# Focal Length of a Diverging Lens

1. Mar 18, 2009

### dcramps

1. The problem statement, all variables and given/known data
An object that is 6.00 cm tall is placed 40.0 cm in front of a diverging lens.
The magnitude of the focal length of the lens is 20.0 cm. Find the image position and size. Is the
image real or virtual? Upright or inverted?

2. Relevant equations
/

3. The attempt at a solution
I am confused on "the magnitude of the focal length of the lens is 20.0cm" part. Does that mean the principle and the secondary focal lengths are 20.0cm each, or is the entire focal length (primary and secondary) 20.0cm?

2. Mar 18, 2009

### rl.bhat

I am confused on "the magnitude of the focal length of the lens is 20.0cm"
It means that focal length may be positive or negative depending on the sign convention.
And in the diverging lens image is virtual.

3. Mar 18, 2009

### dcramps

So does that mean primary and secondary are both 20.0cm though? Your response didn't really answer my question :[

4. Mar 18, 2009

### dcramps

Oh, wait, it did...I think. Since it means it can be either negative or positive, then the primary could be the positive, secondary could be the negative, thus each focal length has an absolute value of 20.0cm. Right?

5. Mar 19, 2009

### Lambduh

I'm not sure what you mean by primary and secondary focal lengths. A thick lens (most of the time we talk about thin lenses and can use the thin lens formula: 1/f = 1/s + 1/s') effectively has two focal lengths from the front and the back of the lens but since only one value is given and no other information(Radii of curvature and such) about the optic i would assume you want to use the thin lens formula.

Now using the thin lens formula: http://en.wikipedia.org/wiki/Thin_lens

you want to put in a negative number for focal length because you are using a negative (diverging/concave) lens. Knowing the object distance you can solve for the image distance which will be virtual.

edit: concave not convex

Last edited: Mar 19, 2009
6. Mar 19, 2009

### dcramps

Yes, it is a thin lens. I guess I should have mentioned that.

What I mean by primary and secondary focal lengths is exactly that...primary focal length on the same side of the lens as the object, and secondary on the other side. See attached diagram.

http://www.roflz.us/Untitled.jpg [Broken]

My confusion was that I wasn't sure if the 20.0cm focal length given spanned both the primary and secondary, making each of them 10.0cm, or was 20.0cm for both. I know the image is virtual/upright/smaller than the object..that isn't what I was asking. The only thing I was confused on was the focal length being 10 or 20. :[

Last edited by a moderator: May 4, 2017
7. Mar 19, 2009

### Lambduh

Hmm i can't see the attachment yet. So i'm guessing you're deriving the focal length from the image you posted then?

Check this out it will help you i think. If you don't have mathematica installed you can install the player and use it that way.

http://demonstrations.wolfram.com/RayTracingWithLenses/

8. Mar 19, 2009

### dcramps

I have updated my post. Didn't realise there was admin approval required for the attachments here.

9. Mar 19, 2009

### Lambduh

For the focal length it would only be from the center of the lens out. So 20cm is from the "primary" position to the center of the lens. Note that it's actually -20cm when using the thin lens formula to solve for the image distance.

10. Mar 19, 2009

### dcramps

Excellent. That's what I wanted to know! Thanks for all the help. :)

11. Mar 19, 2009

### Lambduh

No problem:) did you figure out the virtual image size?