Lens Problem, Help from Anyone Thank you, I appreciate it very much

[physicsbhelp]

[SOLVED] Lens Problem, Help from Anyone! Thank you, I appreciate it very much!

Homework Statement

An object is placed 30 mm in front of a lens. An image of the object is located 90 mm behind the lens.
a) What is the focal length of the lens?
b) On an axis (the x-axis) draw the lens at position x=0. draw at least two rays and locate the image to show the situation described above.

Homework Equations

1/f = 1/Do + 1/Di

The Attempt at a Solution

if i use -90mm then f= -45.

 

[dlgoff]

"... i am not sure of is whether or not i should say -90mm because it says behind the lens."
Distances in the gaussian form to the thin lens equation you gave are real positive values.

"...but since i got 22.5 as the focal, my drawing isint coming out right."
Remember, the focal length is the distance in front or behind where rays from a point source will converge. In your problem, they are wanting you to show how the arrow will look at 90mm, if I understand correctly.

 

[physicsbhelp]

but don't you draw the rays thorugh the focal point?

 

[dlgoff]

A picture is worth a thousand words and since this old computer is limited I would like you to look at http://en.wikipedia.org/wiki/Lens_(optics)#Imaging_properties"

Note where the focal points are. Hope this helps.

Regards

 

[physicsbhelp]

actually i think that picture just confused me more. but thanks for tyring to help.

 

[physicsbhelp]

wait so would the resulting image be real or virtual?

 

[dlgoff]

Sorry the link caused confusion but the pic is exactly what you have here.

Edit: removed quote about virtual images.

 

[physicsbhelp]

what do u mean by "Edit: removed quote about virtual images."

 

[dlgoff]

This is what I was wanting to quote in the previous post:

"...if an object is placed at a distance S1 along the axis in front of a positive lens of focal length f, a screen placed at a distance S2 behind the lens will have an image of the object projected onto it, as long as S1 > f. This is the principle behind photography. The image in this case is known as a real image."

 

[physicsbhelp]

oooooooooooo okay i understand you now, thank you,
but is my answer to part a) right?

 

[dlgoff]

Yep. Glad I could help.

 

[physicsbhelp]

so 22.5 is right?
and thanks

 

[dlgoff]

Yes. 22.5mm. And since this is image distance is grater than this focal length, the image is real.