Question

An object of height 4 cm is placed on the principal
axis at a distance of 20 cm in front of a convex
lens of focal length 30 cm. Find the distance of
image from the lens, its height and nature
​

Answer 1

Answer:

distance of image is -60cm, Height is 12cm and nature it is virtual and erect.

Explanation:

step by step explanation.

u= -20cm

F = 30cm

v= ?

1/f = 1/v - 1/u

1/30 = 1/v - 1/-20

1/30-1/20=1/v

take LCM then, we get

v= -60cm

M=v/u = height of image/Height of object

-60/-20=height of image/4

height of image= 4×3=12cm