GR0177 #11
Posted: 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!!!
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!!!