Question

An object is placed at a distance of 30 cm from a cocave lens of focal length 15 what is the position of the. Object

Answer 1

U=-30cm   f= -15cm 

1/f=1/v-1/u

1/v=1/f+1/u

on substituting  v=-10cm

m=v/u

m=-10/-30

m=1/3=.33

so the magnification is positive so it is a virtual erect image .

m isless  than 1 so the image is  diminished .

and v is negative so the image is formed on the same side of the lens.

Hope so it helps u