Geometric optics - why inverted image from thin lens

[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"?

 

[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

 

[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...?

 

[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.

 

[Mårten]

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.

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?

 

[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).

 

[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

 

[Mårten]

In physical optics, the lens (more properly, the exit pupil of the lens) acts as a low-pass filter.

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).

 

[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!

 

[Mårten]

Aha, in other words, finer details in the object get lost in the image. Simply, the image having lower resolution?

Found some on Wikipedia here, as a reference:

http://en.wikipedia.org/wiki/Spatial_frequency
http://en.wikipedia.org/wiki/Fourier_optics

 

[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.

 

[Mårten]

Thanks for clarifying on the Fourier transform thing. Interesting there, the pinhole.

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

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?

 

[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.

 

[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...

 

[tiny-tim]

… 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."

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.

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

Which part(s) of the Library article do you find confusing?

 

[Redbelly98]

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

 

[Redbelly98]

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 d_o is negative.

Hope that helps.

 

[tiny-tim]

… the 3rd image shows a "virtual object" where d_o is negative.

Nice one!

 

[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?

https://www.physicsforums.com/mgc_gloss/148/img_3.gif

 

[Hurkyl]

But how can I see the virtual object?

Stand on the other side of the lens.

 

[Redbelly98]

Hi Mårten,

What I had in mind when I said "virtual object" is that the incoming rays are converging toward some common point located beyond the lens. I think we can all agree that d_o is negative in this case, even if we don't agree on the "virtual object" terminology.

This situation arises in multi-lens systems (eg. cameras), where one lens forms an image that's located beyond the next lens in the system. Calling this object distance negative, and using the focal length, we can be consistent in calculating where the image of that next lens is located.


Hmmm, now I'm getting paranoid that I've invented some non-standard jargon. But a google search on

"virtual object" lens​

shows that others do use that term the same way I do:


From http://en.allexperts.com/q/Physics-1358/few-Questions-Lenses.htm

A virtual object means than light is converging when it enters the lens.

From http://physics.bu.edu/py106/notes/Lenses.html

Note that in certain cases involving more than one lens the object distance can be negative. This occurs when the image from the first lens lies on the far side of the second lens; that image is the object for the second lens, and is called a virtual object.

Regards,

Mark

 

[Mårten]

This situation arises in multi-lens systems (eg. cameras), where one lens forms an image that's located beyond the next lens in the system. Calling this object distance negative, and using the focal length, we can be consistent in calculating where the image of that next lens is located.

Okey, thank you. I think even my tardy brain understands now...
I think my confusion arised from thinking about the virtual object more as an object than as something virtual. It's not really an object, I'd say now. My way to see it now, is to imagine the real object going further and further until it reaches infinity where the rays get parallel; then the object "goes beyond" infinity to minus infinity and from there on, the rays are not diverging from the real object anymore, but rather converging to the virtual object on the minus side.

Stand on the other side of the lens.

I wouldn't agree on that. If you mean, standing on the left side in the figure above, that would be to interpret the real image as the new object sending out rays to the left, and the virtual object to the right in the figure would then rather be a virtual image. I believe that a virtual object can never be seen, cause it never sends out any rays. It's just an abstract entity that makes our calculations work.

 

[tiny-tim]

to infinity … and beyond!

My way to see it now, is to imagine the real object going further and further until it reaches infinity where the rays get parallel; then the object "goes beyond" infinity to minus infinity and from there on, the rays are not diverging from the real object anymore, but rather converging to the virtual object on the minus side.

I like that!

Reminds me of an ellipse turning into a hyperbola ​

 

[Mårten]

I like that!

Reminds me of an ellipse turning into a hyperbola ​

Hm, interesting! How do you mean? The equation for an ellipse is
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
and for a hyperbola is
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.$$
To make them same, b^2 must turn negative, which is not possible... unless b is complex! Is that what you mean?

Talking about complex numbers. This also reminds me of a thing in complex analysis (a class which I've not taken), where I heard that you imagine infinity and -infinity to reach each other in a sort of infinitely big circle. Is that right?

 

[tiny-tim]

Hm, interesting! How do you mean?

Hi Mårten!

Take an ellipse. Fix one focus, and drag the other focus away (keeping the same directrix).

When the prodigal focus reaches infinity, you have a parabola …

now let the prodigal focus come back from infinity, on the other side of the directrix …

you have a hyperbola.

Now rejoice, and kill the fatted calf! ​

You get the same thing if you produce the figures with a plane cutting a conic … swivel the plane round, keeping the "apex" of the conic in the same place, and you see the ellipse turning into a parabola and then a hyperbola.

And you can also do it using caustic curves (of light), but I can't remember how off-hand.

Talking about complex numbers. This also reminds me of a thing in complex analysis (a class which I've not taken), where I heard that you imagine infinity and -infinity to reach each other in a sort of infinitely big circle. Is that right?

Yes … if you use the inversion z goes to 1/z, then the origin goes to the whole of the "circle at infinity", so if you define that circle to be one point, then the inversion is one-to-one.

I think this is a an exampe of a general technique (which, once again , I can't remember the details of ).

might be something to do with conformal maping or projective geometry …

 

[Mårten]

Hi tiny-tim!

Thank's for your reply! I'll have to contemplate over this for a while...