I was doing this experiment. Let's look at this image:

enter image description here

Suppose we get the imaginary object at dv by convergence lens. And this object is like a real object for the concave lens. Then how do we get a real object on the screen? It is known that only virtual objects can be made on the screen by convex lenses, and the image should be before the real object and after the lens.

According to TLE (Thin lens equation) we can find the focal point, however i'm confused about what is real and imaginary here.

Also, I had tried hard to draw ray diagram for this one. No success in forming an imaginary image in dv or real image at dr.


1 Answer 1


If the concave lens was not there and the object was at a distance greater that its focal length the convex lens would form a real image.

enter image description here

Introducing a concave lens results in the incoming rays from the convex lens being refracted as shown in the diagram above.
The refracted rays are still convergent and so form an image $I$ at a distance $q_2$, your $d_v$, from the the concave lens.
After passing through the convex lens where the rays would have met is at a distance $p_2$, your $d_R$, from the concave lens and that can be thought of as a virtual object for the concave lens.

  • $\begingroup$ Thank you, only 2 things - I think you confused dv and dr. And also, I think the object for the concave lens is acctually I. And the virtual image is at point p2. $\endgroup$ Aug 24, 2016 at 12:42
  • $\begingroup$ @TheCapacitor You can certainly look at it your way but my way is equally valid because of the reversibility of light. Try it both ways and you should get the same value for the focal length of the concave lens. $\endgroup$
    – Farcher
    Aug 24, 2016 at 14:23

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