# Homework Help: Lens Problem, Help from Anyone! Thank you, I appreciate it very much!

1. Jan 16, 2008

### physicsbhelp

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

1. The problem statement, all variables and given/known data

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 lenght 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.

2. Relevant equations

1/f = 1/Do + 1/Di

3. The attempt at a solution

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

Last edited: Jan 16, 2008
2. Jan 16, 2008

### 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.

3. Jan 16, 2008

### physicsbhelp

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

4. Jan 16, 2008

### dlgoff

Last edited by a moderator: Apr 23, 2017
5. Jan 16, 2008

### physicsbhelp

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

6. Jan 16, 2008

### physicsbhelp

wait so would the resulting image be real or virtual?

7. Jan 16, 2008

### dlgoff

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

Edit: removed quote about virtual images.

Last edited: Jan 16, 2008
8. Jan 16, 2008

### physicsbhelp

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

9. Jan 16, 2008

### 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."

10. Jan 16, 2008

### physicsbhelp

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

11. Jan 16, 2008

### dlgoff

Yep. Glad I could help.

12. Jan 16, 2008

### physicsbhelp

so 22.5 is right?
and thanks

13. Jan 16, 2008

### dlgoff

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