Tip:
Highlight text to annotate it
X
Let's set this up properly here, we have
again a wall with two slits
and the distance between two slits we are going to call d
and then way over here, we would have a screen
some distance L apart and we are interested in, from the center, at some point that's y away from the center
which also forms a certain angle between here
what is the magnitude of the interference pattern at that point
over here we have the laser beam coming through and so we have spherical waves coming through
we have one beam that comes like this, and one beam that goes like that
and if we are talking about interference, once again, we have to talk about the Δphase between the two of them
so reminder: we can look at kΔx or Δr in our case, ...
now in most cases, when we deal with double slit experiments, it's coming from the same light
the laser beam on this down here is the same beam of light
because it's going so fast, we are not really going to change one vs. the other
so this is usually zero and this is zero because it's from the same thing
the only thing that's different would be the Δx
and that Δx
is going to be
this distance right here
it's getting a little messy, so let me redraw that with that blown up a little bit
once again you have the wall here
this distance is d
we have one beam of light going that way
one beam go that way
hitting something way far down the line
and we will call that θ
couple approximations that we do, we first of all say
θ is small, once again
or that's to say that
...
...
if that's the case, that lets us say
these are roughly parallel
and these two angles are going to be roughly the same as the θ there
therefore
the only difference in your distances
would be this here
and that's approximately perpendicular as well
with that, that Δx is then going to be given by
... because doing a little geometry
this θ is the same as this θ here, so making this triangle
so for constructive
we recall, we need
...
...
...
...
...
and then destructive
...
and so rewriting that a little neater, looking at the constructive case
we have 2nπ interger multiple of 2π
...
...
... and of course, k we replace with ....
so swing that across, we have
...
...
and that's going to give us the θ at which we will get constructive interference
similarly
we can also do for destructive
we need odd multiples of π, so we have ...
...
...
...
remember though, this only works if θ is small, so your screen has to be nice and far away
By θ small, we need θ to be
say less than about 0.1 rad, about 6°, that would typically give you pretty good approximation
one more approximation that we do
if θ is small, we can further say that sinθ = θ = tanθ
and what is tanθ, well tanθ is
y/L right here, so we can directly relate
y and L with that so ...
therefore, we have
a further set of equation that says we have ...
...
so usually we would like to solve for y
so ...
...
...
...
similarly, this is for constructive
excuse me for the mess, and for destructive
we have ...
and that's where you have your
those are the y for your maximum fringe or bright fringe, and then your minimum or dark fringe
so noticing that the bright fringes are evenly spaced apart, they are going up by integer multiples
what you see on screen is something like this
back here, you have your double slit
and then comes along
your two slits, and then if you have your screen right here, what you would see is you have
at θ=0 you have a bright band, and then you have a bright band somewhere down here and then you have a bright band
and they are evenly spaced
looks kind of like a comb, in terms of intensity, it goes up and down, up and down, and that what we like to typically draw and see
in real life
you see something more like this, you still see that
there's the evenly spaced fringes, bright fringes and dark fringes, but there is something else going on, and that has to do
with the physical size of each individual slit and we'll look at that
in the lectures, but first let's do a couple examples on just the double slit by itself