Snell's law example 2
Snell's Law Example 2. Created by Sal Khan.
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- So we have the distance away where the light will actually hit the bottom of the pool, but my question is: what distance from the observer will the light appear to hit the bottom of the pool?(22 votes)
- Basic Pythagoras Theorem, you know the distance of the observer to the bottom of the pool which is (1.7+3) and you found the distance of the of the laser beam which is
So you can connect the observes point to the laser point and that will be your 11.2m hypothenuse. So Hypothenuse square = (11.2m)squared + (4.7m) squared.
Answer: 125.44 + 22.09 = 147.53(4 votes)
- Hi Sal, I am a sonography student and the physics of sound is extremely important to our practice (for obvious reasons). We are learning a lot about refraction and Snell's law currently. Could you give some examples of Snell's law relating to sound traveling through different densities of tissue mediums as opposed to light traveling through different mediums? The average propagation speed of sound through soft tissues (which is the majority of our practice) is 1540m/s or 1.54mm/microsecond. Lung is 500m/s, blood is 1560m/s, adipose is 1450m/s, muscle is 1600m/s, tendon is 1700m/s and bone is 3500m/s. Thanks for your help if you decide to make some more videos! These videos are great!(8 votes)
- Any medium which causes the speed of a wave to change will result in refraction, whether it be light, sound, water, etc... This is due to the boundary condition that the energy of the wave be conserved when the wave transitions from one medium to another.(4 votes)
- Why does speed of light slow down when it passes in a medium, be it glass or water no medium in specific..??(0 votes)
- Due to increase/decrease in the interaction between the particles present in the medium..!!(13 votes)
- Why is sine function taken in snell's law?(4 votes)
- This is because the operation needed to solve the triangle created by the normal and ray of light is opposite over hypotenuse which is performed by the function sine.
Good question(4 votes)
- I'm having difficulty in understanding Snell's Law. If anyone can explain it in simple words, It will be very nice.(1 vote)
- Whenever the speed of light changes, the direction the light is moving will change.(4 votes)
- Why is snell's law the way it is? I mean how did snell bought up his law?(2 votes)
That is a derivation ^ of it. See if you Understand it.(2 votes)
- Just to confirm, the correct answer with the right number of significant figures would be 11.2m right?(2 votes)
- Yes. At10:20, Sal says that using significant figures 11.18 would round to 11.2, implicitly applying the additive rules for significant figures to respect the location of the decimal point.(2 votes)
- what will happen when a handful salt is put into a glass of water?
is there any change in it?(1 vote)
- The salt will increase the optical density of water.(3 votes)
- Why is there an exponent in the square root? don't they cancel out each other? (referring to2:30of the video)(2 votes)
- That will only happen if it would have been division and the base would have been same. For example: if it would have been 81^2 / 81^2 the exponents would have canceled out each other 81^(2-2) = 81. If it would have been 81^a/81^b where a is not equal to b then it would have been 81^(a-b).(1 vote)
- are there videos related to refraction in glass slab..?(2 votes)
- they are dealing with refraction between any two medium may be denser (or) rarer and optical(or)
water medium etc(1 vote)
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.