Converging Lens Focusing: Finding the Optimal Lens Position

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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 a screen is positioned 30 cm from an object. Participants are exploring how to determine the optimal position of the lens to achieve a focused image.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the lens formula and the relationship between object distance (do) and image distance (di). There are attempts to set up simultaneous equations based on the given distances and the focal length. Some participants question whether logical reasoning can replace formal equations.

Discussion Status

The discussion is active, with participants sharing various approaches to the problem. Some have proposed specific values for do and di, while others are exploring the implications of their assumptions. There is a recognition of multiple potential combinations for do and di that satisfy the conditions of the problem.

Contextual Notes

Participants note that the sum of do and di must equal 30 cm, and there is uncertainty about how to derive the correct values from the equations presented. The discussion includes considerations of whether certain distances yield focused images.

bananan
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I'm having a little problem with this question.. my brain is too small.

"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?"

I've started with the formula

1/do + 1/di = 1/F

1/do + 1/di = 1/10

At this point I don't have a distance for do or di, because they're relative to the lens, and the lens position is the variable. Can someone push me in the right direction?
 
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Look like you need to set up two simulataneous equations. You're written one down and you need to make use of 'A screen is placed 30 cm from an object' to find the other.
 
How do I do that? Okay, could i solve this logically without using any equations?

I'm thinking, if the focal length is 10, and the distance is 30, then having the object at 1F and the screen at -2F would produce a focused image on the screen wouldn't it? Therefore the lens would be 10 cm right of the object and 20 cm left of the screen. Does that work? You've probably guess that I am terrible at this by now.
 
Well the distance from the object to the lens is the object distance do and the distance from the lens to the screen is the image distance di, if the image from the object to the scree is 30cm what does this tell us about the sum of di and do.
 
Max Eilerson said:
Well the distance from the object to the lens is the object distance do and the distance from the lens to the screen is the image distance di, if the image from the object to the scree is 30cm what does this tell us about the sum of di and do.

They equal 30. But something's not clicking, I don't know what to do with that.
 
di + do = 30 (1)
and
1/do + 1/di = 1/10 (2)

Surely you can now solve for di and do from that.
 
The best I can figure to do here, is assume do is 15 and di is 15 (half of 30)

Which sortof yields the right answer

1/15 + 1/15 = 0.13333

But I can also use assume di is 10 and do is 20

1/20 + 1/10 = .15

1/F = .1

So both are pretty close to .1 is the closer (first) one the right one?

All I know is that do and di will be 2 numbers that sum 30, but there's many combinations, and I don't know the method of getting the right one.
 
Last edited:
I think I've figured it out, just by looking at the problem and thinking logically.

do = 20 and di =-10 right? I don't know how to /calculate/ to get that though
 
I've sort of led you round the houses here. The image is focused therefore the distance from the lens to the screen is?
 
  • #10
F? :P

Would 2F be in focus as well, out of curiousity?
 
  • #11
no it would be out of focus :).
 
  • #12
Okay, that simplifies things considerably, thanks man. :)
 

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