GR0177 #11

[maxyar]

Ok, when I draw the light rays I get the answer very easily. However, if I try to solve this problem using the concepts of real images, virtual images, and so forth I start losing it. Do I need to understand which one is the virtual image and which one isn't ?

As far as I can tell there is no 2nd image because the 1st image is being intercepted by the 2nd lens. All the solutions I have seen so far use the thin lens formula twice invoking the negative virtual image--- but I don't understand this at all.

as i said before if I just look at the light rays it is clear that they converge in between the first real image and the 2nd lens so the only possible answer is A)

but I am concerned that I may need to understand this sort of problem more w/ the virtual image vocabulary in other problems


SO CONFUSING ARGH!!!

[maxyar]

ok so I found the answer on grephysics.net that makes some sense:

"for virtual objects (objects) the distance is taken as negative by convention"

seems like, indeed, the first lens image is a virtual object for the second lens.
but I can't find any source of this by google search. Id like to see more discussion of virtual object somewhere.

[physicsworks]

This is actually a well known sign convention for lenses. You can read the details in Hecht's Optics (Fourth Edition, Chapter 5). Also, summarizing table can be found here.