Question

an object 3 cm high is placed perpendicular to the principal axis of a concave lens of focal length 15 CM the image is formed at a distance of 10 cm from the lens calculate distance at which object is placed V size and nature of image formed

Answer 1

nature of the image:
1)virtual,erect
2) diminished

Answer 2

Answer:

object distance (u) = 30cm

image height = 1cm

Explanation:

height of the object(ho)= 3 cm

focal length(f) = 15cm

image distance (v)= 10cm

1/v - 1/u = 1/ f

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

-1/u = 1/15-1/10

-1/u = 2-3/30

-1/u = - 1/30

1/u = 1/30

u= 30cm (ans)

m= v/u = hi/ ho

v/u=hi/ho

10/30=hi/3

1/3=hi/3

by cross multiplication

3=3hi

hi=3/3

hi=1 cm (ans)