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# plz solve the question with explanation

Asked by futureisbright051101 11th April 2018, 12:18 AM
Let us consider image is real

(1/f) = (1/v) + (1/u) ................(1)

magnification m = -v/u = -2; hence v = 2u

substitute for v in (1), we get (1/f) = [ 1/(2u) ] + (1/u) ; we get u = (3/2)f  and v = 3f;

hence v-u = (3/2)f

let us consder image is vrtual

magnification m = -v/u = 2 ; hence v =-2u;

substitute for v in (1), we get (1/f) = [ 1/(-2u) ] + (1/u) ; we get u = (1/2)f  and v = -f;

hence v-u = -(3/2)f;

magnitude of the distance between the object and its doubly magnified image is (3/2)f, whether the image is real or virtual
Answered by Expert 16th April 2018, 3:27 PM
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