A car’s convex rearview mirror has a radius of curvature equal to 15 m.
What is the location of the image dI that is formed by an object that is 14 m from the mirror? Follow the sign convention.
It is convex so the focal length is negative and half the radius = -7.5 m
Using 1/di + 1/d0 = 1/f we get 1/di = -1/7.5 - 1/14
di = - 4.88 m which means it is virtual and upright
What is the magnification of the image that is formed by an object that is 14 m from the mirror?
The magnification = -di/do = 4.88/14 = 0.349
