WEBVTT
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This is College Physics
Answers with Shaun Dychko.
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An oil slick is on top of some water
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and the oil has a thickness
of 120 nanometers we're told.
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It has an index of refraction
of 1.4 and it's on top of water
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which has an index refraction of 1.3.
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The light ray of white light which means
it consists of all the different colors
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mixed together comes in
perpendicular to the oil slick
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but I've drawn it on an angle just so
we can see the two reflected rays.
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And then I can distinguish
between them in the drawing.
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If I were to draw it straight down
then I'd have line straight down
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and another line straight up
and they would all overlap.
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That would be confusing looking.
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So we imagine that these rays are
actually coming straight down
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and then reflecting straight
back up and mixing together.
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Now we need to know
what is the phase change
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for each of these reflected rays here
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and the phase changes measured
in number of wavelengths.
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So it's travelling through air initially
which has an index refraction of one
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and when this ray number one bounces
off the top of the oil surface
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it will have a phase change
of half of a wavelength.
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And that's because it's
going from a low index of
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1.0 to a material of high index of 1.4 .
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And so reflecting off of an
interface from low to high,
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causes a phase change of half a wavelength.
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And then we have to figure out
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what is the phase shift of
this second Ray number two.
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Now when there's a reflection
off of this interface
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there will be no phase change due to
reflection because it's starting.
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In this material with an
index refraction of 1.4
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and bouncing off an interface
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with a material that has
a lower index of refraction
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it's going from a high index and reflecting
off of interface with a low index material
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produces no phase change as a
result of that reflection.
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So we do not have a lambda
over two term here.
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This, by the way, is the way I write
phase change with this symbol *Phi*.
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And this is *Phi 1* which is
the phase change of ray one.
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So the phase change for Ray
two is going to be just
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due to the additional
path length of this ray
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through the thickness of the
oil once and then twice.
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And how many wavelengths
will it be shifted.
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Well it's going to be it's going
through the thickness twice
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and so it's going to be
travelling two times a thickness
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and we're going to divide
that by the wavelength
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but I have a subscript *n* here
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because it's the wavelength in this oil
material that we're concerned with.
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So how many wavelengths is
this total thickness of *2t*,
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this total path length additional
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and so that's *2t* divided by the
wavelength in the material.
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So the wavelength in the material is
the wavelength in an air or a vacuum
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divided by the index of
refraction of the material.
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And since we're dividing by this we're
going to multiply by it's reciprocal.
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So we're multiplying by *n* over *lambda*
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so we have the phase change for *Ray two*
is two times the thickness of the oil
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times Index of refraction of oil
divided by the wavelength in air
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times the wavelength in air
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and these cancel giving us a phase shift
of two times index refraction of the oil
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times the thickness.
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So constructive interference occurs
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when you have the total phase shift is
some integer multiplied by the wavelength
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and so we can add these two
phase shifts together.
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And by the way the question asks
us what color will the oil b.
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Well it'll be whatever color
constructively interference
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because constructive interference increases
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the amplitude of that particular wavelength
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and it will be the one
that is more noticeable.
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So the total phase shift is gonna
be the sum of the phase shifts
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from each of these rays one and two.
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And that has to equal *m lambda*
for constructive interference.
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And then we substitute for each of these.
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So that's lambda over two phase shift
from the reflection at the top of the oil
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plus two times index of refraction
of oil times oil thickness,
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phase shift for the second Ray that
bounces off the oil water interface.
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And that equals *m lambda* and
then we get rid of fractions
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because fractions are a bit messy looking
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and we multiply everything by two
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and we get lambda plus
*4nt* equals to *m lambda* .
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And then subtract lambda from both sides
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and we get and then factor out the
lambda as well on this right side now
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and then switch sides around.
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So we have lambda times bracket
*2m* minus one equals *4nt*
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and then divide both sides by *2m* minus one
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and so here is the wavelength that will
experience constructive interference.
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It's *4nt* over *2m* minus one
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and because this number *m* is
an integer that we can choose.
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You'll notice that there
are many wavelengths
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that will have constructive interference
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but the one we're concerned with
is the one that is visible.
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And so we're going to try
different values of *m* .
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Let's try one first of all.
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So we have four times index
of refraction of oil
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1.4 times the thickness of the oil slick
120 nanometres divided by two times *m*
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or to choosing to be one minus
one which gives 672 nanometres
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and that is a lucky guess of *m* equals one
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because this is visible and
this is the color red.