Thin Lens Formula: Deriving 1/f = 1/S_O + 1/S_I

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Discussion Overview

The discussion revolves around the derivation and interpretation of the thin lens formula, specifically the relationship expressed as 1/f = 1/S_O + 1/S_I. Participants explore the conditions under which this formula is valid, particularly focusing on the implications of object and image distances being at infinity.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants assert that the thin lens equation is general, while the equation 1/f = 1/S_O + 1/S_I applies specifically when either the object or image is at infinity.
  • Others argue that the right-hand side of the thin lens equation is a constant, and thus the image distance can define the focal length.
  • A participant claims that while 1/f = 1/S_I holds when S_O is at infinity, it does not prove the general case of 1/f = 1/S_O + 1/S_I.
  • Some participants agree that many authors extend the validity of 1/f = 1/S_O + 1/S_I beyond the case of infinity, which they contest.
  • There is a suggestion that defining the focal length as the image distance when the object distance is at infinity leads to confusion about the general applicability of the formula.
  • One participant acknowledges a misunderstanding regarding the relationship between the thin lens equation and the focal length, indicating a shift in their perspective.

Areas of Agreement / Disagreement

Participants express disagreement regarding the general applicability of the thin lens formula. While some accept the traditional derivation, others challenge its extension to cases where object and image distances are not at infinity. The discussion remains unresolved on this point.

Contextual Notes

Participants note that the focal length is defined in relation to the object distance being at infinity, which raises questions about the conditions under which the thin lens formula can be applied. There are also references to potential misunderstandings in interpreting the relationships between the variables involved.

Who May Find This Useful

This discussion may be of interest to students and educators in optics, particularly those exploring the nuances of lens formulas and their derivations.

Niles
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Hi

The thin lens equation we know is given by

<br /> \frac{1}{S_O} + \frac{1}{S_I} = (n-1)(\frac{1}{R_1}-\frac{1}{R_2})<br />

assuming light is incident from air. The "usual" way to derive the thin lens formula (aka the Gaussian Lens Formula) is to say that when looking at the system when either the image of object is at infinity, then we introduce the focal length by

<br /> \frac{1}{f} = (n-1)(\frac{1}{R_1}-\frac{1}{R_2})<br />

From this it is usually stated that then 1/f = 1/SO + 1/SI. This I don't agree with, since the thin lens equation at the top is general, however the second equation stated is when either the object or image is at infinity. From this one can't state 1/f = 1/SO + 1/SI, as done in e.g. Hecht.

What is the correct argument?Niles.
 
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Niles said:
This I don't agree with, since the thin lens equation at the top is general, however the second equation stated is when either the object or image is at infinity.
But the right hand side of that equation is a constant. If you put parallel rays in, the image distance defines the focal length.
 
Doc Al said:
But the right hand side of that equation is a constant. If you put parallel rays in, the image distance defines the focal length.

Yes, that is a proof showing that f = SO = SI. However, it doesn't prove that 1/f = 1/SO + 1/SI, but that is what most authors state that it does.
 
Niles said:
Yes, that is a proof showing that f = SO = SI. However, it doesn't prove that 1/f = 1/SO + 1/SI, but that is what most authors state that it does.
Sure it proves it. (It's really just the definition of focal length.) 1/f = 1/SI when SO = ∞, which equals the constant expression on the right hand side of the thin lens equation. Thus f is fixed.
 
Last edited:
Doc Al said:
Sure it proves it. (It's really just the definition of focal length.) 1/f = 1/SI when SO = ∞, which equals the constant expression on the right hand side of the thin lens equation. Thus f is fixed.

It proves that 1/f = 1/SO + 1/SI assuming that either SO or SI is at infinity. However, many authors say that 1/f = 1/SO + 1/SI is also valid when SO and SI are not at infinity, in which case we can use it to find the image distance given some arbitrary object distance. This latter case I don't agree with.
 
Niles said:
It proves that 1/f = 1/SO + 1/SI assuming that either SO or SI is at infinity.
No. First you start with the thin lens formula, which it seems you accept. Then you define f as the image distance when SO is at infinity; f is a constant for the lens, and 1/f is equal to the right hand side of your original thin lens equation. Thus you can rewrite the thin lens equation in terms of f.
However, many authors say that 1/f = 1/SO + 1/SI is also valid when SO and SI are not at infinity, in which case we can use it to find the image distance given some arbitrary object distance. This latter case I don't agree with.
If you still don't agree, start by defining f.
 
Doc Al said:
No. First you start with the thin lens formula, which it seems you accept. Then you define f as the image distance when SO is at infinity; f is a constant for the lens, and 1/f is equal to the right hand side of your original thin lens equation (the following is added by Niles, not Doc Al) when the object is at infinity.

I added something to your quote (the bolded part), which is where we disagree. The focal length is defined as the image distance when the object distance is at infinity, so 1/f is only equal to the LHS of the thin lens equation, when SO is at infinity.
 
Last edited:
Niles said:
I added something to your quote (the bolded part), which is where we disagree. The focal length is defined as the image distance when the object distance is at infinity, so 1/f is only equal to the LHS of the thin lens equation, when SO is at infinity.

That is just the definition of f, surely. What is the problem with that? f is a very convenient parameter for a lens, which why we specify lenses in terms of focal length. Can you propose a better way?

Have you tried doing what Doc Al suggested?
 
Ahhh, now I see it. Great! Thanks for being patient with me.

Best wishes to both of you,
Niles.
 
  • #10
Niles said:
I added something to your quote (the bolded part), which is where we disagree.
Careful about adding something to a quote box, as it makes it look like I said it.
The focal length is defined as the image distance when the object distance is at infinity, so 1/f is only equal to the LHS of the thin lens equation, when SO is at infinity.
Let me ask you this: If I give you n, R1, and R2 for a lens, can you calculate the focal length? How would you do that?
 
  • #11
Doc Al said:
Careful about adding something to a quote box, as it makes it look like I said it.

Yeah, sorry about that. I changed it.


Doc Al said:
Let me ask you this: If I give you n, R1, and R2 for a lens, can you calculate the focal length? How would you do that?

Of course I can. I set SO to be at infinity, and since SI = f in this case (by definition), I just use the RHS of the thin lens equation to find 1/f. But the RHS *is* SO + SI, which is the part I missed. But I see it now.

Thanks.

Best wishes,
Niles.
 
  • #12
Cool. :smile:
 

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