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An object 25 cm high is placed in front ...

An object `25` cm high is placed in front of a convex lens of focal length `30` cm . If the height of image formed is `50` cm , find the distance between the object and the image (real and virtual)?

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As object is in front of the lens , it is real and as `h_(1) = 25 "cm" f = 30 "cm" , h_(2) = -50"cm" , m= (h_(2))/(h_(1)) = (-50)/(25) = -2`
`m=(f)/(f+u) implies -2 = (30)/(30+u) implies u=-45 "cm" implies "m" = (v)/(u) implies -2 = (v)/(-45) implies v=90"cm"`

As in this situation object and image are on opposite sides of lens, the distance between object and image `d_(1) = u +v = 45 +90 = 135 "cm" . If the image is erect (i.e., virtual)
`m= (f)/(f+u) implies 2 (30)/(30 + u) implies u=-15 "cm"implies "m" = -(v)/(u) implies 2 = (-v)/(-15) implies v = 30 "cm"`
As in the situation both image and object are in front of the lens , the distance between object and image
`d_(2) = v-u = 30-15 = 15 "cm"`.
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