Understanding Virtual Objects and Negative Do Values in Lenses

[CaneAA]

I would appreciate some clarification on working with lenses and virtual objects (when the object distance do is negative).

I'm not sure what happens when the do is negative and you get a positive value for di. On which side of the lens does the image go?

Normally, for a lens, if di is positive, it is a "real image" and goes on the side of the lens opposite the object; and if di is negative, it is a "virtual image" and goes on the same side of the lens as the object. But what are the rules for when you start out with a virtual object?

Thanks!

 

[ehild]

The problem of virtual object arises in an optical device with combination of lenses/mirrors. The object is put in front of the device. Assume the combination of two lenses. A virtual object means that before the real image by the first lens is formed, a second lens is put in. This second lens refracts the light rays again and their intersection defines the place of the new image. If this new image can appear on a screen it is real. That means it is at the opposite side of the two-lens device than the real object. With respect to this lens, do is negative and di is positive.

ehild

 

[CaneAA]

Thank you :)