Physics - Mirrors & Lenses
posted by Anonymous .
When standing 1.25 m in front of an amusement park mirror, you notice that your image is three times taller. What is the radius of curvature of the mirror?
Physics - Mirrors & Lenses -
I will assume it is an upright image, behind the mirror. The distance to the image is di = -3 do because of the 3x magnification, and do = 1.25 m. The minus sign is necessary for upright virtual images.
The focal length f is given by the equation
1/do + 1/di = 1/f
1/1.25 -1/3.75 = 2/3.75 = 1/f
Therefore f = 1.87 m.
The radius of curvature is 2f, or 3.75 m. It is a concave mirror; otherwise f would be negative.