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

[Niles]

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/S_O + 1/S_I. 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/S_O + 1/S_I, as done in e.g. Hecht.

What is the correct argument?Niles.

 

[Doc Al]

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.

 

[Niles]

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 = S_O = S_I. However, it doesn't prove that 1/f = 1/S_O + 1/S_I, but that is what most authors state that it does.

 

[Doc Al]

Yes, that is a proof showing that f = S_O = S_I. However, it doesn't prove that 1/f = 1/S_O + 1/S_I, 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/S_I when S_O = ∞, which equals the constant expression on the right hand side of the thin lens equation. Thus f is fixed.

 

[Niles]

Sure it proves it. (It's really just the definition of focal length.) 1/f = 1/S_I when S_O = ∞, 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/S_O + 1/S_I assuming that either S_O or S_I is at infinity. However, many authors say that 1/f = 1/S_O + 1/S_I is also valid when S_O and S_I 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.

 

[Doc Al]

It proves that 1/f = 1/S_O + 1/S_I assuming that either S_O or S_I 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 S_O 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/S_O + 1/S_I is also valid when S_O and S_I 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.

 

[Niles]

No. First you start with the thin lens formula, which it seems you accept. Then you define f as the image distance when S_O 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 S_O is at infinity.

 

[sophiecentaur]

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 S_O 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?

 

[Niles]

Ahhh, now I see it. Great! Thanks for being patient with me.

Best wishes to both of you,
Niles.

 

[Doc Al]

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 S_O 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?

 

[Niles]

Careful about adding something to a quote box, as it makes it look like I said it.

Yeah, sorry about that. I changed it.

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

Thanks.

Best wishes,
Niles.

 

[Doc Al]

Cool.