Hi,

I have seen a bunch of solutions for this problem using the thin lense equation. But I don't understand.

The image from the first lens serves as the object for the second lens, but the image falls on the focal point of the secone lens. If there is an object that is sitting on the focal point of a thin converging lens then isn't the spot where the object will not cast an image?

If you draw a ray diagram the rays don't converge anywhere, and if you use the thin lens equation and define positive to be in the other direction, then you get 1/f=1/do+1/di and 1/f=1/10 and 1/do=1/10 hence 1/di=0 and the image is infinitly far away!

please help me conceptually, I understand that if you flip the coordinates around that 1/do=-1/10 and then you get the right answer, but it should be equivalent to change the sign of the axis and I don't see how the light rays converge to form an image.

-Adam

## PGRE0177 problem#11 - I'm getting no image

- WhoaNonstop
**Posts:**853**Joined:**Mon Sep 21, 2009 1:31 am

### Re: PGRE0177 problem#11 - I'm getting no image

Well, first lens:

1/f = 1/do + 1/di

1/20 = 1/40 + 1/di

2/40 - 1/40 = 1/di

1/40 = 1/di

di = 40.

Where does this place the image? When the image is behind the lens instead of in front of it, how do our values change for the next equation?

-Riley

1/f = 1/do + 1/di

1/20 = 1/40 + 1/di

2/40 - 1/40 = 1/di

1/40 = 1/di

di = 40.

Where does this place the image? When the image is behind the lens instead of in front of it, how do our values change for the next equation?

-Riley