Image Distance with two concave lenses

[Jende]

Homework Statement

An object 1.25 cm tall is placed 100 cm in front of a convex lens with a focal length of magnitude 50 cm. A concave lens with a focal length of magnitude 20 cm is placed 90 cm beyond the first lens. Where is the final image located?

Homework Equations

(1/s)+(1/s')=1/f
s= object distance from lens
s'= image distance from lens
f= focal length

The Attempt at a Solution

I wasn't sure where to start so I tried find the image distance for the first lens.
(1/100cm)+(1/s')=(1/50cm) => (1/100cm)-(1/50cm)=(-1/100cm)=(1/s') => s'= -100

After this i got stuck because I wasn't sure what to do when the image distance is behind the second lens.

Thanks in advance.

 

[ehild]

Homework Statement

An object 1.25 cm tall is placed 100 cm in front of a convex lens with a focal length of magnitude 50 cm. A concave lens with a focal length of magnitude 20 cm is placed 90 cm beyond the first lens. Where is the final image located?

Homework Equations

(1/s)+(1/s')=1/f
s= object distance from lens
s'= image distance from lens
f= focal length

The Attempt at a Solution

I wasn't sure where to start so I tried find the image distance for the first lens.
(1/100cm)+(1/s')=(1/50cm) => (1/100cm)-(1/50cm)=(-1/100cm)=(-1/s') => s'= +100

You have sign errors here.

After this i got stuck because I wasn't sure what to do when the image distance is behind the second lens.

Thanks in advance.

Take the object distance negative.

 

[James R]

The image formed by the first lens acts as the object for the second lens. However, when you apply the thin-lens equation for the second lens you'll need to make sure that the object distance is the distance measured from the second lens, which you can do by taking into account the distance between the two lenses.

Be careful of the signs. As ehild says, you have an error in your first calculation.

 

[andrevdh]

Being behind he second lens this image is then a virtual object for the second lens. The light rays are still entering the second lens, but they seem to originate from a object located on the "wrong" side of the lens. The rays will therefore not form the object and is thus a virtual object for the second lens.

 

[Jende]

Ok thanks a lot. This was giving me a lot of frustration.