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Questijon 91otsThe focal distancc for a convex lens 80 cm: What kind of Image produced when the object distance is greater than 160 cm?inverted and sqalletInvcrted ...

Question

Questijon 91otsThe focal distancc for a convex lens 80 cm: What kind of Image produced when the object distance is greater than 160 cm?inverted and sqalletInvcrted Jnd hgeruprighi and sraller

Questijon 9 1ots The focal distancc for a convex lens 80 cm: What kind of Image produced when the object distance is greater than 160 cm? inverted and sqallet Invcrted Jnd hger uprighi and sraller



Answers

An object is placed $80 \mathrm{~cm}$ from a screen. (a) At what point from the object should a converging lens with a focal length of $20 \mathrm{~cm}$ be placed so that it will produce a sharp image on the screen? (b) What is the image's magnification?

In this problem, you're given a lot of information about the image. It tells you that it's really it's inverted, that it's one point. It's in the meters tall and that it is located some 1.0.4 centimeters from the lips. So we can first a solve for the distance of the object How far the object is from the lens. To do this, we can use the thin lens equation, which is one of her dio plus one of three d I equals one of our F and we want to solve four d Oh so we could go ahead and move D I to the other side and then to sulphur d o. I do need to inverse the entire fraction. So this becomes one over 1/6 0.8 minus one over d. I just 10.4. And then when you get the inverse of that, you get a distance of the object as 19.6 centimeters. Now that you know that the distance of the object, you can use the magnification equation, which is H I over h o equals negative deep I over d o. Now the fact that it's holding it was inverted tells us that that 1.8 is actually a negative value. Or you could absolute value both of the sides and then just work with the ratios. Either way, we'll get the same answer. But I'm gonna go ahead and use that negative. So we are asked to solve for each o. So to do that, we not only wanted by itself, we also want it at the top of the fraction. So I'm gonna go ahead and multiply h o on both sides, and then I'm going to get a job by itself. To do that, I'm gonna bring Dio up d I down and then I'm leaving h I write stuff again. This h I is a well, hold it at a second. So this becomes 19.6 up on top, divided by a negative 10.4. Now that negative is really just part of the fractions not attached to d I. And then each I is and negative 1.8. You could have absolute value this and you would've gotten the same answer. So 19.6, divided by 10.4 times in one we eat gets you ahead of the object of 3.4 centimeters. So looking at this, if you have the height of the object is 3.4 centimeters and the height of the images 1.8. This was a reduction.

All right, so we have to find ah, distance and height of the object. So we're given image distance on Were Given Focal Inc So we have that one of her focal ing is one of our an object, a senseless woman for image distance. So one of her image distance are rather one of the object distance we want here is equal to one over F minus one over X prime ad s prime will be negative, 17 centimeters. And that's because it's on the right side of the lens. Nah, as they are right side of lens. And so it's a conversion lands. And if you're on the right side of this converging lens, your value you will be a negative distance. And that's just the saint invention we use. So it's a negative there. So we put those in for going to 12 centimeters minus one over minus 17 centimeters of the first sister. Plus here of the values and so s the image. This object distances seven centimeters from the lens on then and then we can figure out magnification from that. It's negative image, distance over object distance. So that's minus minus 17. That's a plus 17 over seven. So magnification is plus 2.4 uh, plus 2.4 times. And so and so this is also equal to image height over object. Hi. So, object height is just image height of the magnification. And so that's Tio. Uh, oops. That's 0.8 centimeters. It's eight millimeters over 2.4. So its point object height as 0.36 centimeters. Whoops, that's ah, 4.34 centimeters or 3.4 millimeters. So since M is greater than zero Ah, this image is direct and not inverted on DH in this instance s prime is negative and s is a CE prime is negative and s is positive. This implies that the image an object are on opposite sides So object will be crowd mission object on opposite side so object will be on the left and image will be on the right

This is a problem involving a diverging lens. You're asked to find the, um, image distance que given certain object distances p in the problem. Give five different distances along with at least one solution. Then you can use the same method for solving all five. So you're gonna want to settle for a Cuban the end. So the several different ways you could do this plug in the numbers first in the algebra. I'm gonna go ahead and do the algebra first on the equation and then, uh, use that to get the solution for at least one. So I'm gonna sell for Q The the lens equation. Eso first I subtracting one of repeat the other side common denominator on the left hand side would be FP. And so you're going to have p over FP minus, um f over FP, uh equals whenever cute or P minus F over FP equals one of a cube and then you're in verse. Both sides when you get F p over a P minus, f equals Q. So I'm gonna use that equation you're given after the problem you're given, As I said, the various p values. So I'm gonna go ahead and plug in for the first one here. F negative eight times five p over P five minus minus eight is gonna equal que and so you get negative. 40 over. This becomes plus positive over 13 equals Q or when you do the division there. Q is equal to three 0.8 centimeters equals Q. And so that would be the answer for the five centimeter distance. Now use the same approach with all the other answers. Um, on when I did that, I got the following numerical answers. By the way, this should be negative is the negative sign is still there. So negative 3.8 So for the second one and I solved it, I got negative for 0.0 centimeters for the 14 centimeter object distance. I got negative 5.9 centimeters for the 16. I got negative 5.33 centimeters and for the 20 I got negative 5.71 centimeters. And so these air the answers for each one of those object distances. If you use the same approach as I did for the first one, it also asks him that problem about the UM it says in each case described the images as riel or virtual operating burden, and so on. Um, because it's a diverging lens, there's only one set of possibilities. Diverging lens will always produce images that are virtual upright and diminished in size. Alright, diminished. And so those are the answers for that part of the question. There's also a part B for part B. You're asked to find the, um, uh, image height for each one of these Not for each one of these, but for two of the scenarios. So you have to know that h i image height over h o. The object tight is equal to negative que overpay. And you're asked to do this for two different values. One when P is five and 11 p is 20. So you're gonna solve rage. I I'll do this for the five centimeter position first, So h I is what I'm looking for. They tell you the image height is four centimeters equals, um, for at five centimeters we had up here this answer S R Q. Value is negative of negative 3.8 divided by P is five. When you do the math on this and self h I, you get the image. Height is equal to 2.46 centimeters. So that's the answer for that part. Then finally asked me to do the same thing at 20 centimeters. So h I over four. 27 m. We have this value up here so negative of negative. 5.71 divided by 20. You solve that for H I during the algebra and you get 1.14 centimeters. So that is the answer for the second part of part B.

In this particular problem, it is given that does image sizes. You know 0.75 of the object size, So magnification is hi does image over height off object equals 0.75 but that also comes out. Toby Images distance over object distance. In this particular problem, the object distance is given by the image suggestions. I'm sorry, the images distances given by 24 centimeter, so the object distance will be 24 centimeter, divided by 0.75 so that will come out to be 32 centimeters. So in order to find the focal point of the dance, so you use the land situation. One of work ethic was one over object distance plus one over Amos Distance, which is 1/32 plus 1/24. So focal then should be won over 32. Thus one of 24 in verse off that which is 32 inverse plus 24 inverse inverse off that is 13.7 to 1 centimeter


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