An upright object is placed at a distance of `40cm` in front of a convergent less of focal length `20cm`. A convergent mirror of focal length `10cm` is placed at a distance of `60cm` on the other side of the lens. The position and size of the final image will be:

Options

A. 40 cm from the convergent lens, twice the size of the object
B. 20 cm from the convergent mirror, twice the size of the object
C. 20 cm from the convergent mirror, same size as the object
D. 40 cm from the convergent mirror, same size as the object

Answer

D

Explanation

(1) For first refraction by lens
`4=-40cm =-2f implies v+2f =+40cm`
`m=v/u=-1`
(2) For reflection by mirror
`u=-(60-20)=-40cm=-2f`
`impliesv=-2f=-40cm`
`m=-v/u=-1`
(3) For second refraction by lens
u = - (60-20)=-40 cm = -2f
`impliesv=2f=40cmimplies m=v/u=-1`
`implies` image is at same position as object `m_("net") =(-1).(-1).(-1)=-1`
`implies ` image is of same size and inverted .