What is the Distance 'L' in the Mirror and Lens Optical System?

[chris_avfc]

Homework Statement

Optical System:
Biconcave lens and convex mirror separated by distance 'L'.
Object placed 0.40m in front of the lens.
Final image is .12m from the vertex of the mirror

Lens curvature radii: 0.10m and 0.20m
Mirror curvature radii: .12m

Calculate the distance 'L'

Homework Equations

1/S + 1/S' = 1/f = (n2-n1)/n1 . (1/R1 - 1/R2)

1/S + 1/S' = -2/R

The Attempt at a Solution

Calculated where the lens creates an image using the first equation and got an answer of -0.20m, seen as the image would be virtual due to it being a divergent lens.
Then calculated where the mirror thought the object was using the second formula rearranged for S and got 0.086m.

Does that not mean the mirror is closer than the lenses?EDIT: Actually got the sign wrong got -0.086m, so a virtual image, so now I have two virtual images, so I'm lost.

 

[jbsteele]

A:I think you can ignore the lens, because it creates a virtual image. In this case the mirror will be placed at the same position of the lens (the object will be positioned in between the mirror and the lens).So the distance L is 0.40-0.12=0.28 m.