in this video we're given a crown glass lens that's an index of refraction of 1.52
and we've applied an anti-reflective coating to it that's fluorite with an index of refraction
of 1.43 and this time we're given the thickness of the thin film that's 89.0 nanometers and we're
trying to calculate the wavelengths of light that will interfere constructively and destructively
through this film just to clarify what we mean by finding the wavelengths we're expected to convert
those back to the wavelengths they would be in the air which is pretty much the same as they would be
in vacuum in the third part of the question we take the entire lens and immerse it in oil
and this creates a bit of a new situation because the oil has a higher index of refraction than
the fluorite that creates a slow to fast to slow transition and this forces us to go back to first
principles and figure out the wavelengths that will interfere constructively and destructively
because we never found a formula for this slow to fast to slow situation so to start things out
we're asked to compute any visible wavelengths of light so we're restricting ourselves to only those
that land between 400 and 700 nanometers roughly whose reflections interfere destructively when
white light is incident on this fluorite layer so there's our incoming ray coming in at right angles
and we end up with two reflections one off of the top surface of the fluoride
and just to clarify we're assuming that it's air out here which would have an
index of refraction of about one and that first reflection is phase shifted i'll just write ps
because it's at an interface where the light is slowing down in other
words i'm going from a lower index of refraction to a higher index of refraction
the part of the light that penetrates all the way through the fluorite to the crown glass
some of that is going to be reflected off of the crown glass and because again the index of
refraction is increasing at that interface the reflected wave is going to be phase shifted by
the reflection and then there's the additional possibility of a phase shift occurring because
of the extra travel distance for that wave the thickness is given to us as 89 nanometers and
the additional travel distance is twice that much because the wave has to go through the fluorite
on the way down and on the way back up in part one we're looking for destructive interference
and because both of these waves are phase shifted i need my extra travel distance to undo the phase
shift on the second reflection to get these waves back out of phase well the first way this can
happen is if the extra travel distance 2t is equal to half a wavelength of the light in fluorite
and that wavelength of light in the fluorite is notated as lambda prime solving for lambda
prime i get 4t i can plug in my thickness of 89.0 nanometers and i get 356 nanometers
for a wavelength of light interfering destructively now remember that's the wavelength
of the light within the fluorite and i need to convert back to what it would be in vacuum well we
know that lambda prime is the wavelength in vacuum over the index of refraction for the materials so
that's lambda over 1.43 and we multiply it by 1.43 obtaining a visible wavelength of 509 nanometers
so in our reflected light that 509 nanometer light is going to be missing and in fact you get partial
destructive interference in a neighborhood of that 509 nanometers and this is a good anti-reflective
coating because it's basically carving out kind of in the middle of the visible spectrum and it
should make it so you get less reflection off this lens than you otherwise would now we're
going to look for another wavelength that interferes destructively and the next one
still requires 2t to get us a half a wavelength a phase shift and that happens if we fit in
one and a half wavelengths or three halves lambda prime in a distance of twice the thickness so
that additional integer wavelength doesn't cause any problem as long as we're left with an extra
one-half of a wavelength to cause a phase shift by way of extra travel distance we're going to
get both of these reflections to be out of phase when they emerge from the fluorite so before
i do any calculations i want to note that we're definitely talking about a shorter wavelength now
because one and a half wavelengths fit into twice the thickness whereas before it was only halfway
in length so as we look for these higher order destructive interferences they must be shorter
wavelengths solving for lambda prime that's four t over three so that's four times 89 nanometers
over three and for lambda prime i get 119 nanometers again that's equal to the wavelength
in a vacuum over the index of refraction so taking 119 times 1.43 i get 170 nanometers for my next
destructively interfering light now that's a wavelength shorter than the shortest visible
wavelength of about 400 nanometers so this is actually ultraviolet light and it's not really
what's been requested here because we only want the visible wavelengths and if i continue to look
for higher order destructive interferences those wavelengths are increasingly shorter
because i'm fitting more wiggles on that 2t travel distance so i'm guaranteed that there are no more
visible wavelengths of light that are going to interfere destructively
through this thin film in the second part we're asked for any visible wavelengths of light whose
reflections interfere constructively when white light is incident at right angles on this film
and again we have a phase shift for the light at each of the reflections
so the reflections don't put the waves out of phase if our extra travel distance doesn't mess
with the phase of the second reflection then we'll get constructive interference
and the first one happens when the extra travel distance is equal to exactly one wavelength
so that it doesn't mess with the phase of that reflection and i get lambda prime is equal to 2t
that's 178 nanometers and again that's the wavelength and vacuum over the index
of refraction for the material i get 255 nanometers and that's going to be the longest
wavelength of light that interferes constructively through this thin film
because if i look at my next order constructive interference that's when two wavelengths fit into
twice the thickness of the film in other words i'm talking about a wavelength that's half as much
so given that this first one is ultraviolet already beyond the visible spectrum the
second one is going to be even shorter so it's also going to be ultraviolet just for
completeness i'll work out the calculation real quick the wavelength is 89 nanometers
within the fluorite and the wavelength in a vacuum is 127 nanometers so definitely getting shorter
and definitely ultraviolet so neither one of those is a visible wavelength so the answer
is that there are no visible wavelengths that interfere constructively through this thin film
now in part three i want to know how the answers change if the lens is immersed in oil and this
oil has an index of refraction of 1.5 making it larger than the index for the fluorite
so there's my incoming light wave there's my first reflection there's my second reflection and what
changes is that at the first reflection that's a transition from a slower material to a faster
material the index of refraction went down and there's no phase shift caused by that reflection
the second reflection is still phase shifted because it's a transition from a
faster material to a slower material so to begin with maybe i'll do the destructive interference
because one reflection is phase shifted and the other isn't we're going to get destructive
interference if our travel distance doesn't mess with the phase of that second reflection
so that would be when the travel distance is an integer number of wavelengths within the fluorite
the first one would be when 2t is equal to lambda prime the second time this occurs is when 2t
is equal to 2 lambda prime so all that's happened here is that i'm going to get
destructive interference in the same places where i got constructive interference before
i put this thing in the oil that first wavelength we've already done the math is 255 nanometers and
the second one is 127 nanometers and they just keep getting smaller so there are no visible
wavelengths of light that are going to interfere destructively once i immerse this thing in oil
what about my constructive interferences because one of these reflections has no phase shift and
the other is phase shifted i'll get constructive interference if my extra travel distance for the
second reflection puts that reflection a half wavelength out of phase the first time that
happens is when two t is equal to lambda prime over two and again i notice that i've done these
calculations before it's just that that was the calculation for destructive interference when the
lens wasn't in oil so two t equals lambda prime over two the way that turned out lambda was 509
nanometers so now my 509 nanometer light is going to interfere constructively instead of
destructively again the next time this happens is when 2t is equal to one and a half wavelengths or
three halves lambda prime and the way that turned out was a wavelength of 170 nanometers and again
that's ultraviolet so basically you immerse it in oil and all the answers from the first two parts
just reverse i get only one interference that's happening in the visible spectrum and it's a
constructive interference on 509 nanometer lights exactly the same wavelength that was destructive
before we put it in the oil if you find the physics content on zack's lab helpful click on the
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