Question

Question #4b0ff

Answer 1

`f=4 cm`
I think that this is a converging lens creating a virtual image bigger than the object (a diverging one would produce an erect image but smaller than the object). The image is erect (I suppose having the same orientation of the object) and placed at 12 cm from the lens.
Graphically:

(Picture source: http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/raydiag.html)

I´ll use the following equations:
1) Distances: `1/(o)+1/(i)=1/f`
2) Magnification: `M=(h')/h=-(i)/(o)`

I start using the Magnification: `M=(h')/h=-(i)/(o)` to find the distance of the object, `o`:

Where:
`h'=2cm>0` is the image height positive (erect);
`h=0.5>0` is the object height (positive as well);
`i=-12 cm` is the image distance (on the left or in front of the lens);
`o=?`

`(2)/(0.5)=-(-12)/(o)`
`o=3cm`

Now I use the focal length formula: `1/(o)+1/(i)=1/f` to find the focal length, `f`:
Where:
`o=3cm`;
`i=-12cm`;
`f=?` is the focal length.
Giving:
`1/3+1/-12=1/f`
and `f=4cm`

Note:
In this exercise I used the following conventions:

(Picture source: www.physics.uc.edu)

To decide which configuration to use in the relative positioning of object, image and lens I used reference diagrams such as:

(Picture source: www.physics.uc.edu)
Where I can deduce the characteristics of the system (magnification, orientation, etc.).