Snell's law example 2

- [Voiceover] 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, it gets refracted inward 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 wanna figure out, what I wanna figure out is how far away does the this point hit? So what is this distance right over here? What is, what is this distance right over there. 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 could figure out, I can figure out this part. I'll try and 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 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 water, 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, 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. And if we wanna solve for x, x is going to be the positive square root of this, 'cause we only care about positive distances. X is going to be equal to the principal root of 8.1 squared minus 1.7 squared number. So this is equal to 7., 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 angle of, the incident angle is and then angle of refraction is. So I've dropped a perpendicular to the interface, or to the surface. So our angle, our incident angle is this angle, 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 wanna 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, this is air. So it's gonna be index of refraction for air times the sine of theta one. Times the sine of theta one, 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. For the water, and we'll put the values in at the next step. Times the sine of theta two. times the sine of our refraction angle, sine of theta two. Now we know, we can figure out these, these ends from this table right over here. I actually got this problem from CK12.org's FlexBook as well, at least the image for the problem. And so if we wanna solve for theta two, or if we know theta two, we can 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 two, we'll be able to solve for this. All right, well we'll think about either way. Actually we'll just solve for this angle and then if we know this angle, then we're able, 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 one is. So let's put in, let's put in all of the values. Our index of refraction of air is 1.00029, so let me put that in there, that's this 1.00029. So 1.00029 times the sine of theta. And you say, oh well 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 if opposite over hypotenuse. So if you have this angle here, let's make it part of a, let's make it part of a right triangle. So if you make that as part of a right triangle, opposite over hypotenuse is 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 one is going to be the opposite of the angle, opposite over the hypotenuse. That's just, comes from the definition of sine. So it's going to be times. So this part right over here, sine of theta one. We don't even have to know what theta one is. It's going to be 7.92. 7.92 over, over 8.1, 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 two. And so if we wanna solve for sine of theta two, 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, 7.92 over 8.1. And we're also going to divide by 1.33. So we're also dividing by, we are also dividing by 1.33. That is going to be equal to this sine of theta two. That is going to be equal to sine of theta two. So let's figure out what that is. 0.735 is equal to the sine of theta two. Now we can take the inverse sine of both sides of this equation to solve for theta two. So we get theta two is equal to, let's just take the inverse sine of this value, is 47, 47. 34 degrees. So we were able to figure out what theta two is, 47.34 degrees. So now we just have to use a little bit of trigonometry to actually figure out, to actually figure out this distance over here. Now what trig ratio involves... So we know this angle, we wanna figure out its opposite side, we wanna figure out the opposite side to that angle, and we know the adjacent side, we know that this right here is three. So what trig identity deals with opposite and adjacent. Well tangent, toa. Tangent is opposite over adjacent. So we know that the tangent, the tangent of this angle right over here, of 47.34 degrees is going to be equal to this opposite side over here. So let me just call that, I'll call that y. Is going to equal y over our adjacent side, and that's just three meters. Or if we wanna solve for y, or if we wanna solve for y, we multiple both sides of this equation by three. You get three times the tangent of 47.34 degrees is equal to y. This distance right here, y, 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, plus our answer right now. So we get about 11.18, or maybe if we wanna really round or get the same number of significant digits that we started maybe 11.2 meters, I'll just say 11.18 meters. So this is, this right here is the distance that we wanted to figure out is the point on the bottom of the pool where the actual laser pointer pointer actually hits the surface of the pool will be 11.1, 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 was a little bit more involved of a 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 of the information for the sine of that angle. You could actually figure out that angle now, now that you know it's 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.