# Concave Lenses algebraically

1. Mar 24, 2017

### Matt M

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

An object is in the focal length of a convex lens. As the object is moved from the focus towards the lens, what happens to the object?
2. Relevant equations
1/f=1/d0+1/di

3. The attempt at a solution
I know how to solve this problem using ray diagrams, and the solution is:

The image increases in size and moves farther from the lens

I am trying to conclude this algebraically instead of just with a ray diagram. Just choosing numbers though, I let f=5 and do=4. Plugging this into the thin lens equation, I find that d0=-20, showing a virtual image on the same side of the lens as the object. However, then when I move the image closer and let f=5 and d0=3, the thin lens equation gives that d0=-7.5. Hence it seems to me that as the object is moving closer, it is still past the focal point, but moving closer to the lens as opposed to farther. Why is the numerical example leading to different results than the ray diagram? Thanks in advance!

2. Mar 25, 2017

### stevendaryl

Staff Emeritus
Because you drew the wrong ray diagram?

The result given by the lens equation is intuitively right. If you have a magnifying glass, and you're looking at something small (say, an ant) through the lens, then
1. The image is larger than the object.
2. The image appears to be on the same side of the lens as the object (so it's a virtual image).
3. As you move the object closer, the image appears to get closer, as well.
That's exactly what the lens equation tells you. So what ray diagram is telling you something different?