Lenses- magnification adn focal length

[PhysicsChick8]

Homework Statement

Hey,,
I was wondering if anyone could help me with some physics work i am doing as i am completely stuck.
Is there supposed to be any relationship with the magnification and focal length of lenses and if so what is it?
I did an experiment and measure the magnification of lenses different distances from the lenses and also calculated the focal length at these points. Now i am stuck on how to write all this up and what graphs to draw.
Any help would be much appreciated :s
xxx

Homework Equations

I used 1/v-1/u=1/f for focal length
and size of image/ size of object for magnification

The Attempt at a Solution

 

[tiny-tim]

Welcome to PF!

Hi PhysicsChick8! Welcome to PF!

See the PF Library entry on lens ​

 

[PhysicsChick8]

thanks i read that but I'm kinda very bad at physics and need it explained in very simple terms if possible without too many confusing equations xx

 

[tiny-tim]

As the Library says …

The image is magnified (with respect to the size of the object) by the factor m = -d_i/d_o

and …

d_o is positive for a real object and negative for a virtual object (See Images section).
d_i is positive for a real image and negative for a virtual image (See Images section).

and …

For the common situation where the object and image are real (d_o and d_i both positive), the magnification is negative indicating an inverted image with respect to the object.

 

[PhysicsChick8]

so if i were plotting stuff on a graph what would i plot? x

 

[tiny-tim]

Well, what are you trying to show?

 

[PhysicsChick8]

thats the problem i don't know,,, we had to choose an experiment and i choose the magnification and power of lenses as it seemed easy. Now i have to write it up and draw graphs but I am not sure what of as i don't really know what is linked. I really am useless at this if you haven't guessed already x

 

[tiny-tim]

I suppose the obvious graphs would be d_i against d_o, and m (= -d_i/d_o) against d_o.

Try drawing them, and see what you get.

 

[PhysicsChick8]

thank you :D x