let's talk about single slit
interference now if I were you I'd
already be upset and a little mad single
slit interference interference wave
interference is by definition multiple
waves overlapping at a single point so
how could a single slit ever produce
multiple waves that could overlap I mean
when we had a double slit if I put a
barrier in here and we have a double
slit at least then okay I send in my
wave it gets over to here there's a
small hole we know what waves do at a
small hole they diffract which is to say
they spread out at least with a double
slit you would have two waves spreading
out now they can overlap interference
but for a single slit how are we ever
going to get this well I never really
told you what why do the wave spread out
at a hole why does the fraction happen
at all why when waves encounter a hole
do they spread out and the answer to
this question is the key to single slit
interference and the the answer to why
they spread out at a hole is something
called Huygens principle and I can't say
it this is a Dutch physicist scientist
who figured this out Huygens principle
and I apologize right now to all the
Dutch people out there I'm butchering
this name Huygens principle easier to
spell than to say what he said he
figured out something ingenious he
figured out this if you've got a wave
coming in these wave fronts number these
wave fronts are like Peaks and in
between them are the troughs or the
valleys if you've got a wave front
coming in propagating this way you could
say yeah that wave front moves from here
to there that's what it does or he
realized with a wave you could treat
every point on this wave as a source of
another wave that spreads out
spherically if in the forwards direction
this wave spreads out spherically this
point here he said that a wave front you
can think of as an infinite source of
waves each point is the source of
another wave and you're thinking this is
horribly complicated what what kind of
messes
is going to give you well if you add
this up these are going to interfere
with each other constructive destructive
in a way that just gives you this same
wavefront right back this is crazy but
true if you let every point on this way
be another wave source it will just add
up to another wavefront here you'll just
get the same thing back and this is the
key to understanding why diffraction
happens is because the wave spread the
wave was already diffracting so to speak
it was already doing diffraction every
point on here was doing diffraction it's
just it always added up with the other
waves around it at every other point and
gave you the same wave back
but when there's a barrier when there's
something in the way these here can't
rejoin up with their buddies you just
get this one here spreading you know and
then this one down here spreads out all
the rest of these get blocked now that
these are blocked they're not going to
get to interfere constructively and
destructively with these points here and
so what do you see when it hits the hole
you just see this thing spreading out so
it was always diffracting so to speak we
just didn't notice it because it always
added up when you've got a hole or a
barrier that's when we actually notice
it and this is the key to single slit
interference because if I get rid of all
that if we imagine our wave coming in
here like this
well this waves going to hit here every
point it's the source of another wave so
this point is going to start spreading
out this point is going to start
spreading out when we have a single slit
we really have infinitely many sources
of waves here and since some of them are
blocked
we could see an interference pattern
over here on the wall because these can
interact and interfere with each other
what interference pattern are we going
to see well on the wall over here we see
a big ol bright spot right in the middle
and if I were guessing I would have
thought that was it
big old bright spot because you shine a
light through a small hole single hole
you'll get a big bright spot there the
weird thing is this jumps back up goes
to a minimum
zero point then it jumps back up and
then it comes back up again and you get
this these are going to be not very
pronounced these aren't very pronounced
you get a big bright spot in the middle
these are relatively weak compared to
other interference patterns that we've
looked at and down here jumps up a
little bit again over and over here
so this is the pattern you see how can
we get this how do we analyze it that's
we're going to try to figure out to
figure that out okay well this is a I
said there's infinitely many sources
here with when this wave gets to here
that would take a long time to draw I'm
going to draw eight so let's say there's
one two eight sources let's just imagine
there's eight here to make this a little
bit easier to think about and the weird
part is that this jumps back up here so
let's look at this minimum right here
let's look at this point where it goes
to zero this destructive point so the
wave from this topmost point this way
from the topmost uppermost point has to
travel a certain distance to get there
I'm going to also look at the fifth one
down this one that's basically halfway
how about these two if these two
interfere destructively the argument I'm
going to make is if these two interfere
destructively all the rest of them are
going to have to interfere destructively
why well
we not play this game let's draw our
right angle line here there we go and so
we know that okay if these are going to
interfere destructively this is the
extra path length this extra path length
this second wave this lower middle wave
has to travel has to be what if I want
destructive over here it's got to be a
half wavelength three halves wavelength
5 halves wavelength that's how much it's
got to be in order to be destructive
so if this is the first point let's just
say that's one half of a wavelength and
what's the relationship between the
angle that this is at on the wall
compared to the center line well we
already figured that out remember that
relationship was d sine theta equals the
path length difference between these
that we derived the screen had to be
very far away compared
the width of the hole but that that
relationship still applies what would D
be in this case now we have to be pretty
careful you have to be careful because
this hole has a certain width we'll call
that with W so if this hole has a
certain width W how far apart are these
these are not W apart these are W over
two apart and so what's the relationship
here for the path length difference
between these two well if they're W over
two apart
I'd have d sine-theta the path length
difference so D would be W over two
times sine of the angle that this makes
to this point on the wall and if their
path length difference is lambda over
two then that would be destructive so
equals lambda over two and this is a
little weird already because look I can
cancel off the twos and what do I get I
get the W the width of the entire width
of the slit times sine of theta equals
lambda this is giving me destructive
remember before all the points that were
integer wavelengths were giving me
constructive this time it's giving me a
destructive point over here and the
reason is we played this game where W is
the whole width these are only W over
two apart that two cancels with that -
okay but I really proved that this whole
that they should all be destructive yet
this is just for these two we've got
infinitely many more in here how are we
ever going to show that if these two
cancel the rest of them cancel well
we'll just pair them off look at this
now imagine you come down one I go to
this one I consider this wave that makes
it over to here and the next wave down
from this other one here so I move this
one down a smidge and I move this one
down a smidgen and I imagine these two
waves traveling a certain distance to
get over to this point what relationship
holds between these two I can do the
same thing these are also W over two
apart so this here
is also W over two so I get the same
relationship I get W over two sign up is
that going to be the same angle yeah
it's the same angle same point on the
wall this is really far away so these
approximations are whole where these
lines are supposed to be approximately
parallel because this screen or this
walls very far away compared to the
width that equals well that's going to
be the same thing I got W over two times
sine of the same angle shoot that's got
to equal the same thing that it did up
here if the angle is the same my W over
two is the same that's also going to
equal half a wavelength that's also
going to be destructive these two will
also interfere destructively and I could
keep playing this game I can pick this
point here over to here and the next one
down these two would have to be
destructive I can pair them off and keep
pairing them off I get destructive for
all of them
I can annihilate all of them by pairing
them off and finding a partner that's
destructive to that one and so this
really is a destructive point this point
over here all the light is gone
completely annihilated gives you
destructive so the short of it is that
this relationship here this relationship
that W the slit width times sine of
theta the angle same angle we've always
been defining it as equals integer
wavelengths this time got to be careful
though this time this gives you
destructive points not the constructive
points it was always constructive before
this gives you destructive points now
and you might be upset you might say
hold on a minute we only proved this for
this was just for N equals 1 or M equals
one one way funny even prove this for
anything besides M equals one well you
can just as easily show that 3 lambda
over 2 would also give destructive or 5
lambda over 2 that would give us all the
odd integers here so M and here can be
it can't be zero we'll talk about that
in a minute it could be 1 2 3 4 5 and so
on 1 we already showed 3 you get well if
you made this three-halves wave
that's also destructive that b3
five-halves wavelength the twos are
always canceling so five halves light
wavelength would work what about the
even integers how do we get these well
those come from the fact that I didn't
have to pair these off with the top one
and the middle one that's dividing this
into W over two so pairing them off by
lengths of W over two I can pair them
off I can divide this by any even
integer
I can imagine pairing off instead of
doing the topmost one in the middle one
I can do the topmost one and skip one
down here and so I can pair these off if
I divide this into this distance right
here that distance would be what that be
W over four and so I can imagine pairing
off okay if these two cancel if those
two points cancel then the next one down
so this one here and this one here would
also cancel by the same reasoning and so
I can play the same game now but W over
four would be how I divided I can't
divide it by anything
I can't divide it by three or like two
point five because I always want to pair
these off in twos always twos thats it
my whole plan that's my whole strategy
here is to cancel these in twos and I
can do that by dividing this by any even
integer so W over four would work what
would that give us okay W before the
distance between these times sine theta
equals this just say is the first
one-half of a wavelength well if I solve
this if I move the four over I get W
sine theta equals to lambda so the twos
also give us destructive interference I
could divide by eight that would give us
four once I move it over I can divide by
any even integer any integer here is
going to give us destructive points on
the wall so this would be M equals one
this would be M equals two and so on
upwards so this relationship right here
gives you all the destructive points
how come n equals 0 is not a destructive
point well M equals zero is right in the
middle that's the most constructive
points the brightest spot so M equals
zero is not a destructive point but any
other integer does give you a
destructive point so this is the formula
for the destructive points W is the
entire width of the single slit theta is
the angle the way we normally measure
angle here you imagine a center line
like that imagine a line up to your
point on the wall this angle here would
be theta and M is any integer that's not
zero lambda is the wavelength of the
actual light that you're sending in here
now this just gives you the destructive
points you might wonder hey I'm clever
if the integers are giving us
destructive points then the half
integers should give us the constructive
points if W sine theta equals you know
lambda over 2 or 3 lambda over 2 is this
going to give us constructive points and
a not really so there are some
complications here and if you're
interested in why this does not give the
constructive points I'm going to make
another video watch that one because if
you've been paying close attention you
should be upset about something else too
you should be upset about something
earlier I've said might make it seem
like we can prove this does not happen
with the diffraction grading if you were
paying close attention we proved
quote-unquote that these do not occur
and if you're upset by any of that or
you want to know why the constructive
formula does not exactly give you
constructive points watch that video if
you're happy with what we do know that
this gives you the destructive points on
the wall then you're good