A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is

Options

A. Virtual and at a distance of 16 cm from the mirror
B. Real and at a distance of 16 cm from the mirror
C. Virtual and at a distance of 20 cm from the mirror
D. Real and at a distance of 20 cm from the mirror.

Answer

B

Explanation

Focal length of the convex lens is 15 cm and object is kept at 30 cm , hence at a distance twice the focal length . In this case we should already know that real image will be formed at a distance 30 cm from the lens on the other side shown as `l_1` in figure. Plane mirror is kept at a distance 10 cm from the lens and hence distance of `I_1` from plane mirror will be 20 cm , `I_1` is virtual object for plane mirror and it will create its real image at a distance 20 cm from itself as `I_2` in figure. Now this point `I_2` is 10 cm ahead of lens and hence will serve the purpose of virtual object for it. Now we can apply lens formula to find distance of final image `I_3` from lens (x).
`1/f =1/v =1/u`
`implies (1)/(+15)=1/v-(1)/(+10)`
`implies 1/v=1/15+1/10`
`implies 1/ v = (10+15)/150`
`implies v = (150)/25=6 cm`
Hence distance from te mirror is 16 cm. We can see that final image is real and it a distance 16 cm from the mirror