# Thin lens, optics

Gold Member

## 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?

## Answers and Replies

tiny-tim
Homework Helper
Hi fluidistic!

You're making this far too complicated!

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

Gold Member
Hi fluidistic!

You're making this far too complicated!

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

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!

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

tiny-tim
Homework Helper
Edit: Wait! Why did you choose the expression So-Si rather than minimizing So+Si?

oops!

Gold Member
oops!

But this worked! ahahahaha, I made an error but I reached the result. Wow, amazing. I'll retry. Ahahahah.

Gold Member
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.

tiny-tim
Homework Helper
No, So + Si = So + 1/(1/f - 1/So) = So + fSo/(So - f) …

carry on from there.

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

Gold Member
No, So + Si = So + 1/(1/f - 1/So) = So + fSo/(So - f) …

carry on from there.

(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,

tiny-tim