Solving Optics Lens Problems: 2 Possibilities for Object Placement

[oneplusone]

Hello,

If you have an object, and a bi-convex lense, you are given the focal length, and you are given the magnification factor (M) to be 2, how are there two possibilities to place the object?

One question I was doing asked this, and said:

a) if the image is virtual, and M=2, find the distance.
b) if the image is real, and M=2 find the distance.

Basically, i found one answer easily, and noticed that the absolute value of the difference between my distance and the focal length, equaled the absolute value of the other distance and the focal length.

So is this always the case?

 

[Simon Bridge]

The short answer is because there are two ways to get a magnified image from a converging lens.

To understand this - draw the ray diagrams.
image height is 2units and object height is 1unit (M=2)
put image and object some reasonable distance apart and draw the rays that must connect them ... and you'll find where the lens has to be and what the focal length is.

Repeat for a virtual image.

 

[oneplusone]

So would this be correct analytically? :

------|--------------[C]---------------|-----------

If the distances between the vertical segments, and the focal length [C] are equal, then both will produce an image of the Same size, same distortion, except one will be real, and one will be virtual?

 

[Simon Bridge]

That makes no sense to me, sorry.