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# Young's double slit problem solving

## Video transcript

I think we should look at an example of Young's double-slit let's consider light of wavelength 700 nanometers that would mean this distance right here between Peaks 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 a 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 it's kind of like a shadowy line right there there's our bright spot constructive point how far would 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 that D sine theta remember this is the formula right here D sine theta is the path length difference that's supposed to equal M lambda sometimes you'll see this as n lambda but n reminds me of index of refraction which confuses me so I'm going to write it as M all right so what do we do D what is 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 so I've got 1300 nanometers times the sine of an angle what angle are we going to talk about well what we want to know is the distance this distance here so I'm going to worry about this angle I'm going to worry about the angle from here's my centerline from there to the point I'm concerned with is this first bright spot past the center point so that's the angle I'm concerned with this angle right here equals M what should mV well this is zero should I put zero no because I don't want the angle to the center one 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 nanometer wide piece of information we have to use that no we don't in fact that does not play into here the only time that the 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 have to use that one in our calculation all right so we calculate the angle here we go so I'm going to find sine of theta and I'll get the sine of theta equals 1 is just 1 so I divide both sides by 1,300 I get 700 over 1,300 the nanometers cancels nanometers as long as it I'm in the same units it doesn't matter I'll solve this for theta how do I get theta I've got to use inverse sine of both sides so the inverse sine of sine theta is just theta and the inverse sine of this side gives me 32 point 6 degrees that's what this angle is right here 32 point six degrees but that's not what I was trying to find what I'm trying to find is this distance not this angle so how do we do that well this side this side right here I'm going to call it Delta Y because it looks like it's 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 was three meters away from the screen 3 meted the screen was 3 meters away from the double slit and how do we relate the opposite side to the adjacent side shoot I know how to do that tangent theta equals opposite over adjacent in our office it is Delta Y over three meters in this case so if I solve this for Delta Y I'm going to get Delta y equals x sides by 3 meters times the tangent of theta theta we solve for right here 32 point six degrees and if you multiply all that out you get one point nine two meters that's how big this would be from here Center point to the next bright spot is one point nine two meters that's how you solve this problem you've got to use a little trigonometry once you get your angle you've got to relate it to a distance vertically on the screen and 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 with 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 determined by the laser I fire in here and this wavelength staying the same so this whole sides got to stay the same because M is still 1 this point so what's got to happen to theta if the D goes down and the whole thing has to remain the same the angles got to go up because sine of a bigger angle will give me a bigger number sine of zero zero sine of something bigger than zero gives me something bigger than zero on the bigger the theta the bigger sine theta so as d decreases sine theta has got to go up that's mathematically why that I can show you it 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 and then we come back together and those shadowy lines of constructive kind of ironic they look like shadows but they should bright it's just the way this visually looks you get more and more lines this way they get scrunched together and as you push D closer together as you get smaller they spread apart these bright spots spread apart so in other words if I were to move these distances 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