this video gives a quick introduction to diffraction gratings including the derivation of
the formula for interference maxima then we show a couple simple demonstrations including one with
laser light and one with white light and finally we wrap things up with an example of calculating
the interference maxima for a couple different wavelengths of light passing through a diffraction
grating so a diffraction grating is a film with effectively thousands of slits in it and as light
passes through these gaps each gap acts as a source of circular wavelets so this is really an
application of huygens principle and i'll post a link real quick to where that was first introduced
these wavelets interfere with each other and for whatever wavelength of light we're using
we'll see constructive interference at different angles to the screen as we can see
in our diagram here so along this center line we have wave crests aligned with wave crests
that means they're interfering constructively and we'll see a bright spot on the screen
in this direction and then in this upper diagonal line i can see that the wave
crests are aligning with wave crests again for this particular angle so when we project
this onto a screen we'll see a bright spot coming from that direction and finally in this bottom ray
i have wave crests align with wave press again and when i project this on the screen i'll get
a bright spot in that direction to understand what's going on here we have to make several
diagrams and we're sort of progressively zooming in closer and closer to see what's happening
in the first diagram i'm really zoomed out so the diffraction grating is just this tiny little slit
and i have all these rays coming through there so that represents all the rays coming through
the gaps of the diffraction grating rather than showing them as concentric semi-circles i'm just
showing the direction of each of those sources as a ray and those are interfering constructively
at this particular angle on the screen so i put an angle of theta here a distance between
the center line and the interference maximum of y and the distance to the screen is d in the second
picture we zoom in on the diffraction grating to find the conditions under which interference
is constructive and one important assumption at this point is that the angle is the same
for each one of these rays and that's a really good approximation provided that the distance
to the screen is large compared to the size of the diffraction grating so in the little red circle in
that second diagram i drew a little perpendicular line in there to try to highlight the path length
difference for two adjacent rays the ray that's all the way at the bottom here has to travel just
a little bit farther than the ray that's above it and at the special angle for which that path
length difference is equal to one wavelength for whatever wavelength of light we're using here
those rays are going to arrive in phase with each other at the screen so being off by one wavelength
puts them in phase a crest will be lined up with a crest and a trough of the trough so if
this special angle gives a path length difference of one wavelength between two adjacent rays well
what if i went to the third ray above the bottom that would have a path length difference of one
wavelength relative to the ray just below it so it's going to be in phase with that
and then it would have a path length difference of two wavelengths relative to the ray that
we started with at the very bottom so it's in phase with both of those waves below it
and we can continue this reasoning all the way up the diffraction grading at this special angle
theta where the path length difference for two adjacent slits is equal to one wavelength
every single wavelet emanating from slits in the diffraction grating is interfering constructively
as it turns out this very strongly concentrates the light at this angle even a small deviation
from this special angle means wavelets are arriving enough out of phase from each other
to kill off the vast majority of the intensity of the light so if we zoom in on that little
reddish circle from the second diagram and really blow it up big enough to do some math we get our
third diagram in the third diagram i'm viewing two rays that are coming from adjacent slits
in the diffraction grating my slit separation is labeled with a d so that's just the slit to slit
spacing these rays are coming off at an angle of theta again by assumption that's an angle for an
interference maximum i've drawn the perpendicular into the diagram as a solid white line in order to
measure the path length difference for these two rays the bottom one has to travel just a
little bit farther than the top one and that's given by this little leg of a right triangle
and we also know that when the path length difference is equal to one wavelength
that's when we're going to see constructive interference so we have to work out a little
bit of trig here that angle theta to the bright spot on the screen it adds up to 90 degrees
with this angle so i can call that 90 minus theta but this angle in my right triangle
adds up to 90 degrees with the angle 90 minus theta so if i'm trying to find
that little angle on my right triangle i'm just going to call it question mark for now
when i add it to the angle 90 minus theta i get 90 degrees subtract 90 from both sides add theta
to both sides and i just figured out the angle that i want in my right triangle is actually
the same as theta so that's theta in there now d is the hypotenuse for that right triangle so the
path length difference then can be expressed as d sine theta so i've found a constructive
interference when this condition is met d sine theta is equal to lambda so the path length
difference between two adjacent slits is equal to one wavelength of the light that i'm using
note that this can be generalized because if the path length difference was two lambda between two
adjacent slits then i could move one slit beyond that and i would have a path length difference
of four lambda and i still get constructive interference and six lambda eight lambda et cetera
so i can have multiple locations of constructive interference those are
given by the path length difference being an integer multiple of the wavelength of the light
note that zero should be included as one of the values of n for this formula if i'm talking about
an angle of zero that's the center line of my experiment and the path length difference
of light coming from adjacent slits is 0 at that point so they just all stay in phase and i'll get
a bright spot right along the center so i'm going to write n equals 0 1 2 and so on you're going to
see in practice 1 or maybe 2 interference maxima before you basically go off the screen so in
this slide we're illustrating a couple different things you can do with the diffraction grading in
the first picture i'm just shining red laser light through a diffraction grating that's 1 000 lines
per millimeter and this bright spot right here is the center line for the experiment so that's the n
equals zero maximum and then on the left and right of the diffraction grating i have my n equals 1
maxima so you can take monochromatic light like the light coming from a laser and you can clearly
see these diffraction grading interference maxima at rather large angles away from the center line
however one of the most useful features of a diffraction grating is illustrated in the second
picture in the second picture we're passing white light through a diffraction grating and we see
different angles to the constructive interference of different wavelengths in the mixture
so this violet light is coming to us from the more shallow angles through the diffraction grating
violet light interferes constructively at shallower angles because you don't have to go
as far from the center line to get a one lambda path length difference between adjacent slits
then in here maybe you can see some blue and then some green some yellow and orange and red and this
is the familiar roy g biv just written backwards we're supposed to put in an i for indigo which
is basically blue and i've separated the white light out into all of its constituent wavelengths
so when we study atomic physics we're going to use diffraction gratings and those are either going to
be built into cheap spectroscope instruments or very expensive spectrophotometer instruments that
plug into the computer to separate the wavelengths of light emitted by an excited gas so whatever
device we're using to separate the light the diffraction grading is the piece of technology
at the core of the spectrometer so let's wrap things up with an example in the example i have
white light passing through a diffraction grating with 500 lines per millimeter and projected onto
a screen that's 5 meters distant in part a i want the distance from the center line
of the diffraction pattern to where violet light is interfering constructively and i'm saying this
is 450 nanometer violet lights that's happening right here in part b we want the distance from
the center line to the red light that i'm saying is 635 nanometers so again notice that it takes a
shallower angle to get constructive interference out of a shorter wavelength light like violet
because you don't have to tilt as far to get a one wavelength path length difference between adjacent
slits so i get this violet fringe coming first and then blue and then green and then yellow and then
orange and then red red is the longest wavelength so i have to go to a rather steep angle to get
a one wavelength path length difference between adjacent slits so one last thing to note about the
diagram before we start is that i put a little white rectangle right in the center so you're
going to see a white spot right there because the path length difference is zero no matter what the
wavelength of light is so you still have a mixture of all the wavelengths and you have white light
right at the center so for each of these questions i'm going to apply my interference maxima formula
d sine theta equals n lambda and these are first order maxima so the first maximum that's off
the center so n is equal to one d is the slit separation from slit to slit and that has to
be measured in meters so we have a little bit of work to do we know we have 500 lines every
millimeter how many meters is it between lines so d is going to be one millimeter
for every 500 lines in other words .001 meters for every 500 lines and when i smash the numbers i get
2 times 10 to the negative 6 meters per line to streamline things a little bit i can solve for
theta in general and then i'll just apply that to part a and part b so first you divide by d
then you invert the sine function
so i get inverse sine of lambda over d now to answer part a i have an angle to this interference
of inverse sine lambda over d my wavelength is 450 nanometers so 450 times 10 to the negative 9.
my slit separation is 2 times 10 to the negative 6 meters notice that the units of meters cancel here
and keeping three sig figs i get 13.0 degrees for that violet fringe when i separate light through
this diffraction grating now we were asked for the actual distance y between the center line
and that interference maximum so we're going to have to do just a little bit of trig here
if you've seen the videos on one slit and two slit diffraction patterns
we were assuming the angles were small sometimes well it's usually not the case with diffraction
gratings so 13 degrees is not a small enough angle to make some of the approximations you might be
used to making so we're going to stay exact here i have 13 degrees for the angle i have 5 meters
for the distance to the screen and i'm looking for y so i could say that tangent of 13 degrees
is y over 5 in other words y is 5 tangent 13 and i get a distance of 1.15 meters
in part b i want to calculate that same distance y but for the red fringe in the pattern plug in
the wavelength of the red light plug in the slit spacing and keeping three sig figs i get
18 and a half degrees for this one now there's no need to draw the triangle again i know that
i'm going to get 5 times the tangent of 18.5 for the y value this time and i get 1.67 meters
for that red fringe so if you're able to generate a narrow bright white beam of light and run it
through this diffraction grating you'll project a rainbow on the wall that runs all the way
from 1.15 meters off the center to 1.67 meters off the center always starting with the violet
light closest to the center line because again for shorter wavelengths of light you don't have to go
to as extreme of an angle to get a one wavelength path length difference between adjacent slits
if you find the physics content on zack's lab helpful click on the zaxlab logo on the right
to browse playlists and subscribe to the channel i produce over 100 new videos per
month and subscribing is the easiest way to find new content thanks for watching