Physics, asked by vikrantsingh12, 1 year ago

A lens of focal length 12, cm produced a virtual image.The size of images is 1/3 times the size of the object.what kind of lens it is ? determine yhe position of the object and the image

Answers

Answered by AnkitaSahni
7

Given :

A lens of focal length 12 cm which produces virtual image of magnification 1/3 times

To Find :

Type of lens and position of image & position of object

Solution :

•It is given that magnification prouduced by given lens 1/3

magnification =

Height of image / height of object

also, magnification =distance of

image/distance of object

1/3 = v/u

u = 3v _______(1)

•Also , By lens Formula

1/v - 1/u = 1/f

1/v -1/3v = 1/12 ( From (1) )

(3-1)/3v = 1/12

2/3v = 1/12

v = 8 cm

u = 24 cm

•Hence , position of image & object are 8 cm & 24 cm respectively .

Lens used is concave lens

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