaced 1300 nm apart, and the video demonstrates that the measurement is from the center of one slit to the center of the other. This is not the same as saying the object between the slits is 1300 nm, and it would in fact be 1100 nm.(12 votes)

- How can we find distance between 1st and 2nd bright dot ??plzzz help ..(1 vote)
- First find distance between bright spots 0 and 2 then distance between bright spots 0 and 1 and then subtract them(19 votes)

- Hello, I'm kind of a nube, forgive me. I've been doing the experiment at home and can't seem to figure out what kind of measuring device i can use to collapse the wave and make it act like a particle. I would like to preform this for my friends and family to inspire some curiosity. Is this part of the experiment beyond my DIY capabilities?(4 votes)
- OK. So I have thought of a way to DO it. But a) you can not Prove its happening and b) the experiment / demonstration is a bit 'unexciting'.

However; here goes

First, if you want to show wavefunction collapse, then you need to start with a wave-like thing and then change into something with a specific momentum, position, energy... etc. (ie NOT delocalised).

OK, so how do we 'prove' that we have a wave-like - thing'? Well, if we can diffract it, then it shows us that we have a wave. (Diffraction is one of the 'tests' of wave nature of a thing)

Then if we want the wave function to collapse, we need to 'observe' the thing. ie measure its position, energy momentum etc.

OK, so how to do it at home?? Well, I did say it was not exciting....

the answer is to squint !!

I mean if you look at a light source; for example a street lamp outside at night or a star or any bright-ish light. and then squint. I mean like almost close your eyes. You will see the light 'spreading out'. It makes a star-like shape in your eye. Try it n see. This is due to the light 'diffracting' as it goes past your eyelid.

So, now we have showed that the light going into your eye is a wave. And therefore has a wavefunction.

THEN, the light passes through your eye (jelly bits and lens) and then lands on your retina at the back of your eye. How do we know it lands there? Because you see it.

So now you know the exact position of the light. Exactly on your retina. You also know that the energy required to stimulate your rods and cones in your retina will come from the photon that 'formed' on your retina at the time of observation. Now, no longer a wave but 'collapsed' to a measurable thing.

I did say it was not an exciting experiment and there is no proof of wavefunction collapse but it is a demonstration :) and worthy of discussion.

Great question. Let me know what you think

Well done(5 votes)

- at3:05, since d=distance from the centres of the two slits, wouldnt we add 1300 to 0.5*slit length to find the distance between slits 'd'? So the correct answer would be 1300+(0.5*200*2)=1300+200=1500 nm.(3 votes)
- What happens when the distance from the screen increases? Do the distance between the dots increase?(3 votes)
- Why did you say that sin=tan? this rule is only applicable for small angles (less than 10)

I used the rule Δy= Lλ/d

and I got 1.62

that is different from your approximated answer by 0.30(3 votes) - To calculate the distance between the slits, do we take the distance from the center of the slits to the other center or simply from one end to another? Also, if I use theta = wavelength/ b, then i get theta to be 3.5. Why is this wrong?(1 vote)
- The distance is from one intensity maximum to the next intensity maximum.(3 votes)

- Then what is this .....Xn= n(Lamda)(D)/d?(2 votes)
- My guess is Xn is the distance from the central maxima to the nth maxima (the central maxima being n=0).

From the diagram:

tan(θ)=Xn/D where D is the distance to the screen

and

sin(θ)=n(lambda)/d

For**very small angles**, we can say that tan(θ)=sin(θ)

So: Xn/D=n(lambda)/d

Xn =n(lambda)(D)/d

However, in this question, it is not a "very small angle"

The angle is bigger than 30 degrees! So we first had to find θ from sin(θ) before finding tan(θ) instead of simply equating sin(θ) and tan(θ).(2 votes)

- What would be the practical application of this equation and experiment?(1 vote)
- This equation is also used for investigating crystal structure by passing light through a crystal. The input values are screen distance and a coherent monochromatic source of light. The derived quantity is slit width, which can provide insights into the crystal structure.(2 votes)

- What happens if the wavelength is larger than the spacing between the slits? Inverse sin can't process numbers larger than one.(2 votes)
- interference will not take place if the wavelength is larger than the spacing between the slits. It is necessäry condition that the width of the slit should have a value of the order same as that of wavelength of the wave.(1 vote)

## Video transcript

- [Voiceover] I think we should look at an example of Young's Double Slit. Let's consider the light of
wavelength 700 nanometers. That would mean this distance right here between pix is 700 nanometers apart shines through a double slit whose
holes are 200 nanometers wide. That means from here to
there is 200 nanometers and they're spaced 1300 nanometers apart. That means from the center of one to the center of the
other is 1300 nanometers. If the screen is placed three meters away, here's our screen, and
it is three meters away. Then what would be the distance
from the central bright spot on the screen to the next bright spot? The central bright spot is going to be, well, it's in the center. You can follow this line. Look at it, it's kind
of like a shadowy line. Right there, there's our
bright spot constructive point. How far will it be from that point vertically to our next one? Our next one is right here. You can see these lines of
constructive interference. This one's about right here. So the question is, how far
is this distance right here? How do we figure this out? Well, we're going to use
the equation we found which is to say d sin theta. Remember, this is the formula right here. D sin theta is the path link difference. That's supposed to equal m lambda. Sometimes you'll see this as n lambda but n reminds me of index over
fraction which confuses me so I'm going to write that as m. All right, so what do we do? D, what is the d? We got all these numbers in here. D is defined to be the
distance between the holes. So d in this case is this 1300 nanometers. I got 1300 nanometers
times the sin of an angle. What angle are we going to talk about? Well, what we want to know
is this distance here. So I'm going to worry about this angle. I'm going to worry about the angle from, here's my center line, from there to the point
I'm concerned with is this first bright spot
pass the center point. That's the angle I'm concerned with. This angle right here. Equals m, what should m be? Well, this is zero. Should I put zero? No, because I don't want
the angle to the center. I want the angle to the
first one over here. So this is m equals one. The first order bright
spot constructive point. Times the wavelength,
what's the wavelength? The wavelength of the light
we said was 700 nanometers. Now you might be
wondering, "Wait a minute. "What about this 200 nanometers
wide piece of information? "Do we have to use that?" No, we don't. In fact, that does not play in here. The only time that this
spacing is important. It's not going to change your calculation. You just need the spacing to be small, small enough that you
get enough diffraction, that you get a wide enough
angle of diffraction that these two waves will
overlap significantly enough that they'll create the
interference pattern that you want to see over here
to a degree that's visible. Okay, but we don't need it. We only know how to use
that one in our calculation. All right, so we calculate the angle. Here we go. I'm going to find sin of theta. I'd get the sin of theta
equals, one is just once, so I'll divide both sides by 1300. I get 700 over 1300. The nanometers cancels nanometers. As long as I'm in the same
units, it doesn't matter. I'll solve this for theta. How do I get theta? I got to use inverse sin of both sides. So the inverse sin of
sin theta is just theta. The inverse sin of this
side gives me 32.6 degrees. That's what this angle is right here. 32.6 degrees but that's not
what I was trying to find. What I'm trying to find is
this distance, not this angle. How do we do that? Well, this side, this side
right here, I'll call it delta y because it looks like a vertical distance. This is the opposite side to that angle. That's the opposite side. We know the adjacent side. This adjacent side we were told. This three meters away from the screen. The screen was three meters
away from the double slit. How do we relate the opposite
side to the adjacent side? Sure, I know how to do that tangent theta equals opposite over
adjacent and our opposite is delta y over three meters in this case. If I solve this for
delta y, I'm going to get delta y equals, multiply
both sides by three meters times the tangent of theta. Theta we solved for
right here, 32.6 degrees. If you multiply all that
out, you get 1.92 meters. That's how big this would
be from here, center point, to the next bright spot is 1.92 meters. That's how you solve this problem. You got to use a little trigonometry. Once you get your angle,
you got to relate it to a distance vertically on the screen. This is a common problem
using Young's Double Slit. I will say one more thing. Oftentimes, a popular question,
a follow-up question is, what would happen if we reduce the distance between the slits? What would happen if we take this distance between slits and we make it smaller? We scrunch these together. Would this angle get bigger or smaller? Well mathematically,
let's just look at it. If the distance over here goes down, in our formula, if d goes down. Notice I'm not changing the wavelength. That's the term by the
laser I fire in here. This wavelength staying the same. So this whole side has
got to stay the same because m is still one, this point. What's going to happen to
theta if the d goes down and the whole thing
has to remain the same? The angle's got to go up because sin of a bigger angle
will give me a bigger number. Sin of zero is zero. Sin of something bigger than zero gives me something bigger than zero. The bigger the theta,
the bigger sin theta. So as d decreases, sin
theta has got to go up. That's mathematically why
that I can show you in here. Check this out. Let's just take this. Let's take this here and I'm going to move this whole thing down
and watch what happens. Can you see the shadowy lines spread out? See how they're spreading out? Then we come back together and those shadowy lines have constructive. It's kind of ironic. They look like shadows
but they should be bright. It's just the way this visually looks. We get more and more lines. This way, they get squashed together. As you push d closer
together, they get smaller. They spread apart. The bright spots spread apart. So in other words, if I
were to move these distances and the slits closer
together, you would see these bright spots get
farther and farther away from each other on the screen. So that's an application
of Young's Double Slit. Good luck.