a. Using `1/v-1/u=1/f`
1/v=1/u+1/f=1-(-3.6cm)+1/(4.0cm)`
or `v=-36cm`
b. Linear magnification `=v/u`
`9-36cm)/(-3.6cm)=10`
c. If th object placed at a distance `u_0 form the lens, the angle subtended by the object on thelens is `beta=h/u_0` where h is the height of the object. The maximum angle subtended on the uN/Aided eye is `alpha=h/D`
Thus, the angular magnification is
`m=beta/alpha=D/u_0=(25cm)/(3.6cm)7.0`