Correct Answer - B
For `M_(1)`
`v=(uf)/(u-f) =(-15xx(-10))/(-15-(-10))=-30 cm`
For `M_(2) u=10 cm`
`therefore v=(10xx(-10))/(10-(-10))=-5 cm`
magnification `m=(-v)/(u)=-((-5)/(10))=(1)/(2)`
so, distance of image from `CD=(1)/(2)xx3=(3)/(2) cm`
`therefore` distance of image from `AB=3-(3)/(2)=(3)/(2) cm`