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Question

an object 3cm high is placed prependicularto the principal axis of concave lens of focal length 15cm.the image is form at a distance of 10cm from the lens.find the distance at which object is placed and size and nature of image.?

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Solution

Focal length (f) = -20 cm

Here,

Object Distance (u) = ?

ho = 3 cm

Image Distance, v = -10 cm

now according to lens formula,

1/f =1/v - 1/u

1/u = 1/v - 1/f

This gives,

u = -30 cm

Magnification,

m = hi/ho= v/u

m = 10/30 = hi/ho

hi =1/3 x 3=1 cm

Image is virtual , erect and smaller.


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