A common method to solve for the image formed due to a two lens system consisting of two thin-lenses separated by a distance is:

enter image description here

  1. Locate the (intermediate) image formed by the first lens, ignoring the second one
  2. Use this image as the object for the second lens to get the final image

However, how do we solve for the final image if the intermediate image is formed beyond the second lens? My textbook follows the Cartesian sign convention for optics. So, according to the book, to solve for the final image, we take the object distance for the second lens as positive instead of negative.

However, what is physically the reason for making this change of sign? Why does the lens formula even work in this case?


The physics behind can be explained by invoking the idea of real and virtual objects (or images). If light rays from a point or location in space physically go out in all possible directions (or diverge, as they indeed do from point $O$ in the figure), the object at that point is real. Similarly, if light rays physically converge, the image produced is real. This means the final image formed at point $I$ is real.

It also implies that a real image would have been formed at point $I_1$ if the lens $L_2$ had not been present. However, the presence of the lens $L_2$ makes the light rays emanate virtually from point $I_1$ (which is on the right hand side of the reference lens). In other words, there is now a virtual object at point $I_1$ and the light from this object interacts with $L_2$ to form a real image.

  • $\begingroup$ But if light emanates virtually from $I_1$ then it means that some light goes back into the second lens. However, that is clearly not possible if we trace the light going from $O$. Neither does the light appear to emanate from $I_1$. $\endgroup$ – Gerard Sep 30 '14 at 9:52
  • $\begingroup$ Maybe one notion was not clarified: As you go from left to right, you consider one optical component at a time. That is to say, at $I_1$ when you consider that the real image produced by $L_1$ = virtual object for $L_2$, you can then forget $L_1$ completely. $\endgroup$ – jayann Sep 30 '14 at 11:21
  • $\begingroup$ Yes, but that is precisely my question. Why can we forget $L_1$? I can visualize the situation when the virtual object is between the two lenses, but I just don't understand how an image can act as an object when it is formed beyond the second lens. $\endgroup$ – Gerard Sep 30 '14 at 11:25
  • $\begingroup$ At the end of the day, information is carried by the physical transmission of light. The notion of virtual images/objects mainly serves to ease the overall visualization of this flow while sign conventions allow the correct calculation of distances in each step. Also, the one at a time procedure holds for any complex assembly of lenses. $\endgroup$ – jayann Sep 30 '14 at 11:25
  • $\begingroup$ Because the role played by the first lens together with the first (real) object has been completely taken into account. This one at a time procedure even holds for any complex optical assembly such as a cascade of $n$ lenses. The response of the $k^{th}$ lens in the cascade is influenced by lenses $1$ to $k-1$ but once we know this response, we can use that as an input to find the response of the $(k+1)^{th}$ lens while again ignoring all other lenses. $\endgroup$ – jayann Sep 30 '14 at 11:32

The intermediate object you describe is a virtual object.

You should be familiar with the concept of a virtual image. A virtual image doesn't exist because it cannot be displayed on a screen, however the light rays follow paths as if they had come from the image. A virtual image is (generally) formed on a different side of the lens to a real image, so we mark its position by an opposite sign for $v$.

A virtual object is conceptually similar. A virtual object doesn't exist, but the paths the light rays follow are the same as if the object did exist and the light rays were coming from it. A virtual object is (generally) on the other side of the lens to a real image, so we mark its position by an opposite sign for $u$.

  • $\begingroup$ So, what is the difference between a virtual object and a virtual image? Could you add some references to articles on 'virtual objects'? $\endgroup$ – Gerard Sep 30 '14 at 9:53
  • $\begingroup$ @Gerard any introductory optics text will do. $\endgroup$ – Carl Witthoft Sep 30 '14 at 11:40

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