A physics student wants toobtain a real and virtual image that each is 3 times as large as the original object she has a concave mirror with a 0.50 m radius of curvature. what object distance will she require to obtain the real image?
1/do + 1/di = 1/f
The Attempt at a Solution
m = 3 = -di / do 3do = -di
I made do=d di=3d so...
1/d + 1/3d =1/0.25
[1/d + 1/3d =1/0.25]3d
2d + 1 = 3d/0.25
[2d + 1 = 3d/0.25]0.25
0.5d + 0.25 = 3d
d= .10m m =do huh? I know I did something wrong
based upon my ray diagram do= .32m ish
and if that was the do then there would not be a real image formed becaues the object is inside of the focal point so only a virtual image would be formed right?
since m = 3 = -di / do 3do = -di
I tried to make di negative and that didnt turn out either I am not sure what I have done wrong