# At what distance is the object from the mirror?

• dmolson
In summary, the problem involves an object placed in front of a convex mirror with a 65 cm radius of curvature. The virtual image formed is half the size of the object, and the question asks for the distance of the object from the mirror. Using the mirror equation 1/p + 1/q = 1/f, with p representing the object distance and q representing the image distance, the correct equation should be 1/p + 2/p = 1/f. Solving for p, the object distance is found to be 32.5 cm.

## Homework Statement

An object is placed in front of a convex mirror with a 65 cm radius of curvature. A virtual image half the size of the object is formed. At what distance is the object from the mirror?

1/p + 1/q = 1/f

## The Attempt at a Solution

I have tried this problem several times different ways and cannot get the correct solution. I tried it this way.

focal length = Radius of curvature/2 = -32.5 cm (negative because convex mirror
virtual image = 1/2 object = -q
object = + p

I obtained 16.25 using equation 1/p -1/32.5 = -1/32.5, but it is wrong, as well as -16.25. Any help would be great.

"virtual image = 1/2 object = -q"

You didn't use this correctly in your calculation. The question says the image is half the size of the object, so that means q = -0.5p. You can use that in your equation. I don't understand why you have -1/32.5 twice there.

dmolson said:
focal length = Radius of curvature/2 = -32.5 cm (negative because convex mirror
virtual image = 1/2 object = -q
object = + p
You're on the right track here, but you didn't follow through. Hint: Write the image distance (q) in terms of the object distance (p). Then apply that to the mirror equation.

I obtained 16.25 using equation 1/p -1/32.5 = -1/32.5, but it is wrong, as well as -16.25.
Not sure what you are doing here. Redo this given my advice above. (And realize that a negative value of p would imply that the object was inside the mirror!)

(Looks like hage567 beat me too it while I was daydreaming.)

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