Diverging Lens: Find Object Given Real Image & Size

[lha08]

Homework Statement

The focal length of a diverging lens is -20 cm. Locate the object, given that the image is real, erect, and 150% of the size of the object.

Homework Equations

(1/f)=(1/do)+(1/di)
M=-(di/do)

The Attempt at a Solution

i'm not really sure how a diverging lens forms a real image (i thought they could only from virtual since most of the images are on the same side) and a size that is larger than the original object. At first, I thought that the image would be inverted (negative magnification) since a virtual image would have a positive magnification (image is upright)...

 

[mgb_phys]

For a diverging lens revealed the image of the object is virtual, upright, reduced in size and located on the same side of the lens as the object.

You can form a real image from a diverging lens - but only by feeding it a converging beam from another lens (eg in a retrofocus wide angle camera lens)