Is it possible to get a real image of virtual objects, if so please explain with examples and some real life situations?

Look at my assertion "rays will not pass through virtual objects so how it can form real image", is this possible, please help.

  • 1
    $\begingroup$ Give us an example of virtual object, what you mean is not clear to me. $\endgroup$
    – DelCrosB
    Commented Nov 24, 2016 at 16:10
  • $\begingroup$ @DelCrosB like images by a concave lens, when it treated as objects. $\endgroup$
    – mnulb
    Commented Nov 24, 2016 at 16:12
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    $\begingroup$ Every time you use a magnifying glass, the lens in your eye accepts light that appears to come from a virtual image, and forms a real image on your retina... $\endgroup$
    – DJohnM
    Commented Nov 24, 2016 at 19:25
  • $\begingroup$ @DJohnM that should be written as an answer. $\endgroup$ Commented Jun 21, 2018 at 1:36

3 Answers 3


Suppose one has a two lens system with the first one being a converging lens and the second being a diverging lens. One can choose distances and focal length so that in the end the image created by the converging lens is located behind the diverging lens. Now for diverging lens the focal length is negative and object distance for diverging lens will be negative since the object is behind the diverging lens. Let us now look at the lens equation we have $$1/f = 1/d_i +1/d_o\implies d_i = \frac{d_o f}{d_o-f} $$. In order to get a real image the image distance must be positive. The numerator see must be positive since both quantities are negative. And the denominator is positive if the magnitude of the focal length of diverging lens is greater than the virtual object distance. This gives us a real image.

In short we got a virtual object for the diverging lens but finally got a real image with respect to the original object. But we could also have a two lens systems consisting just of converging lenses and i.e the diverging lens is replaced by the converging lens. We see in this case that focal length can be less than or greater than object distance but we finally get a real image with respect to original object.

I am not sure if this directly answers your question but it does use a virtual object to produce a real image at least with respect to original object which is real.


Yes it is possible. The best example would be :- 1) Suppose a condition in which you are looking at you image in a plane mirror. The mirror forms a virtual image of your, now when you are looking at the mirror on your image then a a real image of the mirror image is formed on your retina. 2) Suppose you are standing in front of a plane mirror holding a camera in your hand now when you capture the photo office looking in the mirror, the real image of the main images formed in the camera which is captured.

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    $\begingroup$ There are several grammatical errors or typos in this answer, eg "photo office". Please fix it so that readers can understand what you are trying to say. $\endgroup$
    – PM 2Ring
    Commented Feb 21, 2019 at 12:32

For a virtual object, the rays do not ontersect at a point in front of the mirror. So we take a concave mirror and adjust itsuch the image formed by the concave mirror appears to be behind the plane mirror(although the image will not be formed as the rays are blocked by the plane mirror). But for the plane mirror, it will seem like the rays do not intersect anywhere in front of the mirror and thus making it a 'virtual object'

{See the image for a better visualization} (https://i.sstatic.net/dcJKv.jpg)


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