predictcalculate-an-obiect-is-located-to-the-left-of-a-convexlens-whose-focal-length-is-34-mathrmcm-

Question

Predict/Calculate An obiect is located to the left of a convexlens whose focal length is 34 $\mathrm{cm}$ . The magnification produced by the lens
is $m=3.0 .$ (a) To increase the magnification to 4.0 , should the object
be moved closerto the lens or farther away? Explain. (b) Calculate the
distance through which the object should be moved.

William Affel · Accepted Answer

All right. So for this question, we have the Comdex lens of focal length of 34 centimeters on object distance that is unknown. That&#x27;s creating an image of unknown distance. Except we do know that the magnification is initially three we&#x27;re trying to do is raise it to four. And what we want to know is that if we&#x27;re raising magnification from 3 to 4, do we have to move the object closer to or further from the lens? And we want to find that out initially, just analytically, without any calculation. And then we&#x27;re gonna find that, actually, by how much we actually have to move the the object. So in order, find it analytically were to write down our lens equation. One over f is gonna be equal to one over Dio plus one over D I I&#x27;m gonna write down a magnification equation Magnification. It&#x27;s gonna equal three initially, which is equal to negative. D I over dio So what every one of these equations initially tell us? Well, the first thing you figure out is that the eye is virtual. It&#x27;s not really That&#x27;s because our magnification is positive. If our magnification were negative, then we have a real image, but it&#x27;s positive and we have a convex lens. So we have a virtual image. Next you want to know. Is it because we have a virtual image and we have a convex lens? The distance of the object must be within the focal length. That&#x27;s the only way that we can create a virtual image of the convex lens. So there two things we can find out just by looking at our equations. The next thing we could find out is we have to analyze. We realize we&#x27;re trying to raise the magnification from 3 to 4. We&#x27;re trying like this number here bigger. So in order to do that, we must obviously have to make the numerator of this fraction bigger. We&#x27;re trying to increase d. I essentially make d I a greater and negative number. So how we&#x27;re gonna do that? We&#x27;re Look at our lens equation. We&#x27;re in. Rearrange it actually were to say one over f minus the object distance one over the object distance. Excuse me as an equal to one over the image distance, so this should show us exactly what we have to do in order to make the image distance a greater negative number. All right, so in order to make the image isn&#x27;t a greater negative number, we need to make its inverse a smaller negative number. Try and follow my logic here. The M assistance has to be greater. So it&#x27;s inverse. Must be smaller because this is a difference here in order to make the smallest number possible or a smaller number as faras. This inverse goes the universe of D I that we need to make the inverse of F and the inverse of dio as close together as possible so that their difference is very small. Does that make sense? We need to make the inverse of F in the universe of Dio very close together numerically so the difference is small to create a small inverse for one over d I. So if we&#x27;re doing that, we&#x27;re trying to make this smaller than it already is. We must get dio closer to deep tow f than it already is. We know Dios on the near side of F come when compared to the lens, so you must be moving away from the lens. So the answer for a is away from once, However, what we want to know next is by how much are we moving away from the lens? And so we need to do some math in order to figure that out. First math that we have to do is you have to set up our lens equation. Except now we&#x27;re going to plug in the numbers of me now. So I would say one over half is just gonna be won over 34 centimeters. That&#x27;s going to be equal to one over Dio plus one over D I. We can use our relationship of the magnification being equal to negative D I over Dio. In order to create a new relationship between d i. India, we can multiply my deal on both sides and say that three times dio people to negative d i If we plug that in to our lens equation, we can actually solve for a value of Dios. We can say 1/34. It&#x27;s gonna be able to one over Dio minus 1/3 dio which should give you 2/3 dio Now if we do this correctly will say 34 it&#x27;s going to be equal to three D over two imploding and or a calculator. We should get an initial object distance of 22 0.67 centimeters. Now that&#x27;s where object is initially placed. In order, Teoh create a magnification of three. We could find out where our images using this number and our lens equation, but we&#x27;re not going to cause the images not of importance in this question, were only trying to figure out why, how much we change the distance of the object from the lens. So that&#x27;s our initial distance of the object from the lens. Let&#x27;s find our new distance when our magnification isn&#x27;t three anymore. But it&#x27;s now four that&#x27;s gonna change up our math just a little bit. We&#x27;re still going to say 1/34 is equal to one over Dio, except now it&#x27;s gonna be subtracted by 1/4 Dio and that will give us 3/4 Dio you then say 34 is equal to four dio over three and I&#x27;m gonna make this a dio Prime causes are new dio so dio prime should be equal to 25.5 centimeters and this is the new distance of our objects in order to raise the magnification from 3 to 4. Now, this isn&#x27;t our answer or answers the difference between these two. So we still need to find that out. Where I say Dio prime minus dio is gonna be equal to 2.8 centimeters approximately. And we already answered part A. Which means that we know that we&#x27;re moving 2.8 centimeters away from the lens and this is our answer.

Adnan Gill · Answer

so far our party to change and from four points year old to negative four points year old object should be moved. Thank you away from thence because as the object gets away from lens Andy, the image will become work it and riel now for part B. So initially we know that magnification is equal to 4.0 and magnification is equal to negative. D I over deal. So if we say four is equal to meditate the eye over deal, then we can say that the I is equal to negative four people. So using the Finland&#x27;s Formula One over the old plus one over t, I is equal to one over f. We&#x27;re trying to solve this for a while. So what were the oh minus 1/4 0 is equal to one over f and that gives me 3/4. Deal is equal to one over f and that would give you the old is equal to 3/2 over four. So with Mac magnification 4.0, the distance off the object from the land should be three f over four. Now we will do the same thing for him is equal to negative for. So now we have em is equal to negative for 20 year old, which leads that negative for is equal to negative the eye over deal. And that would mean that the I will not equal to four d oath. So using the mainland&#x27;s Formula One over the old plus 1/4 Dio isn&#x27;t going to one over F, and that is five over four. Depot is equal to one over F, and that gives deal is equal to five f or four. So the object distant through which the object moves distance object moves is equal to find f over four minus three F 44 which will be equal to F over to. And the question tells us that F is equal to 30 centimeters, so 30 centimeters divided too close to 15 centimeters, or X equal to 0.15 meters.

Nathan Silvano · Answer

Okay, so and this problem we have a student with to magnify glass. Correct? Yeah. The student has two lenses and the first lens as a focal end off five centimeters. And the second chance as a focal length off 30 dirty ST emitters. So five centimeters and 13 70 metres. Right. Well, there first itin off this problem. We need to explain which off this lances we&#x27;ll provide, um, the greater magnification. Okay, So first of all, we know that the magnification gration for magnification just one plus and divided by f So they lens there has they&#x27;re smaller. Focal lens is going to provide the greater magnification. So a smaller focal lens is F one. So following this theory, F one should be the greater magnification. And what is the second thing we need to calculate? And here, the second item sporting here the second ison. We need actually to calculated the magnification and we know that the near point distance is 25 20 meters. So to the glass one that&#x27;s portend blue. A glass one. We have magnification one off, one plus, but five divided by five. This is just going to be six to the glass too. Think less true. We have magnification to with one less hopes. One plus 25 divided by 13. This is going to be a good to point knife, and we prove that our answer was correct.

Nathan Silvano · Answer

Okay, so in this problem, we have two lenses. The first lens has a focal length off to a point six. Same team enters, and the 2nd 1 has a focal lens off 20.4. Same team enters. And in their first itin of this problem, we have to predict how you we would arrange this lens to have a magnify image of the moon. Okay, so remember that this is the first item I think a Remember that to get her magnify image, we have the folk Oh, objective. Divided by the focal. I piss and always remember that the focal objective needs to be bigger. Then the focal eyepiece. That&#x27;s how we built a telescope. So since we have in this problem that I have to is greater than F one, we can say for sure that the telescope has objective equals the F two, and we have the focal I p&#x27;s You closed the F one and that&#x27;s the answer off the first item. The second item. We have to calculate the magnification. Okay, so we already know how to calculate. It&#x27;s just dividing objective by eyepiece. So we&#x27;re gonna get here 20 point four divided by 2.6. This is going to be equal to seven point 85. And that&#x27;s the magnification off this telescope. The turn itin his detergent itin. In this item, we have to rearrange the lens to Butte a microscope. And in the microscope, we have the inverse of the telescope we have that the objective needs to be smaller, then the eyepiece. So here we have objective equals F one and the piece it close F too. You just need to remember that the microscope works the opposite way off the telescope. Okay. Ah, Now we have to calculate the distance off the image off this off this microscope. So what&#x27;s pushing here? The distance off the image we know using the feelings of creation. The distance of the images. Just one divided by the objective miners one divided by zero, all to the power off minus one. So we&#x27;re gonna get the I equals one divided by 2.6 miners. One divided by two points. Nine oh, to the power off miners. One, This is equals two 25 saying team enters and now we can calculate the Soto magnification. Actually, the last thing we need to calculate the magnification. So let&#x27;s circulate in here. It&#x27;s Toto magnification defined by miners. Do I times and divided by the focal objective and the focal I piss. So this is just miners 25 times 25 divided by 2.6 times 20.4. So this is minus 12. Same team enters. I think you manage 12. This is the magnification. We don&#x27;t have unities. So the magnification is just miners 12 and that&#x27;s the final answer.

William Affel · Answer

All right. So for this question, we have two lenses at our disposal, one lens of a focal length of positive, 40 centimeters and another lens of focal length of negatives. 40 centimeters are questioning. They wanna answer is if we&#x27;re trying to project an image onto a wall that is some distance away from the lens Which lens should be used? And this is a really easy question. Answer. Because all we have to figure out is, uh, which lenses con Ah, vex. Excuse me. Do you want the converging lens? We want the ones that will be able to create a real image. Diverging lenses are able to create real lenses. There are real images. They&#x27;re only able create virtual images. So the positive 40 centimetre one is the con vex lends, and the negative 40 centimetre one is the con cave ones. And we know this because when we set up our lens of quite equation, conclave lenses always have negative focal lengths. So you want the one with the positive focal length the convex lens, because that&#x27;ll be able to create a real image that we can project onto a wall. So we want the one with the positive 40 centimetre with lens deposited for Senator your full length, that is our ones. So now that we know we want this lens, next time you want to know is where do we place the lens and we&#x27;re to create an image that has a magnification factor of two. So it doesn&#x27;t mean we have a magnification of to. It means that we&#x27;re increasing the images of size by a factor of two, whether it&#x27;s real or virtual. Now, we already know that we want to create a riel image, so we know where modifications will be negative. So I would say our modifications gonna be negative to create a real image with a mask, a factor of two, and that&#x27;s going to equal to the negative image distance over the object distance. Now we can drop the negatives from the two and the you have negative image distance and say two is equal to humans. Distance over the object distance, using the fact that our focal lengths 40 centimeters we set up our lens equation over 40 centimeters is gonna be equal to one over dio most one over d i to solve for either D or D I can use this system of equations here. So what we want to solve four is ah di I The distance of the image since that will tell us where we want to put our lens. So I want to say to Dio is gonna be equal d I and we&#x27;re in a substitute out dio for 1/2 d I was a dio it&#x27;s gonna be 1/2 d I So we&#x27;ll say 1/40 is gonna be equal to two over d I That&#x27;s just me substituting in 1/2 d I infer one over Dio plus one over the I don&#x27;t give us three over d I. So now we can say that 40. It&#x27;s gonna be equal to D I over three in the image distance most thusly be 120 centimeters. So this is where replace our lens because it 120 centimeters away from the wall that were to create an image that has a magnification factor of negative too. And that&#x27;s our final answer

Nathan Silvano · Answer

Okay. So in this problem we have a librarian that has a near point off 0.5 and meters and a five point off five meters. Okay? And we know that he cannot see distance closer to 0.5 our distance farther off five meters. He can only see between this range. And we need to design a by focal glasses with two different refractive powers in order to correct both his near sightedness and farsightedness. So let&#x27;s begin the problem. The first item off the problem. We need to tell what kind of lens we&#x27;re going to use to correct his near sightedness. Well, near sightedness is when the person can only see close objects. You can only he cannot see distance objects. So to direct this kind of problem, we need Die Virgin lens to correct his near Sinus because he reproduce image of a distant object. Ah, he&#x27;s uncorrected Farpoint. And to correct his farsightedness, we&#x27;re going to need a comb virgin lens because the correct the converging lens I used to correct the vision off people who can on Lee see distance objects and cannot see clothes closer objects. So now let&#x27;s calculate that the refraction, the reflective power, the reflective power off each one of those lenses. So let&#x27;s see, he ordered to correct the Farpoint. The actual Farpoint of this person is five meters Now. We need a Farpoint so this person can be able to see objects at infinity. And we know that the refractive power is just one divided by F and the actual Farpoint of this person when he produced the image is at five meters. But remember that distance off the image always need to be related to the glasses and not to the ice. Five meters is related to the ice, so we need to subtract the distance of the glasses off 0.2 Now we have that reflective power is just one divided by the zero plus one divided by the I. This is going to be one divided by infinity, plus one divided by miners. Because this is this is of the image owner converts converging type of lens five miles, you know point 02 So finally we have the reflective power off the off. The first lance is going to be miners zero point two Diop tres. Now that&#x27;s calculated the reflective power or the converging lens to correct the near point. We also have one divided by F one, divided by zero plus one divided by the I. The zero is ready. Images now farm. So is the near point. The standard near point is 25 centimeters, or 0.25 meters. Again, we need to subtract the distance off the glasses. Zero points 02 and the distance off the image is just 0.5 minor, 0.2 So putting the values for Miller we have 0.25 mile, 0.2 minus one divided by minus 0.5 miners. You know, points, you know to and this gives us are effective. Our sports in here. Refractive power fraud ical virgin lands off positive 2.3 Diop tres And that&#x27;s the final right.

Problem

Predict/Calculate You have two lenses at your dis…

Problem 73 Hard Difficulty

Answer

(a) see work
(b) see work

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