Answer:
magnification will be -0.025
Explanation:
We have given the radius of curvature = 12 cm
And object distance = 3 m
So focal length [tex]f=\frac{R}{2}=\frac{12}{2}=6cm[/tex]
Now for mirror we know that [tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]
So [tex]\frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}[/tex]
[tex]16.66-0.333=\frac{1}{v}[/tex]
v = 0.750 m
Now magnification of the mirror is [tex]m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025[/tex]
