Geometric optics - why inverted image from thin lens

tiny-tim

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

[tex]\frac{1}{s} + \frac{1}{s'} = \frac{1}{f}[/tex]

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

Hurkyl

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-1...ons-Lenses.htm

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

Regards,

Mark

Mårten

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.

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

I like that!

Reminds me of an ellipse turning into a hyperbola

Mårten

Hm, interesting! How do you mean? The equation for an ellipse is
[tex]\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1[/tex]
and for a hyperbola is
[tex]\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.[/tex]
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

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.
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...