# Geometric optics - why inverted image from thin lens

1. Jan 11, 2009

### Mårten

Three questions:

1) Anyone have a pedagogic answer to why the image formed by a thin lens is inverted (i.e. upside-down)? I realize that to get a focused image, the lens has to converge the rays one way or another, so then, eventually, the rays have to cross the optic axis, which in turn leads to the upside down image. But couldn't the image be formed before the rays cross the optic axis? Maybe not, because there it cannot be focused, but why can't it be focused there?

2) Is it possible to construct a lens (or any single transparent material) that actually forms images which are not upside down? (Note: Several lenses in series are not allowed.)

3) That the image seems to have to be inverted really puzzles me - is there some deep fundamentality in nature which predicts this inversion behaviour of an optical system like a lens? Or is it more of "that's just the way it is"?

2. Jan 11, 2009

### mgb_phys

Simplest way is probably to consider a pinhole camera - one with no lens.
If you want a straight line from the object to go through the pinhole and hit the film it will have to cross the axis and so end up on the opposite side = upside down and left-right
If you wanted the image to be the right war round each ray would have to bend when it went through the pinhole, and each would have to bend a different amount depending where on the object it came from - there's no way it can know to bend if there is nothing at the pinhole or knowhow much to bend to form an image

3. Jan 12, 2009

### Mårten

Ah, thank you! That pinhole camera thing, really opened up a new little world for me.

The pinhole and the ray, is like the pivot axis for the lever. Moving one side down, moves the other side up. Moving the light source down, moves the image behind the pinhole up. That's maybe one underlying principle, that is implemented in the lever and the pinhole ray...?

4. Jan 12, 2009

### mgb_phys

The pinhole is just a filter - it only allows one ray from each point in the object to reach the image. (Actually a small bundle of rays for a finite size pinhole of course)

If you know the maths a Fourier transform makes this even clearer, mathematically a lens is just a filter.

5. Jan 13, 2009

### Mårten

Hm, that sounds interesting! I've taken a class in Fourier transforms. Please explain, how is the lens a filter? A low-pass filter? What does it really filter?

Btw, in that lens article in the PF library, it says that the distance to the object could be negative, and then it's called a virtual object. My textbook also says that (using s and s' as notations), that in geometric optics in general, s, the object distance, is negative if not on the same side as the incoming light. How could it not be on the same side? Aren't we just talking about virtual images here? Maybe, the question is: what's the difference between a virtual object and a virtual image?

6. Jan 13, 2009

### Andy Resnick

I have not heard of the term "virtual object". In geometric optics, the world is divided into "object space" and "image space" by the lens, and a lens is said to perform a 1:1 map of object space onto image space. The side with the physical object is "object space", and the other side is "image space". If the image is in fact in image space then it's a "real image", while if the image is in object space, it's a 'virtual iamge'.

It's not clear what the advantage is to the nomenclature, but there it is.

In physical optics, the lens (more properly, the exit pupil of the lens) acts as a low-pass filter. In most systems, the image is considered as a convolution of the object with the 'point spread functon' of the lens, which is determined by the numerical aperture (and a few other characteristics).

7. Jan 13, 2009

### mgb_phys

If you form a virtual image, you could call that a virtual object for the next element
I suppose the advantage is that it keeps the 'object-lens-image' terms in order

8. Jan 13, 2009

### Mårten

But what is actually filtered away? High frequencies of light? If so, it doesn't make sense to me, since I thought that the image behind a lens, had the same colours as the object (i.e., frequencies of light are not altered).

9. Jan 13, 2009

### mgb_phys

Spatial frequencies (essentially levels of detail in the image) nothing to do with light
Sorry - that was probably confusing if you don't know the maths, but incredibly revealing if you do!

10. Jan 13, 2009

### Mårten

11. Jan 13, 2009

### mgb_phys

Yes, it relates the detail in the image to the size of the lens.

It's also useful to explain why a pinhole works - most people are puzzled by how a hole ie =nothing, can act as a lens. But if you picture the pinhole (or lens) limiting where a ray from the object can hit an image then it becomes clearer.
Fourier is just the mathematical treatment of this process.

12. Jan 13, 2009

### Mårten

Thanks for clarifying on the fourier transform thing. Interesting there, the pinhole.
But does it make sense to talk about a negative object distance? In what situations is it negative, if one assumes that positive means being on the same side as the incoming light (as my textbook, Young-Freedman, University Physics, says)? It's only when you talk about this virtual object then, or?

Last edited: Jan 13, 2009
13. Jan 13, 2009

### mgb_phys

As long as you know what you mean!

In eg. some lenses ,the front elements produce a virtual image in front of the lens.
The rear elements then re-imige this virtual image to create the real image (on the film)
Rather than saying 're-image the virtual image onto the image' you could say the lens creates an image of the virtual object.
It isn't common terminology but I can see why it might be useful.

14. Jan 16, 2009

### Mårten

Sorry for late reply - have been away for awhile...

Okey, I understand this virtual object concept now. But still (sorry for being tiresome), I cannot see why any object (not even a virtual object as far as I can see) anytime could have a negative distance. Young-Freedman, University Physics (9th ed, p. 1087), says the following, and the part after the semicolon feels not correct:

"Sign rule for the object distance: When the object is on the same side of the reflecting or refracting surface as the incoming light, the object distance s is positive; otherwise, it is negative."

I may comment this on the lens article in the PF library as well, that it is a little bit confusing...

15. Jan 16, 2009

### tiny-tim

Hi Mårten!

If the incoming light is converging, then the object will be on the opposite side, and the object will be virtual.

This can happen, for example, in a telescope if the light from the primary lens is still converging before it hits the secondary lens.
Which part(s) of the Library article do you find confusing?

16. Jan 16, 2009

### Redbelly98

Staff Emeritus
Mårten, I just saw your note on the lens article, and then this thread. I'll try to get a figure into the article, perhaps later tonight, to show what is meant by a negative object distance.

(I'm the one who wrote the lens article.)

Regards,

Mark

17. Jan 17, 2009

### Redbelly98

Staff Emeritus
I've added 3 images to the lens library article.

You'll have to click on the image to see it, but the 3rd image shows a "virtual object" where do is negative.

Hope that helps.

18. Jan 17, 2009

### tiny-tim

Nice one! :tongue2:

19. Jan 19, 2009

### Mårten

Thank you so much for patience and help, and for the nice pics in the lens article!

But still... I thought I understood what virtual object was first, when mgb_phys and I discussed this earlier in this thread. But now I think (s)he meant one meaning of virtual object, and you other guys are talking about another meaning of virtual object - is that so do you think?

Now, to the meaning of virtual object displayed in the pics in the lens article: What is this virtual object really? It seems even more "virtual" than a virtual image, since a virtual image I can at least see (or my mind believes it sees it). But how can I see the virtual object? Or is it just some formal auxiliary object, so that for a figure like the one we're talking about (with the virtual object in - I've enclosed it below), the virtual object satisfies the lens equation

$$\frac{1}{s} + \frac{1}{s'} = \frac{1}{f}$$

making it possible to for instance calculate the lens' focal length - is that the purpose with virtual object?

Last edited by a moderator: Apr 24, 2017 at 10:27 AM
20. Jan 19, 2009

### Hurkyl

Staff Emeritus
Stand on the other side of the lens.