Question

A 4 cm tall object is placed perpendicular to the principal axis. of a converging lens of focal length 24 cm.the object distance is 16 cm. find the position size and nature of image formed .

Answer 1

ho = 4 cm

f =24 cm

u = -16 cm

1/v = 1/f +1/u

1/v = 1/24 + 1/-16 =-48 cm

now, m = v/u

m= -48/-16=3

m is also = hi / ho

so , 4 = hi/3

so 12 cm =hi

so image is erect and enlarged and image is formed at same side of lens

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Answer 2

Height of the object, h = +4cm

Focal length (convex lens) = +24cm

Object distance, u = -16cm

By lens formula: 1/v - 1/u = 1/f

=> 1/v = 1/f + 1/u

=> 1/v = 1/24 + 1/(-16)

=> 1/v = 1/24 - 1/16

=> 1/v = -1/48 (taking LCM)

So, V = -48 cm

Magnification, m = v/u

=> m = -48/-16

=> m = +3

Now, m = h’/h (where, h’ is the height of the image)

=> 3 = h’/4

=> h’ = +12 cm

Thus, the image is enlarged and erect