# Another sign convention issue (optics)

1. Mar 21, 2012

### chitturp

I read through all of the previous sign convention posts on this forum (and googled a lot and checked a bunch of textbooks) before posting this question. I don't understand the sign convention properly for geometric optics. There are two conventions that I've understood so far, and I've got a few questions... One convention is that anything measured in the direction of incident light is positive, and against the direction of incident light is negative (I guess we'd be measuring from the object to the mirror/lens initially, and then from the lens itself after the light hits it?).

Doubt 1) The thing about this convention... I understand it for mirrors, but I'm confused about how it works for lenses? There are two foci, so which one do we take into account? If we took the one on the same side as the object, wouldn't it be negative? And the opposite side focus would be a positive distance?? So which one?

The other convention I noticed was this -- on a particular website: "Real object and real image distances are given positive sign while virtual object and virtual image distances are given negative sign"

Doubt 2) What is a virtual object?

Huge doubt 3) I tried to use both conventions to solve a problem with two convex lenses next to each other ... I was able to get the distance of the image from the first lens using either sign convention, but then I got stuck with the second one. The way the problem was, the object was on the left side of the first lens, and the image from that ended up on the right side of the second lens. Now -- I thought that using sign convention # 1, couldn't I just consider that (the image of the first lens) a whole new object for the second lens (therefore with positive distance) and then the image distance (virtual) would also be positive because it's on the opposite side of the lens, and the focus would also be positive since it's a convex lens?... I figured that the "new" incident light could come from the right side of the second lens... (but should that really make a difference, because then object distance, image distance, and focus would all be negative since measured towards the left).

Using sign convention #2, the image of the first lens would be a real image, right? (this is where I'm not sure what a virtual object is...) -- So then the new object for the second lens would be "real", so the distance would be positive, and the image distance should be "negative" because it would be virtual, and the focus would be positive because it's a convex lens...

BUT
It turns out that for the second lens, the object distance was negative, the image distance was positive, and the focus was positive (according to the answers). How did this happen!?

Formula used was 1/u + 1/v = 1/f, with signs substituted in.

What sort of sign convention is used?... I'm really confused... I'm sorry this question is so long, but I'd really appreciate if someone could help!

2. Mar 21, 2012

### Philip Wood

I'll just chip in one thing, if I may. A virtual object of a lens is a point to which rays approaching the lens would converge if the lens weren't there. So if light is going left-to-right, the virtual object would be to the right of the lens.

So, in your example of the two juxtaposed lenses, the first lens would form a real image. This real image is the virtual object for the second lens.

Last edited: Mar 21, 2012
3. Mar 21, 2012

### chitturp

There's still something a bit confusing about that.. Wouldn't the new image now be a virtual one?.. so the sign convention would be negative for that? But in the original problem. the "virtual object" for the second lens has a negative sign and the final image was given a positive sign.

Last edited: Mar 21, 2012