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The center of a cube that measures Zcm Zcm X Zcm (a volume of Scm is [3cm from convex lens of focal length ]Oem a. What is the volume Of the image? b. If the center...

Question

The center of a cube that measures Zcm Zcm X Zcm (a volume of Scm is [3cm from convex lens of focal length ]Oem a. What is the volume Of the image? b. If the center is also moving away Irom the lens ata rate of Imls_ at what rate is the image moving?

The center of a cube that measures Zcm Zcm X Zcm (a volume of Scm is [3cm from convex lens of focal length ]Oem a. What is the volume Of the image? b. If the center is also moving away Irom the lens ata rate of Imls_ at what rate is the image moving?



Answers

The front surface of a glass cube $5.00 \mathrm{~cm}$ on each side is placed a distance of $30.0 \mathrm{~cm}$ in front of a converging mirror that has a focal length of $20.0 \mathrm{~cm} .$ (a) Where is the image of the front and back surface of the cube located, and what are the image characteristics? (b) Is the image of the cube still a cube?

Hello Thing this problem were asked to find the volume off a cube. Given that one reference system is moving relative to the other in the X direction with the speed off B. So in the initial frame, let's go. It s we have this coordinate system where we have the X s X y zed and the length off the cube, The cube sitting at the origin with one with one with one corner of the origin and the we can read up the volume where we can read up the sides of the edges as the coordinates mart on these on these axes. And so since the frame is not moving, all of these appear to be a distance. A or all of them appear to be in the eighth coordinate and on the given access. So the volume is just a multiple of these readings, and so that turns out to be a huge now were asked to find the same quantity in another frame that is moving relative to this frame. So what we know is that as prime, this other frame is moving with respect to the initial frame ass with a speed of V in the X direction. So that means that in this new coordinate system we're gonna have a new coordinates we're gonna have ex prime y Prime is a prime. And given that the lengths So given that we're moving in the X direction, we know that the X coordinates are going to be different from X in the S frame. So what that tells us is that things in the export are going to contract with respect to what we see as when we look at s prime. So, in fact, the way it's gonna contract is that ex prime Theo ex prime readings on this coordinate system or in this coordinate system are gonna be the readings off. Act off the X coordinates in the S or the initial coordinate system times the square root of more minus B squared overseas word. Now, this is just the relativistic length contraction factor. And so that's what you have to use here. Um, and this is purely because we're moving in the X direction. Okay, so if we were moving in and why or zed or some complicated direction, then there would be other factors, too. But in this case, because It's so simple. The under coordinates do not change it also. Why? Prime status. White crime is the same as why Zach Crime is the same Zet. So the volume is gonna be again just the product of these readings. And so a prime which we read on the X axis and a times a what you read in the Y and Z axes respectively. Eso If you multiply this together, you notice that a prime which is the reading on the ex prime access He's gonna be eight times the square with tough one minus We squared over C squared times a square which is the product of the Y ends everything's and so multiplying everything. Together you get the volume is equal to a cube times dis relativistic contraction factor of the square root of one minus resp

This's Chapter thirty seven. Problem number forty six o. There is a cube sitting on the S free. So assume that it's exercising right of X Y Z, this is arrest Frayn and there is another cream here as prime frame. His prime frame is moving in the positive extraction. Let's close X prime. Why prime and see prime in depositor extraction with the speed of you. And is you know, the volume. He fear writing it on the s frame. The volume of this Q is given to US. Sides are given to us as a so it's going to be the length alone extraction a landfill on why and length along busy. So we're gonna have a cute for the wall you. Now let's see what the volume would look like for an observer that is sitting on the S prime frame. So the the on ly length that is gonna get shortened is going to be the length alone direction off this motion, which is the extraction and the other to the length along why NC directions are going to remain constant. So they're going to be a right. So for writing the volume has it seen in the s prime frame. We're going to put the length. Is it seen from the, uh, along the extraction? So let's call it in hell and then wire directions. Still still a Because the relative motion is not along the wide direction. Right? And time's a along Z directions for these two are going to remain unchanged. However, the length in the extraction is going to be different. How different? Well, we can write down the the L equals not over gamma equation that gives you the the new length in terms ofthe the proper length, right? Is it seen from the S rain? This is the proper length. So then they give Affect her Here is, you know, is one over spirit of one minus. You squared over C squared in this case, and now then we need to do is to reverse this because GM maize and denominator So l is going to be equal to one minus use for it over c squared times, Al Not right. And if you remember, I'll not was a so i'll not. This term is actually a So you're l is going to be spurred a one minus use words over C squared times, eh? Now all we need to do is to put it in the volume equation. Then we have spurred of one minus use. Where over C squared Time's a times a squared so combined into one minus square, root off one minus view, scared over C squared times a huge he is going to be the volume as it's measured by an observer on best price free.

So we have in convex mirror of folk of magnitude off focal length, eight centimeters, um, and where we also have an object and we want to know where Where is this object located? For an image that has I'm an image distance That's 1/3 of the object distance. We want to know things about whether or not is really a virtual or upright and inverted. So the first thing you wanna note here is that we have a convex mirror, convex mirrors. Look, it looks like this in our in our problems here. In a previous video, I mentioned con convex mirrors. They always create virtual images, and they're a radius of curvature, and their focal wings are always going to be behind. And that's part of the reason why they create virtual images just because they're virtual. Their images were shown behind the mirror due to rays of light that appear to be coming from and, um, But when we talk about convex mirrors, the main thing that we want to talk about here is that the radius of curvature and the focal length they're going to be behind the mirror. And in this textbook, whatever is behind the mirror is negative. Whatever is in front of the mirror is positive. That's what we have here. For all the questions, this is our convention. Um, and since we're given that are magnitude of of focal length, which will just frightened here is Ah, we're given the magnitude of this is eight centimeters. We can use this information to determine the sign of our focal length are focal length is the distance from the mirror to the focal point F located here, um, which is half of rays of curvature. That's always the case for whether or not it's a convex or concave mirror. So f since we since its con cave our Sorry, it's convex. Ah, I'll say convex here. And our focal in for convex mirror is going to be less than zero. We can deduce, and that is going to be minus eight centimeters. So that's what we have for our focal length now. We also were given this information on army image and object. Distant image distance Que is 1/3 hearth object. Distance P. That's her magnification. We don't know the sign of this as well. Um, So what were you need to do you need to find the sign of these two things here? So when we talk about images for convex mirrors, um, we're actually what we talk about. Object distance. An object is going to be placed in front of the mirror. That's always the case, Um, at least for these type cups types of problems. Sometimes an image is formed and you have another piece of equipment, and the image is going to be, ah, the new object. And that might be a negative distance, but usually the object is going to be placed in front of the mirror, and it's always positive in that. In that sense, soapy is going to be positive. And that's usually the case. So he is greater than zero. Well, since we have, um, this value here we want we want to see what happens to the image when you when you show this object in front of a con convex Yeah, you always put in we get The best way to do this is to make a ray diagram. So to make a ray diagram, you have one ray of light that's parallel to the principal access, which is this darker line here. This is called the principal axis, and we always draw one rail by this parallel in that, actually for convex, Maribel diverge. It will spread apart, and it will go away from the mirror like this. And what happens is that there's a ray of light that appears to be coming from behind the mirror, and it connects to this point, so it's a virtual ray of light. Now let's say we had another ray of light hit the principal axis. You can really draw any second ray of light because what will happen to it will hit this, and it will diverge as well. But then this will prepare to be showing coming from this direction here, and they have your image that's always formed when two rays of light intersect. And since these rays of Einar Virtual, this is a virtual image for convex, for convex ah mirrors. And as you can see, this image is behind the mirror, so it's negative. So he was going to be lesson zero. So we have three things now that we learned we have really learned the sign of the focal length, which is minus eight centimeters. We learned that object distances. Air always almost always. Zero r Sorry. Sorry. I almost always greater than zero. There also was positive and we learned that for convex mirrors. I'll just put this year convex. Your image is going to be less than zero. It's gonna be negative. So we can say here, Uh, since these two signs are opposite, he is positive and Q was negative. Que Yeah, let's keep it this way. Que is going to be negative 1/3. And we deduce that from physics from seeing that con context mirrors due to the way that the image is formed is always going to be behind the mirror. The images of the convex mirror always gonna be behind this marriage That would be virtual and since is behind the mirror, it's wanna have a the distance for the for the image to the mirror is gonna be the opposite sign of the distance from the object. So we have this. We can use our magnification equation. Magnification is, and Miss Smith drank it. This crap here, right, so are our equation for magnification is I m equal to minus que overpay. We know, um, that Q overpay from here is going to be minus 1/3. So let's just write it like this. We have minus que overpay for the magnification, but Q over P is going to be a minus 1/3. So we actually have minus a. Mine is 1/3 which is this positive? 1/3 and old is right in decimal notation with just 0.33 repeating. That's actually our question for B. And, um, what's the men magnification of our object or are our image? And that's actually positive 1/3 Uh, well, we want to fund for a though. Using our magnification is, uh, the distance from the object to the mirror. So where do we place the object to create an image that has 1/3 the magnitude of the object distance? Ah, So here here, we'll do this. We'll just say that we're using the mirror equation. We'll just say one overpay plus one over Q. Does he go one over f We know the signs for each of these now, so we can substitute que for minus 1/3 p. And that will give us one overpay, uh, minus 1/3 p. I'm sorry. That's actually three over minus three over P, which is equal to one over half. And this gives us minus two overpay pickles too long graph, which also is just saying that P is equal to, um minus two times F. So since we know what asset is after is minus eight, we get the object distance as positive 16 centimeters. And that makes sense because our our object is supposed to be positive, given what we thought of before. Our object is going to be a positive, because that's usually what it is for these optics problems. The object is going to be set in front of the mirror and whatever is in front of the mirror. It's a positive value I have seen on this sort of number line here, so that makes sense. That was a good verification of our of our situation. So we have the object distance. That's the the answer for party. What's the distance from the mirror for our given image? Um, we answered Part B, which was always a magnification of this image. We found that out to be positive 1/3 and we need to answer now whether or not the images upright or inverted century know the sign of our magnification, and we were able to find the sign due to, um, are signs for the image and object distance. We found this out by drawing a ray diagram and seeing that images are always show up in virtual images are always shown behind the convex mirror. And you know, objects are always set in front of the mirror, so they always have a positive, uh, object distance. But since the images set behind the merits and negative image, distance is always negative since it's set behind the mirror. And that's our convention. So we found out the sign, and we were given that the image distance is 1/3 of the object distance. We weren't given the sign, but we were able to find out that they have to be opposite. They're opposite signs for convex marriage, so Q is equal to minus 1/3 p. That was what we found out. He was equal to negative one therapy. And since we know the former love for the magnification we set, um, we've we basically said minus que overpay is you go to minus minus 1/3 which is equal to a positive 1/3 and we got that from here. So, um, when we have this positive value, that means our image is upright. So a positive value. Sorry, but this correct here, my name is will be upright since we have a positive value for our magnification. So, um, using of the ray diagram, we found out this information that we needed to solve foots this problem. Um, so, yeah, that's how you saw this, This problem.

Hi given in this problem there is a go on give or convex lens which may be shown like this. So for this convicts the surface, the radius of curvature Arvin for the concave surface, the radius of curvature are too and these radio are given us are one is equal to 6.0 centimetre and are too is equal to 10.0 centimetre. So this is for convex surface and this is for the concave surface. The object is stands in front of this lens. It's given us and we know object instance, he's always taken as negative. That is S is equal to minus 35.0 centimetre and image distance on the other side of these lens. The image is real. So this is on the other side of the lens and its distance will be taken as positive. That is positive 50 centimetre. Now, in the first part of the problem, we have to find a focal length of such uh lens. So using team lens equation, which is a relation among the object, is stands the industry stands and the focal length of the lens and it says one by S. Dash, The image distance minus one by S. The object distance is equal to one by F. Now, plugging in unknown values here, the fokker land will be given by one by S. Dash, which is 50 minus one by S, which is minus 35. So it comes out to be one by 50 plus one by 35 and then taking an L. C. Um This one by F comes out to be 15 in 2 35 and in enumerated, this is 35 plus 50. So finally this focal length here comes out to be 20.6 centimetre positive. So this is the answer for the first part of this given problem. The focal length of giving con cable convex lens. Now in the second part of the problem, we have to find the index of refraction of the material of rich. This concave convex lens is formed so to find it. We use lens makers formula which is again a relation for the focal land. But in terms of index of refraction and radius of curvature of both the sides and it says one by F is equal to n minus one. Record one by our one minus one by are too. So the focal length we know this is one by 20.6 is equal to end, which is missing and minus one for our one disease six centimeter for our two this is 10 centimetre. So yet it comes out to be one by 21. 6 is equal to and minus one and taking an L. C. M. Party, it comes out to be five minus tree, so this is to buy 30 which will remain 15 in the denominator. This is n minus one by 15. Hence and minus one will be given by 15 by 20.6. It will come out to be 0.73 so finally, the index of refraction of the lens material is 1.73 which is the answer for the the second part of the same problem. Thank you.


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