The correct option is A 1 mm behind the mirror and 0.08 mm high
The ball bearing surface will act as a convex mirror, its diameter,D=0.4cm,so,f=D4=0.1cm
Also, according to the sign convention,
u=−20 cm.
Now applying mirror formula,
1u+1v=1f
Or,
1v=1f−1u.
Substituting the values;
1v=10.1−1−20=20+0.12,
We get,
1v=10.05cm−1.
Hence, v=+0.1 cm or 1 mm behind the mirror, or oposite to the side the object (A positive value of v means image lies behind the mirror).
Now, for image height:
hih0=−vu,
or,hi+1.6=−+0.1−20
Therefore, hi=0.08 mm, is the height of the image.
Also, it is positive meaning that the image is errect, and the image is diminished, as expected!