Use the worked example above to help you solve this problem. A converging lens o

Question

Use the worked example above to help you solve this problem. A converging lens of focal length 9.9 cm forms images of an object situated at various distances. If the object is placed 29.7 cm from the lens, locate the image, state whether it's real or virtual, and find its magnification. (If either of the quantities evaluate to infinity, type INFINITY.) Repeat the problem when the object is at 9.9 cm. Repeat again when the object is 4.95 cm from the lens. Suppose the image of an object is upright and magnified 1.79 times when the object is placed 15.8 cm from a lens. Find the location of the image and the focal length of the lens.

Explanation / Answer

priactice it

f = 9.9 cm

a) object distance, u = 29.7 cm

let v is the image distance.

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/9.9 - 1/29.7

v = 14.85 cm (real)

M = -v/u

= -14.85/29.7

= -0.5

b) object distance, u = 9.9 cm

let v is the image distance.

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/9.9 - 1/9.9

v = ininite (real)

M = -v/u

-infinite

c) object distance, u = 4.95 cm

let v is the image distance.

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/9.9 - 1/4.95

v = -9.9 cm (virtual)

M = -v/u

= -(-9.9)/4.95

= +2

excericise

given, m = 1.79

and u = 15.8 cm

we know, m = -v/u

v = -m*u

= -1.79*15.8

= -28.282 cm (image distance)

use, 1/f = 1/u + 1/v

1/f = 1/15.8 + 1/(-28.282)

f = 35.8 cm

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