Virtual Image Location for Negative Lens System?

[mrlevis]

Homework Statement

S is a lighted object, height=1.
what is the final image that is created by the system?

Homework Equations

The Attempt at a Solution

i just don't understand how do i find the location of a virtual image created by a negative lens when the rays converge to a point beyond its focal point

thnks for helpers.

 

[Doc Al]

Why can't you just use the thin lens formula?

 

[mrlevis]

you mean this?:
1/f=1/U+1/V

i used this for the lens 1, and found that V, image distance, is 12, which means it is located 6 to the negative's right, 2 after the focal point.

So how can i use the equation for the second lens? U is supposed to be an object, not converging rays.

 

[Doc Al]

you mean this?:
1/f=1/U+1/V

i used this for the lens 1, and found that V, image distance, is 12, which means it is located 6 to the negative's right, 2 after the focal point.

All good.

So how can i use the equation for the second lens? U is supposed to be an object, not converging rays.

The image from the first lens becomes the object for the second. But since it's past the second lens, it must be treated as a virtual object. Just use the usual sign convention: If an object before the lens has a positive object distance, one after the lens would have a negative object distance.

 

[mrlevis]

oh... you right.
i don't know, for some reason i didnt think it would work.

but can you explain why is that? is there a simple proof to it?

 

[Doc Al]

I suggest that you review the derivation of the thin lens formula.

 

[mrlevis]

ok, thanks for help