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How far from the mirror must an object be placed to create an upright image three times the...

A concave mirror has a 50 cm radius of curvature. How far from the mirror must an object be placed to create an upright image three times the height of the object?
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Answer #1

Since the image is going to be 3x the size of the original, we know that it must be a virtual image.

Also since the image is going to be 3x larger, the image must also be 3x farther away than the object (it's a concave mirror).

If s is the object distance and s' is the image distance, we know that s'=-3s.

*Note that s' is negative: this is because the image must be upright, which means m must be positive in the equation for magnification m=-s'/s*

Since focal length (f) is given by f=2/R, and the miror equation states that (1/s)+(1/s')=1/f, we can substitute in for f and get:

(1/s)+(1/s')=(2/R)

Also, since we found that s'=-3s, we can further substitute to get:

(1/3s)+(1/s)=(2/R)

Simplifying, we get that (2/3s)=(2/R)

Plug in 50cm for R, and you get that s=16.6666666666

source: Taking calc-based physics 2, and had the exact same homework question
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