in this video we're looking at one slit diffraction for a light wave so on the
left we have a bunch of parallel wave fronts coming into this slit
i've shown the slit as rather wide for purposes of illustration and the key to understanding
what's going on here is to use huygens principle and i'll post a link back to the video where
we introduced huygens principle real quick huygens principle says that any wavefront can
be represented as a bunch of superimposed point sources so that's what i've shown inside the slit
that little piece of the wavefront that survives to pass through the slit can be represented as a
bunch of little point sources stacked on each other each of those emitting a circular wave
in the forward direction so this creates this whole interference pattern of semicircular waves
each set of those concentric about one of these little point sources
well all those waves start to interfere and where their crests align with each other
and their troughs align with each other you end up with constructive interference
where you have a mix of crests on top of troughs you end up with destructive interference and
in this diagram we only had enough space to capture what's called the central maximum for
the diffraction pattern so mathematically that's represented by this red curve
my maximum intensity is right along the center line and then it dies off and i end up with a
dead spot right here and a dead spot right here now this pattern actually continues
farther out into space and you'll get many many diffraction maxima and minima off to the sides
so what's going on here along the center line is that if i look at all these little circular waves
created by my point sources their crests are overlapping crests almost perfectly on the center
line the alignment starts to get off as i get away from that center line so the intensity of light
that i'm seeing is diminishing as i get away from that center line by the time i get to these green
lines this upper one here and the lower one here i just have this terrible mix of wave crests sitting
on top of troughs and vice versa and on average i'm getting destructive interference and i get my
first dark spot on the screen so this blob on the right is an actual photograph i took this morning
of a diffraction pattern from passing red laser light through a single slit it was so bright in
the center that my phone just interpreted it as white but it's really a bright red
and you can see that brightness diminishing and reaching some kind of minimum value
right about here and right about here where the interference of all these circular wave fronts
becomes destructive now at even greater angles you're going to get constructive
interference again but it's only partial none of them are as bright as that central blob
so our interference maxima aside from this one right down the center are never going
to be perfect maxima they're always partial constructive interference and they diminish
in intensity pretty fast as you get away from the center line however we are able to calculate the
exact angles of the interference minima that's where you get perfect destructive interference
between all the semi-circular waves emanating from these point sources so to get a handle on things
we look first at this very zoomed out picture on top and i have a bunch of rays emanating from the
slit each of those has a little point source as its origin and if i look off in this direction
theta i'm pointing to the first diffraction minimum off the center line now i showed that
as having a distance from the center line of y and the screen is shown to have a distance of
capital d from the slit to figure out what's going on we need to zoom in so that's what we did in the
second picture we've zoomed in on the slit all of these rays represent light originating from the
different point sources that are superimposed to create the wave front that's in that slit the slit
has a width of a it's important that we're able to assume all of these rays have the same angle theta
with respect to the center line and as long as the distance to the screen is very large compared
to the width of the slit this is a really good approximation and what i've done in the picture
is label the path length difference between two of these rays i have one right at the top of the slit
and one at the halfway point along the slit something very special happens if that path
length difference is half a wavelength for the light that we're using in this case
the top ray and the one in the middle the two that are labeled as green in this picture are going to
interfere destructively when they arrive at the screen they're off by half a wavelength so a crest
from one will overlap a trough from the other one and they'll cancel each other out so you can
think of those two as just being gone now but look what happens if i take a ray second from the top
and then one below the middle ray those have exactly the same separation distance
i get exactly the same path length difference the lower one having to travel a half wavelength
longer than the upper one and those two cancel out then i move on to the next one
second from the top second from the middle those have exactly the same separation
distance therefore they have exactly the same path length difference the bottom one
having to travel a half wavelength farther than the one on top and those two cancel
so in this special case where taking half the slit width in other words a over 2 at this angle theta
gives me a path length difference equal to one half of a wavelength all of my little wavelets
cancel pairwise covering the entire slit and i know that i'm going to get a dark spot
in my diffraction pattern well all we have to do now is express that path length difference in
terms of the parameters of the problem and i had to get my triangle really small to fit everything
into the picture here but there's the path length difference i'm making it white right now
a over 2 is the hypotenuse of this tiny little right triangle and i still need the angle of it
right here so let's just redraw the triangle so we can zoom in a little bit to get the trig done and
again the lower ray has just a little bit farther to travel before it gets to the screen that's the
path length difference if that's equal to one half of the wavelength i know i'm going to get a dark
spot on the screen i need to put in a reference to the horizontal here and a reference to the angle
theta so theta adds up with this angle to give me 90 degrees this is a 90 degree angle right here
and that means this angle is 90 minus theta but that 90 minus theta adds up with the angle we're
interested in to give me 90 degrees like this so 90 minus theta plus the angle of interest
is going to be 90 degrees i subtract 90 from both sides add theta to both sides and i find out the
angle of interest is actually the same as that angle theta that's the angle between the center
line for our experiment and the location of the first diffraction minimum all right
so that's actually theta in there a over 2 is the hypotenuse of this triangle and that means
a over 2 times the sine of theta that gives me the opposite side here that's the path length
difference so we arrive at a formula for our first diffraction minimum and that's a over 2 sine theta
that's equal to the path length difference and we get a minimum if that's one half of a wavelength
and i can clean that up by multiplying by two and i get a sine theta equals lambda well what about
the second one where this comes from is just using a similar trick to what we already did
we said if the path length difference between two rays that are separated by half the width of
the slit is one half of a wavelength then they can cancel pairwise well the next steepest angle where
a pairwise cancellation becomes possible is when the slit is cut into four pieces instead of two
so we'll go back to our original picture and we'll say suppose i look at this pair
of rays i'll go ahead and recolor the second one that i'm focused on here and those have
a path length difference of lambda over two then those two will cancel out the two below them will
cancel out and so on and when you're done doing that cancellation you end up with half the slit
deleted those have all interfered destructively then i moved to the lower half of the slit
and the separation between this ray and one that's below it by distance of a over four
would also have a path length difference of lambda over two those two would interfere destructively
and then if i pick the rays below each of those two those interfere destructively and so on until
i've covered the lower half of the slit so if the slit can be evenly divided into four pieces
and you get a path length difference of lambda over 2 across that a over 4
distance then again we can pairwise destructively interfere all these rays and i get a dark spot on
the screen now for the same wavelength of light and the same slit width it takes a bigger angle
to get that path length difference over a quarter of the width of the slit but there's no need to
draw an elaborate second picture i know i get my second diffraction min when a over 4 sine theta
so we do the same trigonometry just with an a over 4 right here instead of an a over 2.
so when a over 4 sine theta the path length difference is equal to 1 half of a wavelength
and then we're able to do twice as much pairwise annihilation of these rays
to completely delete all the light from that slit cleaning things up a little bit i get a sine theta
equals two lambda the next minimum occurs if we can divide the slit evenly into six pieces
so that two rays separated by a over six are out of phase by lambda over two
and then you can completely cover the slit once again with pairwise annihilating rays
so to generalize here if i have the a over six case that would just give me a 3 lambda on the
right hand side to generalize i see the pattern here i just have n lambda on the right hand side
and here's my formula for diffraction minima with a slit width of a i see dark spots at
these angles where a sine theta is equal to an integer number of wavelengths for whatever the
light is that we're using now there's a useful way of reworking this in terms of the parameters
of the experiment what we usually measure in the lab is actually this y value the actual
distance between the center of the pattern and then the first dark spot second dark spot
and so on so we have to use an approximation here that as long as the angle is very small
and typically we're dealing with fractions of degrees here so that's definitely true
the sine of theta is approximately equal to the tangent of theta but the tangent of this
angle theta is just y that's the opposite side over d so i can rework this formula
replacing sine theta with y over d and i get a y over capital d is equal to n lambda and
there's our formula for the diffraction minima in terms of the actual distance
between the center of the pattern and the nth minimum on either side finally we'll practice
using our formula by solving a simple example so in this example i have 633
nanometer red laser light it's passing through a single slit with a width of 0.15 millimeters
and the screen that we're projecting the pattern on is 5 meters away that's the capital d
i want the angle and the distance from the central maximum for the first two diffraction minima
so i don't mean the first diffraction minimum on the left and right of the central bulge i
mean give me the size of the angle to the first two on the same side of the central
bulge so for the first part of this we use the formula in terms of angle and our first
diffraction minimum happens when n is equal to one i plug in my slit width in the right units
that's meters plug in my n equals one and the wavelength of the light again in the right units
nanometers are 10 to the negative 9. into 3 sig figs i get 0.242 degrees to that first dark spot
then i can answer by using the formula in terms of the actual y the distance between the central max
and the first diffraction minimum solving for y in general here just to save a little bit
of time down the road i multiply by big d on both sides and divide by a and these are the locations
of my diffraction minima all right let's sub in n equals 1 sub in our wavelength
distance to the screen slit width in meters and writing it in centimeters to 3 sig figs i get 2.11
centimeters off the center line for that first dark spot now we go to the n equals 2 case just
replace m with 2 in the formula everything else is the same and i get an angle of 0.484
degrees now notice this is twice the angle we got for the first diffraction minimum and the
reason why is these angular changes are so small that the sine function is approximately linear
if you start going to really large values of n you're going to see a deviation from this
linearity but typically we don't do that in the lab the same should be approximately true
for the actual distance between the center line and the second diffraction minimum plug
in n equals 2 plug in the wavelength distance to the screen slit width and i get 4.22 centimeters
for the distance between the center line of our diffraction pattern and the second dark spot
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