We call an image virtual if it is made by intersection of extension of diverging rays. And real if it is made by intersection of converging rays.

Assume a real image is made somewhere. Now if we place an optical device before the intersection point, that image now is like a virtual object for the new device.

We can say a virtual object is like a defeated real image. Rays weren't able to intersect normally, instead, they go through another device.

Now, assume a virtual image. Can this kind of image act as a virtual image for another optical device?

  • $\begingroup$ no the definition of virtual is that the rays do not pass through the image location, for a real image they do. $\endgroup$ – JMLCarter Jan 26 '17 at 13:03
  • $\begingroup$ if you place a device before in front of the real image, that image is destroyed. Unless it is a mirror or lens or otherwise allows the light to continue to propagate, in which case the image can become a virtual object as you state. $\endgroup$ – JMLCarter Jan 26 '17 at 13:06
  • $\begingroup$ Yes a virtual image can be used to create another virtual image or even a real image. $\endgroup$ – JMLCarter Jan 26 '17 at 13:07

I am doing just what the OP wonders as I type this.

The lens inside my eye is exactly the correct power to take an object at infinity and form a perfect real, inverted image on the retina of my eye. So I can drive without glasses, and I could answer SE questions using a giant video screen.

Unfortunately, my monitor is small, only about 1 metre away, and quite out of focus.

So I place a converging lens a few inches in front of my eye. This lens forms a virtual image of my monitor at infinity, so the lens of my eye can focus it on my retina. The virtual image of the screen serves as a real object for the lens in my eye.

Consider a person who has surgery to correct for cataracts, and through some strange screw-up has a diverging lens inserted in their eye.

To correct this error, the victim would use a very powerful converging lens in their glasses. This would form a real inverted image somewhere in the victim's eye, between the lens of the eye and the retina.

From a ray point of few, the diverging lens would "unbend" these rays just enough to move the image to the retina.

From a lens formula point of view, the real image formed by the glasses would be a virtual object for the eye lens. The sign convention for the particular version of the lens formula would apply, and the result would be a real inverted image on the retina...


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