A lens of focal length 12.0cm forms an upright image three times the size of a real object. The distance between the object and the images is
an upright image is a virtual image and it is expected to be on the same side as the object for a lens but not for a mirror.
magnification M = \(\frac{v}{u}\) = \(\frac{v}{2}\) = 3 ⇒⇒ v = 3u ( image is virtual = -3u)
using \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{}{v}\) ⇒⇒ \(\frac{1}{u}\) - \(\frac{1}{3u}\) = \(\frac{1}{12}\)
\(\frac{3 -1}{3u}\) = \(\frac{1}{12}\)
cross multiplying
24 = 3u = u = 8cm but v = 3u = 24cm
virtual image and object are at the side of the lens,
Hence, the distance between the object and the image = 24 - 8 = 16cm.
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