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A piece of thin spherical shell that has a radius of curvature of 106 cm is silvered on both sides. The concave side of the piece forms a real image 79.5 cm from the mirror. The piece is then turned around so that its convex side faces the object. The piece is moved so that the image is now 35 cm from the piece on the concave side. How far was the mirror moved? Was it moved toward or away from the object?

I'm not sure if the shell is the mirror or not, but this is what I have done:

I took the radius of curvature divided by 2 to find the focal length. $f=53$. Then using $1/f=1/s+1/s'$:

$1/53-1/79.5=1/s$, $s=159$.

Now that I have $s$, I started working on flipping it to the convex side, which makes $f=-53$. I am not sure how to continue to find $s$ from here.

In the end I believe taking the first object distance, 159, minus the second object distance should give me the distance the mirror has been shifted.

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    $\begingroup$ Hi MathMan08. Welcome to Physics.SE. This site deals with conceptual Physics Q&A. We don't encourage homework questions that doesn't involve any sort of work done by the author (which is you) and asks other users to solve the problem. If you think you could clarify your question, add what you've done along with your question. We're ready to help you. If you aren't clear, Please have a look at our homework policy for more info. After improving the post, flag it for moderator attention. $\endgroup$ Commented Apr 23, 2013 at 1:51

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You find $s$ in the same way you found it for concave mirror, just keep in mind that for the convex mirror the image is always virtual(as explained nicely in the Wikipedia article on curved mirrors, it cannot be projected on a surface, unlike the real image), so in this case (by convention) $s'=-35cm$ and $f=-53cm$.

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    $\begingroup$ Thanks. I figured it out earlier. Then subtracting the two distances got me the answer I wanted. $\endgroup$
    – anon
    Commented Apr 23, 2013 at 17:09

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