This is the first lecture in a two lecture series on the double slit diffraction equation. In this
first lecture I'm going to be proving the equation that predicts the double slit diffraction pattern,
and in the second video I'll be using the equation to solve some problems and give you some examples
of how to use it. There's a specific way of observing the effects of diffraction in light
by passing the light through a barrier with two slits cut out and observing the pattern the light
forms on a second barrier beyond the first. The pattern shows clear maxima, which are points of
constructive interference, and minima, which are points of destructive interference. So when light
passes through a material with two small openings cut through it, a diffraction pattern appears
where the light spreads out in all directions and the two waves of light coming in from the
two slits interact with each other in certain ways. There are certain points where those waves
constructively interfere to create maxima and certain places where they destructively interfere
to create minima, and if we place a barrier a certain distance away from those two openings
and observe how the light looks on the barrier, we'll see the pattern that I put on the right
where you can see there are places where there's a lot of constructive interference where the light
is very bright, and then between those places are points of destructive interference. So you've got
these points of light separated by points of darkness where the constructive and destructive
interference is happening respectively. So in this video we're going to talk about how to
predict exactly how this pattern will look, but before we talk about that we need to talk about
the idea of path difference of waves. The path difference is just the difference in the length
of the total path of one of the waves of light and the other. You'll remember that constructive
interference occurs when two waves are perfectly in phase and destructive interference occurs when
two waves are perfectly out of phase, so if the crests of one wave line up with the crests of
another the two waves will enter into constructive interference with each other to create a single
overall larger wave, and if the crest of one wave line up with the troughs of the other wave,
that's when destructive interference occurs, and the waves cancel each other out. We can also talk
about constructive and destructive interference in terms of the path difference of waves. I'm
going to show you how that works. You can see here as an example that these two waves are in
constructive interference. I'm going to play that one more time. Because they're lined up
in such a way when they're hitting the wall that their crests line up and their troughs line up,
these two waves are in constructive interference, and if I look at the difference in their path I
can see that the difference in their path is exactly 1 full wavelength of the wave. 1 wave
is traveling for 1 full wavelength longer than the other. So because the difference in their
paths is exactly one full wavelength, they line up perfectly to create this constructive interference
pattern. We can now try to figure out where the next constructive interference pattern will occur
if we change the path difference. So if we make this path longer I can see that the next place
the perfectly constructive interference will occur right here is when the path difference is exactly
2 wavelengths long, and if I do this one more time I can see that there's one more point that
I can find where the constructive interfere and that's where the path difference is three
wavelengths long so this is proving the rule that says that constructive interference will occur if
the path difference is exactly one integer (one regular number) multiplied by the wavelength of
light or n times lambda (n being an integer and lambda being the wavelength) so the only time
constructive interference occurs here is when you have a whole number multiplied by the wavelength
of path difference. As long as you have that, as long as you have complete wave lengths of
path difference, constructive interference is going to occur, so that's where that rule comes
from. I can now line these waves up to create perfectly destructive interference. You can see
that the crest of one wave perfectly lines up with the trough of the other, so that results
in perfectly destructive interference, and this is going to occur if the path difference is exactly
an integer plus 0.5 multiplied by the wavelength. So here I can see that this path difference is
exactly half of one full wavelength of the wave, and that's producing destructive interfere. If
I extend this to the next point where perfectly destructive interference occurs, I can see that
the new path length is 1.5 times the wavelength, and if I do that again to the next point where
perfectly destructive interference occurs I can see that that's 2.5 times the wavelength. So the
pattern here is that destructive interference will occur for every integer plus 0.5 multiplied by the
wavelength. So just looking back here, first integer I'm using is 0 so 0 plus 0.5 is just
0.5 times lambda, and then the next integer is 1, so 1 plus 0.5 is 1.5 lambda, then going to 2,
so the next point of destructive interference occurs at 2.5 times lambda. So if the waves are
otherwise perfectly the same, the path difference can tell you whether constructive or destructive
interference will occur. In an actual double slit experiment the two slits are extremely close and
the wall the light is projected on is very far away. For the proof of this equation I'm going to
exaggerate the distance between the slits so the proof is easier to visualize. So again the goal
of this equation is to try to predict exactly when constructive interference will occur on the wall,
and when that happens a dot of light appears. When perfectly destructive interference occurs no light
appears. So in an actual experiment you would see the pattern on the left, and just to make
that a little bit clearer I've made a diagram of what's going on there on the right. Okay so now
let's come up with an equation to predict exactly where each point of constructive interference
will occur. Constructive interference is going to occur whenever the path difference is an integer
multiplied by the wavelength. So we can try to figure out different points where that will be
true, and one point where that will definitely be true is the point on the wall that's in the exact
center of the two sources of light, because at that point the path difference is just 0.
They've both taken the exact same length of path to get to that one point, so if there's no path
difference constructive interference is going to occur because that's like using the integer 0 in
the constructive interference equation, and so a point of constructive interference will always
appear at the exact midpoint of the two slits in the double slit experiment. So now we need
to figure out where the next point of constructive interference will occur and how far away that will
be from the center point. That's the goal of this video: to understand exactly where those
points will appear. Now that we have that first point of constructive interference the goal of
this video is to understand where the next point of constructive interference is going to occur.
That's what the equation does that we're looking to get. So before we do that I'm just going to
define some terms. I'm going to call the distance from the diffraction grating to the wall capital D
I'm going to call the distance between the two slits lowercase d, and I'm going to call the
distance between two points of constructive interference S. So to understand where the
next point of constructive interference is going to occur, we need to understand the
path difference between these two lines, like how much longer is one line in comparison to another,
and one way of finding the path difference is to draw an isosceles triangle over the two paths,
because two sides of an isosceles triangle are the same. So if I superimpose this triangle over
my system and draw this new green line down here, marking off the edge of that triangle everything
past that green line is going to be the path difference between the two beams of light. So
in order for that point on the wall to be a point of constructive interference, that path difference
length has to be equal to an integer multiplied by the wavelength of light. So again just a reminder
the reason why we want that path difference to be an integer times lambda is because that is when
constructive interference occurs. So as long as that path difference is an integer times lambda,
constructive interference will occur on the wall, and we will find our next point of constructive
interference. I'm going to draw this blue line in the middle just to show the midpoint of the two
beams of light. That will come in handy later. And just to visualize what's happening here,
if I imagine this beam of light going toward the wall and I can see I've made it so that the path
difference from one beam to the other is exactly one full wavelength, like that green line is
marking off exactly one full wavelength of light, perfectly constructive interference occurs at the
point of the wall. You can see that two crests are occurring at the exact point of the wall,
leading to constructive interference. Let's see if we can find more relationships between these
variables. In an actual double slit experiment, the slits are extremely close together compared
to the distance to the wall, so because they're extremely close together that means that the two
rays leave the slits at roughly the same angle, because they only need to angle toward each
other a tiny little bit over that huge expanse of space in order to eventually meet up at a point
on the wall. So for this part of the proof we're basically going to pretend that those two rays
have the exact same angle, and if the two rays are the same angle the line marking their path
difference to the wall makes a 90 degree angle with them. This is because if these two things are
effectively the same, we could theoretically just kind of rotate this down like this, because again
they have the same angle compared to each other, so they could also have an angle of 0, and at 0 it
becomes very clear that that green line is making a 90-degree angle with the rays, so we can say
that the green line is making an approximately 90 degree angle with the two rays. I'm going to use
the angle on the top of this resulting triangle to highlight a relationship where I can see that
the path difference is the opposite side of the right triangle and the space between the slits
is the hypotenuse. So I can write an equation that says that sine of that theta, that angle,
is equal to the path difference n times lambda over the distance between the slits D. So I'm
going to make a note of that and use that later on in this proof. There's a way of proving
geometrically that the blue line that I've drawn has the same angle as the green line. I'm going to
skip that part of the proof right now just because I do want to get to the main part of the equation,
but we can assume that those two angles there are the same. And you'll notice that that blue line is
actually the hypotenuse of a right triangle made with s and D as the two bases. You'll notice that
the blue line is actually the hypotenuse of a right triangle made with sides s and D as well,
so I can define tangent of theta as being equal to s over capital D because the difference in the
angle between the two beams of light is actually extremely small. The angle in the experiment is
also extremely small. This is important because for extremely small angles sine is approximately
equal to tangent because cosine for very small angles is roughly equal to 1, and tangent is
equal to sine over cosine, which here will be about the same as sine over 1. So we can say
that sine and tangent are roughly equal for these very small angles which we're dealing with in this
experiment. So that means that the two different equations that we got are actually equal to each
other as well, so we can connect them like this and get a single relationship for the distance
between constructive interference that has nothing to do with the angle. So what this equation is
saying is that the distance between points of constructive interference is equal to the
distance from the slits to the wall multiplied by an integer (which just tells you how many points
of constructive interference you are away from the center) multiplied by the wavelength of light
divided by the distance between the two slits in the double slit experiment. That's how you
prove the double slit equation. In the next video I'll show you how to use it to solve problems.