A 4cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 24 cm. the distance of the object from the lens is 16 cm. Find the position, size and nature of the image formed using the lens formula.

Explanation

For convex lens
`f=+24 cm, u=-16 cm`
Using lens formula
`(1)/(f)=(1)/(v)-(1)/(u)`
`(1)/(24)=(1)/(v)-(1)/(-16)=(1)/(v)+(1)/(16)`
`(1)/(v)=(1)/(24)-(1)/(16)=(2-3)/(48)=(-1)/(48)`
`v=-48 cm`
Now `= h_(1)4cm`
So magnification, `m=(h_(2))/(h_(1))=(v)/(u)`
`h_(2)=(v)/(u)xx h_(1)=(-48)/(-16)xx4`
= 12 cm.
Position of image is formed at a distance of 48 cm from the optical of the lens on the same side of the object. It is indicates by the negative sign size of image. it is three times the size of object i.e., 12 cm.
Nature of image : positive sign in the image height indicates that image is virtual and erect.