a-jeweler-examines-a-diamond-with-a-magnifying-glass-if-the-near-point-distance-of-the-jeweler-is-22

Question

A jeweler examines a diamond with a magnifying glass. If the
near-point distance of the jeweler is $22.8 \mathrm{cm},$ and the focal length
of the magnifying glass is $7.70 \mathrm{cm},$ find the angular magnification when the diamond is held at the focal point of the magnifier,
Assume the magnifying glass is directly in front of the jeweler's eyes.

Nathan Silvano · Accepted Answer

So in order to solve this problem, we just need to know that when a Diamandis place of the focal point of a magnifying bless the magnified image will be created at infinity. So knowing this, we know that equation we need to use. It&#x27;s just an divided by F, and the problem gives us what is the near point and what is Defoe Kel lent off the magnifying glass. So the new point the problem tells us, is 22 0.8 centimeters and the focal lens. The problem tells us that a seven point seven think team enters. So the magnification, the angler magnification just going to be to a point 96 to dis glass.

Adnan Gill · Answer

so be airport that for the Children, er the year point and is could do 72 centimeters. And we also know that the object is placed four centimeters from the lens and we have to find the angular magnification. Angular magnification is approximately equal to one over F minus one or D I times of the year point now one over F minus one over D. I is actually one over deal and we can see that very clearly. One over deal plus one over g I is equal to one over f task Italians formula. So if I were to subtract one over t I from both sides, that would give me one over. Deal is equal to one were F minus one over d I. So I can rewriting magnification. Angular magnification formula has won over t o times and so this would be equal to 1/4 times 72 the answer would be it

Nathan Silvano · Answer

Okay, So in this problem, we have a similar problem off the problem. 55. But this time the diamond is held on a distance that&#x27;s put to hear the distance off the object or five point 59. Same team enters, and we know. What else do you get? We know that the focal length off is magnifying less is seven 0.7. Think emitters. We know that the near point of this glass near distance is 22 0.8 ST emitters, and I think this is it. What else s o? What the problem wants to discover the problem, wants to discover a magnification, the angular magnification off this glass. And we know by definition that the younger magnification is just the ratio off the and go off the image divided by the Anglo off the object. Okay, so we do not have this to end goes in here. So we have to find a way to discover them first. So then we can calculate the magnification. And how are we going to discover these angles? We know the for definition. The Anglo off the object is defined as their height off the object divided by oops, there near point business that&#x27;s put in in here, okay? And the angular, the angle off the image is just the height off the image divided by the distance modeler. Distance off the weather images form okay, so again we have a problem because we do not have neither the height of the object need of the height off the image or the distance off the image. So let&#x27;s begin with the distance off the image. We know since we have the distance off the object and the focal lens, we can use the feelings equation and calculate the distance off. The image is just one divided by F miners, one divided by the zero oh, to the power off my notes. One. It is going to be too warm, divided by 7.7 minus one, divided by five point 59 Oh, this the power of minus one off the calculate this, we&#x27;re going to have minors. 20 0.4. State team enters Now that we have the distance off the image, we can remember in Chapter 26 that there is a way, really, with correlate the height in the distance off the object, which is Let&#x27;s put to hear the equation is just h. I equals miners. He I divided by the zero, multiplied by 80 So if we use this equation, we can discover that they let&#x27;s put in here the height off the image. Just going to be miners. Miners 20.4 divided by the D zero, which is five point 59. Let&#x27;s rewrite. This goes, though. Okay, again, we have my news minus 20.4, divided by five point 59 this actually 51 59 80 because we do not have 80 Yes, well, so this is just going to be three point 65 80 So now we have a correlation between the height of the image and the height off the object. So let&#x27;s calculate the let&#x27;s remember, we cannot calculate the angle off the image, which is going to be angle off the image she&#x27;s going to be. H. I. Divided by modular off age off the I, this is going to be three point 65 80 divided by 20.4, and the angle off the object is just 80 divided by 22.8 So now finally, we can calculate a magnification. The magnification, which is that&#x27;s a flash divided by Peter. He&#x27;s just three points. 65 age zero. Oops, that&#x27;s Montana. Send Caller 80 divided by 20 points. Four. Divided by age zero. Divided by 22.8. This is a eight. So now we can cross the ate cereal and the angler magnification is just going to be four 0.0 eight. And that&#x27;s define our answer to this problem. Thanks for watching.

Ryan Kutayiah · Answer

in Question fifty six. We&#x27;re told that a thin lenses to be used as a magnifier and the lens has a focal length of six centimeters. We&#x27;re asked to find the magnification, assuming that the object is at the focal point, the angular magnification. So we&#x27;LL use Equation thirty four point two two Ondas the holds for when you have an object at the focal point, and this gives us thie and get her magnification. So that tells us that it&#x27;s twenty five. San Amir&#x27;s divided by the focal length of the lens that you&#x27;re using in this, uh, we have F, which is six hundred years, so we have twenty five centimeters over six centimeters, which gives us an angular magnification of four point one seven in Part B. We&#x27;re told that an image is formed at the near point, said that as prime does twenty five centimeters and let me go ahead right away, right? But that&#x27;s negative. So when you have a magnifier, the the image is is a virtual image. So we have ask. Prime is a negative number, and the lens I distance, or the distance of the leads to the eye, is very small so we can neglect it and were asked to find essentially the object distance. So really, in the question they&#x27;re asking you out. Closure you Khun of bringing? No, because can the object be brought to the lens? So that&#x27;s that&#x27;s the object distance s that were asked to find here. And you can use Equation thirty four point one six which is one over s plus one over ass Prime is equal to one over f and we can just solve for death which is s prime f over Ask prime minus f, both of which we know as as prime and F so as prime is minus twenty five centimeters apus six centimeters and then we have minus twenty five centimeters, minus six centimeters and that gives us four point eight four seven years.

Adnan Gill · Answer

So for our it, we are poor that the image is 25 centimeters from his eye and it&#x27;s a virtual image. So the I is equal to negative 25 centimeters, and this also is his near points of the near point is 25 centimeters Long live. We&#x27;re told that the focal length off the magnifying glass is using is 9.50 centimeters, and we are asked to find Dio. So the Taelons formula says one over Dio plus one over the I is equal to one over F or one over. Eagle is equal to one over F minus Wonder work D I. So that&#x27;s going to be won over 9.5 minus one over negative 25. So solving this for D o we get deal is equal to 6.88 centimeters. Now for Part B, we want to find the angular magnification, which is equal to one over F minus one over D I times the near point and one over F minus one over the eyes actually would do one over deal. So you can also say that angular magnification is one or two your times, and so this would be 1/6 0.88 times 25 so the magnification is equal to 3.60.

Netra Sharma · Answer

So this problem we have the focal length of the magnifier is five centimeter. So in part A When the object is placed at the focal length or a focal point, the eye is relaxed and the image will form at the infinity and the angler magnification. In this case, if you follow the textbook in the angler, magnification and excuse will be 25 centimeter over f. Okay, so this is very straightforward. 25/5, which is going to be five now for part B. When the image is formed at the normal near point, which is cuticles negative 25 centimeter. This is when the angular magnification is actually going to be a maximum. And in this case, the angler magnification is going to be M. Max because one plus 25 centimeter over f and this is going to be one plus. You can just follow from part a 25 centimeter over F five and this is gonna be a six. So the maximum magnification is six. So for Parsi now for maximum magnification using cuticles negative 25 centimeter. And we know the focal length is five. Senator, we can just write down p, which is the object distance as que times half over Q minus F and this is going to be 25 negative 25 times focal length, which is five over negative 25 minus five. This will give me a 4.17 centimeter, which is the object distance in this case.

Problem

In Problem $55,$ find the angular magnification w…

Problem 55 Easy Difficulty

Answer

2.96

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