Object distance, u=−5 m (always negative)
Radius of curvature, R=3 m
Focal length, f=32=1.5 m
Using the mirror formula,
1u+1v=1f
1−5+1v=11.5
1v=11.5−1−5
1v=11.5+15=5+1.57.5=6.57.5⇒v=1.15 m
For finding the magnification,
m=−vu=−1.15−5=0.23
Therefore, the image is formed on the opposite side of the object. Hence, the image is virtual and since the magnification is positive, the image is erect. The value of magnification is less than 1 ⟹ the image is diminished
The nature of the image is virtual, erect, and diminished.