Find image distance multiple optics.

[viperassasin]

Homework Statement

As shown in the figure View Figure the candle is at the center of curvature of the concave mirror, whose focal length is 10.0 {\rm cm}. The converging lens has a focal length of 32.0 {\rm cm} and is 85.0 {\rm cm} to the right of the candle. The candle is viewed looking through the lens from the right. The lens forms two images of the candle. The first is formed by light passing directly through the lens. The second image is formed from the light that goes from the candle to the mirror, is reflected, and then passes through the lens.

http://session.masteringphysics.com/problemAsset/1043040/4/YF-34-094.jpg

Homework Equations

1/s+1/s'=1/f

The Attempt at a Solution

I have already found the first image to be a distance of 51.3 cm I am now looking for the second image.
Trying to figure out what I put for the object distance to the concave mirror or do I not worry about it.

 

[EngageEngage]

Since the object is at the center of curvature, you know that the image formed from the mirror will be at the same center of curvature, only the image will be inverted. Now, use this image as the object for the lens -- you should find that the image distances are the same for both situations. However, in one case the object will be inverted with respect to the original object, and the other will be upright.

 

[Moetasim]

I would approach this problem by using the equation 1/s + 1/s' = 1/f to determine the image distance for the second image formed by the light that goes through the mirror. In this equation, s represents the object distance and s' represents the image distance. Since the candle is located at the center of curvature of the concave mirror, the object distance can be assumed to be infinity. Therefore, the equation becomes 1/∞ + 1/s' = 1/10.0 cm. Simplifying this, we get 1/s' = 1/10.0 cm, which means that the image distance for the second image is also 10.0 cm. This means that the second image is formed at the same location as the first image.