# Two Lenses In A Row

## Homework Statement

Two lenses, one converging with focal length 20.0cm and one diverging with focal length -10.0cm, are placed 25.cm apart. An object is placed 60.0cm in front of the converging lense. Determine the position of the final image.

## Homework Equations

$$\frac{1}{s_i} + \frac{1}{s_o} = \frac{1}{f}$$

## The Attempt at a Solution

I understand that the idea behind this problem is that the image of the first lense becomes the object for the second lense. So the first thing I did was to calculate the position of the image of the first lense.
$$s_i=\frac{1}{\frac{1}{f} - \frac{1}{s_o}}=\frac{1}{\frac{1}{20} - \frac{1}{60}}=30cm$$
Now the part I am stuck on is that according to this, the image of the first lense is behind the second lense (the separation of the two lenses is only 25cm). I'm not sure where to go next.

Would it be sensible to say that you treat the image of the first lense as sort of a "virtual" object... Meaning that you have to look through the second lense towards the first image to see it. My problem with this is that if you look through the second lense is such a way, then the light rays are going away from you and will never hit your eyes.

What do you think?

## Answers and Replies

Kurdt
Staff Emeritus
Science Advisor
Gold Member
You're going about it the right way. You should be able to work out the object distance for the second lens in terms of their separation and the image distance from the first lens. Then you can work out the resulting image distance in terms of the focal lengths the separation and the object distance from the first lens.

Hmm, so it does not matter that the rays from the first lens go through the second lens and then form the first image?

the image that forms from the first lens will be a virtual object for the next lens. you need to make your sign convention follow along with this. Dont worry about the lenses in combination, just worry about them separately.

Ah thanks!