Thin Lens Equation - How to Solve for the Position of 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 10 cm, where participants are trying to determine the appropriate position of the lens relative to an object and a screen placed 30 cm apart to achieve a focused image.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationships between object distance (do), image distance (di), and the focal length, referencing the thin lens equation. There are attempts to eliminate one variable using the other equation but with varying levels of success. Some participants express frustration over the complexity of the problem and the need for a systematic approach.

Discussion Status

The discussion is ongoing, with some participants offering guidance on manipulating equations, while others express confusion and seek clarification on the mathematical processes involved. There is no clear consensus on the method to solve the problem, and multiple interpretations of the setup are being explored.

Contextual Notes

There is a noted change in the problem parameters, with a new total distance of 60 cm introduced, which may affect the approach to finding a solution. Participants are also questioning the implications of object placement and its effect on image formation.

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Homework Statement



3. A converging lens has a focal length of 10 cm. A screen is placed 30 cm from an object. Where should the lens be placed, in relation to the object, to produce a focused image?

Homework Equations



1/do + 1/di = 1f

The Attempt at a Solution



Do can not be greater than 2F, because then the image appears between F and 2F on the other side of the lens. From 2F to 2F is 40cm, the distance we want is 30cm.

Do cannot be equal to 2F because then the image appears at 2F on the other side. 2F to 2F in this case is 40cm. So its not possible.

Do cannot be between f (10cm) and 2f or the image will appear beyond 2f.
F to 2F is 30cm, so the object can't be beyond F and the image can't be beyond 2F when there’s only 30cm between them.

Do cannot be F because no image is formed when do is at F.

Therefore, do must be less than F. However, this is not possible because the lens must be between the screen and the object, as specified by the question, and do being less than f results in the image appearing on the same side of the lens as the object.If I actually attempt it mathematically it turns into a sadistic guessing game.

di + do = 30
1/do + 1/di = 1/10

How many combinations of 2 numbers sum 30? Am I expected to draw a ray diagram for all of them until I find a combo that works, or is there a formulaic way of approaching this that's going over my head?

The problem being... do and di are defined in relation to the position of the lens. In this case, the position of the lens is the variable.
 
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You've got the right equations use the first one to eliminate di or do from the second one. Then work out di or do, and hence find the othervalue.
 
Yes.. I guess the real problem is I've been trying to do that for 4 hours with no success which is why I posted here.
 
Of course you could just think how far from a lens a focused image is :).
 
Can you tell me how to do it mathematically? I need to know how to solve this type of problem.
 
use the first one to eliminate di or do from the second one

Please explain how to do this?
 
Max Eilerson said:
Of course you could just think how far from a lens a focused image is :).

Doesnt it depend on where the object is?
 
I suspect that your getting confused with signs. You do know how to cross multiiply right?

[tex]\frac{1}{d_i} - \frac{1}{30 - d_i} = \frac{1}{f}[/tex]
 
I dunno.. That latex graphic thing didnt seem to work.

There was a problem with the question though. Its been changed to di+do=60cm

So could you teach me how to evaluate this?

1/di + 1/(60-di) = 1/10
 
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