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Have small open top tank Into Which | could pour some water. The edges are all 2Ocm long: Ata top edge mount laser so that it points below the horizontal at 45"...

Question

Have small open top tank Into Which | could pour some water. The edges are all 2Ocm long: Ata top edge mount laser so that it points below the horizontal at 45". This means that the beam would hit the bottom of the tank right at the far side now pour water ( nwiter 33 ) to depth of 1Ocm into the tank; how far from the far side does the laser beam hit the bottom of the tank? Be sure t0 drawa good picturel Submit an Image of your solutlon below;

have small open top tank Into Which | could pour some water. The edges are all 2Ocm long: Ata top edge mount laser so that it points below the horizontal at 45". This means that the beam would hit the bottom of the tank right at the far side now pour water ( nwiter 33 ) to depth of 1Ocm into the tank; how far from the far side does the laser beam hit the bottom of the tank? Be sure t0 drawa good picturel Submit an Image of your solutlon below;



Answers

An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the air-water interface at a 45° angle relative to the vertical. You see a bright spot where the beam hits the bottom of the aquarium. How much water (in terms of height) should you add to the tank so the bright spot on the bottom moves 5.0 cm?

Okay. So, as far as I can tell, there's a couple different approaches and even ways that you can address the solution of this particular question. And so what we have here is the surface of a pool, Okay? And the pool is pool is 1.5 m deep, and you're standing somewhere here at the edge of the pool. You're gonna shine a laser. The laser is going to strike the pool. Uh huh. 5 m from where you're standing. Okay, so we have a triangle like so and then the lights going to continue onwards. And we want to know how far away we actually don't know how long the pool is. So erase this edge here. We want to know this distance from where the light strikes the bottom of the pool to where you're standing. How far away is that? Okay. And so what we're given were given this 5 m distance from where it enters the pool and were given the pools 1.5 m deep. What we don't know is how high, How high above the surface of the pool we are. So we're going to call this, um x three, right here. Let's see what clocks and the data. The data. So that the angle, um, they don't want and at which it strikes the pool. We don't know in our data to right here our data to so are refracted angle. Once we enter the water, we also don't know. And of course, we're trying to solve this total distance, but let's call the distance from where it hits the pool two where it hits the bottom of the pool X one. That's X one. Okay. And what's actually on the noise, x three. It's called sex, too. Sex too. Okay, so, um, using are equation and one sign data one equals end to sign data to If, in theory, we can find what data to is, then we can solve for X one. Using are triangle here, like so, um, but we have too many variables. Too many variables, um, that we don't know because we know and one is going to be one. We're in air, so we're just gonna have signed Data one, and that's going to be equal to 1.33 since we're entering the water. Sign Data two. Okay. And we know that we're told in the question that the person is standing very low to the water and shining the light almost parallel, almost parallel to the surface of the water. So we do know that fate of one is going to be small. It's going to be very small. This angle here is very itty bitty. And what we're told, one thing that we know is that for small Thetas, small Thetas sign data is going to be approximately equal to tangent of theta approximately equal to just data for really small data's. So we could I could simplify this further by saying that well, sine theta one we know is gonna be so small that theta one is just going to be equal to 1.33 sign data to okay, And remember, we want to try to find what x one is, so we can then add it to the 5 m to find this total distance here. Okay, so the next expression that I can come up with because I know that if this is theta two, you know, to, um we know that the height of this triangle this triangle hearing is 1.5 m So then we can set up the expression data to is going to be equal to the tan. The inverse tangent of opposites or 1.5 m over adjacent data. Um, sorry over X one. Like so. And so now we have an expression for data to and so we can plug this back into this equation here. Okay, so we'll get set to one is equal to 1.33 10 in verse, 1.5 m over X one. Okay. And we can multiply by tam date on both sides, so we'll end up with tan Data. One equals 1.33 times 1.5 is about to two 2 m, divided by X one. Sorry. Divided by yeah, times. Okay. And then solving for X one solving for X one, we get X one equals 2 m over 10. Theater one. Okay. And so then our total distance from how far away from us at this point here, will the, um okay. Um so x one equals 2 m over a tan theta one. And we want to know how far from us where we're standing is the laser at the bottom. So um, we can then say we can simplify this further by saying that Well, we know the tangent of the theater for a very small angle. Once again, 3 to 1 we know is really small. It is approximately equal to fate. Uh, so then we get X one. Next one equals 2 m over 3 to 1, and our total distance is going to be X one plus 5 m. So that gives us an expression of 5 m plus 2 m over data one. And if they do, one is, say really close to zero, approximately zero. Then that would be about 7 m away. Um, but remember, we're doing a bunch of approximations here, so let's leave it like so for small theaters. So this is Maastricht. This is for four small theater ones, okay?

In this problem. On the topic of ray optics, we have shown a figure where one corner of a rectangular boxes filled with water and the laser beam starts 10 centimeters from said A of this side of this container. It enters the water at Position X, and we want to know if the laser beam will refract back into the air through side B or reflect from side. Be back into the water. We want to determine this angle of refraction or reflection, and then repeat for X is now 25 centimeters that we want to find the minimum value of X, for which the laser beam passes through side B and emerges into the air. So we'll use the ray model of light and the law of refraction, and we'll assume the laser beam is a ray of light. We will set up the geometry as follows now from this geometry of the diagram and outside a. We can see that the 10 of the angle phi is equal to 10 centimeters, divided by 15 centimeters, which means that fire is the Oct 10 of 10 by 15, which gives us the angle five to be 33 point 69 degrees. So this means that the angle of incidence at that side a vita A is equal to 90 degrees minus five minus 33 point 69 degrees, which gives us this angle of incidence to be 56 point 31 degrees. Now we can use Snell's law at side A and using smells lower, we get the defective index of air times sign Delta Air must equal to the effective index of water times sign pita water. Hey, it means that this angle vita w A is equal to the oxide of the effective in the index of one times the sine 56 point 31 degrees divided by the effective index of water. 1.33 gives us this angle vita A for water to be 38 0.73 degrees. Now this ray of light will strike side B and the angle of incidents at this water air boundary is data what b and this is equal to 90 degrees minus the, uh, W A, which we calculated above. So this is 51 0.3 degrees Now. The critical angle for the water air boundary is due to see and the critical angle. DTSC is equal to the ox sign of three factor index of air and a over the refractive index of water and W. And so this is the ox sign of one divided by 1.33 and so the critical angle is 48.8 degrees now, because the angle vita Water B is larger than the critical angle, the to see this ray will experience total internal reflection. So after striking side B today will completely reflect back into the water. Now for Part B, we want to repeat the calculations for Part A, but now we change X 2 25 centimeters, and so from the diagram outside A, we get five to be 21.8 degrees, similar to how we've done it both the to A to be 68 point two degrees. And we can use Snell's law again like above and using Snell's law at the air water boundary, we get data W A. To be 44 point 28 degrees and following from their TTWB his 45 point 72 degrees. Now we can see that the to WN side B is less than the critical angle which calculated above DTC. So this means that the ray will be reflected into the air, so there's no total internal reflection. In this case. There will be refraction, so changing extra 25 centimeters will allow the ray to diffract into the air. Now the angle of refraction we can calculate as follows. So in a sign, the to A B is equal to the reflective index of water and water again using stones law times sine the to what b, which means that the angle of refraction data a at B is equal to the ox. Sign after effective index of water. 1.33 times sine of the angle, 41 point 25 degrees divided by effective index of air, which is one and so well, actually, this angle is 40 5.72 25.7 two degrees. And so we calculate we get Peter B. The angle of reflection to be 72 0.2 degrees Now for part C of the problem, we want to find the minimum value of X, for which the laser beam will pass through side B and emerge into the air. So we'll use the critical angle for the air water boundary that we found in part A. And we get the, uh, W A. To be 90 degrees minus 48 point 75 degrees, which is 41 0.25 degrees, and we again use Snell's law. Using Snell's law, we get the refractive index of air time. Sign Peter for a is equal to the effective index of water. I'm sign Peter W. A. Which means the angle of incidence Peter A is the ox ein of one 0.33 times sine of 41.25 degrees over the fact of index of A, which is one so calculating. We get the angle of incidents to be 61 0.27 degrees now, the minimum value of X, for which the laser beam passes through side B and emerges into the air and refracted, is calculated as follows. We know that 10 of the angle phi will be equal to 10 centimeters over X, and so therefore we re arrange X is 10 centimeters divided by 10 or five, which is 28.73 degrees. We get this from above since we have to. We know five is equal to 90 degrees minus 61.27 degrees, which is where we get 28 0.7 three degrees and so calculating. We get the value for X to be 18.2 centimetres.

For Number 59 there was a 4.1 meter deep pull of water and of ray of late. Guess comes in here an angle 30 degrees in the normal. Yeah, it's going to hit the bottom of poem, hit this mirror and come back out and were testifying the distance between where the right enters the pool and whether it will use the pool. Of course, we know that when light goes in, it's going to get refracted and like, goes slower and water. So it's gonna get Vince wore the normal, so it's not going to go straight. Is going events war the normal? You gonna bend it to here to my mirror? And then, of course, when light hits a mirror If I was a love reflection workers in same angle, it comes out of the same angle trying to make that about the same angle. That looks pretty decent iffing. So what we're trying to find it's gonna be this distance. I'm gonna make it pink. Good. Well, I'm gonna use my snows long to find the angle refraction. Here's the angle of incidence. This would be the angle refraction. So Snell's law of this air. So that's kind of index of refraction of one point. Well, times the angle of incidence, which is 30 degrees. Let it go. What comes? The next medium is water. So this 1.33 is the index refraction, and I want to find that angle. So I divided by 1.33 and let me took inverse sign and they go with that angle, I was 22.1 degrees. So it's gonna be this angle right here. Well, you figure I'm gonna minute for him. This this is like, that's my normal get that little better. I made two right triangles here, So I'm gonna look at this right triangle right here, and I'm going to find the site of the strengthened call that eggs. There's different ways I could have done the math you, but I'm going to recognize that I just found this angle. And that angle is the same as this angle because their ultimate interiors. So I'm looking at this skinny right triangle right here, and I know this side of the triangle, cause that's the depth of the water will point for. So that's the adjacent site. And I'm trying to find the opposite side, so I'm going to use tangent. Tangent of the angle, which I just found to be 22.1, is equal to the opposite side, which is X over my adjacent side, which is one point for the depth of the water. And for that I get my exercise point 566 But that's an answer. The question, because remember, my ex is just from here to here. But, um, if it came in a 22.1, it's gonna bounce up 22.1. So this trying over here is identical. So my distance here would be just twice what? Exes. So I'm gonna won't play that by two, and I get that it's 1.13 meters away.

Here. I've made a teacher off What is happening? This submarine is located at point as the top off the building is located at point B. Age is the height of the building. De is the depth off the submarine El is the horizontal distance beating the submarine and the shore. L one is the horizontal distance between the point birthing beam strikes the shore face off water and the shore l two is the horizontal distance between the point where the beam strikes the surface of water and the submarine. You might notice that l one plus l two is equal to L From this equation we obtain that l two musical toe l minus one. Let's know at this triangle tangent off the angle batter is equal to the opposite side divided by the ingestion site. So tangent Betta is equal to D over l two better. In its turn is equal toe arc tangent Off D divided by l two all four I is the anger off incidents. Often I is equal toe 90 degrees minus angle. Better off our is the Anglo Free Fraction angle Gamma is equal to 90 degrees minus alpha are according Toa snails lower off our is equal toe arc sine over there. A show off refractive indices off water and air multiplied by the sign over the angle off incidents. The angle off the incidence is equal toe scientist degrees minus better. Let's substitute this expression into the equation for Alfa are the refractive index. Off air is almost one. So if you can give ride this expression as arc sign off NW multiplied by the sign off 90 degrees minus better but better is equal toe arc tangent off D over l two. So let's substitute the expression for Betta into the expression for our for our Finally the obtained this expression for Alfa are now I remind you that l two is equal toe l minus l one Let's substitute the expression for l two into this expression for Alfa. Are we obtain that Alfa R is equal toe arc sine often w time sign off 90 degrees minus our dungeon off D over L minus a one. Let's now consider these triangle from this triangle dungeon Comma is equal toe age divided by one. Let's multiply both sides of this equation by everyone, Age equals l one multiplied by Dungeon Gamma. Now let's recall that gamma is equal to 90 degrees minus often are. Let's substitute the expression for gamma into this equation. Age equals L Juan, multiplied by dungeon 90 degrees minus Alfa are but also our is equal to this expression. Let's substitute the expression for off our in place off Alfa are so the obtain age equals L one times tangent off 90 degrees minus this expression age is now expressed in terms off the known quantities. I have used my car cool later to substitute the numbers. De is equal to 115 meters. L is equal to 325 meters. L one is equal to 205 meters. The refractive index off water is equal toe 1.333 The answer is 57.9 meters. Please note that the final result is required toe have three significant figures


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