Magnification of an Image formed by a converging lens

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Homework Help Overview

The discussion revolves around a problem involving a converging lens with a focal length of 14.2 cm, which is used to form a virtual image of a coin located 30.1 cm away. Participants are examining the calculation of magnification based on the lens formula and the characteristics of virtual images.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of the lens formula and the implications of forming a virtual image, questioning the signs of the distances involved. There is mention of alternative methods, such as ray diagrams, for visualizing the problem.

Discussion Status

Some participants have provided guidance regarding the importance of sign conventions in the calculations. There is acknowledgment of the original poster's uncertainty about their solution, and a participant notes a successful adjustment after considering the signs.

Contextual Notes

Participants highlight the need to account for the specific rules regarding the signs of distances in lens equations, particularly when dealing with virtual images formed by converging lenses.

fordy314
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Homework Statement


A converging lens with a focal length of 14.2 cm is used to inspect a coin. The lens forms a virtual image 30.1 cm away. What is the magnification?


Homework Equations


(1/Si) + (1/So) = (1/f)
M = (hi/ho) = (-Si/So)


The Attempt at a Solution


1/So = 1/f - 1/Si

1/(1/So) = So = 26.88cm

M = (-Si/So) = -30.1/26.88 = 1.12


I'm pretty sure that I have the solution and answer correct, but I just wanted to double check. This homework is online and the site is telling me that I have the answer wrong and I just wanted to make sure that I wasn't missing anything before I bother my teacher about it. As of right now I think it's just an error in the code of the site, but I'm just double checking. Thanks.
 
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fordy314 said:

Homework Statement


A converging lens with a focal length of 14.2 cm is used to inspect a coin. The lens forms a virtual image 30.1 cm away. What is the magnification?


Have you accounted for the fact that it's a virtual image?
 
fordy314 said:

Homework Statement


A converging lens with a focal length of 14.2 cm is used to inspect a coin. The lens forms a virtual image 30.1 cm away. What is the magnification?


Homework Equations


(1/Si) + (1/So) = (1/f)
M = (hi/ho) = (-Si/So)


The Attempt at a Solution


1/So = 1/f - 1/Si

1/(1/So) = So = 26.88cm

M = (-Si/So) = -30.1/26.88 = 1.12


I'm pretty sure that I have the solution and answer correct, but I just wanted to double check. This homework is online and the site is telling me that I have the answer wrong and I just wanted to make sure that I wasn't missing anything before I bother my teacher about it. As of right now I think it's just an error in the code of the site, but I'm just double checking. Thanks.

This object [the coin] is placed inside the focus [giving a virtual image] so some/all of those distances may be negative.
I am never confident with that (1/Si) + (1/So) = (1/f) formula, so I always use a couple of alternates, and a ray diagram.
 
Check your notes or your textbook. There are specific rules for deciding on the signs of the various quantities.
 
TSny said:
Check your notes or your textbook. There are specific rules for deciding on the signs of the various quantities.

Ah, right, I forgot about the signs. Looks like the summer off has made me a little rusty. I just accounted for the signs and got the right answer. Thanks for your help.
 

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