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Problem 2: Refractionsubmerged scuba diver looks Up toward the calm surface of a freshwater lake and notes that the sun appears to be 209 from the vertical. a) At w...

Question

Problem 2: Refractionsubmerged scuba diver looks Up toward the calm surface of a freshwater lake and notes that the sun appears to be 209 from the vertical. a) At what angle would he see the sun were he diving in a sugar solution? (nwaler 1.33 nsukat Sol = 1.49)b) The diver directs a laser beam towards the surface. At what angle with respect to the vertical will the laser beam undergo complete reflection?

Problem 2: Refraction submerged scuba diver looks Up toward the calm surface of a freshwater lake and notes that the sun appears to be 209 from the vertical. a) At what angle would he see the sun were he diving in a sugar solution? (nwaler 1.33 nsukat Sol = 1.49) b) The diver directs a laser beam towards the surface. At what angle with respect to the vertical will the laser beam undergo complete reflection?



Answers

(II) An aquarium filled with water has flat glass sides whose index of refraction is 1.54. A beam of light from outside the aquarium strikes the glass at a 43.5$^\circ$ angle to the perpendicular (Fig. 23-56). What is the angle of this light ray when it enters (a) the glass, and then (b) the water? (c) What would be
the refracted angle if the ray entered the water directly?

Okay, so we're doing chocolate. 32 Problem 44. So this other you haven't Aquarian with water with class sides. So years are glass and here's our water. And here's our air. So what we have here is we have light country here on call data color stated one, and data one is 43.5 degrees. So oops, am in glass until this is 156 and water is on 0.33 Okay, so first for part pay, it asks, What is the angle of this light ray when it endures class? Okay, So for this British use medals law in one sign, data one equals into sign data too. So for this, we have in one is air. So it's just one so sign of 43 court 54 people, five degrees. And this is in glass. Six sign they did to solve this is stated to he calls inverse sine of one over 1.6 sign of 43 45 degrees so we can put that in our calculator. It we should see this is 26 2 degrees cool. So for part, being around, we want to figure out What is the angle interests of water? So for this we have in glass sign of 26.2 degrees. That's our new angle, and this is in water terms. Sign of the tube. So now I think the two is the end. For some sign of 1.56 over 1.33 sign of 26 2 degrees. We plugged that in our calculator. We shouldn't see year. We should have data to of 31.2 degrees on the part. See? See asked us what would be the refracted angle. It's the ray into the water directly. So this is no the glass, which means in on sign they won two sons to it at our snows on again. So sign of 43 points, five degrees. We're entering straight into the water. That class song that they too stuff We saw this. We know that ADA to must be the inverse sine of one over 133 times the sign of 43.5 degrees. Then we can plug that in a calculator and we shouldn't see is exactly 31.2 degrees, which is exactly this. So it doesn't matter. We enter the light first or just come directly in the water, it's gonna be at the same angle story.

So we have an interface between water and air, and a ring off light meets the normal at some angle. Taylor one, which is unknown. But once it goes into the air and water, the two is equal to 28 degrees. So the refractive index off air What 10 1 is equal to one and the refractive, and that's a water and two is equal to one point beneath everything. Snell's losses and one sign off fatal one is equal to end to sign off there, too. So one times side off Failla one is equal to 1.333 times. Sign off 28 degrees. Sophia one is equal to signing worse off 1.333 Sign off 28 degrees that comes out to be equal. Do 38.7 degrees.

Okay, in this problem, we have an aquarium, and it's with water. And it has, ah, last sides, and their index of refraction is 1.52 So let's write that first given down that and G is equal to 1.52 Let's go ahead and look at the figure while we're at it, and I'll pause the video while I pull it up. Okay, got it. So we have, um, you have glass, and then we have water. We have light entering on. This angle is 43.5 degrees. Let's double check. Yeah, great. And we want to get the angle of the ray when it enters the glass and then the water. So first ones right down the index of reflect fraction of water. So that's N W. And that value is 1.33 So, um, let's find what the angle is when it goes into the glass. So here is the glass, and here's the water. So for that, we want to use Snell's law so we could say and she sign. Beta G is equal to an air. Ah, sign of data in the air. And then this is of course one. So if we want to solve for theta G, we want to divide both sides by N G. And then take the arc sine of both sides. So with that, we can save data G is equal to the ark or the inverse sign of an A, which is one over N g multiplied by the sine of the angle and the air. And that data air is 43.5. Smoke. Go ahead and write that out expressively. Great. So now we have a formula and we have everything we're looking for in the formula. So I just need to plug it into a calculator and so on the do art sign of 1/1 0.52 times of sign of 43.5 degrees and, um, convert that to three is so that is 26.9 degrees, and then we also want the angle in water. So we just found this angle and that angle is right here, a state, a g. And now we want what's going on in water. So, um, we can use the same formula except for we want to use n g sign data G is equal. Teoh and water sign beta off water and we can do the same thing and just solved for science data Water. Um, so it's the same math. So you with different variable names. So you end up with data. W is equal to the inverse sign of N G. Divided by and w times the sign of the, uh, Tom G great. And now I can plug this into a calculator. So one by the by of 1.33 and then the sign of 26.9 degrees, understand a double chocolate and accept a fraction of water. I think I remember it was 1.33 but I want to double check. Yeah, it's 1.33 Great. So for this angle, I get 19.9 degrees. All right, what's I realized? I made a mistake in the calculation, so I needed to be using the index of refraction of glass and water on. I think I was using air and water. So let me revise that copulation. I'm getting the arc sine of 1.52 provided by one point three times with sign of ankle and the glass, which is 26.9. Decree is all right, so with that, I get 31.1 degrees. So all raised this and write that down. Thank you. All right. Okay. And then we want to know what would the angle b if it entered the water directly. So to do that, what we would do is, um, an air sign. Sita in the air is equal to and water sign data in the water. However, we know that on a sign, data A is equal to n g signed David G. And so then we know that that would be the same thing of solving the above equation. And so we would get the same answer. So without doing a calculation, we know that the answer would be the same.

The angle at which the light reflected is completely polarized is known as a brief stirring, which Can Burton is that it? B is 1/4 and two derived by N one in two isn't affecting nips off water. We can write that Toby is equal to sin, please in worse off Tang in Warsaw, friend to drive and one so substituting the values we have reflecting its off water, which is 1.3 t divided by inflicting dicks off and which is one so simplifying this expression gives us the angle off 3.1 degrees.


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