Question

A lens of focal length 12cm forms an erect image, three times the size of the object. The distance the object and image is:​

Answer 1

the distance of the object is 4cm and image is -6

Answer 2

Magnification , 

$m = \frac{v}{u}$

here m =3

=> 3u = v

Lens formula : 

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

=

$\frac{1}{3u} - \frac{1}{u} = \frac{1}{12}$

-2/3u = 1/12

⇒ u= − 8cm 

8cm v = 3×−8 = − 24cm 

Distance between object and image

 =u−v = − 8− (−24)

= −8+24

= 16cm

hope its helpful