Why does a positive lens focus two images at different lengths?

[MechEngrStdnt]

Missed this part of the lab and I need to complete my lab write up.

We have a light source at one end of a track, and at the other end a screen. there is a positive lens in between them. When we set the screen at a distance away from the light source, let's say one meter, and slide the lens on the track connecting them, starting from the light source and moving towards the screen, at two distinct locations we get a focussed image. The first image, closest to the light source, is roughly 4X the size actual size of the image, while other is roughly 1/4X the size. My question is, why is this? what are those lengths? are the the focal length? and how can I show these two lengths mathematically?

Thanks!

 

[rude man]

You have two equations to solve simultaneously.
The first one relates the focal distance to the object and image distances.
The second represents the constraint on your setup. What is that constraint?

 

[MechEngrStdnt]

The focal length of the positive lens was supposed to be 10cm, the distance between the light and projected image was 1 meter, is this was you need?

 

[dauto]

The distance between the lens and the object (light source) represented by p and the distance between the lens and the image (screen) represented by q satisfy the equation $$\frac{1}{p}+\frac{1}{q} = \frac{1}{f},$$ where f is the focal length of the lens. In the experiment you described they also satisfy $$p+q=L,$$ where L is the fixed distance between the light source and the screen - one meter according with your description. These two equations have two possible solutions - the two situations where a focused image was obtained. Combine that with the fact that the magnification M of a simple set up like that can be calculated by $$M=\frac{-q}{p},$$ and you'll have all you need to understand your experiment.

 

[MechEngrStdnt]

awesome, makes sense. Thanks to you both!