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>> In this video, we're going
to take a closer look at
diffraction and interference.
We've looked at different
sorts of waves.
Waves on a string,
waves in the air, light waves,
and water waves.
All of these waves have
two things in common.
They bend around barriers,
and when they have overlap,
they combine into a new wave.
When a wave bends
around a barrier,
that's called diffraction.
And when waves mix together,
that's called interference.
All waves can diffract
and interfere.
One way to see diffraction
and interference in action
is to send a wave
through a narrow slit.
And look at the wave pattern
that appears on the other side.
When you send a wave through
a single slit onto a screen,
you get a pattern of
bright areas on the screen
called maxima.
That alternates with areas
of dark called minima.
Why does sending a wave
through a single slit
cause this pattern?
Diffraction allows the wave
to cover a much larger
area of the screen
than you would expect.
Interference occurs because
each point of a wave behaves
as a separate wave source.
This is known as
the Huygens Principle.
The yellow dots in this image
represent those individual
points called wavelets.
This means that these
independently behaving waves
will interact on the screen.
Some in phase with one another,
producing maxima,
and some out of phase,
producing minima.
The peaks and valleys
appear so regularly,
we can fit them
into an equation.
When the width of the slit
is much smaller than
the distance to the screen
and the wavelength of the wave
passing through the slit
is much smaller than
the width of the slit,
the wave doesn't bend much
after passing through
the single slit.
When this is the case,
we can approximate
the position of dark spots,
or minima, on the screen
using this equation.
Y sub min equals
L times lambda,
times m, divided by a.
Y sub min is the distance
on the screen from
the intensity minimum
being considered to
the center of the main peak.
L is the distance from
the slit to the screen.
Lambda is the wavelength
of the incoming wave,
and m is a whole number
called the order number,
telling which minimum
we are referring to.
M can be positive or negative
depending on whether
you are looking at minima
to the left or right
of the central maximum.
Finally, 'a' is the width
of the slit the wave
passes through.
This is the equation for minima
from a single slit.
On the other hand,
if our setup has two slits,
the equation for
intensity maxima,
rather than minima, is the same,
where y sub max
equals L times lambda,
times m, divided by b.
All the variables in this
equation are the same as those
in the single slit equation
except for b,
which is the distance
between the two slits.
We'll use both the single
and double slit relationships
to study wave effects in action.
For our first example, consider
the single slit set up.
Our wavelength is one meter.
The slit width is 150 meters.
And the backdrop is
2,300 meters behind the slit.
How far above the center
of the screen is the first
minimum point in the pattern?
We'll use the single slit
equation we looked at above
to answer this question.
The distance of the first
minimum point from
the center line, y sub min,
equals the distance from
the slit to the backdrop,
times the wavelength
divided by slit width,
all multiplied by m.
But what is m?
Since we're asked about
the distance of the first
minimum point,
we know that m is 1
in this case.
We're solving for y sub min.
Let's insert the values we know
for the other parameters.
L is the distance between
the slit and the screen.
That's given as 2,300 meters.
Lambda, the wavelength
of our wave is one meter.
And a, the width of the slit,
we're told, is 150 meters.
So, y sub min equals
2,300 meters times 1 meter,
times 1,
divided by 150 meters,
which equals 15.3 meters.
That means the first minimum
in the wave interference pattern
on the screen is 15.3 meters
above the pattern's center.
For an interference pattern
made with sound waves,
this minimum would
be a dead spot.
The absence of sound.
Now we'll look at an example
with a light wave.
But remember, it's the same
dynamics with any type of wave.
In this case, we have
two slits instead of one.
So we'll use the double slit
equation we looked at earlier.
Green laser light with
a wavelength of 545 nanometers
shines on a surface
with two slits spaced
at 0.5 millimeters apart.
The diffracted light waves mix
and land on a screen
behind the slits.
Looking at the pattern of
bright and dark spots created,
take a ruler and measure
the distance from
the center of the pattern
to the second bright spot
above it.
The distance is
6.0 millimeters.
What is the distance between
the slits and the screen?
We want to solve for
the distance, L.
So let's rearrange the equation
so it says L equals something.
Multiplying both sides
of the equation by b,
and dividing both sides
by m times lambda,
we see the equation now reads
L equals Y sub m, times b,
divided by m, times lambda.
Now if we can find out
and fill in the other
parts of the equation,
we'll be able to solve for L.
Y sub max is the
distance we measured,
the position of the second
bright spot above the center
of the pattern.
6.0 millimeters,
or 0.006 meters.
B is the distance in meters
between the two slits.
That's 0.5 millimeters,
or 5 times 10 to the
negative fourth meters.
Now, what's m?
Remember, m is a whole number
that tells us just which
bright spot we're working with.
The question stated that
we measured the second
bright spot above the
interference pattern center.
That means m equals positive 2
in this case.
And, finally,
the wavelength lambda,
we're given that value.
It's 545 nanometers,
which is 5.45 times 10 to
the negative seventh meters.
With all these numbers
figured out, we enter them
into the equation
to solve for L.
L equals 0.006 meters,
times 5.0, times 10 to
the negative fourth meters,
divided by the quantity 2,
times 5.45,
times 10 to the negative
seventh meters,
which equals 2.8 meters.
That's how far the screen
needs to be behind the slits,
in order for a pattern like
the one we measured to appear.
For this third
practice question,
let's keep our light waves
and double slit,
but switch up a few things
about the setup.
In this situation,
we shine a red laser with
a wavelength of 560 nanometers
on the two slits.
We don't know how far apart
the slits are from one another,
but we do know
they are 1.5 meters away
from the screen behind them.
We see another interference
pattern on the screen.
And this time,
measure the position of
the third dark spot
below the center point.
Measuring a distance
of 7.5 millimeters.
We want to figure out,
what's the distance
between the two slits?
Here's a start.
We'll rely on the same
relationship as in
the last example.
Knowing that, press pause
and take a minute
to see if you can
answer this question.
We'll start back
when you're ready.
Okay, did you find the
distance between the slits
is 0.28 millimeters?
If so, great work.
If not, let's solve it together.
Once we know we'll use
the double slit equation,
we want to figure out what
exactly we'd like to solve for.
This question asks for the
distance between the two slits.
That's b.
So we'll rearrange
the double slit equation
so it says b equals something.
Multiplying both sides by b
and dividing both sides
by y sub min,
we get b equals m
times L times lambda,
divided by y sub min.
Now, we'll want to figure out
all the variables on the
right side of the equation.
We are looking for the third
dark spot below the center.
The equation we have been
using for double slits,
has m as a positive
or negative integer,
plus or minus one,
plus or minus two,
and so-forth.
This is true for when
we are looking for the
distance to a bright spot.
But what if we have
or are looking for the distance
to a dark spot to a minimum?
In this case,
the third dark spot,
or third minimum
below the central maximum,
lies at a position where
m is equal to minus 2.5.
L, the distance between
the slits and the screen
is given to be 1.5 meters.
Lambda, the wavelength of the
red laser, is 560 nanometers,
or 5.6 times 10 to
the negative seven meters.
And y sub min, the position
of the third dark spot,
we measured to be
7.5 millimeters below
the central maximum,
or negative 0.0075 meters.
Let's insert all these values
into the equation for b.
So b equals minus 2.5,
times 1.5 meters, times 5.65,
times 10 to the negative
seventh meters,
all divided by
negative 0.0075 meters,
which equals 0.28 millimeters.
That's how far apart
the two slits are that would
create this particular
wave interference pattern.
That's a closer look at
diffraction and interference.
For more "Physics in Motion",
go to our homepage,
where the entire series
is available to you.
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