Current time:0:00Total duration:10:44

0 energy points

# Snell's law example 1

Video transcript

As promised, let's do a couple
of simple Snell's law examples. So let's say, that
I have two media-- I guess the plural of mediums. So let's say I have
air right here. And then right here
is the surface. Let me do that in a
more appropriate color. That is the surface
of the water. And I know that I have a light
ray, coming in with an incident angle of-- so relative to the
perpendicular-- 35 degrees. And what I want to know is
what the angle of refraction will be. So it will refract a little bit. It will bend inwards
a little bit, since this outside's going to
be in the air a little longer, if you buy into my car
travelling into the mud analogy. So it will then
bend a little bit. And I want to figure out
what this new angle will be. I want to figure out
the angle of refraction. I'll call that theta 2. What is this? This is just straight
up applying Snell's law. And I'm going to use the
version using the refraction indices, since we
have a table here from the ck12.org FlexBook
on the refraction indices-- and you can go get it
for free if you like. And that just tells
us that the refraction index for the
first medium-- that is air-- times the sine of the
incident angle, in this case is 35 degrees, is going to
be equal to the refraction index for water, times the
sine of this angle right over here-- times
the sine of theta 2. And we know what the refraction
index for air and for water is, and then we just have
to solve for theta 2. So let's just do that. The refraction index for
air is this number right over here, 1.00029. So it's going to be, there's
three 0's, 1.00029 times the sine of 35 degrees, is going
to be equal to the refraction index for water, which is 1.33. So it's 1.33 times
sine of theta 2. Now we can divide both sides
of this equation by 1.33. On this side, we're just left
with the sine of theta 2. On the left-hand side, let's
get our calculator out for this. So let me get the
handy calculator. And so we want to
calculate-- and I made sure my calculator is in degree
mode-- 1.00029 times the sine of 35
degrees, so that's the numerator of
this expression right here-- the green part--
that's 0.5737 divided by 1.33. I'm just dividing by
the numerator here. When you just
divide this answer, it means your last answer. That's the numerator up here
divided by that denominator. And so I get 0.4314. I'll just round a little bit. So I'll get-- I'll
switch colors-- 0.4314 is equal to
sine of theta 2. And now to solve
for theta, you just have to take the inverse
sine of both sides of this. This doesn't mean sine
to the negative 1. You could also use the arcsine. The sine inverse
of 0.4314 is going to be equal to-- the inverse
sine of sine is just the angle itself or, I guess
when we're dealing with angles in a
normal range, it's going to be the angle itself. And that's going to be the
case with this right over here. And if any of that
is confusing you might want to review the
videos on the inverse sine and the inverse
cosine, and they're in the Trigonometry playlist. But we can very
easily figure out the inverse sine for
this right over here. You literally, you
have sine here, when you press Second
you get the inverse sine. So it's the inverse
sine, or the arcsine of that number right over there. And instead of
retyping it, I can just put Second, and then Answer. So I'm taking the inverse
sine of that number. And that will give me an angle. And I get 25.55, or I'll
round it, 25.6 degrees. So this theta 2,
is equal to 25.6, or I'll say approximately
equal to some 25.6 degrees. So Snell's law goes
with our little car driving in to the mud analogy. It's going to be
a narrow degree. It's going to come inwards a
little bit closer to vertical. And theta 2 is equal
to 25.6 degrees. And you could do the other way. Let's do another example. Let's say that we have,
just to make things simpler, that I have some
surface right over here. So this is some
unknown material. And we're traveling in space,
we're on the space shuttle, and so this is a vacuum. Or pretty darn
close to a vacuum. And I have light coming in at
some angle, just like that. Let me drop a vertical. So it's coming in at some angle. Actually, let me
make it interesting. Let me make the light go
from the slower medium to the faster medium, just
because the last time we went from the faster
to the slower. So it's in a vacuum. So let's say I have some
light traveling like this. And once again, just to get
the "get" of whether it's going to bend inward
or bend outward, the left side is going
to get out first, so is going to
travel faster first. So it will bend inwards when it
goes into the faster material. So this is some unknown. This is some unknown material,
where light travels slower. And let's say we were able
to measure, the angles. So let me drop a
vertical right here. And so let's say that this
right here, is 30 degrees. And let's say we're
able to measure the angle of refraction. And the angle of
refraction over here is, let's say that
this is 40 degrees. So given that we're able
to measure the incident angle, and the
angle of refraction, can we figure out the refraction
index for this material? Or even better, can we
figure out the speed of light in that material? So let's figure out the
refraction index first. So we know the refraction index
for this questionable material times the sine of
30 degrees is going to be equal to the refraction
index for a vacuum. Well, that's just the
ratio of the speed of light in the vacuum to the
speed of light in the vacuum. So that's just going to be 1. This is the same thing
as n for a vacuum-- and I'll just write a 1 there--
times the sine of 40 degrees. Or If we wanted to solve for
this unknown refraction index, we just divide both sides of
the equation by sine of 30. So our unknown
refraction index is going to be-- this is
just the sine of 40 degrees-- over this-- over
the sine of 30 degrees. So we can get our
handy calculator out. And so we have the
sine of 40 divided by the sine of 30 degrees. Make sure you're in degree
mode, if you try this. And you get, let's
just round it, 1.29. So this is approximately equal
to-- so our unknown refraction index for our material
is equal to 1.29. So we were able to figure out
the unknown refraction index. And we can actually
use this to figure out the velocity of light
in this material. Because remember, this
unknown refraction index is equal to the
velocity of light in a vacuum, which is 300
million meters per second, divided by the velocity in this
material, the unknown material. So we know that 1.29 is equal
to the velocity of light in a vacuum. So we could write
300 million meters per second, divided by
the unknown velocity in this material. I'll put a question mark. And so we can
multiply both sides times our unknown velocity. I'm running out of
space over here, I have other stuff
written over here. So I could multiply
both sides by this v and I'll get 1.29 times
this v with a question mark, is going to be equal to 300
million meters per second. And then I could divide
both sides by 1.29. v question mark is going to be
this whole thing, 300 million divided by 1.29. Or another way to
think of it is, light travels 1.29
times faster in a vacuum than it does in this
material right over here. But let's figure
out it's velocity. So in this material,
light will travel a slow-- so the 300 million
divided by 1.29. Light will travel a super slow
232 million meters per second. So this is approximately,
just to round off, 232 million meters per second. And if we had to guess
what this material is, let's see-- I just
made up these numbers-- but let's see if there's
something that has a refraction index close to 1.29. So that's pretty
close to 1.29 here. So maybe this is some
type of interface with water in a vacuum,
where the water somehow isn't actually
evaporating because of the lack of pressure. Or maybe it's some
other material. Let's keep it that
way, maybe it's some type of solid material. But anyway, those
were, hopefully, two fairly straightforward
Snell's law problems. In the next video, I'll do a
slightly more involved one.