Object Distance From Converging Lens for 1% Focal Length Difference

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Homework Help Overview

The discussion revolves around determining the object distance from a converging lens, specifically when the difference between the image distance and the focal length is set at 1% of the focal length. The subject area is optics, focusing on the thin lens equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to formulate the problem using the thin lens equation but expresses uncertainty about the correctness of their equation and how to isolate the object distance. Other participants engage by clarifying the meaning of "difference" and suggesting alternative expressions for the image distance.

Discussion Status

Participants are actively discussing the setup of the problem and exploring different interpretations of the relationships between the object distance and image distance. Some guidance has been offered regarding the interpretation of terms and the formulation of the equations, but there is no explicit consensus on a final approach yet.

Contextual Notes

There is a focus on ensuring the correct interpretation of the terms used in the problem, particularly regarding the definitions of image distance and object distance. The original poster's equation has been questioned, and adjustments to terminology have been suggested.

spoonthrower
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How far away from a converging lens must an object be ( in terms of focal length, f ) so that the difference between the image distance and the focal lengh is 1% of the focal length?

I tried to translate the words into an equation based on the thin lens equation but i don't know if it is right or how to solve it for x in terms of f: tell me if I am right:

1/xf + 1/(f+.01f) = 1/f

please help! thanks.
 
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"Difference" means subtraction.

f-di=.01f so di=?

then, find do.
 
Seems right to me. You could also go the other way 0.99f.
 
Oops. di will be greater than f, so you were right there. di=1.01f .

and do, call it "x" or "do" but not "xf"
 

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