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

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SUMMARY

The minimum distance between two conjugate points for a positive thin lens is established as 4f. This conclusion is derived from the lens formula, 1/f = 1/S_o + 1/S_i, and the relationship between the object distance (S_o) and image distance (S_i). The discussion emphasizes the importance of differentiating the function S_i - S_o to find the minimum distance, confirming that the inequality S_o + S_i ≥ 4f holds true when analyzed correctly.

PREREQUISITES
  • Understanding of the lens formula: 1/f = 1/S_o + 1/S_i
  • Basic knowledge of optics, specifically thin lens behavior
  • Familiarity with differentiation in calculus
  • Concept of conjugate points in optics
NEXT STEPS
  • Explore the derivation of the lens maker's equation for biconvex lenses
  • Study the application of calculus in optimizing functions related to optics
  • Investigate the implications of varying object distances on image formation
  • Learn about different types of lenses and their focal properties
USEFUL FOR

Students studying optics, physics educators, and anyone interested in the mathematical principles governing lens behavior and image formation.

fluidistic
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Homework Statement


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

Homework Equations


\frac{1}{f}=\frac{1}{S_o}+\frac{1}{S_i}.

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 \frac{1}{f}=(n_1-n_0)\left ( \frac{2}{R} \right ).So I must show that S_0+S_1 \geq 4f.
Using these 2 formulae, I reach that the inequation holds if and only if S_0+S_i \leq 2R 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?
 
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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 \frac{1}{f}=\frac{1}{S_o}+\frac{1}{S_i} 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. S_0+S_i=\frac{S_0S_i}{f}. 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:
 

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