Prove 4f is the Min. Distance Between Conjugate Points for Thin Lens

Click For Summary

Homework Help Overview

The discussion revolves around proving that the minimum distance between two conjugate points (real object and image) for a positive thin lens is 4f. The subject area involves optics, specifically the behavior of thin lenses and their focal properties.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore various mathematical approaches to relate object distance (S_o) and image distance (S_i) using the lens formula. Some question the choice of expressions to minimize, while others suggest differentiating functions to find minimum values.

Discussion Status

There is an ongoing exploration of different methods to approach the problem. Some participants have made progress by differentiating expressions, while others express confusion about the next steps in their reasoning. No consensus has been reached, but several productive lines of inquiry are being pursued.

Contextual Notes

Participants are navigating through the complexities of the problem, with some expressing uncertainty about the assumptions made regarding the lens type and the mathematical expressions used. There is also mention of potential errors in reasoning that may affect the outcome.

fluidistic
Gold Member
Messages
3,934
Reaction score
286

Homework Statement


Show that the minimum distance between 2 conjugate points (real object and image) for a positive thin lens is 4f.

Homework Equations


[tex]\frac{1}{f}=\frac{1}{S_o}+\frac{1}{S_i}[/tex].

The Attempt at a Solution


I assumed the lens to be biconvex (though I know that I can't. There are so many types of positive lens...).
So I get that [tex]\frac{1}{f}=(n_1-n_0)\left ( \frac{2}{R} \right )[/tex].So I must show that [tex]S_0+S_1 \geq 4f[/tex].
Using these 2 formulae, I reach that the inequation holds if and only if [tex]S_0+S_i \leq 2R[/tex] where R is the curvature radius of the thin lens. I'm stuck here. Are there any other equation I should use? Or am I in the right direction?
 
Physics news on Phys.org
Hi fluidistic! :smile:

You're making this far too complicated!

Just write Si - So as a function of So, and differentiate. :wink:
 
tiny-tim said:
Hi fluidistic! :smile:

You're making this far too complicated!

Just write Si - So as a function of So, and differentiate. :wink:

Wow, this worked. Awsome! So I only needed the formula [tex]\frac{1}{f}=\frac{1}{S_o}+\frac{1}{S_i}[/tex] and your nice idea!:biggrin:

Edit: Wait! Why did you choose the expression So-Si rather than minimizing So+Si?
 
fluidistic said:
Edit: Wait! Why did you choose the expression So-Si rather than minimizing So+Si?

oops! :rolleyes:
 
tiny-tim said:
oops! :rolleyes:

But this worked! ahahahaha, I made an error but I reached the result. Wow, amazing. I'll retry. Ahahahah.
 
Have you solved the problem? I'm getting stuck, I just don't reach anything. [tex]S_0+S_i=\frac{S_0S_i}{f}[/tex]. I have to minimize this function.
 
No, So + Si = So + 1/(1/f - 1/So) = So + fSo/(So - f) …

carry on from there. :smile:

(and I'm off to bed :zzz:)
 
tiny-tim said:
No, So + Si = So + 1/(1/f - 1/So) = So + fSo/(So - f) …

carry on from there. :smile:

(and I'm off to bed :zzz:)


Worked! Thanks a lot.
 
It is still unclear to me where to go from here. What should I be differentiating with respects to?

Regards,
Adam
 
  • #10
Hi Adam! :smile:
Titans86 said:
What should I be differentiating with respects to?

hmm … there's only one variable in the formula …
tiny-tim said:
No, So + Si = So + 1/(1/f - 1/So) = So + fSo/(So - f)

… so i suppose you'd better differentiate wrt that! :wink:
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K