An object of size `3.0 cm` is placed `14 cm` in front of a concave lens of focal length `21 cm`. Describe the image produced by the lens. What happens if the object is moved further from the lens ?

Explanation

Size of the object, h1 = 3cm
Object distance, u=`-14cm`
Focal length of the concave lens, `f=-21cm`
Image distance =v
According to the lens formula, we have the relation:
`1/v-1/u=1/f`
`1/v=-1/21-1/14= (-2-3)/(42)=-5/42`
`therefore v=-42/5 = -8.4 cm`
Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.
the magnification of the image is given as:
`m=("image height"(h_(2)))/("Object height"(h_(2))) = v/u`
`therefore h_(2) = -8.4/-14 xx 3 = 0.6 xx 3=1.8cm`
Hence, the height of the image is 1.8 cm.
if the object is moved further away from the lens, the virtual image will move towards the focus of the lens, but not beyont it. The size of the image will decrease with the increase in the object distance.