I am currently studying the textbook Modern Optical Engineering, fourth edition, by Warren Smith. The textbook presents the following images and explanation:

In Fig. 1.9 the action of a concave lens is sketched. In this case the lens is thicker at the edge and thus retards the wave front more at the edge than at the center and increases the divergence. After passing through the lens, the wave front appears to have originated from the neighborhood of point $P'$, which is the image of point $P$ formed by the lens. In this case, however, it would be futile to place a screen at $P'$ and expect to find a concentration of light; all that would be observed would be the general illumination produced by the light emanating from $P$. This type of image is called a virtual image to distinguish it from the type of image diagramed in Fig. 1.8, which is called a real image. Thus a virtual image may be observed directly or may serve as a source to be reimaged by a subsequent lens system, but it cannot be produced on a screen. The terms “real” and “virtual” also may be applied to rays, where “virtual” applies to the extended part of a real ray.

enter image description here

enter image description here

It is this part that confuses me:

Thus a virtual image may be observed directly or may serve as a source to be reimaged by a subsequent lens system, but it cannot be produced on a screen.

I thought the entire point was that the virtual image cannot be observed directly (or, actually, observed at all). So I'm wondering what the author means by this part?

I would greatly appreciate it if people would please take the time to clarify this.

  • $\begingroup$ The light will still reach your eyes, right? So you can still directly observe it. $\endgroup$ Commented Feb 12, 2020 at 2:57
  • $\begingroup$ @AaronStevens But then why won't a screen at $P'$ capture it? It seems that there is a subtle distinction here between, in the author's words, "a concentration of light" and "general illumination"? Can you please elaborate on specifically what my misunderstanding is? $\endgroup$ Commented Feb 12, 2020 at 3:07
  • $\begingroup$ Because light isn't converging at the virtual imager, it just appears to be coming from where the virtual image is. Contrast this with a real image where light actually converges at where the image is located. $\endgroup$ Commented Feb 12, 2020 at 4:21
  • $\begingroup$ @AaronStevens That all makes sense to me, and is actually what I'm arguing: If that's all true, then what does the author mean by "a virtual image may be observed directly or may serve as a source to be reimaged by a subsequent lens system"? How could it be that a virtual image may be observed directly or may serve as a source to be reimaged by a subsequent lens system, but that it cannot be produced on a screen? These two statements seem contradictory? $\endgroup$ Commented Feb 12, 2020 at 4:28

2 Answers 2


You can directly observe virtual images. All you have to do is look through a diverging lens. People who are nearsighted and wear glasses to correct their vision do this all of the time.

The virtual image is, well, virtual. You could argue that it is a trick: it is just the location where light appears to be coming from. In other words, a virtual image looks just like if the object was at the image position (magnified appropriately). This is why we can directly observe them or why they can be imaged by other systems; they are indistinguishable from just regular sources of light.

This is also why you cannot see them on a screen. If you put the screen at the light source and then look at the screen then you aren't going to see anything. You need light to converge onto the screen in order for an image to be on the screen. Light diverges away from virtual images, so they cannot be imaged.

  • $\begingroup$ Ahh, ok, I understand now. So the virtual image is more similar to the object than it is to an actual image, in the sense that you cannot image it onto a screen, as you can the actual image, but it acts as the object in a different location (at the location of the virtual image). Am I understanding it correctly? $\endgroup$ Commented Feb 12, 2020 at 4:45
  • 1
    $\begingroup$ @ThePointer Kind of. If by "actual image" you mean a real image, then yes. The key point though is that light only diverges away from a virtual image, so there is no point in space where you could put something that the light converges at in order to capture an image. $\endgroup$ Commented Feb 12, 2020 at 17:24
  • $\begingroup$ Yes, that's what I meant. $\endgroup$ Commented Feb 12, 2020 at 17:35

Your retina is just a screen. A real image needs to be projected onto it in order for you to see it. So in this sense, you can only 'directly' observe real images.

However, don't forget that your eye also has a lens in front of the retina. This lens can transform a virtual image into a real one that is projected onto your retina. In this sense, you can 'directly' observe a virtual image.

So both can be considered correct. Only real images can be 'directly' observed in the sense that only they can be projected onto a screen. But your eye can 'directly' observe virtual images as well, because it's not just a screen - it is a lens in front of a screen.

(The same reasoning can be applied to any camera, of course, not just our natural cameras.)

  • $\begingroup$ Another way of looking at it: screens only 'see' real images, optical elements like lenses and mirrors can transform any object or image, real or virtual. $\endgroup$ Commented Feb 12, 2020 at 11:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.