GR0177 #11

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maxyar
Posts: 6
Joined: Sun Apr 24, 2011 1:03 am

GR0177 #11

Post by maxyar » Thu Apr 28, 2011 4:52 am

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
Posts: 6
Joined: Sun Apr 24, 2011 1:03 am

Re: GR0177 #11

Post by maxyar » Thu Apr 28, 2011 5:06 am

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
Posts: 80
Joined: Tue Oct 12, 2010 8:00 am

Re: GR0177 #11

Post by physicsworks » Thu Apr 28, 2011 12:38 pm

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.



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