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GR0177 #12
Problem
 GREPhysics.NET Official Solution Alternate Solutions
This problem is still being typed.
Optics$\Rightarrow$}Mirror Formula

The mirror formula is identical in form to the lensmaker's formula in Problem 11. The sign convention varies for the image distance. If an image distance is on the inside of the mirror, then it is taken as negative.

Thus, $1/f=1/d_o+1/d_i\Rightarrow 1/d_i=1/f-1/d_o$. From the diagram, one deduces that $d_o. Thus, one finds that $d_i$ has to be negative. The virtual image has to be inside the mirror. Only choice (V) shows this.

Alternate Solutions
 ramparts2009-10-05 12:29:10 Just look at it qualitatively - an object at the focal point will cause the rays to be parallel. An object closer than the focal point will then cause the rays to diverge, making the image appear to be behind the mirror.Reply to this comment
 gmeheretu2012-05-07 15:35:09 This is the only position where the object has to be placed in a concave mirror to get a magnified virtual image at position V.Reply to this comment ramparts2009-10-05 12:29:10 Just look at it qualitatively - an object at the focal point will cause the rays to be parallel. An object closer than the focal point will then cause the rays to diverge, making the image appear to be behind the mirror.Reply to this comment gn0m0n2008-11-05 23:00:50 Hm. So is the convention for both mirrors and lenses that virtual images have negative distances? Thanks, by the way, to all.Reply to this comment grae3132007-11-01 16:21:18 A little ray tracing also gets the job done.Reply to this comment

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