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A lens placed at a distance of `20cm`from a object produces of virtual Image `2//3`the size of the object.Find the position of the image,kind of lens and its focal length.

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Virtual image means,I is positive and it is given that I=`(2)/(3)O`Thus,
`m=+(2)/(3)`
Further because`u=-20cm`(given),using
`m=(f)/(f+u)` ltrbgtwe get,`(2)/(3)=(f)/(f+(-20))`
or`f=-40cm`
The `f` is negative,thus the lens is a concave lens. Again using
`m=(v)/(u)`
we get`(2)/(3)=(v)/(-20)`
or`v=-(2)/(3)`
`=-1.33cm`
The virtual image is on the same side of the object.
(can you draw the ray diagram for the above problem?Try)

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