**PhD., University of Maryland**

Published author

Jonathan is a published author and recently completed a book on physics and applied mathematics.

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Jonathan Osbourne

**PhD., University of Maryland**

Published author

Jonathan is a published author and recently completed a book on physics and applied mathematics.

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**Double slit experiment**, also known as the Young's experiment, was conduct by Thomas Young in 1803. This experiment confirms the wave nature of light, which is wave behaves as a wave, not a particle. **Double slit experiment** showed that two distances were formed by the slit. Constructive interference occurs when *d(sin of the angle) = n(wavelength)* and instructive interference occurs when *d(sin of the angle)=(n+1/2)(wavelength)*.

So let's talk about the double slit experiment this is an extremely important experiment that was first performed by Thomas Young in 1803 and it was the first experiment that gave conclusive evidence that light behaves as a wave and it started really eroding all of Newton's ideas about light behaving as a particle which then James later on anyway. But for right now we're looking at light as a wave and what we're going to do is we're going to have a screen here and we're going to poke 2 holes in this obstacle. We're going to send light through it and then we've got a screen a distance capital D away and we want to look at the pattern that we get and what we end up seeing is something that looks like this. So we have a central maximum where it's very bright then we got dark spot no light at all, then another maximum dark spot maximum, dark spot maximum. And this was not at all expected if there were just particles going through because how are going to get a dark spot. So the way that we understand it is by using Huygens theory of wave interference. So when you look you've got some light that's going to come from here and going to go over and meet at the screen here and some light that's coming from the top slit and it'll come up and hit the screen here.

Now these two bits of light are going to interfere with one another, so the question is why are they different? Well here's the idea. The light from the bottom slit has to travel a little bit farther than the light from the top slit to get over to this point. If that difference in distance is exactly one wavelenght then that means that we'll have constructive interference because it'll just go an additional wavelenght and then they'll both meet and they'll do the same thing. But if that additional distance is exactly one half of a wavelenght, then this will go like that and then be overlapping with the next one and they'll just cancel out exactly and that's what gives us the dark spots. Alright now we can determine this approximately this difference in distance by drawing the following triangle right here. So we draw that down and we say the red and the green line are about the same distance and then we've got this little bit of excess from the bottom slit. So if we blow up this triangle what we see is that this delta r, this difference in distance is just equal to d times the sine of theta.

Alright now what's d? d is the distance between the 2 slits lower case d. What's theta? Theta is the angle that I need to go at to get to this distance y at my screen. Alright so this sine theta equals an integral number of wavelenghts so that's n equals 1, n equals 2, n equals 3, 4, 5 not n equals 2.7 okay so an integral number of d sine theta is n wavelenghts plus a half wavelenght now I've got destructive interference and that's going to give me a dark spot. So if we plot the intensity that we see, as a function of angle we'll get bright spot, dark spot, bright, dark, bright, dark. And it'll just go on like that basically uniformly until the angles get too big. Alright when the angles are small, it really is the same size and it just goes on. Alright so let's go ahead and do some problems with this. Not difficult to do but takes a little bit of work.

Alright so the first thing I want to say is, if the wavelenght is much, much smaller than this distance d between the 2 slits which is almost always the case especially with light experiment, because the wavelenght of visible light is really small. Then theta is going to be extremely small and so then we can use this wonderful approximation and say that the sine of theta is approximately y divided by the distance to the screen d. Alright so then that allows us to write y is approximately an integer multiple of the wavelenght divided by the distance between the 2 slits times the distance to the screen. Alright let's go ahead and look at this example, we've got 2 slits they're 0.1 millimeters apart so that's lower case d alright. Determine the distance from the central maximum to the next maximum so that means I'm looking for bright spots so that means I need n to be an integer. Right so we're going to take n equals 1, if the light has wavelength 530 nanometers, and the screen is 2 meters. So this is my lambda, this is my big distance and the word next here says n equals 1.

Alright so that's the idea, so all we need to do is determine y so y equals approximately we've got 1 and then it's 530 nanometers divided by now this distance d is 0.1 millimeters but I've got nanometers up here. So let's try to fix that, what we're going to do is we're going to express this first in terms of meters and then we'll express in terms of nanometers. So this is 0.1 millimeters so it's 10 to the minus 4 meters so that's 10 to the minus 4, 10 to the 9 nanometer. So it's 10 to the 9-4 is 5 nanometers alright so that's lambda over d and then we got times 2 meters. And if you go through this analysis, what you'll find is that it's 1.06 centimeters to the central maximum, to the next maximum from the central maximum. Alright what if we're looking to the first dark spot? Well the first dark spot appears when d sine theta is equal to half of the wavelenght. Because now we'll be destructively interfering alright so if I just cut the wavelenght in half that will give me the answer. So all I have to do is divide this by 2 so this answer will be 0.53 centimeters.

Alright let's do the next one, this is actually what the double slit experiment is used for now to measure wavelenghts. So now we're given the distance between bright bands. We're using some sort of laser and we can see the distance between the bright bands and we measure it as 0.7 centimeters. So this is y alright, so d, it gives us d it gives us capital D too and now we want to know the wavelenght lambda. Alright we're going to use again n equals 1 because we're looking at bright bands that are right next to each other. So that means that we went from one of them to the next n equals 1. Alright so we'll say lambda is equal to d times y over capital D just like that. Again I want to do everything in SI units alright so I'll change this 0.7 centimeters to 7 times 10 to the minus 3 that's y. I'll change this 0.2 millimeters to 2 times 10 to the minus 4 so that's my d and then I'll divide by capital d which is 3 and this will be in meters and if you do that Math you'll end up finding that it's 467 nanometers. So it's going to be bit of like a bluish, green aquamarine color. Alright more blue.

But anyway so this actually allows us to measure the wavelenght I mean this is a really short distance, you can't even see that really but I can measure that wavelenght by using much larger distances 0.7 centimeters and 0.2 millimeters using this idea of wave interference and that's the double slit experiment.