GR0177 #11

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

GR0177 #11

Postby 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

Postby 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

Postby 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.




Return to “Problems, Solutions, and Discussion”

Who is online

Users browsing this forum: No registered users and 1 guest