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STUDENT: Rainbows-- while
we may not notice them,

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there are rainbows all around us
as we go about our daily lives.

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They are hidden
in the reflection

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of soap bubbles,
the shine on a CD,

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and even in oily
puddles on the street.

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In this video, we'll explore
the phenomenon of light wave

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interference and how
it creates the colors

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we see on the surfaces
of thin films.

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Since soap and oil
are usually colorless,

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why do they have iridescence?

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Let's start with the
laws of light reflection

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and refraction, then peruse
through some visual simulations

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of these fundamental principles.

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When waves travel
through space and hit

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an interface or a
surface, some of the wave

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reflects off the surface
and the rest refracts,

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continuing through the new
medium at a different angle.

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When the refracted wave hits
another surface, part of it

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reflects back out of
the medium and combines

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with the first wave or
interferes with the first wave.

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Since the first and
second reflected

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waves travel different
paths, the second wave

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acquires a phase shift in
comparison to the first wave.

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When it adds to the first
wave, this phase shift

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may cause the two
waves to completely

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cancel each other out, resulting
in destructive interference.

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Or if the phases are
perfectly aligned,

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the resultant wave has twice
the amplitude, resulting

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in constructive interference.

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Notice that the resultant wave
has destructive interference

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when the phase shift is about
180 degrees or pi radians.

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And it reaches the
maximum amplitude,

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having constructive
interference,

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when the phase shift is a
multiple of 2 pi radians.

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For light traveling from
one medium or material

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to another medium, for
example, from air to water,

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Snell's law of refraction,
given by this equation here,

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states that the angle at which
light travels through the two

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different mediums is
proportional to the velocities

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of light through the two
mediums, V1 and V2, which

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is inversely proportional to the
refractive indices, N2 and N1.

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This equation can be
simplified to N1 sine theta 1

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equals N2 sine theta 2.

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Snell's law can be
rearranged to represent

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N1, the refractive index
of the first medium,

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in terms of the incident light
angles of the two materials

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and the refractive
index of the thin film.

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This trick will be useful in
some of our later calculations.

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Remember how waves with
shifted phases interfere?

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In this case, the phase shift
is a difference in path length

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that the first reflected
wave and the second travel.

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Geometrically, the
path difference

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between the wave reflected off
the top surface and the wave

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reflected off the bottom
surface of a thin film

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is given by this
equation here, where

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AB, BC, and AD refer to the
line segments in this diagram.

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Using some trigonometry,
the line segment lengths

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can be calculated in terms
of the refractive indices

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and the refracted light angles.

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Now, using Snell floor
to substitute for N1,

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the path difference
between the two

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reflected waves can be expressed
in terms of the thin film

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medium only, only depending
on N2 and theta 2.

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That means for constructive
interference between the two

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waves, the phase
shifts has to be 0

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and the path
difference has to be

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an integer multiple
of the incident lights

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wavelength, lambda.

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When the medium a
wave is traveling into

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has a refractive
index that is greater

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than the index for the medium
that the wave is coming

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from, that is, N2
is greater than N1,

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there is one extra caveat.

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The reflected wave has a
180-degree phase shift.

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In that case, the condition
for constructive interference

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is that the path difference
must be a half integer

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multiple of the incident
light's wavelength.

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For a typical thin film,
which has a higher refractive

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index than its surroundings,
which is usually air,

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this is a condition for
constructive interference

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between waves reflected off
the top and bottom surfaces

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of the film.

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This simulation shows a
wave traveling through air,

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then reflecting off the top and
bottom surfaces of a thin film.

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The bold reflected
waves represents

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the summation of the
waves reflected off

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the top and the
bottom on the film,

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and should grow
largest when there

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is constructive interference
and the two waves' phases align

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line.

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I can change four different
parameters in this situation--

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the thickness of the film,
D, the refractive index

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of the film, the angle
of the incident light

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hitting the surface of
a film, and finally,

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the wavelength of the
light hitting the film.

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Changing all of these parameters
changes the resultant wave that

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reflects back out of the film.

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You might be wondering why
changing the wavelength

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makes the wave different colors.

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That's because the
visible light spectrum

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can be represented by waves of
different wavelengths ranging

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from three 80 nanometers
to 750 nanometers.

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The human eye has
light receptors,

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or cones, that
actually only perceive

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three colors from the
visible spectrum-- red,

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green, and blue.

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The entire visible
spectrum people

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see is really a superposition
of the waves of these three

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colors, which mix
in different ratios

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to provide all the other
colors of the rainbow.

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The light that humans see
making rainbow swirly patterns

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on the surface of soap bubbles
or in the oil sheen on water

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has a minimum wavelength
of 380 nanometers

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and a maximum wavelength of 750.

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Red has a wavelength of about
700 nanometers, green 546,

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and blue 436.

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The strength of each
of these colors,

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or the intensity,
that people perceive

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is proportional to the square
of the wave's amplitude.

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Since the amplitude represents
the energy, E, of a wave,

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intensity is also
a representation

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of the wave's energy density.

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The more intense a wave,
the larger amplitude

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it has, and the more its color
dominates our perception.

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The total color that
we see reflected off

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a point on the film
surface is a combination

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of these three
waves proportional

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to their intensities.

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For example, for
a film thickness

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of 200 nanometers with
refractive index 1.5

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and incident angle
45, we can see

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that there is more blue in
this color than green or red.

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If we try changing
the film thickness to,

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let's say, 400 nanometers,
the refractive index to 2,

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and keep the incident
light angle at 45,

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we can see that green would
be the dominating color.

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For thin films which
have a greater refractive

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index than the air, we can
show the color reflected off

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a thin film as a function
of the film's thickness.

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Likewise, we can see how the
reflective wavelength colors

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vary with the incident
light wave angle.

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Here are two plots
Mathematica has

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generated for a thin film
with refractive index 1.5,

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first with a constant
incident angle of 45 degrees

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and second with a constant
thickness of 200 nanometers.

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Notice that with
varying thicknesses,

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the color shifts
alternate between

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reddish, greenish,
and bluish colors,

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while for varying
angle, the spectrum

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is a more gradual gradient.

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Let's envision a flat thin film
of oil floating on a perfectly

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flat surface of water.

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The refractive indices
of oil, water, and air

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are respectively about
1.5, 1.33, and 1.

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As light reflects off
the top of the oil,

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there is a 180-degree
phase shift

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since the refractive
index of oil

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is greater than the
refractive index of air.

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But there isn't a phase shift
when the wave reflects off

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the bottom interface between
the oil and the water

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because water's refractive
index is less than oil's.

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This interactive
graphic shows how

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an oil firm with
varying thicknesses

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reflects light of a point
source at varying heights

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above the oil's surface.

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We can increase the
height of the light

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to get this iridescence pattern.

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Or we can increase
the film's thickness.

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We can see that with
increasing the light height,

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there is a wider range
of incident light angles,

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whereas when we move
the film thickness,

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there is a wider range of
colors that are produced

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on the surface of the film.

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In reality, the
surface of a puddle

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really isn't completely flat.

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This interactive graphic shows
thin film interference of oil

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on rippling water.

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Again, if we increase the
thickness of the oil film,

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there is a wider
range of colors.

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We can also change the x, y,
and z positions of the light

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and see how the iridescence
pattern changes.

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For a spherical soap bubble,
the refractive index of so

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varies with the
recipe, but it's always

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greater than the
refractive index of air.

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That means that there is also
a phase shift of 180 degrees

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when light hits the outer
surface of the bubble,

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but not when it reflects
off the inner surface.

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This simulation shows the effect
of multiple point source lights

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on a soap bubble.

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Note that in real life, light
sources around a syllable

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usually aren't
just point sources

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and the bubble isn't an
even thickness throughout,

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as is assumed in
this simulation.

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Hence, the color
patterns that we

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see on surfaces of soap
bundles are usually

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far more complex than what
we see in the simulation.

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Special thanks to the 3016 staff
for their help and guidance

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on this project and thanks
for watching my video.