# Homework Help: Help with light and lens problem

1. Feb 5, 2006

### ACLerok

Here is the problem:

A converging lens with a focal length of 12.0 cm forms a virtual image 8.00 mm tall, 17.0 cm to the right of the lens.

I am supposed to find the position of the object. I tried using the lens eq. 1/s + 1/s' = 1/f where s is the distance of the object from the lens, s' is the dist of the image from the lens, and f is the focal point. I got s to be 40.82 cm but I'm being told it's wrong. Is there something im missing here? Any tips and points would be great.

Thanks!!

2. Feb 5, 2006

### George Jones

Staff Emeritus
Careful with the signs!

For a converging lens, where must an object be for a virtual image to be produced.

Regards,
George

3. Feb 5, 2006

### ACLerok

to the right of the lens correct? does this change the equation in some way?

4. Feb 5, 2006

### George Jones

Staff Emeritus
And where with respect to the focal point?

George

5. Feb 5, 2006

### ACLerok

Well my obvious guess would just be (s' - f) centimeters to the right of the focal point..

6. Feb 5, 2006

### George Jones

Staff Emeritus
In order for a converging lens to produce a virtual image, the object must be between the focal point and then lens. I have been trying to get you to see that your answer of 40.8 cm can't be correct. I am reasonably certain that you have made a sign mistake, but I used a different sign convention (and thus different formulae) last semster, and I can't remember, off the top of my head, the other sign convention, so I'm not quite sure where you're going wrong.

When I use the convention with which I am familiar, I can:

1) get an answer that is between the focal point and the lens;

2) get 40.8 cm when I make a sign mistake.

Regards,
George

7. Feb 5, 2006

### nrqed

What sign did you use for f and for s'? That`s your mistake. f is positive for a converging lens. What is the sign for s' for a *virtual* image?

Pat