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As promised, let's do a couple
of simple Snell's law examples.

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So let's say, that
I have two media-- I

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guess the plural of mediums.

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So let's say I have
air right here.

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And then right here
is the surface.

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Let me do that in a
more appropriate color.

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That is the surface
of the water.

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And I know that I have a light
ray, coming in with an incident

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angle of-- so relative to the
perpendicular-- 35 degrees.

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And what I want to know is
what the angle of refraction

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will be.

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So it will refract a little bit.

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It will bend inwards
a little bit,

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since this outside's going to
be in the air a little longer,

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if you buy into my car
travelling into the mud

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analogy.

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So it will then
bend a little bit.

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And I want to figure out
what this new angle will be.

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I want to figure out
the angle of refraction.

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I'll call that theta 2.

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What is this?

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This is just straight
up applying Snell's law.

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And I'm going to use the
version using the refraction

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indices, since we
have a table here

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from the ck12.org FlexBook
on the refraction indices--

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and you can go get it
for free if you like.

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And that just tells
us that the refraction

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index for the
first medium-- that

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is air-- times the sine of the
incident angle, in this case

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is 35 degrees, is going to
be equal to the refraction

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index for water, times the
sine of this angle right

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over here-- times
the sine of theta 2.

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And we know what the refraction
index for air and for water is,

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and then we just have
to solve for theta 2.

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So let's just do that.

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The refraction index for
air is this number right

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over here, 1.00029.

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So it's going to be, there's
three 0's, 1.00029 times

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the sine of 35 degrees, is going
to be equal to the refraction

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index for water, which is 1.33.

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So it's 1.33 times
sine of theta 2.

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Now we can divide both sides
of this equation by 1.33.

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On this side, we're just left
with the sine of theta 2.

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On the left-hand side, let's
get our calculator out for this.

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So let me get the
handy calculator.

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And so we want to
calculate-- and I made sure

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my calculator is in degree
mode-- 1.00029 times

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the sine of 35
degrees, so that's

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the numerator of
this expression right

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here-- the green part--
that's 0.5737 divided by 1.33.

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I'm just dividing by
the numerator here.

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When you just
divide this answer,

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it means your last answer.

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That's the numerator up here
divided by that denominator.

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And so I get 0.4314.

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I'll just round a little bit.

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So I'll get-- I'll
switch colors--

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0.4314 is equal to
sine of theta 2.

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And now to solve
for theta, you just

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have to take the inverse
sine of both sides of this.

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This doesn't mean sine
to the negative 1.

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You could also use the arcsine.

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The sine inverse
of 0.4314 is going

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to be equal to-- the inverse
sine of sine is just the angle

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itself or, I guess
when we're dealing

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with angles in a
normal range, it's

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going to be the angle itself.

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And that's going to be the
case with this right over here.

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And if any of that
is confusing you

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might want to review the
videos on the inverse sine

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and the inverse
cosine, and they're

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in the Trigonometry playlist.

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But we can very
easily figure out

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the inverse sine for
this right over here.

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You literally, you
have sine here,

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when you press Second
you get the inverse sine.

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So it's the inverse
sine, or the arcsine

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of that number right over there.

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And instead of
retyping it, I can just

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put Second, and then Answer.

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So I'm taking the inverse
sine of that number.

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And that will give me an angle.

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And I get 25.55, or I'll
round it, 25.6 degrees.

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So this theta 2,
is equal to 25.6,

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or I'll say approximately
equal to some 25.6 degrees.

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So Snell's law goes
with our little car

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driving in to the mud analogy.

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It's going to be
a narrow degree.

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It's going to come inwards a
little bit closer to vertical.

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And theta 2 is equal
to 25.6 degrees.

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And you could do the other way.

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Let's do another example.

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Let's say that we have,
just to make things simpler,

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that I have some
surface right over here.

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So this is some
unknown material.

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And we're traveling in space,
we're on the space shuttle,

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and so this is a vacuum.

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Or pretty darn
close to a vacuum.

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And I have light coming in at
some angle, just like that.

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Let me drop a vertical.

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So it's coming in at some angle.

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Actually, let me
make it interesting.

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Let me make the light go
from the slower medium

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to the faster medium, just
because the last time we

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went from the faster
to the slower.

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So it's in a vacuum.

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So let's say I have some
light traveling like this.

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And once again, just to get
the "get" of whether it's

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going to bend inward
or bend outward,

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the left side is going
to get out first,

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so is going to
travel faster first.

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So it will bend inwards when it
goes into the faster material.

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So this is some unknown.

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This is some unknown material,
where light travels slower.

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And let's say we were able
to measure, the angles.

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So let me drop a
vertical right here.

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And so let's say that this
right here, is 30 degrees.

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And let's say we're
able to measure

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the angle of refraction.

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And the angle of
refraction over here

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is, let's say that
this is 40 degrees.

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So given that we're able
to measure the incident

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angle, and the
angle of refraction,

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can we figure out the refraction
index for this material?

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Or even better, can we
figure out the speed of light

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in that material?

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So let's figure out the
refraction index first.

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So we know the refraction index
for this questionable material

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times the sine of
30 degrees is going

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to be equal to the refraction
index for a vacuum.

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Well, that's just the
ratio of the speed

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of light in the vacuum to the
speed of light in the vacuum.

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So that's just going to be 1.

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This is the same thing
as n for a vacuum--

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and I'll just write a 1 there--
times the sine of 40 degrees.

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Or If we wanted to solve for
this unknown refraction index,

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we just divide both sides of
the equation by sine of 30.

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So our unknown
refraction index is

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going to be-- this is
just the sine of 40

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degrees-- over this-- over
the sine of 30 degrees.

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So we can get our
handy calculator out.

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And so we have the
sine of 40 divided

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by the sine of 30 degrees.

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Make sure you're in degree
mode, if you try this.

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And you get, let's
just round it, 1.29.

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So this is approximately equal
to-- so our unknown refraction

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index for our material
is equal to 1.29.

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So we were able to figure out
the unknown refraction index.

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And we can actually
use this to figure out

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the velocity of light
in this material.

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Because remember, this
unknown refraction index

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is equal to the
velocity of light

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in a vacuum, which is 300
million meters per second,

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divided by the velocity in this
material, the unknown material.

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So we know that 1.29 is equal
to the velocity of light

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in a vacuum.

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So we could write
300 million meters

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per second, divided by
the unknown velocity

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in this material.

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I'll put a question mark.

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And so we can
multiply both sides

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times our unknown velocity.

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I'm running out of
space over here,

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I have other stuff
written over here.

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So I could multiply
both sides by this v

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and I'll get 1.29 times
this v with a question mark,

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is going to be equal to 300
million meters per second.

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And then I could divide
both sides by 1.29.

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v question mark is going to be
this whole thing, 300 million

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divided by 1.29.

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Or another way to
think of it is,

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light travels 1.29
times faster in a vacuum

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than it does in this
material right over here.

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But let's figure
out it's velocity.

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So in this material,
light will travel a slow--

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so the 300 million
divided by 1.29.

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Light will travel a super slow
232 million meters per second.

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So this is approximately,
just to round off,

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232 million meters per second.

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And if we had to guess
what this material is,

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let's see-- I just
made up these numbers--

192
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but let's see if there's
something that has a refraction

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index close to 1.29.

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So that's pretty
close to 1.29 here.

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So maybe this is some
type of interface

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with water in a vacuum,
where the water somehow

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isn't actually
evaporating because

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of the lack of pressure.

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Or maybe it's some
other material.

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Let's keep it that
way, maybe it's

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some type of solid material.

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But anyway, those
were, hopefully,

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two fairly straightforward
Snell's law problems.

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In the next video, I'll do a
slightly more involved one.