# Calculating Magnification with lens, an object, and a screen.

1. Sep 8, 2014

### jmacmartin

1. The problem statement, all variables and given/known data

A lens with focal length fo = 9 cm forms an image of an object placed a distance d=3.7fo to the left of the lens, onto a screen to the right of the lens. The distance L from object to image is fixed. The lens is now moved toward the screen a distance x until an image is formed again. (a) What is the magnification in the initial case? (b) How far was the lens moved? (c) What is the magnification in the second case? (d) What is the distance L between the object and the image?

2. Relevant equations

1/f = 1/s + 1/s'

and

m = -(s/s')

3. The attempt at a solution

I just figured out part A and am now stuck on part B. I'm not sure how to go about solving it. I read online that "when the lens is moved, the object and image distances are switched." Can someone please explain to me why this is? I can't seem to find any information in my book. Not looking for the answer, but more how to understand what happens as this lens is moved.

Thanks!

Last edited: Sep 8, 2014
2. Sep 9, 2014

### andrevdh

It comes from the thin lens equation:

1/f = 1/s + 1/s'

You can switch the object and image distances and the equation still holds.
So moving the lens until the image distance is equal to the previous object
distance will result in a new object distance equal to the previous image distance.