#### Question

A lens forms a real image 3 cm high of an object 1 cm high. If the separation of object and image is 15 cm, find the focal length of the lens.

#### Solution

Given,

Height of image (h') = - 3 cm (negative because image is real)

Height of object = 1 cm

Let us first define the sign convention:

Since the object distance is always negative, let it be -u.

Since the focal length is positive for a convex lens, let it be +f .

Magnification =`h/h=v/u`

`v/u=-3`

(i) Since object distance (u) is always negative, and the ratio `v/u` here is negative, v must be positive, i.e., it is on the right side of the lens.

Now, we have object distance + image distance = 15

`v+(-u)=15 (using sign convention)`

`-3u-u=15`

`-4u=15`

`u=(-15)/4`

Applying the lens formula:

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

`1/f=1/(-3u)-1/u`

`1/f=(-4)/(3u)`

`f=(3u)/-4=(3xx-15/4)/4=45/16=2.81`cm

Hence, the focal length of the lens is +2.81 cm.