An object of height 5 cm is placed perpendicular to the principal axis of a concave lens of focal length 10 cm. If the distance of the object from the optical centre of the lens is 20 cm, determine the position, nature and size of the image formed using the lens formula.

#### Solution

Given:-

Height of the object = h = 5 cm

Focal length of the concave lens = f = −10 cm

Object distance = u = −20 cm

Using the lens formula, we get

`1/"f"=1/"v"-1/"u"`

`therefore1/"v"=1/"f"+1/"u"=1/(-10)-1/20=(-2-1)/20=(-3)/20`

`therefore"v"=-6.67cm`

Hence, the image is formed 6.67 cm in front of the lens on the same side as the object.

Because v is negative, we can say that the image is virtual.

From the magnification formula for the lens, we get

`"m"=(h')/h="v"/"u"`

`thereforeh'="vh"/"u"=(-6.67xx5)/-20=1.67cm`

Hence, the size of the image is h′ = 1.67 cm.

Because the height of the image is positive and smaller than the height of the object, the image is erect and diminished.

So, we can conclude that the image is virtual, erect and diminished.