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
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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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