Final answer:
The image formed by a convex mirror with magnification of 0.250 is 0.75 meters behind the mirror. The mirror's focal length is -1 meter, and its radius of curvature is -2 meters. These values are for virtual images and distances as it's a convex mirror.
Step-by-step explanation:
The student is dealing with a problem related to convex mirrors and their properties in optics. A shopper standing 3.00 meters away from a convex security mirror sees an image with a magnification of 0.250. We will use the mirror equation and magnification formula to solve for the position of the image, the focal length, and the radius of curvature of the mirror.
To find the image distance (di), we use the magnification formula:
- Magnification (m) = -di/do
Where 'do' is the object distance and 'di' is the image distance, with the negative sign indicating a virtual image. Given m = 0.250 and do = 3.00 m:
- 0.250 = -di/3.00
- di = -3.00 * 0.250
- di = -0.75 m
For the focal length (f) using the mirror equation (1/f = 1/do + 1/di), we get:
- 1/f = 1/3.00 + 1/(-0.75)
- 1/f = (1/3) - (4/3)
- 1/f = -1
- f = -1 m
The radius of curvature (R) is twice the focal length for mirrors:
- R = 2f
- R = 2 * (-1 m)
- R = -2 m
The image is located 0.75 meters behind the mirror, the focal length of the mirror is -1 meter, and the radius of curvature is -2 meters, indicating that all the values pertain to virtual images and distances in a convex mirror.