Question

A convex of focal length 10 cm is placed at a distance of 12 cm from a wall how far from the lens should be an object be placed so as to form its real image on the wall.​

Answer 1

f=10 cm

v=12cm

1/v - 1/u= 1/f

1/u=-1/60

u=-60 cm

the real image on the world 60 cm

Answer 2

Following Sign Convention,

focal length = +10cm

image distance = +12

Object distance = "unknown"

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

$\frac{1}{12} - \frac{1}{u} = \frac{1}{10}$

$\frac{-1}{u} = \frac{1}{10} - \frac{1}{12}$

$\frac{-1}{u} = \frac{6-5}{60}$

$\frac{-1}{u}=\frac{1}{60}$

$u = (-1) 60\\u = -60cm\\\\\\Object distance = -60cm$

Hope it helps :)