to-engrave-wishes-of-good-luck-on-a-watch-an-engraver-uses-a-magnifier-whose-focal-length-is-875-mat

Question

To engrave wishes of good luck on a watch, an engraver uses a
magnifier whose focal length is 8.75 $\mathrm{cm} .$ If the image formed by
the magnifier is at the engraver's near point of $24.6 \mathrm{cm},$ find (a) the
distance between the watch and the magnifier and (b) the angular magnification of the engraving. Assume the magnifying glass is
directly in front of the engraver's eyes.

Nathan Silvano · Accepted Answer

Okay, so and this problem we have, uh I&#x27;m not gonna find glass with, ah, focal lens off eight point 75 It&#x27;s anti mentors producing a virtual image at a distance off. Let&#x27;s see, 24 0.6 same team enters and their first itin off this problem. We want to calculate it the distance off the object that is the watch to the lens. So the distance off watched the lens. Just if England&#x27;s equation one divided by f this f it&#x27;s really bad one divided by F my noose one divided by D I. All these to the hour of minus one. Okay, so the distance of the object is just going to be one divided by 8.75 minus one divided by drinking 4.6. But remember that distance since this is a visual image, we need to put a minor signing here. So this is all right of minus one. The answer to this is just six point 45 70 meters. So this is the distance off object. Now we have to calculate the angular magnification. So let&#x27;s see the angular magnification we know that is calculated by one plus and divided by F. It&#x27;s happened to my f sturdy, divided by F. So let&#x27;s see. We have here one loss 24 0.6. Divide it by eight point 75 So this needs to be equal to three point 81. So this is there magnification off this.

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.

Nathan Silvano · 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

Dading Chen · Answer

Okay, so for question one is asking us to determine object distance. Okay, So? Well, we know that one over half is equal to one over V plus one for you and we know B, which is the image distance. And it&#x27;s good skimming has snaked 25 point hold centimeters because it&#x27;s on the other side off the lands. Okay, that&#x27;s why is inactive and the focal Ian&#x27;s askew as a point of closing. So we know that the object lessons can be equal to be a over three months. You&#x27;ve replied in the battles will get be just now, get 25 0.0 centimeters times a point. Oh, centimeters. It was just you got to be. He was 25.0, Centerview mirrors minus a point. Oh, sending years and this can give us that you which is the object distance can be able to 6.6 centimeters. Okay, for question be it was asking us what is the Heidel image. Okay, so we know that the formula off a magnification it is, and because violence be over you, which people the HPAI over age is just actual high is finds the height of inch. So we know that, uh, the actual highs along millimeter Okay, so it&#x27;s prime can be equal to minus B over you times age, which can give us minus, uh, 25.0, centimeters. Oh, you. Which is 6.67 years times that actual high, which is one millimetre in this case. I&#x27;m sorry. In this case. Ah, things really easy with minus 25 uh, centimeters. OK. Oh, ah, me If you do. So minus 25 centimeters. Okay. And also, there&#x27;s a minus. Sign here. So there&#x27;s another mind. So you eventually would become possible. And this will give us the height of image is equal to 4.13 man. Here. Thank. And these are the solution for the question. Thank you.

Bruce Edelman · Answer

Okay, so we&#x27;re doing Chapter 34 Problem 57 here. So this is a focal length of a simple magnifier. Is eight centimeters assumed the magnifier? Is that Finland&#x27;s place? Very close to the odd. Okay, so part ay says how far in front of the magnifier should an object the place if the image is formed at the observers and your points, which is 25 centimeters cool and it&#x27;s also in front of the ice, that means you have to find that, um, now to do this, we already says this. Finland&#x27;s equation is balanced, so we know one of her f is one of us one over as prime. If we solve for s, we say this is as pro ing over its prime minus F cinema has put stuff in here. So negative. 25 tons, eight over negative, 25 minus eights. People get in. We should get a positive 6.6 centimeters for the object distance. Cool. So now leads go to part B. Barbie says if the object is one millimetre high, what is the height of the image formed by the night of fire? Okay, so the negative image distance over object distance is the magnification, which is also the image height over the object type. Let&#x27;s rearrange that. And so for image high as negative image distance times object I over object distance. So this is negative. Negative 25 times one millimeter over 6.6 millimeters. And if we plug all that in, we see that this comes out to be for one two, no New Year&#x27;s. So this means we can focus at something closer than what we normally could focus that within 90 fire and week. See it bigger. So yep, that works exactly how we intuitively think a magnifier shouldn&#x27;t work. That was great.

Problem

A jeweler examines a diamond with a magnifying gl…

Problem 54 Easy Difficulty

Answer

a) 6.45 $cm$
b) $3.93$

Related Courses

Physics

Related Topics

Discussion

Top Physics 103 Educators

Liev B.

Zachary M.

Aspen F.

Meghan M.

Physics 103 Courses

Wave Optics - Intro

Multiple Aperture Interference - Overview

Recommended Videos

Watch More Solved Questions in Chapter 27

Video Transcript

James S. Walker

Physics

Liev B.

Zachary M.

Aspen F.

Meghan M.

Wave Optics - Intro

Multiple Aperture Interference - Overview

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

a) 6.45 $cm$b) $3.93$

Physics

Liev B.

Zachary M.

Aspen F.

Meghan M.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Liev B.

Zachary M.

Aspen F.

Meghan M.

a) 6.45 $cm$
b) $3.93$

James S. Walker

Physics