since lens is concave so it will have negative focal length
f=R/2
R=10.7cm=0.107m
so f=0.107/2
f=-0.054m
magnification, m = f / (f-u)
=-0.054/(-0.054-0.189)
m=0.22
since m is positive so image will be virtual image--answer for b
u = f - f /m
=-0.054-(-0.054/0.22)
=0.192m
=19.2cm--answer to a
c)A concave lens always shows a smaller upright virtual image
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