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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 gets a little reflected in or 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 in the pool they tell us is 3 meters deep what I want to figure out what I want to figure out is how far away does 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 can figure out I can figure out this part kind of 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 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 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 meter 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 point 7 squared if we want to solve for 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 eight point one squared minus one point seven squared and let's get our calculator out for that so X is going to be equal to the square root of eight point one eight point one squared minus one point seven squared and I get seven point nine looks about yeah let's just round it seven point nine two so X is about seven point nine two although we could save that number there so we can get an exact number so this is equal to seven point seven point nine two 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 let's think about what the angle of 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 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 want to care about so we know 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 with the 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 in 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 ck-12 org flex book as well or at least the image for the problem and so if we want to all for theta2 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 alright well we'll think about either way actually we'll just thought solve for this angle and then if we know this angle then really we'll 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 let's put in let's put in all of the values our index of refraction of air is one point zero zero zero to nine so let me put that in there so that's this one point zero zero zero to nine so one point zero zero zero to nine 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 so cut OA sine is opposite over hypotenuse so if you have this angle here let's make it as 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 it's the ratio of this side is 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 seven point nine two so the sine of theta one 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 x so this part right over here sine of theta one we don't have to know what theta one is it's going to be seven point nine two seven point nine two over over eight point one over eight point one and that's going to be equal to the index of refraction of water so that's index level is 1.33 so let me do that in a different color so that's going to be two now I wanted to do a different color so that's going to be let me do it in this dark blue so it's going to be one point three three times sine of theta two and so if we want to solve for sine of theta two you just divide both sides of this equation by one point three three so let's do that so I'll do it over here so if you divide both sides by one point three three we get one point zero zero zero two nine times seven point nine two seven point nine two over eight point one and we're also going to divide by one point three three so we're also dividing by we are also dividing by one point three three 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 what that is so let's do that get the calculator out let's get the calculator out so we have one point zero zero zero two nine times seven point nine two well actually I could even say times second answer if we want 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 one point three three that's this right here and then we want to divide by eight point one divided by eight point one and we get that and that's going to be equal to the sine of theta two so that's going to be equal to the sine of theta two so let me write this down so we have zero zero point seven three five is equal to the sine of theta two now we could take the inverse sine of both sides of this equation to solve for theta 2 so we get theta two 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 two being forty seven point three let's say rounded point three four degrees so this is forty seven forty seven point three four degrees so we were able to figure out what theta two is forty seven point three four 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 want to figure out its opposite side we want to 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 edj and well tangent Toa tangent is opposite over adjacent so we know that the tangent the tangent of this angle right over here of forty seven point three four degrees is going to be equal to this opposite side over here so let me just call that I'll call that Y it's going to equal Y over our adjacent side and that's just three meters or if we want to solve for y or if we want to solve for y you multiply both sides of this equation by three you get three times the tangent of forty seven point three four degrees is equal to Y so let's just get the calculator out so three times the tangent of that forty seven point three four degrees I'll use the exact answer three times a tangent of that is three point two five five so this distance right here Y Y and we're at the homestretch Y is equal to three point two five five meters now our question was what is this total distance so it's going to be this distance X plus y plus the three point two five so X was seven point nine two and I'll round here so it's literally just going to be seven point nine two plus our answer just now plus our answer right now so we get about eleven point one eight or maybe if we want to really round or get the same number of significant digits we start maybe eleven point two meters I'll just say eleven point one eight 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 where the where the actual a laser pointer pointer actually hits the surface of the pool will be 11.1 eleven point one eight approximately I'm rounding a little bit meters from from this edge of the pool right there anyway hopefully you found that useful is a little bit more involved in 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 it's sine you could figure out the inverse sign 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