An object 25 cm high is placed infront of a convex lens of focal length 30 cm. If the height of the image formed is 50 cm, find the distance between the object and the image?

Explanation

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"`.