GREPhysics.NET: GR0177 #12

GR0177 #12

Problem

GREPhysics.NET Official Solution

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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 $d_o<f$ . 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

ramparts
2009-10-05 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.

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Comments

gmeheretu 2012-05-07 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
ramparts 2009-10-05 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
gn0m0n 2008-11-05 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
grae313 2007-11-01 16:21:18	A little ray tracing also gets the job done. Reply to this comment

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Hm. So is the convention for both mirrors and lenses that virtual images have negative distances? Thanks, by the way, to all.

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