Object Distance & Magnification: Explained w/Diagram

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Discussion Overview

The discussion revolves around determining the object distance and magnification for a lens system where an object and its real image are 2.4 m apart, given a focal length of 55 cm. Participants seek to understand the relevant formulas and how to apply them to find the required values, including the possibility of virtual images.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • Participants mention the formulas $\frac{1}{{d}_{o}} + \frac{1}{{d}_{i}} = \frac{1}{f}$ and $m = -\frac{{d}_{i}}{{d}_{o}}$ as applicable to the problem.
  • There is a set of equations proposed, including $d_o + d_i = 2.4 \text{ m}$ and $f = 0.55 \text{ m}$, which participants are encouraged to solve.
  • One participant expresses confusion about the relationship between object distance and image distance, initially thinking they should subtract to find the total distance.
  • Another participant clarifies that typically, the object and image are on opposite sides of the lens, leading to their distances adding up, but acknowledges the possibility of a virtual image affecting the sign of the image distance.

Areas of Agreement / Disagreement

Participants generally agree on the formulas to use and the setup of the problem, but there is some confusion regarding the relationship between object and image distances, indicating a lack of consensus on how to interpret the distances in this context.

Contextual Notes

Participants have not resolved the mathematical steps required to find the object distance and magnification, and there are assumptions about the nature of the image (real vs. virtual) that remain unaddressed.

MermaidWonders
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An object and its lens-produced real image are 2.4 m apart. If the lens has 55-cm focal length, what are the possible values for the object distance and magnification?

Can someone please explain this with a diagram of the different possibilities (or, if not, just give a detailed explanation on how one should approach this)?
 
Last edited:
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MermaidWonders said:
An object and its lens-produced real image are 2.4 m apart. If the lens has 55-cm focal length, what are the possible values for the object distance and magnification?

Can someone please explain this with a diagram of the different possibilities (or, if not, just give a detailed explanation on how one should approach this)?

Attempt?
Which formula applies?
 
I like Serena said:
Attempt?
Which formula applies?

Well, formulas $\frac{1}{{d}_{o}} + \frac{1}{{d}_{i}} = \frac{1}{f}$ and $m = -\frac{{d}_{i}}{{d}_{o}}$ probably apply... I just have trouble finding ${d}_{o}$ and ${d}_{i}$ given that the object and its image are 2.4 m apart...
 
MermaidWonders said:
Well, formulas $\frac{1}{{d}_{o}} + \frac{1}{{d}_{i}} = \frac{1}{f}$ and $m = -\frac{{d}_{i}}{{d}_{o}}$ probably apply... I just have trouble finding ${d}_{o}$ and ${d}_{i}$ given that the object and its image are 2.4 m apart...

Good! It means that we have the set of equations:
\begin{cases} d_o+d_i=2.4 \text{ m} \\
f = 0.55 \text{ m}\\
\frac 1{d_o} + \frac 1 {d_i} = \frac 1f \\
m=\left| \frac{d_i}{d_o}\right|
\end{cases}
Can you solve it? (Wondering)
 
I like Serena said:
Good! It means that we have the set of equations:
\begin{cases} d_o+d_i=2.4 \text{ m} \\
f = 0.55 \text{ m}\\
\frac 1{d_o} + \frac 1 {d_i} = \frac 1f \\
m=\left| \frac{d_i}{d_o}\right|
\end{cases}
Can you solve it? (Wondering)

I keep thinking that ${d}_{o} - {d}_{i}$ is 2.4 m... I didn't know they add up to be 2.4 m... Why?
 
Last edited:
MermaidWonders said:
I keep thinking that ${d}_{o} - {d}_{i}$ is 2.4 m... I didn't know they add up to be 2.4 m... Why?

Normally the object and image are at different sides of the lense, meaning their distances add up.
However, it is possible that we have a virtual image that is on the same side of the lense as the object.
In that case we treat the image distance as a negative distance.
Either way, to do the math, we treat them as adding up to a total distance between object and image.
 
I like Serena said:
Normally the object and image are at different sides of the lense, meaning their distances add up.
However, it is possible that we have a virtual image that is on the same side of the lense as the object.
In that case we treat the image distance as a negative distance.
Either way, to do the math, we treat them as adding up to a total distance between object and image.

Oh, I see. I'll try that. :)
 

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