pricey-stamp-a-rare-1918-jenny-stamp-depicting-a-misprinted-upside-down-curtiss-jn-4-jenny-airplane-

Question

Pricey Stamp $A$ rare 1918 "Jenny" stamp, depicting a misprinted,
upside-down Curtiss JN-4 "Jenny" airplane, sold at auction for 
$\$ 525,000 .$ A collector uses a simple magnifying glass to examine the
"Jenny, " obtaining a linear magnification of 2.5 when the stamp is
held 2.76 $\mathrm{cm}$ from the lens. What is the focal length of the magnifying glass?

Nathan Silvano · Accepted Answer

Okay, so this problem lens produces a ritual image with a magnification magnification off 2.5 times over stem that is held at a distance. So we have distance off the object off two point 76. Saying team enters and we want to calculate the focal length off this lens. Okay, so let&#x27;s find out. How are we going to calculated focal? And by definition, the focal length should be one divided by zero plus one divided by the I out the power of minus one. But the problem here is that we do not have that the I So we&#x27;re gonna use the magnification the problem gives us because we know that the magnification is just minus. Oh, dear. I divided by zero. So therefore we can conclude that he I is just miners 2.5 times two point 76 which gives us a photo of minor 6.9. Same team enters. Okay, now that we have distance of the image and in the distance off your object, that&#x27;s calculate focal lens or their focal left is just one divided by 276 That&#x27;s do it again greets again. So we have two point 76 miners, one divided by 6.9 all to the hour of minus one. She gives us 4.6 saying two meters, and that&#x27;s the final answer to this problem thankful.

Dading Chen · Answer

So the first question was asking us, What is the anger Magnification off the device. What? First we know the formula for the angry and magnification physical state. Our prime with Ada is also equal to 25 centimeters over down the Fort Collins. OK. And the focal lenses. People looking at 1.5 centimeters so that angler magnification is equal to and equal to 25 centimeters over 11 point by seven years, which is equal to 2.17 Okay. And the second question was asking us how high should the stand hololens about the stand, Which is means the question is looking for e, uh, object distance. Okay, so since the image is at inability, there is the image. Distance is equal to infinity, and this will give us one over half is equal to one over infinity, plus one over object distance, which is will give us one over ash. It&#x27;s legal to one over you. Okay, remember the focal answers skimmers, gunpoint by seven years. And in this case, it means that you easy equal to have for the kind of this equation here. Okay, so this means that the object this is which is the high in this case, is equal to half. Just also, you go to 11.5 centimeters. Okay, so these are these other solutions questions and thank you.

Julie Farhm · Answer

So in this case were given the magnification of the lens because the stamp is 1.5 zero times larger. Um, van object magnification is positive on 25. The location of the object relative to the lens, which is the object distance, is 2.84 centimeters. They&#x27;re asking us to find the location of the image and the focal length of the lens. The problem just has to find the focal into the lens. But in order to use the lens equation, we need the image location. So we use the formula for magnification. Um, which is negative, que overpay to find the image distance. So 1.5 positive is equal to negative Q over 2.84 So rearranging this equation we can&#x27;t queue is equal to negative 4.267 years. So now we can use the lens a creation, which is one overpay plus. One of her cue is one over the focal ing, so R P is 2.8 for 1/2 0.84 Our Q is negative 4.26 nats equal to one over F. So whenever it&#x27;s numbers like this, where you can&#x27;t find an easy common denominator. I just plug it into my calculator using the reciprocal key, Um, and then take the reciprocal of your answer. So my answer comes out to 8.52 centimeters. This is a positive focal length, which means it&#x27;s a converging lens, which makes sense because an object health close to a converging lens is a magnifying, less so. 8.52 centimeters.

Netra Sharma · Answer

so we have the focal length off a magnifier is seven point point five centimeter. Also, it is given that the virtual images produced by this magnifier at the normal near point, that means the image distance is going to be a negative 25 centimeter. So for part A to find up. Victor says P. We just used the lens equation, which is water or F, because whatever p plus one of acute and from here P is going to be Q times F over Q minus F, and we know the value of Q, which is negative. 25 negative, 25 times 7.5 over negative, 25 minus 7.5. So this is going to give us 5.77 centimeter as the object distance. So to go to part B, when the image of the object is formed at the near point, then the magnification is maximums. So in this case, the magnification is going to be a max, and that&#x27;s going to be cool to one plus 25 centimeter over for the length of the magnifier magnifier. So this is going to give us one plus 25 over 7.5, and this will give us 4.33 which is the angler magnification

Kristela Garcia · Answer

in this problem were asked to calculate the distance of the object for a lends with focal length five centimeters where it&#x27;s producing images at negative infinity, negative 200 centimeters and negative 20 person. So to do that, we can use the thin lens equation, and we want to sell for one over D O. And then we want to go ahead and go back insult for just deal. This means inverse ing everything the entire fraction. And then we&#x27;re gonna be repeating this exact same process for all three situations. So when you have one over infinity, that basically gives you a very, very tiny number, effectively zero. So for this 1st 1 you get a distance of the object of five centimeters to calculate the angular magnification you want to use the near point or the human eye. It&#x27;s about 25 centimeters, so we can use that. It&#x27;s the 25 centimeters divided by five centimeters gets us. But as our magnification for party repeating the same process for Part B again using the same equation and I&#x27;m plugging in a new value of negative 200 I get one over. This gets me in value of 4.88 centimeters as my dio and then magnification for that would be again using that 25 centimeters divided by 4.8. It&#x27;s me magnification of 15.12 and then for our last one gets me a distance of the object of 4.17 centimeters with a magnification of six.

Kristela Garcia · Answer

from You want your magnifying lens? A. Have a magnified a magnification of four and when it is placed when an object displaced 2.5 centimeters in front of the lens. So what does your focal length need to be? Two accomplices. So for this you can use the magnification equation, which is H I over h o equals negative D I over D O. And we can use our only this portion of the equation and this portion of the equation. Since we don&#x27;t have height information and then we can say m equals negative d I over d o so we can bring d o up. And then we can just divide but a negative one or multiplied by making it mathematically the same thing so that we can do negative 3.5 or negative times 3.5 times four gets us a distance of aren&#x27;t image of negative 14 centimeters. This negative just tells me that it is a virtual image that cannot be projected onto a screen, and we can use that information to solve four our focal length by using the Finland&#x27;s equation. So one over D. Ope was one of her D I equals one of her f. Now, when you&#x27;re solving for your focal length, you do want to inverse the right hand side. And while you do that, you need to inverse the entire left outside. You can&#x27;t individually inverse. Um, so then you can plug in your numbers. This becomes 1/3 0.5 plus one over a negative. 14 senators, if you want to cheat, you could do just this bottom portion of the equation first to get you one over 0.214 And then you can inverse that question again and get a vocal links of or quite seven if you round up centimeters.

Problem

Predict/Calculate $A$ Big Eye The largest eye eve…

Problem 89 Medium Difficulty

Answer

4.6 $\mathrm{cm}$

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