Current time:0:00Total duration:11:25

0 energy points

Studying for a test? Prepare with these 3 lessons on Geometric optics.

See 3 lessons

# Snell's law example 2

Video transcript

Let's do a slightly more
involved Snell's law example. So I have this person
over here, sitting at the edge of this pool. And they have a little
laser pointer in their hand and they shine
their laser pointer. So in their hand,
where they shine, it's 1.7 meters above
the surface of the pool. And they shine it so
it travels 8.1 meters to touch the surface
of the water. And then the light
gets refracted inward. It's going to a slower medium. If you think about
the car analogy, the outside tires get to stay
outside a little bit longer, so they move faster. So it gets refracted inward. And then it hits the
bottom of the pool at some point right over here. And the pool, they tell
us, is three meters deep. What I want to figure out is how
far away does this point hit. So what is this distance
right over here? And to figure that
out, I just need to figure out what
this distance is. I need to figure out what this
distance is-- so this distance right over here. And then figure out what
that distance is, and then add them up. So I can figure out
this part-- trying to do it in a different
color-- this part right until we hit the
surface of the water and then figure out this
incremental distance, just like that. And hopefully with a
little trigonometry and maybe a little
bit of Snell's law we'll be able to get there. So let's start on maybe
what's the simplest thing. Let's just figure
out this distance. And it looks like it will
pay off later on, as well. So let's figure out this
distance right over here. So just the distance along
the surface of the water, to where the laser
point actually starts touching the water. And this is just a straight up
Pythagorean theorem problem. This is a right angle. This is the
hypotenuse, over here. So this distance, let's
call this distance x. x squared plus
1.7 meters squared is going to be equal
to 8.1 squared-- just straight up Pythagorean theorem. So x squared plus
1.7 squared is going to be equal to 8.1 squared. Or we could subtract 1.7
squared from both sides. We get x squared is equal to
8.1 squared minus 1.7 squared. If we want to solve
for x, x is going to be the positive
square root of this, because we only care
about positive distances. x is going to be equal to the
principal root of 8.1 squared minus 1.7 squared. And let's get our
calculator out for that. So x is going to be equal to
the square root of 8.1 squared minus 1.7 squared. And I get 7.9-- looks about--
let me just round it, 7.92. So x is about 7.92, although we
could save that number there, so we can get a
more exact number. So this is equal to 7.92. That is x. Now, we just have to figure out
this incremental distance right over here, add that
to this x, and then we know this entire distance. So let's see how we
can think about it. So let's think about what
the incident angle is and then the angle
of refraction is. So I've dropped a
perpendicular to the interface, or to the surface. So our incident angle is
this angle right over here. That is our incident angle. And remember,
Snell's law-- we care about the sine of this angle. Actually, let me just write
down what we want to care about. So we know this is
our incident angle. This is our angle of refraction. We know that the
index of refraction for this medium
out here-- and this is air, so it's going to be the
index of refraction for air-- times the sine of theta 1--
this is just Snell's law, so times our incident
angle, right here-- is going to be equal to
the index of refraction for the water-- and
we'll put the values in the next step-- times
the sine of theta 2-- times the sine of our refraction
angle, sine of theta 2. Now, we know-- we can
figure out these ends from this table right over here. I actually got this problem
from ck12.org's flex book as well or at least the
image for the problem. And so if we want to
solve for theta 2-- or if we know theta 2, we
could then solve for this. And we'll do that with a
little bit of trigonometry. Actually we won't even have to--
if we know the sine of theta 2, we'll be able to solve for this. All right, we'll think
about it either way. Actually, we'll just
solve for this angle, and then if we know
this angle, then we'll be able to use a little
trigonometry to figure out this distance over here. So to solve for that angle,
we can look up these two. And so we just have to
figure out what this is. We need to figure out what
the sine of theta 1 is. So let's put in
all of the values. Our index of refraction
of air is 1.00029. So let me put that in there. So that's this. So 1.00029 times
the sine of theta. And you say, oh, how do we
figure out sine of theta? We don't even know
what that angle is. But remember, this is
basic trigonometry. Remember "soh cah toa." Sine is opposite
over hypotenuse. So if you have this
angle here-- let's make it a part of
a right triangle. So if you make that as part
of a right triangle, opposite over hypotenuse-- it's
the ratio of this side, it's the ratio of that
distance to the hypotenuse. This distance over here
we just figured out, it's the same as this
distance down here. It's x. So this is 7.92. So the sine of theta
1 is going to be the opposite of the angle,
opposite over the hypotenuse. That just comes from
the definition of sine. So it's going to be times-- so
this part right over here, sine of theta 1-- we don't have
to know what theta 1 is. It's going to be 7.92 over 8.1. And that's going to be equal
to the index of refraction of water. So that's index of
water is 1.33-- so let me do that in a different color. So that's going to be-- no, I
wanted to do a different color. So that's going to be-- let
me do it in this dark blue. So that's going to be 1.33
times sine of theta 2. And so, if we want to
solve for sine of theta 2, you just divide both sides
of this equation by 1.33. So let's do that. So I'll do it over here. So if you divide
both sides by 1.33, we get 1.00029
times 7.92 over 8.1, and we're also going
to divide by 1.33. So we're also dividing by 1.33. That is going to be equal
to this sine of theta 2. So let's figure what that is. So let's do that. Get the calculator out. So we have 1.00029 times 7.92. Well, actually I could even
say times second answer, if we want this exact value. That was the last-- so I'm going
to do that-- second answer. So that's the actual
precise, not even rounding. And then we want to divide by
1.33, that's this right here. And then we want to divide
by 8.1, and we get that. And that's going to be equal
to the sine of theta 2. So that's going to be equal
to the sine of theta 2. So let me write this down. So we have 0.735 is equal
to the sine of theta 2. Now, we could take
the inverse sine of both sides of this
equation to solve for theta 2. So we get theta 2 is
equal to-- let's just take the inverse sine of this value. So I take the inverse sine of
the value that we just had, so answer is just
our last answer. And we have theta 2
being 47.3-- let's say rounded-- 47.34 degrees. So this is 47.34 degrees. So we were able to figure out
what theta 2 is, 47.34 degrees. So now we just have to use
a little bit of trigonometry to actually figure out
this distance over here. Now what trig ratio involves--
so we know this angle. We want to figure out its
opposite side, to that angle. And we know the
adjacent side-- we know that this right here is 3. So what trig identity deals
with opposite and adjacent? Well, tangent-- toa. Tangent is opposite
over adjacent. So we know that the tangent
of this angle right over here of 47.34 degrees is going to
be equal to this opposite side over here-- so I'll
call that y-- is going to equal y over
our adjacent side. And that's just 3 meters. Or if we want to solve
for y, we multiply both sides of this
equation by 3. You get 3 times the tangent of
47.34 degrees is equal to y. So let's just get
the calculator out. So 3 times the tangent
of that 47.34 degrees-- I'll use the exact answer--
3 times the tangent of that is 3.255. So this distance right here, y. And we're at the home stretch. y is equal to 3.255 meters. Now, our question was, what
is this total distance? So it's going to be
this distance, x, plus y, plus the 3.25. So x was 7.92. And I'll round here. So it's literally just going
to be 7.92 plus our answer just now. So we get about 11.18. Or maybe if we want to really
round, or get the same number significant digits,
maybe 11.2 meters. I'll just say 11.18 meters. So this right here, the distance
that we wanted to figure out is the point on the bottom of
the pool where the actual laser pointer actually hits
the surface of the pool will be 11.18--
approximately, I'm rounding a little bit-- meters
from this edge of the pool right there. Anyway, hopefully you
found that useful. It's a little bit more involved
than the Snell's law problem. But really the hard part was
just in the trigonometry, recognizing that you didn't
have to know this angle, because you have
all the information for the sine of that angle. You could actually figure
out that angle now. Now that you know its
sine, you could figure out the inverse sine of that. But that's not even necessary. We know the sine of the angle
using basic trigonometry. We can use that and
Snell's law to figure out this angle right here. And then once you
know this angle, use a little bit
more trigonometry to figure out this
incremental distance.