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## Interference of electromagnetic waves

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# Young's double slit problem solving

## 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.

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