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I was reading geometric optics when I came across the concept of virtual objects. I found the concept counter-intuitive, as far as reflection goes. (I perfectly understand the formation of such an object when the light ray is refracted through a glass slab.)

Attaching the definition provided in my book:

Virtual Object defn

My confusion: A mirror forms an image when the rays originate from a certain object. So, how can an image of a virtual object form if the rays do not diverge from a particular point in the first place? What are some real-life examples of such a situation, as depicted by the figure below:

fig

I have checked answers to similar questions but my doubt is not cleared.

Golden_Hawk
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2 Answers2

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So, how can an image of a virtual object form if the rays do not diverge from a particular point in the first place?

The image in your question shows a "virtual object". It does not show an image. You could create a real image from it, and in this case it would do exactly that in front of the mirror, if you finished drawing the rays.

This is just a definition, which is somewhat counterintuative, and it completes the set of real and virtual objects and images, the other three of which are somewhat more intuative.

  • Light diverges from a real object
  • Light appears to diverge from a virtual image (when an optic reflects or refracts it)
  • Light converges on a real image
  • Light appears to converge on a virtual object (except an optic gets on the way and refracts or reflects the light)

For real life examples its somewhat harder to come up with simple examples as a virtual object isnt often much use on its own. So it might be formed as an intermediary in a larger optical system. Or even in this example. If you have an imaging system with a fold mirror at the end, that mirror would create a virtual object, similar to your picture here.

Matt
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What "forms an image" (on your retina) is the lens in your eyes. Neither a real object nor a a flat mirror reflecting a real object can form an image. The virtual object behind the mirror is what we see because all the rays that are reflected by the mirror surface SEEM to be coming from that virtual object (that doesn't exist) if continue them behind the mirror surface.

I believe your diagram is wrong in one detail: it should show rays going in the other direction. This shows the actual optical path of a plane mirror surface: https://www.ck12.org/book/ck-12-physics-concepts-intermediate/r252/section/14.1/

Divergent rays from an object reach us and if we trace them back without the change in direction on the surface, then they seem to converge on the "virtual object" behind the mirror. Except that that's not what we call a virtual object. I defer to somebody else to explain why the convention was chosen that way. It's not obvious to me, either, at this point. I am sure there is a good reason for it.

FlatterMann
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  • The drawing correctly depicts a virtual object. If the rays were flipped it would be a virtual image. – Matt Jun 09 '23 at 17:36
  • @Matt The direction of light is divergent from both an object and a virtual object. It is only convergent on an image focal point. Maybe there is a convention in geometrical optics to reverse the direction of rays (which is perfectly proper because of reciprocity) but for an explanation of light paths for beginners I prefer to show the actual flow of energy. It always diverges from the source. – FlatterMann Jun 09 '23 at 17:39
  • No. Light converges on a real image or a virtual object. It diverges from a real object or virtual image. Those are the (partial) definitions of those terms. In this case the light is diverging from the source that is not pictured and has presumably passed through other optics to converge here. A virtual object is not a source. How could it be? Its virtual. – Matt Jun 09 '23 at 17:47
  • That's what https://dr282zn36sxxg.cloudfront.net/datastreams/f-d%3A430aedeca2255998c5b97149657ef252464e02ee6efa392f1348d22f%2BIMAGE_THUMB_POSTCARD_TINY%2BIMAGE_THUMB_POSTCARD_TINY.1 shows. Are you still looking at the Khan academy image? I posted the wrong link. Sorry. – FlatterMann Jun 09 '23 at 17:58
  • That picture in your link shows a real object and a virtual image. This question and its accompanying picture show/discuss a virtual object. They are not the same. – Matt Jun 09 '23 at 18:17
  • @Matt Maybe it's been too long in ray optics for me. I have to look up the definition of virtual object. In any case, what is depicted there is the real and virtual light path. – FlatterMann Jun 09 '23 at 18:20
  • See here: https://physics.stackexchange.com/q/413109/29170 These (real and virtual objects and images) are defined terms. A virtual object and a virtual image are not the same. Your answer here seems to think they are. What is depicted in this question is a virtual object. It is literally a textbook example. It is not incorrect. – Matt Jun 09 '23 at 18:22
  • @Matt Like All that matters to me as an experimentalist is the actual light path and that never converges on an object. It only ever converges on an image. That ray opticians are defining it differently (to have positive coefficients in the equations?) is one of the counterintuitive aspects of physics theory. That's fine. You are welcome to explain it to the OP why it has been chosen that way. I am not getting in your way. – FlatterMann Jun 09 '23 at 18:28