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Let's do a slightly more involved Snell's Law example
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 1.7 m above the surface of the pool
And they shine it, so it travels 8.1 m to touch the surface of the water
And then the light gets refracted inwards
It's going into 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 inwards and then it hits
the bottom of the pool at some point right over here
and pool they tell us is 3 m deep
What I want to figure out is how far away does this point hit?
So what is this distance right over here?
To figure that out, I need to figure out what this distance is, right over here
and then figure out what that distance is and then add them up
I can figure out this part, this part right until we hit the surface of the water
and then figure out this incremental distance just like that
And hopefully a little bit of trigonometry and maybe Snell's Law will be able to get there
So let's start with 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 along the water
the surface of the water to where the laser point actually starts touching the water
And this is a straight up Pythagorean Theorem problem
This is a right angle
This is the hypotenuse over here. So this distance, let's call it x
X^2 +1.7 m ^2 = 8.1 ^2
just a straight up Pythagorean Theorem
X^2 +1.7 m ^2 = 8.1 ^2
Subtract 1.7 squared from both sides
we get X squared is equal to 8.1 squared -1.7 squared
If you want to solve for x, x = 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 -1.7 squared
Let's get our calculator out for that, so x is going to be equal to
the square root of 8.1 squared -1.7 squared
And I get 7.92
So x is about 7.92. That is x
Now to figure out this incremental distance right over here
add that to this x and then we know this entire distance. Let's see how we can think about it
Let's think about what the incident angle is and what the angle of refraction
I'll drop the perpendicular to the interface or to the surface
So our incident angle is this angle right over here
And remember Snell's Law, we care about the sine of this angle
actually let me just write down what we 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 here--this is air--
the index of refraction for air
times the sine of theta 1--this is just Snell's Law--times our incident angle right here
is going to be equal to the index of refraction for the water
times the sine of theta 2, times the sine of our fraction angle
Now we know we can figure out these n's from this table here
I actually got this problem from ck12.org flexbook as well as the image for the problem
And so if you want to solve for theta 2
or if we know theta 2, we can then solve for this
We'll do that with a little bit of trigonometry
If we know the sine of theta 2, we'll be able to solve for this
Either way. Actually I'll solve for this angle. If we know this angle
then we're able to use a little bit of trigonometry to figure out this distance over here
So to solve for that angle, we have to figure out what the sine of theta 1 is
Let's put in all of the values
Our index of refraction of air 1.00029
times sin theta. Maybe you'll say, how to do that? We don't even know what that angle is
But remember this is basic trigonometry, soh cah toa
Sine is opposite over hypotenuse
If you have this angle here, let's make this part of a right triangle
opposite over hypotenuse is the ratio of this side to the hypotenuse
This distance over here is the same as x, which is 7.92
So the sine of theta 1 is going to be the opposite of the angle over the hypotenuse
That comes from the definition of sine
We don't have to know what theta 1 is
It's going to be 7.92/8.1
And that's going to be equal to the index of refraction of water which is 1.33
times sine of theta 2
If we want to solve for theta 2, we just divide both sides of this equation by 1.33
So if we divide both sides by 1.33, we get 1.00029 times 7.92/8.1 divided by 1.33
That is going to be equal to this sine of theta 2
So let's do that with the calculator
So we have 1.00029 times 7.92--
I can say second answer here. This is the exact value, not even rounding
and then divide it by 1.33 and 8.1
And we get that and 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
Theta 2 is equal to the inverse sine of this value
So I take the inverse sine of our last answer
So this is 47.34 degrees. So we're able to figure out what theta 2 is. 47.34 degrees.
So now we'll just have to use a little bit of trigonometry to actually figure out
to actually figure out this distance over here
So we know this angle
we want to figure its opposite side; we want to figure out the opposite side to that angle
And we know the adjacent side, which is 3
So what trig identity deals with opposite and adjacent?
Tangent, toa. Tangent is opposite over adjacent
is going to be equal to this opposite side over here--I'll call that y--
It's going to equal y over our adjacent side and that's just 3 m
To solve for y, we multiply both sides of this equation by 3
You get three times the tangent of 47.34 degrees is equal to y
Let's 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 is equal to 3.255 m
Now our question was what is this total distance
So it's going to be this distance X plus Y, plus the 3.255
X was 7.92
I'll round here. So it's literally going to be 7.92 + our answer just now
So we get about 11.18 meters
So this right here is the distance that we want to figure out
the point on the bottom of pool where the actual laser pointer hit
the surface of the pool will be approximately 11.18 m from this edge
Anyway, hopefully you found it useful. It's a little bit more involved
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 can actually figure out that angle now
Now that you know it's sine, you can 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 once you know this angle, use a little bit more trigonometry to
figure out this incremental distance