in this video we're computing the
interference and diffraction minima
and maxima for 532 nanometer green laser
light
passing through a pair of slits with a
separation of 0.15 millimeters
with that interference pattern projected
onto a screen three meters away from the
slits
so this big d in the diagram that's
three meters
and really what we're doing here is just
quoting the formulas that we've already
derived
for interference minimum maxima and
diffraction minima and i'll just post
links to those videos
as they come up in part a we're asked
for the location of the first
interference
minimum relative to the central bright
spot on the screen
so the way this formula turned out is d
sine theta
is equal to n plus a half times the
wavelength lambda where n is 0 1
2 and so on that little d is the slit
separation so that's 0.15 millimeters
for us
and we're plugging in n equals zero that
gives us the first minimum
that's where the path length difference
for the two rays
arriving at that point on the screen is
equal to one half of a wavelength
so we're plugging in n equals zero and
we're going to make an approximation
that gives us this value
y the distance between the center line
and that first interference minimum
the approximation we make for small
angles is that sine theta is
approximately equal to tangent theta
which is just y over big d making that
substitution
i get little d y over big d equals one
half lambda
solving for y i get big d lambda over
two little d
now we just substitute the values for
all of these things and get everything
in the right units of meters as we go
nanometers are 10 to the negative nine
meters
point one five millimeters is point zero
zero zero one five
meters and i get point zero zero five
three two meters which i could write as
5.32 millimeters
in part b i want the location of the
first interference
maximum relative to the central bright
spot on the screen
and the way this formula turned out and
again i'll post a link to it as i go
because d sine theta is equal to n
lambda
where n is 0 1 2 and so on now n equals
0 corresponds to an angle of theta
equals 0 here in other words that's the
central maximum that's not what we're
interested in we're interested in the n
equals 1
maximum the first one is off the center
line
and again we make the small angle
approximation the sine theta is
approximately tangent theta which is y
over big d
so we get little d y over big d equals
lambda and we solve for y again
and we sub in the values for all of
these variables
and to get .0106
meters which could be written as 10.6
millimeters finally in part c
we're asked to consider the width of the
slits themselves
each slit has to have some finite width
and we already know that
a slit with a finite width is going to
produce a diffraction pattern
so whenever we look at interference in
the lab we're actually looking at
interference and diffraction
at the same time and we're asked for the
location of the first diffraction
minimum so where the slit width itself
is causing the intensity of the light to
go to zero
this diffraction minimum is going to
suppress the interference pattern
and again i'll just post a link to the
video where this was derived
it turned out that our diffraction
minima occur where a sine theta
that's where a is equal to the slit
width is equal to n lambda
where n is 1 2 3 and so on so we just
plug in the slit width
and again we make the small angle
approximation sine theta is
approximately tangent theta which is y
over d
and we sub in n equals 1 for the first
diffraction minimum
off the center line and we get y equals
lambda d
over the slit with a sub everything in
and the right units
i just tacked on three zeros to turn the
millimeters into meters
into three sig figs i get point zero two
zero
zero meters which i could write as
twenty point zero millimeters
so this is the first point where the
diffraction pattern
is going to suppress the interference
pattern to zero and we'll see our series
of bright dots for interference
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