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The focal length of a concave lens is 30 cm. Where should an object be placed so that its image is `1//10` of its size ?

Answer» `"f = 30 cm, h"_(1)=h, h_(2)=(h)/(10)`
`m=(-"v")/(u)=(h_(2))/(h_(1))`
`(-v)/(u)=(h)/(10h) rArr (-1)/(v)=(10)/(u)`
`(1)/(f)=(1)/(v)+(1)/(u) rArr (1)/(u) =(1)/(f)-(1)/(v) rArr (1)/(u)=(1)/(30)+(10)/(u)`
`(1)/(u)-(10)/(u)=(1)/(30) rArr (1)/(u) (1-10)=(1)/(30) rArr (-9)/(u)=(1)/(30) rArr u = -"270 cm."`


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