An object is placed on the axis of a convex lens. Its real image is formed 80 cm from the object. The magnification is 3. The focal length of the lens is

Given Data:

Magnification produced,$m=-3=\frac{v}{u}$ (since image is real)

Distance of the image from object ,$v-u=80cm$

Step 1: Focal length

From magnification we can find distance of object,

$-3=\frac{v}{u}cm$

From distance of object and image we can say;

$-3=\frac{80+u}{u}$

$-3u=80+u$

$-4u=80$

$u=-20cm$

And distance of image from pole will be,

$v-u=80$

$v-(-20)=80$

$v=60cm$

So, from lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{f}=\frac{1}{60}-\frac{1}{-20}⟹\frac{1}{f}=\frac{1}{60}+\frac{1}{20}$

$\frac{1}{f}=\frac{1+3}{60}=\frac{4}{60}$

$f=\frac{60}{4}=15cm$