Question

As shown in figure. L is half part of an equiconvex lens `(mu=1.5)` whose surface have radius of curvature `R=40cm` and its right surface in silvered. Normal to principal axis, a plane mirror M is placed on the right of lens. A small object O is placed on left of the lens such that there is no parallax between final images formed by the lens and mirror. If transverse length of final image formed by the lens is twice that of image formed by the mirror. Calculate distance a between lens and object and distance `b`.

Answer 1

First calculate focal length of silvered lens.
`(1)/(F)=(1)/(f_l)+(1)/(f_m)+(1)/(f_l)=(2)/(f_l)+(1)/(f_m)`
`=2((3)/(2)-1)((1)/(40)-(1)/(-40))+(1)/((40)/(2))`
`=(1)/(20)+(1)/(20)=(1)/(10)impliesF=10cm`
This silvered lens will behave as a concave mirror of focal length 10 cm.
Image by plane mirror:
Image by silvered lens/concave mirror:
`(1)/(v)+(1)/(-a)=(1)/(-10)implies(1)/(v)=(1)/(a)-(1)/(10)=(10-a)/(10a)`
`v=(10a)/(10-a)`
`m=(I)/(O)=-(v)/(u)rArr2=-((10a)//(10-a))/(-a)`
`10-a=5`
`a=5cm`
`v=(10a)/(10-a)=(10xx5)/(5)=10=a+2bimplies10=5+2b`
`b=2.5cm`