Part of the definition of a virtual image is that it cannot be formed on a screen. I understand this is the case when the screen is right next to the image, since there are no physical rays that can hit the screen. But what I don't understand is why an image can't form on the screen if the screen is located sufficiently far away from the image and/or lens so that rays do physically hit the screen? The 'explanation' usually given is that real rays converge while virtual rays don't, but how is the screen supposed to know if the rays it's seeing actually converged at some point or not? The only apparent difference compared to real rays I can see is that rays for virtual images would have greater angular divergence, which would create an image on the screen, just blurry.

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    $\begingroup$ For virtual images, there are no rays at the point of convergence. $\endgroup$ Sep 7, 2020 at 20:48
  • $\begingroup$ I'm aware, but this doesn't answer my question. $\endgroup$
    – Denn
    Sep 7, 2020 at 21:03
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    $\begingroup$ Why? If there are no rays, there cannot be an image. And if you put the screen on a point out of focus, you can get an image, but it will be out of focus. -- What is your question? $\endgroup$ Sep 7, 2020 at 21:04
  • $\begingroup$ I specifically stated in my question that I'm interested in the case where the screen is located somewhere where rays physically do hit it - so not at the intersection. Can you clarify what you mean by out of focus? This seems to be my hunch as well but I've never seen any mathematical justification. $\endgroup$
    – Denn
    Sep 7, 2020 at 21:07
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    $\begingroup$ FWIW, Looking through an optical system at a "virtual image" is equivalent to looking directly at a scene composed of actual, physical objects. If you simply hold up a white card in front of a scene composed of actual physical objects, would you expect to see picture of those objects form on the screen? $\endgroup$ Sep 8, 2020 at 15:39

2 Answers 2


There seems to be some fundamental confusion here. An image is formed on a screen when light rays emanating from an object converge there. If there is no convergence of rays, then there is no image on a screen.

Think about a portrait located positioned on the left side of the lens. The light emanating from a point on the tip of the nose focuses to the (single) corresponding point on the image. The same is true of all the neighboring points, so there is a one-to-one correspondence between points on the image and points on the object, and the image is clear.

enter image description here

On the other hand, if you position a screen at a different location, then the light emanating from the tip of the portrait's nose will be spread over a whole region of the screen. The light from the neighboring points on the object will overlap, and the result will be a blurred image.

enter image description here

The conclusion is that the calculated image distance is where you will get a clear image; if you put your screen anywhere else, then an image will not form. Now consider what you'd get with a diverging lens.

enter image description here

The blue dotted lines are obtained by tracing the rays on the right hand side backward and pretending the lens wasn't there. The virtual image is the location from which the rays appear to be emanating from the perspective of somebody on the right-hand side of the lens. However, there are no actual light rays which converge there. If you place a screen at the location of the virtual image, can you see why you don't get a nice picture?

enter image description here

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    $\begingroup$ If I understand right, in one sense, it should, perhaps, be called a virtual object, not a virtual image. It's where the object would appear to be (to someone on the right) if the real object and lens weren't there. $\endgroup$
    – TripeHound
    Sep 8, 2020 at 12:46
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    $\begingroup$ @TripeHound That line of thought is a good one. In fact, if you placed a second lens to the right of the diverging lens, then the virtual image created by the first lens would constitute a real object for the second. Furthermore, if you consider a system of two lenses with the first a converging lens, then a real image formed by the first lens can serve as a virtual object for the second. $\endgroup$
    – J. Murray
    Sep 8, 2020 at 15:56

From your statement

would create an image on the screen, just blurry.

I suppose you think of an image as some distribution of light that can be registered by a photo sensor. This is an intuitively obvious definition. But that's not what the technical term "image" means in optics. Citing Wikipedia:

In optics, an image is defined as the collection of focus points of light rays coming from an object.

Image in optics is the term opposed to object, and when an image is formed, it means that we basically have a kind of optical equivalent to the set of points an object is made of—from the point of view of ray propagation. I.e. if you take the rays emitted from an image, and pass them into a well-focused optical instrument, the image will be indistinguishable from an object placed at image's position—as seen by that instrument.


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