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 the image from the lens,its height and nature.
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Answer:
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Explanation:
u = -20
f = 30
1/v-1/u = 1/f
1/v = 1/f+1/u
1/v = 1/30+(-1/20)
1 /v = 1/30-1/20
1/v = 2-3/60
1/v = -1/60
v = -60
m = h'/h = v/u
h' = v/u ×h
= -60/-20×4
= 12 cm
v= -60
h' = 12
nature of image = as image will be formed at the side where object is placed. So, image formed is virtual,erect and magnified
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