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)?

Explanation

We have `|m|=1/O = v/u=50/25=2`
There are two possibilities
(i) If the image is inverted (i. e., real) `m = v/u = -2 or v = -2u`
Let x be the object distance in this case
We have u = - x, v = + 2 x, f = + 30 cm
Using lens formula, we have `1/(2x) - 1/(-x) =1/30` or x=45cm
Hence, the distance between the object and the image = 45 + 90 = 135 cm.
(ii)If the image is erect (i.e., virtual) `m = v/u = 2` or v = 2u
Let `x^’` be the object distance we have `u = x^(’),v=x^(’), f = + 30`
Using the lens formula, we have`1/(-2x) + 1/x^(.) = 1/30 impliesx^(.) = 15ccm`
Hence, the distance between the object and the image = 15 cm.