in this video we investigate the behavior of monochromatic light when it strikes two
thin slits as we've already seen huygens principle is useful for thinking about what happens to light
when it passes through a slit and i'll post a link back to that video real quick in the
huygens principle video we learned that when light passes through a thin slit it can be thought of as
a point source for circular waves in the animation we're viewing right now we see what happens when
wavefronts of light strike two thin slits each slit becomes a point source for circular waves and
those waves propagate outward from the slits in a very interesting way along some lines of sight we
have cress overlapping crests troughs overlapping troughs and these are the directions of
constructive interference between the two sources in between these we have troughs overlapping
crests and these are the directions of destructive interference we see an interference pattern
emerging along the red lines here here along the center and down here we have crests and troughs
that are landing on top of each other those waves reinforce each other and interfere constructively
along the green lines here here here and finally down here we have crests that are lined up right
on top of the trough from the other wave and vice versa that means the waves are going to interfere
destructively so when we project the pattern onto a screen we obtain an intensity pattern
given by this red curve i have a maximum at the very center because that's a direction where the
alignment is perfect and the waves reinforce each other i have a dead spot at this point right here
because that's a longer line of sight where crests land on top of troughs and vice versa and the
interference is destructive and again i get a bright spot along this line of sight where crests
are lined up with crests and troughs lined up with troughs and i have constructive interference
and so on so when you run laser light through two slits you end up seeing a series of dots if
you project it on the wall so i superimposed a photograph i took this morning using red laser
light and unfortunately it was so bright that my phone washed it out to make it look white but
this is what it looks like in real life you have kind of a blob in the center kind of a dead spot
where you're getting destructive interference and then another blob and then another dead spot
pattern is symmetric so the same thing is happening on the other side of the center
i would normally call the one in the middle the central maximum and then the ones that
are the first bright spots off the center i would call the first interference maximum
the two dark spots closest to the central maximum i'd call those the first interference minimum
and then the second dark spot relative to the center line would be the second interference
minimum and so on so to understand what's going on quantitatively i've drawn a couple pictures here
and in the top picture we're zoomed way out i can see that my two slits are very very close
together in this picture and the light from these two slits is aiming towards
the first interference maximum that's off the center so i know that that light is interfering
constructively so the light from the bottom slit clearly has a longer distance to travel here
and the reason the interference is constructive must be that it has exactly one extra wavelength
of distance to travel so that the waves arrive in phase of the screen and interfere constructively
so the path length difference for this first constructive interference must be equal to lambda
whatever the wavelength of the light is i notice in the first picture that i have a y labeled for
the distance between the central max and the first interference max i have the angle theta
relative to the center line for the experiment and i'm told the distance between the slits and
the screen is capital d but to really get into the math here we've got to zoom in on those
slits and that's what we're doing in the second picture i have a slit separation of little d
and it's critical that we're able to assume these two rays coming from the two slits
have the same angle and this will be very close to true provided those slits are really close
together compared to how far away the screen is just to give you a sense for it in the lab we're
normally dealing with a fraction of a millimeter between the slits and at least a couple meters
for the distance between the slits and the screen so the assumption that these rays come out at the
same angle is a very good approximation so again the key physical concept here is that
the bottom ray has a little bit farther to travel than the top ray to get to this
first interference maximum and if that path length difference is equal to one wavelength
the interference will be constructive so the math all comes down to figuring out
how to express that path length difference in terms of the parameters of the experiment
i notice that that path length difference is a leg in a right triangle i'm just going to label the
right angle on the other side here so it doesn't interfere with our drawing so that means the angle
right here is also right angle and the hypotenuse of that triangle is little d the slit separation
but i need to find the angle in this triangle well all it takes is a little bit of trig here i have
theta added to this angle right here giving me 90 degrees because i drew a perpendicular to that
ray in order to find the path length difference and that means this angle is 90 minus theta but
then that 90 minus theta plus the angle that we're interested in this very tiny angle in the triangle
that adds up to 90 degrees so the angle we're interested in plus 90 minus theta is equal to 90.
just hashing out the math real quick the angle we're interested in
plus 90 minus theta is equal to 90. subtract 90 from both sides add theta to both sides and
it turns out the angle we're interested in is also theta and again the path length difference
is given by the opposite side of that right triangle so the side opposite theta and d is
the hypotenuse so my path length difference is going to be d sine theta and i know that
i get a constructive interference if that path length difference is equal to one wavelength
so that the waves arrive in phase at the screen so that's my first constructive interference that's
off the center i also get a constructive right at the center because there is no path length
difference for the rays traveling from the two slits if you're right on the center line
all right of course we want to generalize so i want to figure out how do i express the conditions
for my second constructive interference well that happens if the path length difference is equal to
twice the wavelength so the waves would still arrive in phase and interfere constructively at
the screen and surely this continues for a path length difference of three wavelengths or four or
so i can write down my general formula for the angles that give me constructive interference
so i get d sine theta equals n lambda where n is an integer and i'm going to include 0 for
the central maximum and i'm not going to bother to put a plus or minus on each of these numbers
but i know that these interference maxima happens symmetrically in both directions now there's a
useful modification to make to this formula that's especially convenient when you're in the lab what
you would normally measure is actually y that's the distance between the central interference
maximum and the first one off of that it's the same as the distance between any two adjacent
maxima that's something you can measure with a ruler in the lab so i want to re-express my
formula in terms of y and we're going to use an approximation for this and the approximation is
that sine theta is approximately equal to the tangent of theta provided theta is very small
and we definitely qualify in this case because we're ordinarily talking about fractions of
degrees for these angles well if you buy that then i could replace sine theta with i can see
in the diagram tangent theta is y over d and this allows me to directly talk about
the distance between the center line and the location of these different interference maxima
so i can rework my formula like this and now i've got something that's probably more useful in the
lab the last thing i want to do on this slide is figure out well what are the angles that get
me to these dark spots where i have destructive interference that happens if the waves arrive
out of phase i've already got a general expression for the path length difference from the two slits
based on the angle and so what happens at the first destructive interference is that the
waves arrive out of phase because the path length difference is equal to one half of a wavelength
so i write down path length difference again i can express that as d sine theta equals whatever
the wavelength of the light is over two so this means when a crest from one wave is arriving at
that point on the screen a trough from the other wave is arriving and they interfere destructively
so anything that gets them out of phase by half a wavelength causes destructive interference it
doesn't have to be just one half of a wavelength it could be a half of a wavelength plus a
whole wavelength in other words one and a half wavelengths out of phase and that would be our
second destructive so that happens again at three lambda over two or one and a half wavelengths
out of phase so to generalize this i have my path length difference d sine theta is equal to n plus
a half times lambda where n is equal to 0 1 2 and so on so just plugging in n equals 0 there i would
get a path length difference of a half wavelength that's my first destructive interference plug in
n equals one and i get a path length difference of three lambda over two that's my second destructive
interference and so on finally we can rework this in terms of the actual distances we measure on the
screen those y values by using sine theta is approximately tangent theta which is y
over d so that's pretty much it for the analysis of two slit interference let's wrap things up with
a quick example so in this one we're given the wavelength of the light 633 nanometer red laser
light passing through a pair of slits separated by 0.15 millimeters and the interference pattern
is projected on a wall that's 5 meters distant i want to compute the angle and the approximate
distance in other words the y value between the central maximum and the first and second maxima
and then i want to get the angle and approximate distance between the central maximum and the first
and second minima so really we're just plugging into our formulas to practice a little bit
so first we'll do the constructive interferences so i have a formula giving those locations
as an angle and then a formula giving those locations as a distance from
the center line we'll do our first maximum and i plug in that slit separation of 0.15 millimeters
i've got to put that in the right units and that first maximum occurs when the
two waves are out of phase by exactly one wavelength in other words it's the n equals one
maximum and lambda is 633 nanometers so 633 times 10 to the negative nine so we just divide both
sides by .00015 and invert the sine function make sure you're set in degrees here and i get
0.242 degrees just keeping three sig figs on this alternatively i can solve for the y value
corresponding to that angle now either you can take that angle and just go do the trig from
scratch or i can use my second formula here if i solve that second formula for y that's n lambda
big d the distance to the screen over little d the slit separation and we're talking again about n
equals one in this case the first maximum and this gives me point zero two one one meters it'd be
nicer to write that as two point one centimeters so you'll see that first off center bright spot
about two centimeters away from the center bright spot then we do all the same things for the second
interference maximum the only thing that changes is that we plug in n equals 2 which means our
path length difference is equal to 2 wavelengths of the light and i get an angle of 0.484 degrees
you'll notice that's just twice the angle for the first one and this is true because our angular
changes are so small that we're in a regime where the sine function is practically linear
and then just streamlining the calculation again to get the y value for the second maximum i just
replace the n with a 2 in the formula so i have a 2 times 6 33 times 10 to the negative 9 times
5 all divided by that slit separation 0.00015 and again i end up with twice the value i got before
and that's 4.22 centimeters in this case to the second bright fringe off-center next we handle
our destructive interferences in other words the dark spots that separate the interference
maxima and i have my formula in terms of angle d sine theta equals n plus a half times lambda
where n is equal to zero for the first one one for the second one and so on
or again replacing sine theta with approximately y over d
so to get the angle for my first destructive interference i just plug in n equals zero
into three sig figs i get .121 degrees which you'll notice is half our original angle for the
first maximum in other words this minimum just occurs halfway between the central max and the
first off center max all right then i apply the second formula because i'm interested in finding
y for this dark spot i'll go ahead and solve for y in general that's just big d times n plus a half
times lambda all divided by little d and again we're plugging in n equals zero for the first
interference minimum and i get .0106 meters but it's nicer to look at it in centimeters that's
1.06 centimeters off center again if i compare to where my first maximum was that's about halfway
between the center line and the first maximum i get that dark spot that's 1.06 centimeters
all right then we get our second one and i'm running out of space so i have to streamline
a little bit here all we do is plug in n equals one and that gives me a three halves
in this part of the formula so i get .363 degrees for the angular location of this second dark spot
then when i try to find the actual distance between the center line and the second dark spot
i use this formula where i've solved for y and i'm going to plug in n equals 1 giving
me a 3 halves in those parentheses so i have d that's 5 times 3 halves times the wavelength
all divided by the slit separation in the appropriate units
and writing it in centimeters i get 3.17 centimeters off center for that second dark line
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