Physics lens problem. please, please help!

Converging lens physics question?

A converging lens with a focal length of 6.30 cm forms an image of a 5.00mm -tall real object that is to the left of the lens. The image is 2.00cm tall and erect.
Where are the object and image located?
virtual or real?

i appreciate your help

So first you are given the heights of the object and it's image

some variables i will use that you may or may not know:
s(o)= object distance
s(i)= image distance

Magnification = image height / object height
Magnification is also equal to -s(i)/s(o)

therefore, we can conclude

h(i)/h(o) = -s(i)/s(o)

Now we have two unknowns so we need another equation

So you can use

1/f=1/s(o) + 1/s(i)

You can solve for either image distance or object distance first, doesnt matter which so I'll start with object distance

s(o) = 1/ ( 1/f - 1/s(i) )

Doing some sloppy variable algebra, plug that equation in for s(o) and solve for s(i)

you'll end up with:

s(i) = -h(i)f/h(o) + f

s(i) = -18.9 cm

Since the image distance is negative, it is virtual.

plus s(i) into

1/s(o) = 1/f-1/s(i)

s(o) = 4.725 cm


I am also doing this chapter in my AP physics class so this is good practice, but also not guaranteed 100% right. I hope I was clear enough and helped a little