What Is the Proper Sign Convention for a Virtual Image in Lens Calculations?

[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 I am missing here? Any tips and points would be great.

Thanks!

 

[George Jones]

Careful with the signs!

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

Regards,
George

 

[ACLerok]

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

 

[George Jones]

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

And where with respect to the focal point?

George

 

[ACLerok]

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

 

[George Jones]

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

 

[nrqed]

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 I am missing here? Any tips and points would be great.

Thanks!

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